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University Physics Volume 2 SENIOR CONTRIBUTING AUTHORS WILLIAM MOEBS, FORMERLY OF LOYOLA MARYMOUNT UNIVERSITY SAMUEL J. LING, TRUMAN STATE UNIVERSITY JEFF SANNY, LOYOLA MARYMOUNT UNIVERSITY
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Contents Preface 1 UNIT 1 THERMODYNAMICS CHAPTER 1 Temperature and Heat 5 Introduction 5 1.1 Temperature and Thermal Equilibrium 6 1.2 Thermometers and Temperature Scales 7 1.3 Thermal Expansion 9 1.4 Heat Transfer, Specific Heat, and Calorimetry 16 1.5 Phase Changes 23 1.6 Mechanisms of Heat Transfer 31 Chapter Review 48 CHAPTER 2 The Kinetic Theory of Gases 63 Introduction 63 2.1 Molecular Model of an Ideal Gas 64 2.2 Pressure, Temperature, and RMS Speed 73 2.3 Heat Capacity and Equipartition of Energy 83 2.4 Distribution of Molecular Speeds 88 Chapter Review 93 CHAPTER 3 The First Law of Thermodynamics 103 Introduction 103 3.1 Thermodynamic Systems 104 3.2 Work, Heat, and Internal Energy 106 3.3 First Law of Thermodynamics 110 3.4 Thermodynamic Processes 116 3.5 Heat Capacities of an Ideal Gas 119 3.6 Adiabatic Processes for an Ideal Gas 121 Chapter Review 125 CHAPTER 4 The Second Law of Thermodynamics 135 Introduction 135 4.1 Reversible and Irreversible Processes 136 4.2 Heat Engines 138 4.3 Refrigerators and Heat Pumps 140 4.4 Statements of the Second Law of Thermodynamics 141 4.5 The Carnot Cycle 144 4.6 Entropy 149
4.7 Entropy on a Microscopic Scale 154 Chapter Review 159 UNIT 2 ELECTRICITY AND MAGNETISM CHAPTER 5 Electric Charges and Fields 169 Introduction 169 5.1 Electric Charge 170 5.2 Conductors, Insulators, and Charging by Induction 175 5.3 Coulomb's Law 179 5.4 Electric Field 184 5.5 Calculating Electric Fields of Charge Distributions 190 5.6 Electric Field Lines 198 5.7 Electric Dipoles 202 Chapter Review 205 CHAPTER 6 Gauss's Law 219 Introduction 219 6.1 Electric Flux 220 6.2 Explaining Gauss’s Law 228 6.3 Applying Gauss’s Law 234 6.4 Conductors in Electrostatic Equilibrium 246 Chapter Review 255 CHAPTER 7 Electric Potential 267 Introduction 267 7.1 Electric Potential Energy 268 7.2 Electric Potential and Potential Difference 274 7.3 Calculations of Electric Potential 285 7.4 Determining Field from Potential 296 7.5 Equipotential Surfaces and Conductors 299 7.6 Applications of Electrostatics 307 Chapter Review 312 CHAPTER 8 Capacitance 323 Introduction 323 8.1 Capacitors and Capacitance 323 8.2 Capacitors in Series and in Parallel 333 8.3 Energy Stored in a Capacitor 339 8.4 Capacitor with a Dielectric 342 8.5 Molecular Model of a Dielectric 345 Chapter Review 352 Access for free at openstax.org
CHAPTER 9 Current and Resistance 361 Introduction 361 9.1 Electrical Current 362 9.2 Model of Conduction in Metals 366 9.3 Resistivity and Resistance 371 9.4 Ohm's Law 380 9.5 Electrical Energy and Power 384 9.6 Superconductors 390 Chapter Review 396 CHAPTER 10 Direct-Current Circuits 405 Introduction 405 10.1 Electromotive Force 406 10.2 Resistors in Series and Parallel 413 10.3 Kirchhoff's Rules 425 10.4 Electrical Measuring Instruments 437 10.5 RC Circuits 440 10.6 Household Wiring and Electrical Safety 446 Chapter Review 450 CHAPTER 11 Magnetic Forces and Fields 463 Introduction 463 11.1 Magnetism and Its Historical Discoveries 464 11.2 Magnetic Fields and Lines 466 11.3 Motion of a Charged Particle in a Magnetic Field 470 11.4 Magnetic Force on a Current-Carrying Conductor 474 11.5 Force and Torque on a Current Loop 479 11.6 The Hall Effect 481 11.7 Applications of Magnetic Forces and Fields 484 Chapter Review 489 CHAPTER 12 Sources of Magnetic Fields 501 Introduction 501 12.1 The Biot-Savart Law 502 12.2 Magnetic Field Due to a Thin Straight Wire 505 12.3 Magnetic Force between Two Parallel Currents 509 12.4 Magnetic Field of a Current Loop 511 12.5 Ampère’s Law 514 12.6 Solenoids and Toroids 520 12.7 Magnetism in Matter 525 Chapter Review 532 CHAPTER 13 Electromagnetic Induction 545 Introduction 545
13.1 Faraday’s Law 546 13.2 Lenz's Law 550 13.3 Motional Emf 554 13.4 Induced Electric Fields 562 13.5 Eddy Currents 566 13.6 Electric Generators and Back Emf 570 13.7 Applications of Electromagnetic Induction 576 Chapter Review 578 CHAPTER 14 Inductance 591 Introduction 591 14.1 Mutual Inductance 592 14.2 Self-Inductance and Inductors 595 14.3 Energy in a Magnetic Field 600 14.4 RL Circuits 602 14.5 Oscillations in an LC Circuit 607 14.6 RLC Series Circuits 610 Chapter Review 613 CHAPTER 15 Alternating-Current Circuits 623 Introduction 623 15.1 AC Sources 624 15.2 Simple AC Circuits 625 15.3 RLC Series Circuits with AC 631 15.4 Power in an AC Circuit 636 15.5 Resonance in an AC Circuit 640 15.6 Transformers 645 Chapter Review 649 CHAPTER 16 Electromagnetic Waves 657 Introduction 657 16.1 Maxwell’s Equations and Electromagnetic Waves 658 16.2 Plane Electromagnetic Waves 664 16.3 Energy Carried by Electromagnetic Waves 671 16.4 Momentum and Radiation Pressure 675 16.5 The Electromagnetic Spectrum 680 Chapter Review 687 Appendix A Units 699 Appendix B Conversion Factors 703 Appendix C Fundamental Constants 707 Appendix D Astronomical Data 709 Appendix E Mathematical Formulas 711 Appendix F Chemistry 715 Appendix G The Greek Alphabet 717 Access for free at openstax.org
Answer Key 719 Index 765
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PREFACE Welcome to University Physics, an OpenStax resource. This textbook was written to increase student access to high-quality learning materials, maintaining highest standards of academic rigor at little to no cost. About OpenStax OpenStax is part of Rice University, which is a 501(c)(3) nonprofit charitable corporation. Our mission is to make an amazing education accessible for all. Through our partnerships with philanthropic organizations and our alliance with other educational resource companies, we’re breaking down the most common barriers to learning. Because we believe that everyone should and can have access to knowledge. About OpenStax's resources CustomizationUniversity Physics is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 (CC BY NC-SA) license, which means that you can non- commercially distribute, remix, and build upon the content, as long as you provide attribution to OpenStax and its content contributors, and distribute all derivatives under the same license. Because our books are openly licensed, you are free to use the entire book or pick and choose the sections that are most relevant to the needs of your course. Feel free to remix the content by assigning your students certain chapters and sections in your syllabus in the order that you prefer. You can even provide a direct link in your syllabus to the sections in the web view of your book. For any art that is openly licensed, non-commercial users or organizations may reuse the art as long as they provide the same attribution to its original source. (Commercial entities should contact OpenStax to discuss reuse rights and permissions.) To maximize readability and content flow, some art does not include attribution in the text. If you reuse art from this text that does not have attribution provided, use the following attribution: Copyright Rice University, OpenStax, under CC BY NC-SA 4.0 license. Instructors also have the option of creating a customized version of their OpenStax book. The custom version can be made available to students in low-cost print or digital form through their campus bookstore. Visit your book page on OpenStax.org for more information. Errata All OpenStax textbooks undergo a rigorous review process. However, like any professional-grade textbook, errors sometimes occur. Since our books are web based, we can make updates periodically when deemed pedagogically necessary. If you have a correction to suggest, submit it through the link on your book page on OpenStax.org. Subject matter experts review all errata suggestions. OpenStax is committed to remaining transparent about all updates, so you will also find a list of past errata changes on your book page on OpenStax.org. Format You can access this textbook for free in web view or PDF through OpenStax.org, and for a low cost in print. About University Physics University Physics is designed for the two- or three- semester calculus-based physics course. The text has been developed to meet the scope and sequence of most university physics courses and provides a foundation for a career in mathematics, science, or engineering. The book provides an important opportunity for students to learn the core concepts of physics and understand how those concepts apply to their lives and to the world around them. Due to the comprehensive nature of the material, we are offering the book in three volumes for flexibility and efficiency. Coverage and scope Our University Physics textbook adheres to the scope and sequence of most two- and three-semester physics courses nationwide. We have worked to make physics interesting and accessible to students while maintaining the mathematical rigor inherent in the subject. With this objective in mind, the content of this textbook has been developed and arranged to provide a logical progression from fundamental to more advanced concepts, building upon what students have already learned and emphasizing connections between topics and between theory and applications. The goal of each section is to enable students not just to recognize concepts, but to work with them in ways that will be useful in later courses and future careers. The Preface 1
organization and pedagogical features were developed and vetted with feedback from science educators dedicated to the project. VOLUME I Unit 1: Mechanics Chapter 1: Units and Measurement Chapter 2: Vectors Chapter 3: Motion Along a Straight Line Chapter 4: Motion in Two and Three Dimensions Chapter 5: Newton’s Laws of Motion Chapter 6: Applications of Newton’s Laws Chapter 7: Work and Kinetic Energy Chapter 8: Potential Energy and Conservation of Energy Chapter 9: Linear Momentum and Collisions Chapter 10: Fixed-Axis Rotation Chapter 11: Angular Momentum Chapter 12: Static Equilibrium and Elasticity Chapter 13: Gravitation Chapter 14: Fluid Mechanics Unit 2: Waves and Acoustics Chapter 15: Oscillations Chapter 16: Waves Chapter 17: Sound VOLUME II Unit 1: Thermodynamics Chapter 1: Temperature and Heat Chapter 2: The Kinetic Theory of Gases Chapter 3: The First Law of Thermodynamics Chapter 4: The Second Law of Thermodynamics Unit 2: Electricity and Magnetism Chapter 5: Electric Charges and Fields Chapter 6: Gauss’s Law Chapter 7: Electric Potential Chapter 8: Capacitance Chapter 9: Current and Resistance Chapter 10: Direct-Current Circuits Chapter 11: Magnetic Forces and Fields Chapter 12: Sources of Magnetic Fields Chapter 13: Electromagnetic Induction Chapter 14: Inductance Chapter 15: Alternating-Current Circuits Chapter 16: Electromagnetic Waves VOLUME III Unit 1: Optics Chapter 1: The Nature of Light Chapter 2: Geometric Optics and Image Formation Chapter 3: Interference Chapter 4: Diffraction Unit 2: Modern Physics Chapter 5: Relativity Chapter 6: Photons and Matter Waves Chapter 7: Quantum Mechanics Chapter 8: Atomic Structure Chapter 9: Condensed Matter Physics Chapter 10: Nuclear Physics Chapter 11: Particle Physics and Cosmology Pedagogical foundation Throughout University Physics you will find derivations of concepts that present classical ideas and techniques, as well as modern applications and methods. Most chapters start with observations or experiments that place the material in a context of physical experience. Presentations and explanations rely on years of classroom experience on the part of long-time physics professors, striving for a balance of clarity and rigor that has proven successful with their students. Throughout the text, links enable students to review earlier material and then return to the present discussion, reinforcing connections between topics. Key historical figures and experiments are discussed in the main text (rather than in boxes or sidebars), maintaining a focus on the development of physical intuition. Key ideas, definitions, and equations are highlighted in the text and listed in summary form at the end of each chapter. Examples and chapter- opening images often include contemporary applications from daily life or modern science and engineering that students can relate to, from smart phones to the internet to GPS devices. Assessments that reinforce key concepts In-chapter Examples generally follow a three-part format of Strategy, Solution, and Significance to emphasize how to approach a problem, how to work with the equations, and how to check and generalize the result. Examples are often followed by questions and answers to help reinforce for students the important ideas of the examples. Problem-Solving Strategies in each chapter break down methods of approaching various types of problems into steps students can follow for guidance. The book also includes exercises at the end of each chapter so students can practice what they’ve learned. Conceptual questions do not require calculation but test student learning of the key concepts. Problems categorized by section test student problem-solving skills and the ability to apply 2 Preface Access for free at openstax.org
ideas to practical situations. Additional Problems apply knowledge across the chapter, forcing students to identify what concepts and equations are appropriate for solving given problems. Randomly located throughout the problems are Unreasonable Results exercises that ask students to evaluate the answer to a problem and explain why it is not reasonable and what assumptions made might not be correct. Challenge Problems extend text ideas to interesting but difficult situations. Answers for selected exercises are available in an Answer Key at the end of the book. Answers to Questions in the Book Answers to Unreasonable Results are not provided. All answers to Check Your Understanding are provided in the Answer Key in the book. Odd-numbered answers to Conceptual Questions, Problems, Additional Problems, and Challenge Problems are also provided in the Answer Key. Students can find additional answer information in the Student Solution Guide via the Student Resources page. Answers to even-numbered Conceptual Questions, Problems, Additional Problems, and Challenge Problems are provided only to instructors in the Instructor Answer Guide via the Instructor Resources page. Additional resources Student and instructor resources We’ve compiled additional resources for both students and instructors, including Getting Started Guides, PowerPoint slides, and answer and solution guides for instructors and students. Instructor resources require a verified instructor account, which you can apply for when you log in or create your account on OpenStax.org. Take advantage of these resources to supplement your OpenStax book. Community Hubs OpenStax partners with the Institute for the Study of Knowledge Management in Education (ISKME) to offer Community Hubs on OER Commons – a platform for instructors to share community-created resources that support OpenStax books, free of charge. Through our Community Hubs, instructors can upload their own materials or download resources to use in their own courses, including additional ancillaries, teaching material, multimedia, and relevant course content. We encourage instructors to join the hubs for the subjects most relevant to your teaching and research as an opportunity both to enrich your courses and to engage with other faculty. To reach the Community Hubs, visit www.oercommons.org/hubs/OpenStax. Partner resources OpenStax partners are our allies in the mission to make high-quality learning materials affordable and accessible to students and instructors everywhere. Their tools integrate seamlessly with our OpenStax titles at a low cost. To access the partner resources for your text, visit your book page on OpenStax.org. About the authors Senior contributing authors Samuel J. Ling, Truman State University Dr. Samuel Ling has taught introductory and advanced physics for over 25 years at Truman State University, where he is currently Professor of Physics and the Department Chair. Dr. Ling has two PhDs from Boston University, one in Chemistry and the other in Physics, and he was a Research Fellow at the Indian Institute of Science, Bangalore, before joining Truman. Dr. Ling is also an author of A First Course in Vibrations and Waves, published by Oxford University Press. Dr. Ling has considerable experience with research in Physics Education and has published research on collaborative learning methods in physics teaching. He was awarded a Truman Fellow and a Jepson fellow in recognition of his innovative teaching methods. Dr. Ling’s research publications have spanned Cosmology, Solid State Physics, and Nonlinear Optics. Jeff Sanny, Loyola Marymount University Dr. Jeff Sanny earned a BS in Physics from Harvey Mudd College in 1974 and a PhD in Solid State Physics from the University of California–Los Angeles in 1980. He joined the faculty at Loyola Marymount University in the fall of 1980. During his tenure, he has served as department Chair as well as Associate Dean. Dr. Sanny enjoys teaching introductory physics in particular. He is also passionate about providing students with research experience and has directed an active undergraduate student research group in space physics for many years. William Moebs, Formerly of Loyola Marymount University Dr. William Moebs earned a BS and PhD (1959 and 1965) from the University of Michigan. He then joined their staff as a Research Associate for one year, where he continued his doctoral research in particle physics. In 1966, he accepted an appointment to the Physics Department of Indiana Purdue Fort Wayne (IPFW), where he served as Department Chair from 1971 to Preface 3
1979. In 1979, he moved to Loyola Marymount University (LMU), where he served as Chair of the Physics Department from 1979 to 1986. He retired from LMU in 2000. He has published research in particle physics, chemical kinetics, cell division, atomic physics, and physics teaching. Contributing authors Stephen D. Druger, Northwestern University Alice Kolakowska, University of Memphis David Anderson, Albion College Daniel Bowman, Ferrum College Dedra Demaree, Georgetown University Edw. S. Ginsberg, University of Massachusetts Joseph Trout, Richard Stockton College Kevin Wheelock, Bellevue College David Smith, University of the Virgin Islands Takashi Sato, Kwantlen Polytechnic University Gerald Friedman, Santa Fe Community College Lev Gasparov, University of North Florida Lee LaRue, Paris Junior College Mark Lattery, University of Wisconsin Richard Ludlow, Daniel Webster College Patrick Motl, Indiana University Kokomo Tao Pang, University of Nevada, Las Vegas Kenneth Podolak, Plattsburgh State University Reviewers Salameh Ahmad, Rochester Institute of Technology–Dubai John Aiken, University of Colorado–Boulder Raymond Benge, Terrant County College Gavin Buxton, Robert Morris University Erik Christensen, South Florida State College Clifton Clark, Fort Hays State University Nelson Coates, California Maritime Academy Herve Collin, Kapi’olani Community College Carl Covatto, Arizona State University Alejandro Cozzani, Imperial Valley College Danielle Dalafave, The College of New Jersey Nicholas Darnton, Georgia Institute of Technology Ethan Deneault, University of Tampa Kenneth DeNisco, Harrisburg Area Community College Robert Edmonds, Tarrant County College William Falls, Erie Community College Stanley Forrester, Broward College Umesh Garg, University of Notre Dame Maurizio Giannotti, Barry University Bryan Gibbs, Dallas County Community College Lynn Gillette, Pima Community College–West Campus Mark Giroux, East Tennessee State University Matthew Griffiths, University of New Haven Alfonso Hinojosa, University of Texas–Arlington Steuard Jensen, Alma College David Kagan, University of Massachusetts Sergei Katsev, University of Minnesota–Duluth Gregory Lapicki, East Carolina University Jill Leggett, Florida State College–Jacksonville Alfredo Louro, University of Calgary James Maclaren, Tulane University Ponn Maheswaranathan, Winthrop University Seth Major, Hamilton College Oleg Maksimov, Excelsior College Aristides Marcano, Delaware State University James McDonald, University of Hartford Ralph McGrew, SUNY–Broome Community College Paul Miller, West Virginia University Tamar More, University of Portland Farzaneh Najmabadi, University of Phoenix Richard Olenick, The University of Dallas Christopher Porter, Ohio State University Liza Pujji, Manakau Institute of Technology Baishali Ray, Young Harris University Andrew Robinson, Carleton University Aruvana Roy, Young Harris University Gajendra Tulsian, Daytona State College Adria Updike, Roger Williams University Clark Vangilder, Central Arizona University Steven Wolf, Texas State University Alexander Wurm, Western New England University Lei Zhang, Winston Salem State University Ulrich Zurcher, Cleveland State University 4 Preface Access for free at openstax.org
INTRODUCTION CHAPTER 1 Temperature and Heat 1.1 Temperature and Thermal Equilibrium 1.2 Thermometers and Temperature Scales 1.3 Thermal Expansion 1.4 Heat Transfer, Specific Heat, and Calorimetry 1.5 Phase Changes 1.6 Mechanisms of Heat Transfer Heat and temperature are important concepts for each of us, every day. How we dress in the morning depends on whether the day is hot or cold, and most of what we do requires energy that ultimately comes from the Sun. The study of heat and temperature is part of an area of physics known as thermodynamics. The laws of thermodynamics govern the flow of energy throughout the universe. They are studied in all areas of science and engineering, from chemistry to biology to environmental science. In this chapter, we explore heat and temperature. It is not always easy to distinguish these terms. Heat is the flow of energy from one object to another. This flow of energy is caused by a difference in temperature. The transfer of heat can change temperature, as can work, another kind of energy transfer that is central to thermodynamics. We return to these basic ideas several times throughout the next four chapters, and you will see that they affect everything from the behavior of atoms and molecules to cooking to our weather on Earth to the life cycles of stars. FIGURE 1.1 These snowshoers on Mount Hood in Oregon are enjoying the heat flow and light caused by high temperature. All three mechanisms of heat transfer are relevant to this picture. The heat flowing out of the fire also turns the solid snow to liquid water and vapor. (credit: modification of work by “Mt. Hood Territory”/Flickr) CHAPTER OUTLINE
1.1 Temperature and Thermal Equilibrium LEARNING OBJECTIVES By the end of this section, you will be able to: • Define temperature and describe it qualitatively • Explain thermal equilibrium • Explain the zeroth law of thermodynamics Heat is familiar to all of us. We can feel heat entering our bodies from the summer Sun or from hot coffee or tea after a winter stroll. We can also feel heat leaving our bodies as we feel the chill of night or the cooling effect of sweat after exercise. What is heat? How do we define it and how is it related to temperature? What are the effects of heat and how does it flow from place to place? We will find that, in spite of the richness of the phenomena, a small set of underlying physical principles unites these subjects and ties them to other fields. We start by examining temperature and how to define and measure it. Temperature The concept of temperature has evolved from the common concepts of hot and cold. The scientific definition of temperature explains more than our senses of hot and cold. As you may have already learned, many physical quantities are defined solely in terms of how they are observed or measured, that is, they are defined operationally. Temperature is operationally defined as the quantity of what we measure with a thermometer. As we will see in detail in a later chapter on the kinetic theory of gases, temperature is proportional to the average kinetic energy of translation, a fact that provides a more physical definition. Differences in temperature maintain the transfer of heat, or heat transfer, throughout the universe. Heat transfer is the movement of energy from one place or material to another as a result of a difference in temperature. (You will learn more about heat transfer later in this chapter.) Thermal Equilibrium An important concept related to temperature is thermal equilibrium. Two objects are in thermal equilibrium if they are in close contact that allows either to gain energy from the other, but nevertheless, no net energy is transferred between them, or, even when not in contact, if no net energy is transferred between them when they are brought into contact. If two objects remain in contact for a long time, they typically come to equilibrium. In other words, two objects in thermal equilibrium do not exchange energy. Experimentally, if object A is in equilibrium with object B, and object B is in equilibrium with object C, then (as you may have already guessed) object A is in equilibrium with object C. That statement of transitivity is called the zeroth law of thermodynamics. (The number “zeroth” was suggested by British physicist Ralph Fowler in the 1930s. The first, second, and third laws of thermodynamics were already named and numbered then. The zeroth law had seldom been stated, but it needs to be discussed before the others, so Fowler gave it a smaller number.) Consider the case where A is a thermometer. The zeroth law tells us that if A reads a certain temperature when in equilibrium with B, and it is then placed in contact with C, it will not exchange energy with C; therefore, its temperature reading will remain the same (Figure 1.2). In other words, if two objects are in thermal equilibrium, they have the same temperature. FIGURE 1.2 If thermometer A is in thermal equilibrium with object B, and B is in thermal equilibrium with C, then A is in thermal equilibrium with C. Therefore, the reading on A stays the same when A is moved over to make contact with C. A thermometer measures its own temperature. It is through the concepts of thermal equilibrium and the zeroth law of thermodynamics that we can say that a thermometer measures the temperature of something else, and to make sense of the statement that two objects are at the same temperature. 6 1 • Temperature and Heat Access for free at openstax.org
In the rest of this chapter, we will often refer to “systems” instead of “objects.” As in the chapter on linear momentum and collisions, a system consists of one or more objects—but in thermodynamics, we require a system to be macroscopic, that is, to consist of a huge number (such as ) of molecules. Then we can say that a system is in thermal equilibrium with itself if all parts of it are at the same temperature. (We will return to the definition of a thermodynamic system in the chapter on the first law of thermodynamics.) 1.2 Thermometers and Temperature Scales LEARNING OBJECTIVES By the end of this section, you will be able to: • Describe several different types of thermometers • Convert temperatures between the Celsius, Fahrenheit, and Kelvin scales Any physical property that depends consistently and reproducibly on temperature can be used as the basis of a thermometer. For example, volume increases with temperature for most substances. This property is the basis for the common alcohol thermometer and the original mercury thermometers. Other properties used to measure temperature include electrical resistance, color, and the emission of infrared radiation (Figure 1.3). FIGURE 1.3 Because many physical properties depend on temperature, the variety of thermometers is remarkable. (a) In this common type of thermometer, the alcohol, containing a red dye, expands more rapidly than the glass encasing it. When the thermometer’s temperature increases, the liquid from the bulb is forced into the narrow tube, producing a large change in the length of the column for a small change in temperature. (b) Each of the six squares on this plastic (liquid crystal) thermometer contains a film of a different heat-sensitive liquid crystal material. Below , all six squares are black. When the plastic thermometer is exposed to a temperature of , the first liquid crystal square changes color. When the temperature reaches above , the second liquid crystal square also changes color, and so forth. (c) A firefighter uses a pyrometer to check the temperature of an aircraft carrier’s ventilation system. The pyrometer measures infrared radiation (whose emission varies with temperature) from the vent and quickly produces a temperature readout. Infrared thermometers are also frequently used to measure body temperature by gently placing them in the ear canal. Such thermometers are more accurate than the alcohol thermometers placed under the tongue or in the armpit. (credit b: modification of work by Tess Watson; credit c: modification of work by Lamel J. Hinton, U.S. Navy) Thermometers measure temperature according to well-defined scales of measurement. The three most common temperature scales are Fahrenheit, Celsius, and Kelvin. Temperature scales are created by identifying two reproducible temperatures. The freezing and boiling temperatures of water at standard atmospheric pressure are commonly used. 1.2 • Thermometers and Temperature Scales 7
On the Celsius scale, the freezing point of water is and the boiling point is The unit of temperature on this scale is the degree Celsius . The Fahrenheit scale (still the most frequently used for common purposes in the United States) has the freezing point of water at and the boiling point at Its unit is the degree Fahrenheit ( ). You can see that 100 Celsius degrees span the same range as 180 Fahrenheit degrees. Thus, a temperature difference of one degree on the Celsius scale is 1.8 times as large as a difference of one degree on the Fahrenheit scale, or The definition of temperature in terms of molecular motion suggests that there should be a lowest possible temperature, where the average kinetic energy of molecules is zero (or the minimum allowed by quantum mechanics). Experiments confirm the existence of such a temperature, called absolute zero. An absolute temperature scale is one whose zero point is absolute zero. Such scales are convenient in science because several physical quantities, such as the volume of an ideal gas, are directly related to absolute temperature. The Kelvin scale is the absolute temperature scale that is commonly used in science. The SI temperature unit is thekelvin, which is abbreviated K (not accompanied by a degree sign). Thus 0 K is absolute zero. The freezing and boiling points of water are 273.15 K and 373.15 K, respectively. Therefore, temperature differences are the same in units of kelvins and degrees Celsius, or The relationships between the three common temperature scales are shown in Figure 1.4. Temperatures on these scales can be converted using the equations in Table 1.1. FIGURE 1.4 Relationships between the Fahrenheit, Celsius, and Kelvin temperature scales are shown. The relative sizes of the scales are also shown. To convert from… Use this equation… Celsius to Fahrenheit Fahrenheit to Celsius Celsius to Kelvin Kelvin to Celsius Fahrenheit to Kelvin Kelvin to Fahrenheit TABLE 1.1 Temperature Conversions To convert between Fahrenheit and Kelvin, convert to Celsius as an intermediate step. 8 1 • Temperature and Heat Access for free at openstax.org
EXAMPLE 1.1 Converting between Temperature Scales: Room Temperature “Room temperature” is generally defined in physics to be . (a) What is room temperature in ? (b) What is it in K? Strategy To answer these questions, all we need to do is choose the correct conversion equations and substitute the known values. Solution To convert from to , use the equation Substitute the known value into the equation and solve: Similarly, we find that . The Kelvin scale is part of the SI system of units, so its actual definition is more complicated than the one given above. First, it is not defined in terms of the freezing and boiling points of water, but in terms of the triple point. The triple point is the unique combination of temperature and pressure at which ice, liquid water, and water vapor can coexist stably. As will be discussed in the section on phase changes, the coexistence is achieved by lowering the pressure and consequently the boiling point to reach the freezing point. The triple-point temperature is defined as 273.16 K. This definition has the advantage that although the freezing temperature and boiling temperature of water depend on pressure, there is only one triple-point temperature. Second, even with two points on the scale defined, different thermometers give somewhat different results for other temperatures. Therefore, a standard thermometer is required. Metrologists (experts in the science of measurement) have chosen the constant-volume gas thermometer for this purpose. A vessel of constant volume filled with gas is subjected to temperature changes, and the measured temperature is proportional to the change in pressure. Using “TP” to represent the triple point, The results depend somewhat on the choice of gas, but the less dense the gas in the bulb, the better the results for different gases agree. If the results are extrapolated to zero density, the results agree quite well, with zero pressure corresponding to a temperature of absolute zero. Constant-volume gas thermometers are big and come to equilibrium slowly, so they are used mostly as standards to calibrate other thermometers. INTERACTIVE Visit this site (https://openstax.org/l/21consvolgasth) to learn more about the constant-volume gas thermometer. 1.3 Thermal Expansion LEARNING OBJECTIVES By the end of this section, you will be able to: • Answer qualitative questions about the effects of thermal expansion • Solve problems involving thermal expansion, including those involving thermal stress 1.3 • Thermal Expansion 9
The expansion of alcohol in a thermometer is one of many commonly encountered examples of thermal expansion, which is the change in size or volume of a given system as its temperature changes. The most visible example is the expansion of hot air. When air is heated, it expands and becomes less dense than the surrounding air, which then exerts an (upward) force on the hot air and makes steam and smoke rise, hot air balloons float, and so forth. The same behavior happens in all liquids and gases, driving natural heat transfer upward in homes, oceans, and weather systems, as we will discuss in an upcoming section. Solids also undergo thermal expansion. Railroad tracks and bridges, for example, have expansion joints to allow them to freely expand and contract with temperature changes, as shown in Figure 1.5. FIGURE 1.5 (a) Thermal expansion joints like these in the (b) Trojan Bridge in Prague, Czech Republic allow bridges to change length without buckling. (credit: modification of works by “ŠJů”/Wikimedia Commons) What is the underlying cause of thermal expansion? As previously mentioned, an increase in temperature means an increase in the kinetic energy of individual atoms. In a solid, unlike in a gas, the molecules are held in place by forces from neighboring molecules; as we saw in Oscillations (http://openstax.org/books/university-physics- volume-1/pages/15-introduction), the forces can be modeled as in harmonic springs described by the Lennard- Jones potential. Energy in Simple Harmonic Motion (http://openstax.org/books/university-physics-volume-1/pages/ 15-2-energy-in-simple-harmonic-motion) shows that such potentials are asymmetrical in that the potential energy increases more steeply when the molecules get closer to each other than when they get farther away. Thus, at a given kinetic energy, the distance moved is greater when neighbors move away from each other than when they move toward each other. The result is that increased kinetic energy (increased temperature) increases the average distance between molecules—the substance expands. For most substances under ordinary conditions, it is an excellent approximation that there is no preferred direction (that is, the solid is “isotropic”), and an increase in temperature increases the solid’s size by a certain fraction in each dimension. Therefore, if the solid is free to expand or contract, its proportions stay the same; only its overall size changes. LINEAR THERMAL EXPANSION According to experiments, the dependence of thermal expansion on temperature, substance, and original initial length is summarized in the equation where is the instantaneous change in length per temperature, L is the length, and is the coefficient of 1.1 10 1 • Temperature and Heat Access for free at openstax.org
Table 1.2 lists representative values of the coefficient of linear expansion. As noted earlier, is the same whether it is expressed in units of degrees Celsius or kelvins; thus, may have units of or 1/K with the same value in either case. Approximating as a constant is quite accurate for small changes in temperature and sufficient for most practical purposes, even for large changes in temperature. We examine this approximation more closely in the next example. Material Coefficient of Linear Expansion Coefficient of Volume Expansion Solids Aluminum Brass Copper Gold Iron or steel Invar (nickel-iron alloy) Lead Silver Glass (ordinary) Glass (Pyrex®) Quartz Concrete, brick (approximate) Marble (average) Liquids Ether Ethyl alcohol Gasoline TABLE 1.2 Thermal Expansion Coefficients linear expansion, a material property that varies slightly with temperature. As is nearly constant and also very small, for practical purposes, we use the linear approximation: where is the change in length and is the change in temperature. 1.2 1.3 • Thermal Expansion 11
Material Coefficient of Linear Expansion Coefficient of Volume Expansion Glycerin Mercury Water Gases Air and most other gases at atmospheric pressure TABLE 1.2 Thermal Expansion Coefficients Thermal expansion is exploited in the bimetallic strip (Figure 1.6). This device can be used as a thermometer if the curving strip is attached to a pointer on a scale. It can also be used to automatically close or open a switch at a certain temperature, as in older or analog thermostats. FIGURE 1.6 The curvature of a bimetallic strip depends on temperature. (a) The strip is straight at the starting temperature, where its two components have the same length. (b) At a higher temperature, this strip bends to the right, because the metal on the left has expanded more than the metal on the right. At a lower temperature, the strip would bend to the left. EXAMPLE 1.2 Calculating Linear Thermal Expansion The main span of San Francisco’s Golden Gate Bridge is 1275 m long at its coldest. The bridge is exposed to temperatures ranging from to . What is its change in length between these temperatures? Assume that the bridge is made entirely of steel. Strategy Use the equation for linear thermal expansion to calculate the change in length, . Use the coefficient of linear expansion for steel from Table 1.2, and note that the change in temperature is Solution Substitute all of the known values into the equation to solve for : Significance Although not large compared with the length of the bridge, this change in length is observable. It is generally spread 12 1 • Temperature and Heat Access for free at openstax.org
over many expansion joints so that the expansion at each joint is small. Thermal Expansion in Two and Three Dimensions Unconstrained objects expand in all dimensions, as illustrated in Figure 1.7. That is, their areas and volumes, as well as their lengths, increase with temperature. Because the proportions stay the same, holes and container volumes also get larger with temperature. If you cut a hole in a metal plate, the remaining material will expand exactly as it would if the piece you removed were still in place. The piece would get bigger, so the hole must get bigger too. FIGURE 1.7 In general, objects expand in all directions as temperature increases. In these drawings, the original boundaries of the objects are shown with solid lines, and the expanded boundaries with dashed lines. (a) Area increases because both length and width increase. The area of a circular plug also increases. (b) If the plug is removed, the hole it leaves becomes larger with increasing temperature, just as if the expanding plug were still in place. (c) Volume also increases, because all three dimensions increase. Volume expansion is defined for liquids, but linear and area expansion are not, as a liquid’s changes in linear dimensions and area depend on the shape of its container. Thus, Table 1.2 shows liquids’ values of but not . In general, objects expand with increasing temperature. Water is the most important exception to this rule. Water does expand with increasing temperature (its density decreases) at temperatures greater than . However, it is densest at and expands with decreasing temperature between and ( ), as shown in Figure 1.8. A striking effect of this phenomenon is the freezing of water in a pond. When water near the surface cools down to it is denser than the remaining water and thus sinks to the bottom. This “turnover” leaves a layer of warmer water near the surface, which is then cooled. However, if the temperature in the surface layer drops below , that water is less dense than the water below, and thus stays near the top. As a result, the pond surface can freeze over. The layer of ice insulates the liquid water below it from low air THERMAL EXPANSION IN TWO DIMENSIONS For small temperature changes, the change in area is given by where is the change in area is the change in temperature, and is the coefficient of linear expansion, which varies slightly with temperature. (The derivation of this equation is analogous to that of the more important equation for three dimensions, below.) 1.3 THERMAL EXPANSION IN THREE DIMENSIONS The relationship between volume and temperature is given by , where is the coefficient of volume expansion. As you can show in Exercise 1.60, . This equation is usually written as Note that the values of in Table 1.2 are equal to except for rounding. 1.4 1.3 • Thermal Expansion 13
temperatures. Fish and other aquatic life can survive in water beneath ice, due to this unusual characteristic of water. FIGURE 1.8 This curve shows the density of water as a function of temperature. Note that the thermal expansion at low temperatures is very small. The maximum density at is only greater than the density at , and greater than that at . The decrease of density below occurs because the liquid water approachs the solid crystal form of ice, which contains more empty space than the liquid. EXAMPLE 1.3 Calculating Thermal Expansion Suppose your 60.0-L -gal) steel gasoline tank is full of gas that is cool because it has just been pumped from an underground reservoir. Now, both the tank and the gasoline have a temperature of How much gasoline has spilled by the time they warm to ? Strategy The tank and gasoline increase in volume, but the gasoline increases more, so the amount spilled is the difference in their volume changes. We can use the equation for volume expansion to calculate the change in volume of the gasoline and of the tank. (The gasoline tank can be treated as solid steel.) Solution 1. Use the equation for volume expansion to calculate the increase in volume of the steel tank: 2. The increase in volume of the gasoline is given by this equation: 3. Find the difference in volume to determine the amount spilled as Alternatively, we can combine these three equations into a single equation. (Note that the original volumes are equal.) Significance This amount is significant, particularly for a 60.0-L tank. The effect is so striking because the gasoline and steel expand quickly. The rate of change in thermal properties is discussed later in this chapter. 14 1 • Temperature and Heat Access for free at openstax.org
If you try to cap the tank tightly to prevent overflow, you will find that it leaks anyway, either around the cap or by bursting the tank. Tightly constricting the expanding gas is equivalent to compressing it, and both liquids and solids resist compression with extremely large forces. To avoid rupturing rigid containers, these containers have air gaps, which allow them to expand and contract without stressing them. CHECK YOUR UNDERSTANDING 1.1 Does a given reading on a gasoline gauge indicate more gasoline in cold weather or in hot weather, or does the temperature not matter? Thermal Stress If you change the temperature of an object while preventing it from expanding or contracting, the object is subjected to stress that is compressive if the object would expand in the absence of constraint and tensile if it would contract. This stress resulting from temperature changes is known as thermal stress. It can be quite large and can cause damage. To avoid this stress, engineers may design components so they can expand and contract freely. For instance, in highways, gaps are deliberately left between blocks to prevent thermal stress from developing. When no gaps can be left, engineers must consider thermal stress in their designs. Thus, the reinforcing rods in concrete are made of steel because steel’s coefficient of linear expansion is nearly equal to that of concrete. To calculate the thermal stress in a rod whose ends are both fixed rigidly, we can think of the stress as developing in two steps. First, let the ends be free to expand (or contract) and find the expansion (or contraction). Second, find the stress necessary to compress (or extend) the rod to its original length by the methods you studied in Static Equilibrium and Elasticity (http://openstax.org/books/university-physics-volume-1/pages/12-introduction) on static equilibrium and elasticity. In other words, the of the thermal expansion equals the of the elastic distortion (except that the signs are opposite). EXAMPLE 1.4 Calculating Thermal Stress Concrete blocks are laid out next to each other on a highway without any space between them, so they cannot expand. The construction crew did the work on a winter day when the temperature was . Find the stress in the blocks on a hot summer day when the temperature is . The compressive Young’s modulus of concrete is . Strategy According to the chapter on static equilibrium and elasticity, the stress F/A is given by where Y is the Young’s modulus of the material—concrete, in this case. In thermal expansion, We combine these two equations by noting that the two are equal, as stated above. Because we are not given or A, we can obtain a numerical answer only if they both cancel out. Solution We substitute the thermal-expansion equation into the elasticity equation to get and as we hoped, has canceled and A appears only in F/A, the notation for the quantity we are calculating. 1.3 • Thermal Expansion 15
Now we need only insert the numbers: Significance The ultimate compressive strength of concrete is so the blocks are unlikely to break. However, the ultimate shear strength of concrete is only so some might chip off. CHECK YOUR UNDERSTANDING 1.2 Two objects A and B have the same dimensions and are constrained identically. A is made of a material with a higher thermal expansion coefficient than B. If the objects are heated identically, will A feel a greater stress than B? 1.4 Heat Transfer, Specific Heat, and Calorimetry LEARNING OBJECTIVES By the end of this section, you will be able to: • Explain phenomena involving heat as a form of energy transfer • Solve problems involving heat transfer We have seen in previous chapters that energy is one of the fundamental concepts of physics. Heat is a type of energy transfer that is caused by a temperature difference, and it can change the temperature of an object. As we learned earlier in this chapter, heat transfer is the movement of energy from one place or material to another as a result of a difference in temperature. Heat transfer is fundamental to such everyday activities as home heating and cooking, as well as many industrial processes. It also forms a basis for the topics in the remainder of this chapter. We also introduce the concept of internal energy, which can be increased or decreased by heat transfer. We discuss another way to change the internal energy of a system, namely doing work on it. Thus, we are beginning the study of the relationship of heat and work, which is the basis of engines and refrigerators and the central topic (and origin of the name) of thermodynamics. Internal Energy and Heat A thermal system has internal energy, which is the sum of the microscopic energies of the system. This includes thermal energy, which is associated with the mechanical energies of its molecules and which is proportional to the system’s temperature. As we saw earlier in this chapter, if two objects at different temperatures are brought into contact with each other, energy is transferred from the hotter to the colder object until the bodies reach thermal equilibrium (that is, they are at the same temperature). No work is done by either object because no force acts through a distance (as we discussed in Work and Kinetic Energy (http://openstax.org/books/university-physics- volume-1/pages/7-introduction)). These observations reveal that heat is energy transferred spontaneously due to a temperature difference. Figure 1.9 shows an example of heat transfer. FIGURE 1.9 (a) Here, the soft drink has a higher temperature than the ice, so they are not in thermal equilibrium. (b) When the soft drink and ice are allowed to interact, heat is transferred from the drink to the ice due to the difference in temperatures until they reach the same 16 1 • Temperature and Heat Access for free at openstax.org
temperature, , achieving equilibrium. In fact, since the soft drink and ice are both in contact with the surrounding air and the bench, the ultimate equilibrium temperature will be the same as that of the surroundings. The meaning of “heat” in physics is different from its ordinary meaning. For example, in conversation, we may say “the heat was unbearable,” but in physics, we would say that the temperature was high. Heat is a form of energy flow, whereas temperature is not. Incidentally, humans are sensitive to heat flow rather than to temperature. Since heat is a form of energy, its SI unit is the joule (J). Another common unit of energy often used for heat is the calorie (cal), defined as the energy needed to change the temperature of 1.00 g of water by —specifically, between and , since there is a slight temperature dependence. Also commonly used is the kilocalorie (kcal), which is the energy needed to change the temperature of 1.00 kg of water by . Since mass is most often specified in kilograms, the kilocalorie is convenient. Confusingly, food calories (sometimes called “big calories,” abbreviated Cal) are actually kilocalories, a fact not easily determined from package labeling. Mechanical Equivalent of Heat It is also possible to change the temperature of a substance by doing work, which transfers energy into or out of a system. This realization helped establish that heat is a form of energy. James Prescott Joule (1818–1889) performed many experiments to establish the mechanical equivalent of heat—the work needed to produce the same effects as heat transfer. In the units used for these two quantities, the value for this equivalence is We consider this equation to represent the conversion between two units of energy. (Other numbers that you may see refer to calories defined for temperature ranges other than to .) Figure 1.10 shows one of Joule’s most famous experimental setups for demonstrating that work and heat can produce the same effects and measuring the mechanical equivalent of heat. It helped establish the principle of conservation of energy. Gravitational potential energy ( U) was converted into kinetic energy (K), and then randomized by viscosity and turbulence into increased average kinetic energy of atoms and molecules in the system, producing a temperature increase. Joule’s contributions to thermodynamics were so significant that the SI unit of energy was named after him. FIGURE 1.10 Joule’s experiment established the equivalence of heat and work. As the masses descended, they caused the paddles to do work, , on the water. The result was a temperature increase, , measured by the thermometer. Joule found that was proportional to W and thus determined the mechanical equivalent of heat. Increasing internal energy by heat transfer gives the same result as increasing it by doing work. Therefore, although a system has a well-defined internal energy, we cannot say that it has a certain “heat content” or “work content.” A well-defined quantity that depends only on the current state of the system, rather than on the history of that system, 1.4 • Heat Transfer, Specific Heat, and Calorimetry 17
is known as a state variable. Temperature and internal energy are state variables. To sum up this paragraph, heat and work are not state variables. Incidentally, increasing the internal energy of a system does not necessarily increase its temperature. As we’ll see in the next section, the temperature does not change when a substance changes from one phase to another. An example is the melting of ice, which can be accomplished by adding heat or by doing frictional work, as when an ice cube is rubbed against a rough surface. Temperature Change and Heat Capacity We have noted that heat transfer often causes temperature change. Experiments show that with no phase change and no work done on or by the system, the transferred heat is typically directly proportional to the change in temperature and to the mass of the system, to a good approximation. (Below we show how to handle situations where the approximation is not valid.) The constant of proportionality depends on the substance and its phase, which may be gas, liquid, or solid. We omit discussion of the fourth phase, plasma, because although it is the most common phase in the universe, it is rare and short-lived on Earth. We can understand the experimental facts by noting that the transferred heat is the change in the internal energy, which is the total energy of the molecules. Under typical conditions, the total kinetic energy of the molecules is a constant fraction of the internal energy (for reasons and with exceptions that we’ll see in the next chapter). The average kinetic energy of a molecule is proportional to the absolute temperature. Therefore, the change in internal energy of a system is typically proportional to the change in temperature and to the number of molecules,N. Mathematically, The dependence on the substance results in large part from the different masses of atoms and molecules. We are considering its heat capacity in terms of its mass, but as we will see in the next chapter, in some cases, heat capacities per molecule are similar for different substances. The dependence on substance and phase also results from differences in the potential energy associated with interactions between atoms and molecules. Values of specific heat must generally be measured, because there is no simple way to calculate them precisely. Table 1.3 lists representative values of specific heat for various substances. We see from this table that the specific heat of water is five times that of glass and 10 times that of iron, which means that it takes five times as much heat to raise the temperature of water a given amount as for glass, and 10 times as much as for iron. In fact, water has one of the largest specific heats of any material, which is important for sustaining life on Earth. The specific heats of gases depend on what is maintained constant during the heating—typically either the volume or the pressure. In the table, the first specific heat value for each gas is measured at constant volume, and the second (in parentheses) is measured at constant pressure. We will return to this topic in the chapter on the kinetic theory of gases. HEAT TRANSFER AND TEMPERATURE CHANGE A practical approximation for the relationship between heat transfer and temperature change is: where Q is the symbol for heat transfer (“quantity of heat”), m is the mass of the substance, and is the change in temperature. The symbol c stands for the specific heat (also called “specific heat capacity”) and depends on the material and phase. In the SI system, the specific heat is numerically equal to the amount of heat necessary to change the temperature of kg of mass by . The SI unit for specific heat is or . (Recall that the temperature change is the same in units of kelvin and degrees Celsius.) 1.5 18 1 • Temperature and Heat Access for free at openstax.org
Substances Specific Heat ( c) Solids Aluminum 900 0.215 Asbestos 800 0.19 Concrete, granite (average) 840 0.20 Copper 387 0.0924 Glass 840 0.20 Gold 129 0.0308 Human body (average at ) 3500 0.83 Ice (average, ) 2090 0.50 Iron, steel 452 0.108 Lead 128 0.0305 Silver 235 0.0562 Wood 1700 0.40 Liquids Benzene 1740 0.415 Ethanol 2450 0.586 Glycerin 2410 0.576 Mercury 139 0.0333 Water 4186 1.000 Gases[3] Air (dry) 721 (1015) 0.172 (0.242) Ammonia 1670 (2190) 0.399 (0.523) Carbon dioxide 638 (833) 0.152 (0.199) TABLE 1.3 Specific Heats of Various Substances[1] [1]The values for solids and liquids are at constant volume and , except as noted. [2]These values are identical in units of [3]Specific heats at constant volume and at except as noted, and at 1.00 atm pressure. Values in parentheses are specific heats at a constant pressure of 1.00 atm. 1.4 • Heat Transfer, Specific Heat, and Calorimetry 19
Substances Specific Heat ( c) Nitrogen 739 (1040) 0.177 (0.248) Oxygen 651 (913) 0.156 (0.218) Steam 1520 (2020) 0.363 (0.482) TABLE 1.3 Specific Heats of Various Substances[1] [1]The values for solids and liquids are at constant volume and , except as noted. [2]These values are identical in units of [3]Specific heats at constant volume and at except as noted, and at 1.00 atm pressure. Values in parentheses are specific heats at a constant pressure of 1.00 atm. In general, specific heat also depends on temperature. Thus, a precise definition of c for a substance must be given in terms of an infinitesimal change in temperature. To do this, we note that and replace with d: Except for gases, the temperature and volume dependence of the specific heat of most substances is weak at normal temperatures. Therefore, we will generally take specific heats to be constant at the values given in the table. EXAMPLE 1.5 Calculating the Required Heat A 0.500-kg aluminum pan on a stove and 0.250 L of water in it are heated from to . (a) How much heat is required? What percentage of the heat is used to raise the temperature of (b) the pan and (c) the water? Strategy We can assume that the pan and the water are always at the same temperature. When you put the pan on the stove, the temperature of the water and that of the pan are increased by the same amount. We use the equation for the heat transfer for the given temperature change and mass of water and aluminum. The specific heat values for water and aluminum are given in Table 1.3. Solution 1. Calculate the temperature difference: 2. Calculate the mass of water. Because the density of water is , 1 L of water has a mass of 1 kg, and the mass of 0.250 L of water is . 3. Calculate the heat transferred to the water. Use the specific heat of water in Table 1.3: 4. Calculate the heat transferred to the aluminum. Use the specific heat for aluminum in Table 1.3: 5. Find the total transferred heat: Significance In this example, the heat transferred to the water is more than the aluminum pan. Although the mass of the pan is twice that of the water, the specific heat of water is over four times that of aluminum. Therefore, it takes a bit more than twice as much heat to achieve the given temperature change for the water as for the aluminum pan. Example 1.6 illustrates a temperature rise caused by doing work. (The result is the same as if the same amount of energy had been added with a blowtorch instead of mechanically.) 20 1 • Temperature and Heat Access for free at openstax.org
EXAMPLE 1.6 Calculating the Temperature Increase from the Work Done on a Substance Truck brakes used to control speed on a downhill run do work, converting gravitational potential energy into increased internal energy (higher temperature) of the brake material (Figure 1.11). This conversion prevents the gravitational potential energy from being converted into kinetic energy of the truck. Since the mass of the truck is much greater than that of the brake material absorbing the energy, the temperature increase may occur too fast for sufficient heat to transfer from the brakes to the environment; in other words, the brakes may overheat. FIGURE 1.11 The smoking brakes on a braking truck are visible evidence of the mechanical equivalent of heat. Calculate the temperature increase of 10 kg of brake material with an average specific heat of if the material retains 10% of the energy from a 10,000-kg truck descending 75.0 m (in vertical displacement) at a constant speed. Strategy We calculate the gravitational potential energy ( Mgh) that the entire truck loses in its descent, equate it to the increase in the brakes’ internal energy, and then find the temperature increase produced in the brake material alone. Solution First we calculate the change in gravitational potential energy as the truck goes downhill: Because the kinetic energy of the truck does not change, conservation of energy tells us the lost potential energy is dissipated, and we assume that 10% of it is transferred to internal energy of the brakes, so take . Then we calculate the temperature change from the heat transferred, using where m is the mass of the brake material. Insert the given values to find Significance If the truck had been traveling for some time, then just before the descent, the brake temperature would probably be higher than the ambient temperature. The temperature increase in the descent would likely raise the temperature of the brake material very high, so this technique is not practical. Instead, the truck would use the technique of engine braking. A different idea underlies the recent technology of hybrid and electric cars, where 1.4 • Heat Transfer, Specific Heat, and Calorimetry 21
mechanical energy (kinetic and gravitational potential energy) is converted by the brakes into electrical energy in the battery, a process called regenerative braking. In a common kind of problem, objects at different temperatures are placed in contact with each other but isolated from everything else, and they are allowed to come into equilibrium. A container that prevents heat transfer in or out is called a calorimeter, and the use of a calorimeter to make measurements (typically of heat or specific heat capacity) is called calorimetry. We will use the term “calorimetry problem” to refer to any problem in which the objects concerned are thermally isolated from their surroundings. An important idea in solving calorimetry problems is that during a heat transfer between objects isolated from their surroundings, the heat gained by the colder object must equal the heat lost by the hotter object, due to conservation of energy: We express this idea by writing that the sum of the heats equals zero because the heat gained is usually considered positive; the heat lost, negative. EXAMPLE 1.7 Calculating the Final Temperature in Calorimetry Suppose you pour 0.250 kg of water (about a cup) into a 0.500-kg aluminum pan off the stove with a temperature of . Assume no heat transfer takes place to anything else: The pan is placed on an insulated pad, and heat transfer to the air is neglected in the short time needed to reach equilibrium. Thus, this is a calorimetry problem, even though no isolating container is specified. Also assume that a negligible amount of water boils off. What is the temperature when the water and pan reach thermal equilibrium? Strategy Originally, the pan and water are not in thermal equilibrium: The pan is at a higher temperature than the water. Heat transfer restores thermal equilibrium once the water and pan are in contact; it stops once thermal equilibrium between the pan and the water is achieved. The heat lost by the pan is equal to the heat gained by the water—that is the basic principle of calorimetry. Solution 1. Use the equation for heat transfer to express the heat transferred from the pan in terms of the mass of the pan, the specific heat of aluminum, the initial temperature of the pan, and the final temperature: 2. Express the heat gained by the water in terms of the mass of the water, the specific heat of water, the initial temperature of the water, and the final temperature: 3. Note that and and that as stated above, they must sum to zero: 4. Bring all terms involving on the left hand side and all other terms on the right hand side. Solving for and insert the numerical values: 1.6 22 1 • Temperature and Heat Access for free at openstax.org
Significance Why is the final temperature so much closer to than to ? The reason is that water has a greater specific heat than most common substances and thus undergoes a smaller temperature change for a given heat transfer. A large body of water, such as a lake, requires a large amount of heat to increase its temperature appreciably. This explains why the temperature of a lake stays relatively constant during the day even when the temperature change of the air is large. However, the water temperature does change over longer times (e.g., summer to winter). CHECK YOUR UNDERSTANDING 1.3 If 25 kJ is necessary to raise the temperature of a rock from how much heat is necessary to heat the rock from ? EXAMPLE 1.8 Temperature-Dependent Heat Capacity At low temperatures, the specific heats of solids are typically proportional to . The first understanding of this behavior was due to the Dutch physicist Peter Debye, who in 1912, treated atomic oscillations with the quantum theory that Max Planck had recently used for radiation. For instance, a good approximation for the specific heat of salt, NaCl, is The constant 321 K is called the Debye temperature of NaCl, and the formula works well when Using this formula, how much heat is required to raise the temperature of 24.0 g of NaCl from 5 K to 15 K? Solution Because the heat capacity depends on the temperature, we need to use the equation We solve this equation for Q by integrating both sides: Then we substitute the given values in and evaluate the integral: Significance If we had used the equation and the room-temperature specific heat of salt, we would have gotten a very different value. 1.5 Phase Changes LEARNING OBJECTIVES By the end of this section, you will be able to: • Describe phase transitions and equilibrium between phases • Solve problems involving latent heat • Solve calorimetry problems involving phase changes Phase transitions play an important theoretical and practical role in the study of heat flow. In melting (or “fusion”), a solid turns into a liquid; the opposite process is freezing. In evaporation, a liquid turns into a gas; the opposite process is condensation. 1.5 • Phase Changes 23
A substance melts or freezes at a temperature called its melting point, and boils (evaporates rapidly) or condenses at its boiling point. These temperatures depend on pressure. High pressure favors the denser form, so typically, high pressure raises the melting point and boiling point, and low pressure lowers them. For example, the boiling point of water is at 1.00 atm. At higher pressure, the boiling point is higher, and at lower pressure, it is lower. The main exception is the melting and freezing of water, discussed in the next section. Phase Diagrams The phase of a given substance depends on the pressure and temperature. Thus, plots of pressure versus temperature showing the phase in each region provide considerable insight into thermal properties of substances. Such a pT graph is called a phase diagram. Figure 1.12 shows the phase diagram for water. Using the graph, if you know the pressure and temperature, you can determine the phase of water. The solid curves—boundaries between phases—indicate phase transitions, that is, temperatures and pressures at which the phases coexist. For example, the boiling point of water is at 1.00 atm. As the pressure increases, the boiling temperature rises gradually to at a pressure of 218 atm. A pressure cooker (or even a covered pot) cooks food faster than an open pot, because the water can exist as a liquid at temperatures greater than without all boiling away. (As we’ll see in the next section, liquid water conducts heat better than steam or hot air.) The boiling point curve ends at a certain point called the critical point—that is, a critical temperature, above which the liquid and gas phases cannot be distinguished; the substance is called asupercritical fluid. At sufficiently high pressure above the critical point, the gas has the density of a liquid but does not condense. Carbon dioxide, for example, is supercritical at all temperatures above . Critical pressure is the pressure of the critical point. FIGURE 1.12 The phase diagram (pT graph) for water shows solid (s), liquid (l), and vapor (v) phases. At temperatures and pressure above those of the critical point, there is no distinction between liquid and vapor. Note that the axes are nonlinear and the graph is not to scale. This graph is simplified—it omits several exotic phases of ice at higher pressures. The phase diagram of water is unusual because the melting-point curve has a negative slope, showing that you can melt ice by increasing the pressure. Similarly, the curve between the solid and liquid regions in Figure 1.12 gives the melting temperature at various pressures. For example, the melting point is at 1.00 atm, as expected. Water has the unusual property that ice is less dense than liquid water at the melting point, so at a fixed temperature, you can change the phase from solid (ice) to liquid (water) by increasing the pressure. That is, the melting temperature of ice falls with increased pressure, as the phase diagram shows. For example, when a car is driven over snow, the increased pressure from the tires melts the snowflakes; afterwards, the water refreezes and forms an ice layer. As you learned in the earlier section on thermometers and temperature scales, the triple point is the combination of temperature and pressure at which ice, liquid water, and water vapor can coexist stably—that is, all three phases exist in equilibrium. For water, the triple point occurs at and 611.2 Pa; that is a more accurate 24 1 • Temperature and Heat Access for free at openstax.org
calibration temperature than the melting point of water at 1.00 atm, or . INTERACTIVE View this video (https://openstax.org/l/21triplepoint) to see a substance at its triple point. At pressures below that of the triple point, there is no liquid phase; the substance can exist as either gas or solid. For water, there is no liquid phase at pressures below 0.00600 atm. The phase change from solid to gas is called sublimation. You may have noticed that snow can disappear into thin air without a trace of liquid water, or that ice cubes can disappear in a freezer. Both are examples of sublimation. The reverse also happens: Frost can form on very cold windows without going through the liquid stage. Figure 1.13 shows the result, as well as showing a familiar example of sublimation. Carbon dioxide has no liquid phase at atmospheric pressure. Solid is known as dry ice because instead of melting, it sublimes. Its sublimation temperature at atmospheric pressure is . Certain air fresheners use the sublimation of a solid to spread a perfume around a room. Some solids, such as osmium tetroxide, are so toxic that they must be kept in sealed containers to prevent human exposure to their sublimation- produced vapors. FIGURE 1.13 Direct transitions between solid and vapor are common, sometimes useful, and even beautiful. (a) Dry ice sublimes directly to carbon dioxide gas. The visible “smoke” consists of water droplets that condensed in the air cooled by the dry ice. (b) Frost forms patterns on a very cold window, an example of a solid formed directly from a vapor. (credit a: modification of work by Windell Oskay; credit b: modification of work by Liz West) Equilibrium At the melting temperature, the solid and liquid phases are in equilibrium. If heat is added, some of the solid will melt, and if heat is removed, some of the liquid will freeze. The situation is somewhat more complex for liquid-gas equilibrium. Generally, liquid and gas are in equilibrium at any temperature. We call the gas phase a vapor when it exists at a temperature below the critical temperature, as it does for water at . Liquid in a closed container at a fixed temperature evaporates until the pressure of the gas reaches a certain value, called the vapor pressure, which depends on the gas and the temperature. At this equilibrium, if heat is added, some of the liquid will evaporate, and if heat is removed, some of the gas will condense; molecules either join the liquid or form suspended droplets. If there is not enough liquid for the gas to reach the vapor pressure in the container, all the liquid eventually evaporates. If the vapor pressure of the liquid is greater than the total ambient pressure, including that of any air (or other gas), the liquid evaporates rapidly; in other words, it boils. Thus, the boiling point of a liquid at a given pressure is the temperature at which its vapor pressure equals the ambient pressure. Liquid and gas phases are in equilibrium at the boiling temperature (Figure 1.14). If a substance is in a closed container at the boiling point, then the liquid is boiling and the gas is condensing at the same rate without net change in their amounts. 1.5 • Phase Changes 25
FIGURE 1.14 Equilibrium between liquid and gas at two different boiling points inside a closed container. (a) The rates of boiling and condensation are equal at this combination of temperature and pressure, so the liquid and gas phases are in equilibrium. (b) At a higher temperature, the boiling rate is faster, that is, the rate at which molecules leave the liquid and enter the gas is faster. This increases the number of molecules in the gas, which increases the gas pressure, which in turn increases the rate at which gas molecules condense and enter the liquid. The pressure stops increasing when it reaches the point where the boiling rate and the condensation rate are equal. The gas and liquid are in equilibrium again at this higher temperature and pressure. For water, is the boiling point at 1.00 atm, so water and steam should exist in equilibrium under these conditions. Why does an open pot of water at boil completely away? The gas surrounding an open pot is not pure water: it is mixed with air. If pure water and steam are in a closed container at and 1.00 atm, they will coexist—but with air over the pot, there are fewer water molecules to condense, and water boils away. At any given temperature, including when boiling, vapor is generated at the corresponding saturation vapor pressure. This may be less than the ambient pressure, but part of the ambient pressure is due to air, so the pressure of the vapor above the pot can be less than the pressure of the vapor being generated at that temperature, and evaporation continues. Incidentally, the equilibrium vapor pressure of solids is not zero, a fact that accounts for sublimation. CHECK YOUR UNDERSTANDING 1.4 Explain why a cup of water (or soda) with ice cubes stays at even on a hot summer day. Phase Change and Latent Heat So far, we have discussed heat transfers that cause temperature change. However, in a phase transition, heat transfer does not cause any temperature change. For an example of phase changes, consider the addition of heat to a sample of ice at (Figure 1.15) and atmospheric pressure. The temperature of the ice rises linearly, absorbing heat at a constant rate of until it reaches Once at this temperature, the ice begins to melt and continues until it has all melted, absorbing 333 kJ/kg of heat. The temperature remains constant at during this phase change. Once all the ice has melted, the temperature of the liquid water rises, absorbing heat at a new constant rate of At the water begins to boil. The temperature again remains constant during this phase change while the water absorbs 2256 kJ/kg of heat and turns into steam. When all the liquid has become steam, the temperature rises again, absorbing heat at a rate of . If we started with steam and cooled it to make it condense into liquid water and freeze into ice, the process would exactly reverse, with the temperature again constant during each phase transition. 26 1 • Temperature and Heat Access for free at openstax.org
FIGURE 1.15 Temperature versus heat. The system is constructed so that no vapor evaporates while ice warms to become liquid water, and so that, when vaporization occurs, the vapor remains in the system. The long stretches of constant temperatures at and reflect the large amounts of heat needed to cause melting and vaporization, respectively. Where does the heat added during melting or boiling go, considering that the temperature does not change until the transition is complete? Energy is required to melt a solid, because the attractive forces between the molecules in the solid must be broken apart, so that in the liquid, the molecules can move around at comparable kinetic energies; thus, there is no rise in temperature. Energy is needed to vaporize a liquid for similar reasons. Conversely, work is done by attractive forces when molecules are brought together during freezing and condensation. That energy must be transferred out of the system, usually in the form of heat, to allow the molecules to stay together (Figure 1.18). Thus, condensation occurs in association with cold objects—the glass in Figure 1.16, for example. FIGURE 1.16 Condensation forms on this glass of iced tea because the temperature of the nearby air is reduced. The air cannot hold as much water as it did at room temperature, so water condenses. Energy is released when the water condenses, speeding the melting of the ice in the glass. (credit: Jenny Downing) The energy released when a liquid freezes is used by orange growers when the temperature approaches . Growers spray water on the trees so that the water freezes and heat is released to the growing oranges. This prevents the temperature inside the orange from dropping below freezing, which would damage the fruit (Figure 1.17). 1.5 • Phase Changes 27
FIGURE 1.17 The ice on these trees released large amounts of energy when it froze, helping to prevent the temperature of the trees from dropping below . Water is intentionally sprayed on orchards to help prevent hard frosts. (credit: Hermann Hammer) The energy involved in a phase change depends on the number of bonds or force pairs and their strength. The number of bonds is proportional to the number of molecules and thus to the mass of the sample. The energy per unit mass required to change a substance from the solid phase to the liquid phase, or released when the substance changes from liquid to solid, is known as the heat of fusion. The energy per unit mass required to change a substance from the liquid phase to the vapor phase is known as the heat of vaporization. The strength of the forces depends on the type of molecules. The heat Q absorbed or released in a phase change in a sample of mass m is given by where the latent heat of fusion and latent heat of vaporization are material constants that are determined experimentally. (Latent heats are also called latent heat coefficients and heats of transformation.) These constants are “latent,” or hidden, because in phase changes, energy enters or leaves a system without causing a temperature change in the system, so in effect, the energy is hidden. FIGURE 1.18 (a) Energy is required to partially overcome the attractive forces (modeled as springs) between molecules in a solid to form a liquid. That same energy must be removed from the liquid for freezing to take place. (b) Molecules become separated by large distances when going from liquid to vapor, requiring significant energy to completely overcome molecular attraction. The same energy must be removed from the vapor for condensation to take place. Table 1.4 lists representative values of and in kJ/kg, together with melting and boiling points. Note that in general, . The table shows that the amounts of energy involved in phase changes can easily be comparable to or greater than those involved in temperature changes, as Figure 1.15 and the accompanying discussion also showed. 1.7 1.8 28 1 • Temperature and Heat Access for free at openstax.org
Substance Melting Point kJ/kg kcal/kg Boiling Point kJ/kg kcal/kg Helium[2] 5.23 1.25 20.9 4.99 Hydrogen 58.6 14.0 452 108 Nitrogen 25.5 6.09 201 48.0 Oxygen 13.8 3.30 213 50.9 Ethanol –114 104 24.9 78.3 854 204 Ammonia –75 332 79.3 –33.4 1370 327 Mercury –38.9 11.8 2.82 357 272 65.0 Water 0.00 334 79.8 100.0 2256[3] 539[4] Sulfur 119 38.1 9.10 444.6 326 77.9 Lead 327 24.5 5.85 1750 871 208 Antimony 631 165 39.4 1440 561 134 Aluminum 660 380 90 2450 11400 2720 Silver 961 88.3 21.1 2193 2336 558 Gold 1063 64.5 15.4 2660 1578 377 Copper 1083 134 32.0 2595 5069 1211 Uranium 1133 84 20 3900 1900 454 Tungsten 3410 184 44 5900 4810 1150 TABLE 1.4 Heats of Fusion and Vaporization[1] [1]Values quoted at the normal melting and boiling temperatures at standard atmospheric pressure ( ). [2]Helium has no solid phase at atmospheric pressure. The melting point given is at a pressure of 2.5 MPa.[3] [4] At (body temperature), the heat of vaporization for water is 2430 kJ/kg or 580 kcal/kg. Phase changes can have a strong stabilizing effect on temperatures that are not near the melting and boiling points, since evaporation and condensation occur even at temperatures below the boiling point. For example, temperatures in humid weather rarely fall below the dew point—the temperature where condensation occurs given the concentration of water vapor in the air—because so much heat is released when water vapor condenses. More energy is required to evaporate water below the boiling point than at the boiling point, because the kinetic energy of water molecules at temperatures below is less than that at , so less energy is available from random thermal motions. For example, at body temperature, evaporation of sweat from the skin requires a heat input of 2428 kJ/kg, which is about 10% higher than the latent heat of vaporization at . This heat comes from the skin, and this evaporative cooling effect of sweating helps reduce the body temperature in hot weather. However, high humidity inhibits evaporation, so that body temperature might rise, while unevaporated sweat might be left on your brow. 1.5 • Phase Changes 29
EXAMPLE 1.9 Calculating Final Temperature from Phase Change Three ice cubes are used to chill a soda at with mass . The ice is at and each ice cube has a mass of 6.0 g. Assume that the soda is kept in a foam container so that heat loss can be ignored and that the soda has the same specific heat as water. Find the final temperature when all ice has melted. Strategy The ice cubes are at the melting temperature of Heat is transferred from the soda to the ice for melting. Melting yields water at so more heat is transferred from the soda to this water until the water plus soda system reaches thermal equilibrium. The heat transferred to the ice is The heat transferred by the soda is Since no heat is lost, as in Example 1.7, so that Solve for the unknown quantity : Solution First we identify the known quantities. The mass of ice is and the mass of soda is Then we calculate the final temperature: Significance This example illustrates the large energies involved during a phase change. The mass of ice is about 7% of the mass of the soda but leads to a noticeable change in the temperature of the soda. Although we assumed that the ice was at the freezing temperature, this is unrealistic for ice straight out of a freezer: The typical temperature is . However, this correction makes no significant change from the result we found. Can you explain why? Like solid-liquid and liquid-vapor transitions, direct solid-vapor transitions or sublimations involve heat. The energy transferred is given by the equation , where is the heat of sublimation, analogous to and . The heat of sublimation at a given temperature is equal to the heat of fusion plus the heat of vaporization at that temperature. We can now calculate any number of effects related to temperature and phase change. In each case, it is necessary to identify which temperature and phase changes are taking place. Keep in mind that heat transfer and work can cause both temperature and phase changes. PROBLEM-SOLVING STRATEGY The Effects of Heat Transfer 1. Examine the situation to determine that there is a change in the temperature or phase. Is there heat transfer into or out of the system? When it is not obvious whether a phase change occurs or not, you may wish to first solve the problem as if there were no phase changes, and examine the temperature change obtained. If it is 30 1 • Temperature and Heat Access for free at openstax.org
sufficient to take you past a boiling or melting point, you should then go back and do the problem in steps—temperature change, phase change, subsequent temperature change, and so on. 2. Identify and list all objects that change temperature or phase. 3. Identify exactly what needs to be determined in the problem (identify the unknowns). A written list is useful. 4. Make a list of what is given or what can be inferred from the problem as stated (identify the knowns). If there is a temperature change, the transferred heat depends on the specific heat of the substance (Heat Transfer, Specific Heat, and Calorimetry), and if there is a phase change, the transferred heat depends on the latent heat of the substance (Table 1.4). 5. Solve the appropriate equation for the quantity to be determined (the unknown). 6. Substitute the knowns along with their units into the appropriate equation and obtain numerical solutions complete with units. You may need to do this in steps if there is more than one state to the process, such as a temperature change followed by a phase change. However, in a calorimetry problem, each step corresponds to a term in the single equation . 7. Check the answer to see if it is reasonable. Does it make sense? As an example, be certain that any temperature change does not also cause a phase change that you have not taken into account. CHECK YOUR UNDERSTANDING 1.5 Why does snow often remain even when daytime temperatures are higher than the freezing temperature? 1.6 Mechanisms of Heat Transfer LEARNING OBJECTIVES By the end of this section, you will be able to: • Explain some phenomena that involve conductive, convective, and radiative heat transfer • Solve problems on the relationships between heat transfer, time, and rate of heat transfer • Solve problems using the formulas for conduction and radiation Just as interesting as the effects of heat transfer on a system are the methods by which it occurs. Whenever there is a temperature difference, heat transfer occurs. It may occur rapidly, as through a cooking pan, or slowly, as through the walls of a picnic ice chest. So many processes involve heat transfer that it is hard to imagine a situation where no heat transfer occurs. Yet every heat transfer takes place by only three methods: 1. Conduction is heat transfer through stationary matter by physical contact. (The matter is stationary on a macroscopic scale—we know that thermal motion of the atoms and molecules occurs at any temperature above absolute zero.) Heat transferred from the burner of a stove through the bottom of a pan to food in the pan is transferred by conduction. 2. Convection is the heat transfer by the macroscopic movement of a fluid. This type of transfer takes place in a forced-air furnace and in weather systems, for example. 3. Heat transfer by radiation occurs when microwaves, infrared radiation, visible light, or another form of electromagnetic radiation is emitted or absorbed. An obvious example is the warming of Earth by the Sun. A less obvious example is thermal radiation from the human body. In the illustration at the beginning of this chapter, the fire warms the snowshoers’ faces largely by radiation. Convection carries some heat to them, but most of the air flow from the fire is upward (creating the familiar shape of flames), carrying heat to the food being cooked and into the sky. The snowshoers wear clothes designed with low conductivity to prevent heat flow out of their bodies. In this section, we examine these methods in some detail. Each method has unique and interesting characteristics, but all three have two things in common: there is a net transfer of heat solely because of a temperature difference, and the greater the temperature difference, the faster the heat transfer (Figure 1.19). 1.6 • Mechanisms of Heat Transfer 31
FIGURE 1.19 In a fireplace, heat transfer occurs by all three methods: conduction, convection, and radiation. Radiation is responsible for most of the heat transferred into the room. Heat transfer also occurs through conduction into the room, but much slower. Heat transfer by convection also occurs through cold air entering the room around windows and hot air leaving the room by rising up the chimney. Convection also contributes to heat transfer indirectly by circulating the air in the room and thereby increasing conduction. CHECK YOUR UNDERSTANDING 1.6 Name an example from daily life (different from the text) for each mechanism of heat transfer. Conduction As you walk barefoot across the living room carpet in a cold house and then step onto the kitchen tile floor, your feet feel colder on the tile. This result is intriguing, since the carpet and tile floor are both at the same temperature. The different sensation is explained by the different rates of heat transfer: The heat loss is faster for skin in contact with the tiles than with the carpet, so the sensation of cold is more intense. Some materials conduct thermal energy faster than others. Figure 1.20 shows a material that conducts heat slowly—it is a good thermal insulator, or poor heat conductor—used to reduce heat flow into and out of a house. 32 1 • Temperature and Heat Access for free at openstax.org
FIGURE 1.20 Insulation is used to limit the conduction of heat from the inside to the outside (in winter) and from the outside to the inside (in summer). (credit: Giles Douglas) A molecular picture of heat conduction will help justify the equation that describes it. Figure 1.21 shows molecules in two bodies at different temperatures, and for “hot” and “cold.” The average kinetic energy of a molecule in the hot body is higher than in the colder body. If two molecules collide, energy transfers from the high-energy to the low-energy molecule. In a metal, the picture would also include free valence electrons colliding with each other and with atoms, likewise transferring energy. The cumulative effect of all collisions is a net flux of heat from the hotter body to the colder body. Thus, the rate of heat transfer increases with increasing temperature difference If the temperatures are the same, the net heat transfer rate is zero. Because the number of collisions increases with increasing area, heat conduction is proportional to the cross-sectional area—a second factor in the equation. FIGURE 1.21 Molecules in two bodies at different temperatures have different average kinetic energies. Collisions occurring at the contact surface tend to transfer energy from high-temperature regions to low-temperature regions. In this illustration, a molecule in the lower- temperature region (right side) has low energy before collision, but its energy increases after colliding with a high-energy molecule at the contact surface. In contrast, a molecule in the higher-temperature region (left side) has high energy before collision, but its energy decreases after colliding with a low-energy molecule at the contact surface. A third quantity that affects the conduction rate is the thickness of the material through which heat transfers. Figure 1.6 • Mechanisms of Heat Transfer 33
1.22 shows a slab of material with a higher temperature on the left than on the right. Heat transfers from the left to the right by a series of molecular collisions. The greater the distance between hot and cold, the more time the material takes to transfer the same amount of heat. FIGURE 1.22 Heat conduction occurs through any material, represented here by a rectangular bar, whether window glass or walrus blubber. All four of these quantities appear in a simple equation deduced from and confirmed by experiments. The rate of conductive heat transfer through a slab of material, such as the one in Figure 1.22, is given by where P is the power or rate of heat transfer in watts or in kilocalories per second, A and d are its surface area and thickness, as shown in Figure 1.22, is the temperature difference across the slab, and k is the thermal conductivity of the material. Table 1.5 gives representative values of thermal conductivity. More generally, we can write where x is the coordinate in the direction of heat flow. Since in Figure 1.22, the power and area are constant, dT/dx is constant, and the temperature decreases linearly from to Substance Thermal Conductivity k Diamond 2000 Silver 420 Copper 390 Gold 318 Aluminum 220 Steel iron 80 Steel (stainless) 14 Ice 2.2 TABLE 1.5 Thermal Conductivities of Common Substances Values are given for temperatures near . 1.9 34 1 • Temperature and Heat Access for free at openstax.org
Substance Thermal Conductivity k Glass (average) 0.84 Concrete brick 0.84 Water 0.6 Fatty tissue (without blood) 0.2 Asbestos 0.16 Plasterboard 0.16 Wood 0.08–0.16 Snow (dry) 0.10 Cork 0.042 Glass wool 0.042 Wool 0.04 Polystyrene foam 0.035 Down feathers 0.025 Air 0.023 TABLE 1.5 Thermal Conductivities of Common Substances Values are given for temperatures near . EXAMPLE 1.10 Calculating Heat Transfer through Conduction A polystyrene foam icebox has a total area of and walls with an average thickness of 2.50 cm. The box contains ice, water, and canned beverages at The inside of the box is kept cold by melting ice. How much ice melts in one day if the icebox is kept in the trunk of a car at ? Strategy This question involves both heat for a phase change (melting of ice) and the transfer of heat by conduction. To find the amount of ice melted, we must find the net heat transferred. This value can be obtained by calculating the rate of heat transfer by conduction and multiplying by time. Solution First we identify the knowns. for polystyrene foam; ; ; Then we identify the unknowns. We need to solve for the mass of the ice, m. We also need to solve for the net heat transferred to melt the ice, Q. The rate of heat transfer by conduction is given by 1.6 • Mechanisms of Heat Transfer 35
The heat used to melt the ice is .We insert the known values: Multiplying the rate of heat transfer by the time we obtain We set this equal to the heat transferred to melt the ice, and solve for the mass m: Significance The result of 12 kg, or about 26 lb, is quite large due to the high temperature in the trunk, and even more ice would be required if you add any warm food or beverages. Table 1.5 shows that polystyrene foam is a very poor conductor and thus a good insulator. Other good insulators include fiberglass, wool, and goosedown feathers. Like polystyrene foam, these all contain many small pockets of air, taking advantage of air’s poor thermal conductivity. In developing insulation, the smaller the conductivity k and the larger the thickness d, the better. Thus, the ratio d/k, called the R factor, is large for a good insulator. The rate of conductive heat transfer is inversely proportional to R. R factors are most commonly quoted for household insulation, refrigerators, and the like. Unfortunately, in the United States, R is still in non-metric units of , although the unit usually goes unstated [1 British thermal unit (Btu) is the amount of energy needed to change the temperature of 1.0 lb of water by , which is 1055.1 J]. A couple of representative values are an R factor of 11 for 3.5-inch-thick fiberglass batts (pieces) of insulation and anR factor of 19 for 6.5-inch-thick fiberglass batts (Figure 1.23). In the US, walls are usually insulated with 3.5-inch batts, whereas ceilings are usually insulated with 6.5-inch batts. In cold climates, thicker batts may be used. FIGURE 1.23 The fiberglass batt is used for insulation of walls and ceilings to prevent heat transfer between the inside of the building and the outside environment. (credit: Tracey Nicholls) 36 1 • Temperature and Heat Access for free at openstax.org
Note that in Table 1.5, most of the best thermal conductors—silver, copper, gold, and aluminum—are also the best electrical conductors, because they contain many free electrons that can transport thermal energy. (Diamond, an electrical insulator, conducts heat by atomic vibrations.) Cooking utensils are typically made from good conductors, but the handles of those used on the stove are made from good insulators (bad conductors). EXAMPLE 1.11 Two Conductors End to End A steel rod and an aluminum rod, each of diameter 1.00 cm and length 25.0 cm, are welded end to end. One end of the steel rod is placed in a large tank of boiling water at , while the far end of the aluminum rod is placed in a large tank of water at . The rods are insulated so that no heat escapes from their surfaces. What is the temperature at the joint, and what is the rate of heat conduction through this composite rod? Strategy The heat that enters the steel rod from the boiling water has no place to go but through the steel rod, then through the aluminum rod, to the cold water. Therefore, we can equate the rate of conduction through the steel to the rate of conduction through the aluminum. We repeat the calculation with a second method, in which we use the thermal resistance R of the rod, since it simply adds when two rods are joined end to end. (We will use a similar method in the chapter on direct-current circuits.) Solution 1. Identify the knowns and convert them to SI units. The length of each rod is the cross-sectional area of each rod is the thermal conductivity of aluminum is , the thermal conductivity of steel is , the temperature at the hot end is , and the temperature at the cold end is . 2. Calculate the heat-conduction rate through the steel rod and the heat-conduction rate through the aluminum rod in terms of the unknown temperature T at the joint: 3. Set the two rates equal and solve for the unknown temperature: 4. Calculate either rate: 5. If desired, check your answer by calculating the other rate. Solution 1. Recall that . Now 2. We know that . We also know that and we denote that rate of heat flow by P. Combine the equations: 1.6 • Mechanisms of Heat Transfer 37
Thus, we can simply add R factors. Now, . 3. Find the from the known quantities: and 4. Substitute these values in to find as before. 5. Determine for the aluminum rod (or for the steel rod) and use it to find T at the joint. so , as in Solution . 6. If desired, check by determining for the other rod. Significance In practice, adding R values is common, as in calculating the R value of an insulated wall. In the analogous situation in electronics, the resistance corresponds to R/A in this problem and is additive even when the areas are unequal, as is common in electronics. Our equation for heat conduction can be used only when the areas are equal; otherwise, we would have a problem in three-dimensional heat flow, which is beyond our scope. CHECK YOUR UNDERSTANDING 1.7 How does the rate of heat transfer by conduction change when all spatial dimensions are doubled? Conduction is caused by the random motion of atoms and molecules. As such, it is an ineffective mechanism for heat transport over macroscopic distances and short times. For example, the temperature on Earth would be unbearably cold during the night and extremely hot during the day if heat transport into and through the atmosphere were only through conduction and radiation. Also, car engines would overheat unless there was a more efficient way to remove excess heat from the pistons. The next module discusses the important heat-transfer mechanism in such situations. Convection In convection, thermal energy is carried by the large-scale flow of matter. It can be divided into two types. In forced convection, the flow is driven by fans, pumps, and the like. A simple example is a fan that blows air past you in hot surroundings and cools you by replacing the air heated by your body with cooler air. A more complicated example is the cooling system of a typical car, in which a pump moves coolant through the radiator and engine to cool the engine and a fan blows air to cool the radiator. In free or natural convection, the flow is driven by buoyant forces: hot fluid rises and cold fluid sinks because density decreases as temperature increases. The house in Figure 1.24 is kept warm by natural convection, as is the pot of water on the stove in Figure 1.25. Ocean currents and large-scale atmospheric circulation, which result from the buoyancy of warm air and water, transfer hot air from the tropics toward the poles and cold air from the poles toward the tropics. (Earth’s rotation interacts with those flows, causing the observed eastward flow of air in the temperate zones.) 38 1 • Temperature and Heat Access for free at openstax.org
FIGURE 1.24 Air heated by a so-called gravity furnace expands and rises, forming a convective loop that transfers energy to other parts of the room. As the air is cooled at the ceiling and outside walls, it contracts, eventually becoming denser than room air and sinking to the floor. A properly designed heating system using natural convection, like this one, can heat a home quite efficiently. FIGURE 1.25 Natural convection plays an important role in heat transfer inside this pot of water. Once conducted to the inside, heat transfer to other parts of the pot is mostly by convection. The hotter water expands, decreases in density, and rises to transfer heat to other regions of the water, while colder water sinks to the bottom. This process keeps repeating. INTERACTIVE Natural convection like that of Figure 1.24 and Figure 1.25, but acting on rock in Earth’s mantle, drives plate tectonics (https://openstax.org/l/21platetecton) that are the motions that have shaped Earth’s surface. Convection is usually more complicated than conduction. Beyond noting that the convection rate is often approximately proportional to the temperature difference, we will not do any quantitative work comparable to the formula for conduction. However, we can describe convection qualitatively and relate convection rates to heat and time. Air is a poor conductor, so convection dominates heat transfer by air. Therefore, the amount of available space for airflow determines whether air transfers heat rapidly or slowly. There is little heat transfer in a space filled with air with a small amount of other material that prevents flow. The space between the inside and outside walls of a typical American house, for example, is about 9 cm (3.5 in.)—large enough for convection to work effectively. The addition of wall insulation prevents airflow, so heat loss (or gain) is decreased. On the other hand, the gap between the two panes of a double-paned window is about 1 cm, which largely prevents convection and takes advantage of air’s low conductivity reduce heat loss. Fur, cloth, and fiberglass also take advantage of the low conductivity of air by trapping it in spaces too small to support convection (Figure 1.26). 1.6 • Mechanisms of Heat Transfer 39
FIGURE 1.26 Fur is filled with air, breaking it up into many small pockets. Convection is very slow here, because the loops are so small. The low conductivity of air makes fur a very good lightweight insulator. Some interesting phenomena happen when convection is accompanied by a phase change. The combination allows us to cool off by sweating even if the temperature of the surrounding air exceeds body temperature. Heat from the skin is required for sweat to evaporate from the skin, but without air flow, the air becomes saturated and evaporation stops. Air flow caused by convection replaces the saturated air by dry air and evaporation continues. EXAMPLE 1.12 Calculating the Flow of Mass during Convection The average person produces heat at the rate of about 120 W when at rest. At what rate must water evaporate from the body to get rid of all this energy? (For simplicity, we assume this evaporation occurs when a person is sitting in the shade and surrounding temperatures are the same as skin temperature, eliminating heat transfer by other methods.) Strategy Energy is needed for this phase change ( ). Thus, the energy loss per unit time is We divide both sides of the equation by to find that the mass evaporated per unit time is Solution Insert the value of the latent heat from Table 1.4, . This yields Significance Evaporating about 3 g/min seems reasonable. This would be about 180 g (about 7 oz.) per hour. If the air is very dry, the sweat may evaporate without even being noticed. A significant amount of evaporation also takes place in the lungs and breathing passages. Another important example of the combination of phase change and convection occurs when water evaporates from the oceans. Heat is removed from the ocean when water evaporates. If the water vapor condenses in liquid droplets 40 1 • Temperature and Heat Access for free at openstax.org
as clouds form, possibly far from the ocean, heat is released in the atmosphere. Thus, there is an overall transfer of heat from the ocean to the atmosphere. This process is the driving power behind thunderheads, those great cumulus clouds that rise as much as 20.0 km into the stratosphere (Figure 1.27). Water vapor carried in by convection condenses, releasing tremendous amounts of energy. This energy causes the air to expand and rise to colder altitudes. More condensation occurs in these regions, which in turn drives the cloud even higher. This mechanism is an example of positive feedback, since the process reinforces and accelerates itself. It sometimes produces violent storms, with lightning and hail. The same mechanism drives hurricanes. INTERACTIVE This time-lapse video (https://openstax.org/l/21convthuncurr) shows convection currents in a thunderstorm, including “rolling” motion similar to that of boiling water. FIGURE 1.27 Cumulus clouds are caused by water vapor that rises because of convection. The rise of clouds is driven by a positive feedback mechanism. (credit: “Amada44”/Wikimedia Commons) CHECK YOUR UNDERSTANDING 1.8 Explain why using a fan in the summer feels refreshing. Radiation You can feel the heat transfer from the Sun. The space between Earth and the Sun is largely empty, so the Sun warms us without any possibility of heat transfer by convection or conduction. Similarly, you can sometimes tell that the oven is hot without touching its door or looking inside—it may just warm you as you walk by. In these examples, heat is transferred by radiation (Figure 1.28). That is, the hot body emits electromagnetic waves that are absorbed by the skin. No medium is required for electromagnetic waves to propagate. Different names are used for electromagnetic waves of different wavelengths: radio waves, microwaves, infrared radiation, visible light, ultraviolet radiation, X-rays, and gamma rays. 1.6 • Mechanisms of Heat Transfer 41
FIGURE 1.28 Most of the heat transfer from this fire to the observers occurs through infrared radiation. The visible light, although dramatic, transfers relatively little thermal energy. Convection transfers energy away from the observers as hot air rises, while conduction is negligibly slow here. Skin is very sensitive to infrared radiation, so you can sense the presence of a fire without looking at it directly. (credit: Daniel O’Neil) The energy of electromagnetic radiation varies over a wide range, depending on the wavelength: A shorter wavelength (or higher frequency) corresponds to a higher energy. Because more heat is radiated at higher temperatures, higher temperatures produce more intensity at every wavelength but especially at shorter wavelengths. In visible light, wavelength determines color—red has the longest wavelength and violet the shortest—so a temperature change is accompanied by a color change. For example, an electric heating element on a stove glows from red to orange, while the higher-temperature steel in a blast furnace glows from yellow to white. Infrared radiation is the predominant form radiated by objects cooler than the electric element and the steel. The radiated energy as a function of wavelength depends on its intensity, which is represented in Figure 1.29 by the height of the distribution. (Electromagnetic Waves explains more about the electromagnetic spectrum, and Photons and Matter Waves (http://openstax.org/books/university-physics-volume-3/pages/6-introduction) discusses why the decrease in wavelength corresponds to an increase in energy.) FIGURE 1.29 (a) A graph of the spectrum of electromagnetic waves emitted from an ideal radiator at three different temperatures. The intensity or rate of radiation emission increases dramatically with temperature, and the spectrum shifts down in wavelength toward the visible and ultraviolet parts of the spectrum. The shaded portion denotes the visible part of the spectrum. It is apparent that the shift toward the ultraviolet with temperature makes the visible appearance shift from red to white to blue as temperature increases. (b) Note the variations in color corresponding to variations in flame temperature. The rate of heat transfer by radiation also depends on the object’s color. Black is the most effective, and white is the least effective. On a clear summer day, black asphalt in a parking lot is hotter than adjacent gray sidewalk, because black absorbs better than gray (Figure 1.30). The reverse is also true—black radiates better than gray. Thus, on a clear summer night, the asphalt is colder than the gray sidewalk, because black radiates the energy more rapidly 42 1 • Temperature and Heat Access for free at openstax.org
than gray. A perfectly black object would be an ideal radiator and an ideal absorber, as it would capture all the radiation that falls on it. In contrast, a perfectly white object or a perfect mirror would reflect all radiation, and a perfectly transparent object would transmit it all (Figure 1.31). Such objects would not emit any radiation. Mathematically, the color is represented by the emissivity e. A “blackbody” radiator would have an , whereas a perfect reflector or transmitter would have . For real examples, tungsten light bulb filaments have an e of about 0.5, and carbon black (a material used in printer toner) has an emissivity of about 0.95. FIGURE 1.30 The darker pavement is hotter than the lighter pavement (much more of the ice on the right has melted), although both have been in the sunlight for the same time. The thermal conductivities of the pavements are the same. FIGURE 1.31 A black object is a good absorber and a good radiator, whereas a white, clear, or silver object is a poor absorber and a poor radiator. To see that, consider a silver object and a black object that can exchange heat by radiation and are in thermal equilibrium. We know from experience that they will stay in equilibrium (the result of a principle that will be discussed at length in Second Law of Thermodynamics). For the black object’s temperature to stay constant, it must emit as much radiation as it absorbs, so it must be as good at radiating as absorbing. Similar considerations show that the silver object must radiate as little as it absorbs. Thus, one property, emissivity, controls both radiation and absorption. Finally, the radiated heat is proportional to the object’s surface area, since every part of the surface radiates. If you knock apart the coals of a fire, the radiation increases noticeably due to an increase in radiating surface area. The rate of heat transfer by emitted radiation is described by the Stefan-Boltzmann law of radiation: where is the Stefan-Boltzmann constant, a combination of fundamental constants of nature; A is the surface area of the object; T is its temperature in kelvins, ; and e is the emissivity of the object. The proportionality to the fourth power of the absolute temperature is a remarkably strong temperature 1.6 • Mechanisms of Heat Transfer 43
dependence. It allows the detection of even small temperature variations. Images called thermographs can be used medically to detect regions of abnormally high temperature in the body, perhaps indicative of disease. Similar techniques can be used to detect heat leaks in homes (Figure 1.32), optimize performance of blast furnaces, improve comfort levels in work environments, and even remotely map Earth’s temperature profile. FIGURE 1.32 A thermograph of part of a building shows temperature variations, indicating where heat transfer to the outside is most severe. Windows are a major region of heat transfer to the outside of homes. (credit: US Army) The Stefan-Boltzmann equation needs only slight refinement to deal with a simple case of an object’s absorption of radiation from its surroundings. Assuming that an object with a temperature is surrounded by an environment with uniform temperature , the net rate of heat transfer by radiation is where e is the emissivity of the object alone. In other words, it does not matter whether the surroundings are white, gray, or black: The balance of radiation into and out of the object depends on how well it emits and absorbs radiation. When the quantity is positive (and when the quantity is negative), that is, the net heat transfer is from hot to cold. Before doing an example, we have a complication to discuss: different emissivities at different wavelengths. If the fraction of incident radiation an object reflects is the same at all visible wavelengths, the object is gray; if the fraction depends on the wavelength, the object has some other color. For instance, a red or reddish object reflects red light more strongly than other visible wavelengths. Because it absorbs less red, it radiates less red when hot. Differential reflection and absorption of wavelengths outside the visible range have no effect on what we see, but they may have physically important effects. Skin is a very good absorber and emitter of infrared radiation, having an emissivity of 0.97 in the infrared spectrum. Thus, in spite of the obvious variations in skin color, we are all nearly black in the infrared. This high infrared emissivity is why we can so easily feel radiation on our skin. It is also the basis for the effectiveness of night-vision scopes used by law enforcement and the military to detect human beings. EXAMPLE 1.13 Calculating the Net Heat Transfer of a Person What is the rate of heat transfer by radiation of an unclothed person standing in a dark room whose ambient temperature is ? The person has a normal skin temperature of and a surface area of The emissivity of skin is 0.97 in the infrared, the part of the spectrum where the radiation takes place. 1.10 44 1 • Temperature and Heat Access for free at openstax.org
Strategy We can solve this by using the equation for the rate of radiative heat transfer. Solution Insert the temperature values and , so that Significance This value is a significant rate of heat transfer to the environment (note the minus sign), considering that a person at rest may produce energy at the rate of 125 W and that conduction and convection are also transferring energy to the environment. Indeed, we would probably expect this person to feel cold. Clothing significantly reduces heat transfer to the environment by all mechanisms, because clothing slows down both conduction and convection, and has a lower emissivity (especially if it is light-colored) than skin. The average temperature of Earth is the subject of much current discussion. Earth is in radiative contact with both the Sun and dark space, so we cannot use the equation for an environment at a uniform temperature. Earth receives almost all its energy from radiation of the Sun and reflects some of it back into outer space. Conversely, dark space is very cold, about 3 K, so that Earth radiates energy into the dark sky. The rate of heat transfer from soil and grasses can be so rapid that frost may occur on clear summer evenings, even in warm latitudes. The average temperature of Earth is determined by its energy balance. To a first approximation, it is the temperature at which Earth radiates heat to space as fast as it receives energy from the Sun. An important parameter in calculating the temperature of Earth is its emissivity ( e). On average, it is about 0.65, but calculation of this value is complicated by the great day-to-day variation in the highly reflective cloud coverage. Because clouds have lower emissivity than either oceans or land masses, they emit some of the radiation back to the surface, greatly reducing heat transfer into dark space, just as they greatly reduce heat transfer into the atmosphere during the day. There is negative feedback (in which a change produces an effect that opposes that change) between clouds and heat transfer; higher temperatures evaporate more water to form more clouds, which reflect more radiation back into space, reducing the temperature. The often-mentioned greenhouse effect is directly related to the variation of Earth’s emissivity with wavelength (Figure 1.33). The greenhouse effect is a natural phenomenon responsible for providing temperatures suitable for life on Earth and for making Venus unsuitable for human life. Most of the infrared radiation emitted from Earth is absorbed by carbon dioxide ( ) and water ( ) in the atmosphere and then re-radiated into outer space or back to Earth. Re-radiation back to Earth maintains its surface temperature about higher than it would be if there were no atmosphere. (The glass walls and roof of a greenhouse increase the temperature inside by blocking convective heat losses, not radiative losses.) The greenhouse effect and its causes were first predicted by Eunice Newton Foote after she designed and conducted experiments on heating of different gases. After filling flasks with carbon dioxide, hydrogen, and regular air, then also modifying moisture, she placed them in the sun and carefully measured their heating and, especially, their heat retention. She discovered that the CO2 flask gained the most temperature and held it the longest. After subsequent research, her paper "Circumstances affecting the Heat of the Sun’s Rays" included conclusions that an atmosphere consisting of more carbon dioxide would be hotter resulting from the gas trapping radiation. 1.6 • Mechanisms of Heat Transfer 45
FIGURE 1.33 The greenhouse effect is the name given to the increase of Earth’s temperature due to absorption of radiation in the atmosphere. The atmosphere is transparent to incoming visible radiation and most of the Sun’s infrared. The Earth absorbs that energy and re-emits it. Since Earth’s temperature is much lower than the Sun’s, it re-emits the energy at much longer wavelengths, in the infrared. The atmosphere absorbs much of that infrared radiation and radiates about half of the energy back down, keeping Earth warmer than it would otherwise be. The amount of trapping depends on concentrations of trace gases such as carbon dioxide, and an increase in the concentration of these gases increases Earth’s surface temperature. The greenhouse effect is central to the discussion of global warming due to emission of carbon dioxide and methane (and other greenhouse gases) into Earth’s atmosphere from industry, transportation, and farming. Changes in global climate could lead to more intense storms, precipitation changes (affecting agriculture), reduction in rain forest biodiversity, and rising sea levels. INTERACTIVE You can explore a simulation of the greenhouse effect (https://openstax.org/l/21simgreeneff) that takes the point of view that the atmosphere scatters (redirects) infrared radiation rather than absorbing it and reradiating it. You may want to run the simulation first with no greenhouse gases in the atmosphere and then look at how adding greenhouse gases affects the infrared radiation from the Earth and the Earth’s temperature. PROBLEM-SOLVING STRATEGY Effects of Heat Transfer 1. Examine the situation to determine what type of heat transfer is involved. 2. Identify the type(s) of heat transfer—conduction, convection, or radiation. 3. Identify exactly what needs to be determined in the problem (identify the unknowns). A written list is useful. 4. Make a list of what is given or what can be inferred from the problem as stated (identify the knowns). 5. Solve the appropriate equation for the quantity to be determined (the unknown). 6. For conduction, use the equation . Table 1.5 lists thermal conductivities. For convection, determine the amount of matter moved and the equation , along with or if a substance changes phase. For radiation, the equation gives the net heat transfer rate. 7. Substitute the knowns along with their units into the appropriate equation and obtain numerical solutions 46 1 • Temperature and Heat Access for free at openstax.org
complete with units. 8. Check the answer to see if it is reasonable. Does it make sense? CHECK YOUR UNDERSTANDING 1.9 How much greater is the rate of heat radiation when a body is at the temperature than when it is at the temperature ? 1.6 • Mechanisms of Heat Transfer 47
Chapter Review Key Terms absolute temperature scale scale, such as Kelvin, with a zero point that is absolute zero absolute zero temperature at which the average kinetic energy of molecules is zero calorie (cal) energy needed to change the temperature of 1.00 g of water by calorimeter container that prevents heat transfer in or out calorimetry study of heat transfer inside a container impervious to heat Celsius scale temperature scale in which the freezing point of water is and the boiling point of water is coefficient of linear expansion ( ) material property that gives the change in length, per unit length, per change in temperature; a constant used in the calculation of linear expansion; the coefficient of linear expansion depends to some degree on the temperature of the material coefficient of volume expansion ( ) similar to but gives the change in volume, per unit volume, per change in temperature conduction heat transfer through stationary matter by physical contact convection heat transfer by the macroscopic movement of fluid critical point for a given substance, the combination of temperature and pressure above which the liquid and gas phases are indistinguishable critical pressure pressure at the critical point critical temperature temperature at the critical point degree Celsius ( ) unit on the Celsius temperature scale degree Fahrenheit ( ) unit on the Fahrenheit temperature scale emissivity measure of how well an object radiates Fahrenheit scale temperature scale in which the freezing point of water is and the boiling point of water is greenhouse effect warming of the earth that is due to gases such as carbon dioxide and methane that absorb infrared radiation from Earth’s surface and reradiate it in all directions, thus sending some of it back toward Earth heat energy transferred solely due to a temperature difference heat of fusion energy per unit mass required to change a substance from the solid phase to the liquid phase, or released when the substance changes from liquid to solid heat of sublimation energy per unit mass required to change a substance from the solid phase to the vapor phase heat of vaporization energy per unit mass required to change a substance from the liquid phase to the vapor phase heat transfer movement of energy from one place or material to another as a result of a difference in temperature Kelvin scale (K) temperature scale in which 0 K is the lowest possible temperature, representing absolute zero kilocalorie (kcal) energy needed to change the temperature of 1.00 kg of water between and latent heat coefficient general term for the heats of fusion, vaporization, and sublimation mechanical equivalent of heat work needed to produce the same effects as heat transfer net rate of heat transfer by radiation phase diagram graph of pressure vs. temperature of a particular substance, showing at which pressures and temperatures the phases of the substance occur radiation energy transferred by electromagnetic waves directly as a result of a temperature difference rate of conductive heat transfer rate of heat transfer from one material to another specific heat amount of heat necessary to change the temperature of 1.00 kg of a substance by ; also called “specific heat capacity” Stefan-Boltzmann law of radiation where is the Stefan- Boltzmann constant, A is the surface area of the object, T is the absolute temperature, and e is the emissivity sublimation phase change from solid to gas temperature functionally defined as a quantity measured by a thermometer, which, at least for most of the systems discussed in this chapter, reflects the mechanical energy of particles in the system thermal conductivity property of a material describing its ability to conduct heat thermal equilibrium condition in which heat no longer flows between two objects that are in contact; the two objects have the same temperature thermal expansion change in size or volume of an 48 1 • Chapter Review Access for free at openstax.org
object with change in temperature thermal stress stress caused by thermal expansion or contraction triple point pressure and temperature at which a substance exists in equilibrium as a solid, liquid, and gas vapor gas at a temperature below the critical temperature vapor pressure pressure at which a gas coexists with its solid or liquid phase zeroth law of thermodynamics law that states that if two objects are in thermal equilibrium, and a third object is in thermal equilibrium with one of those objects, it is also in thermal equilibrium with the other object Key Equations Linear thermal expansion Thermal expansion in two dimensions Thermal expansion in three dimensions Heat transfer Transfer of heat in a calorimeter Heat due to phase change (melting and freezing) Heat due to phase change (evaporation and condensation) Rate of conductive heat transfer Net rate of heat transfer by radiation Summary 1.1 Temperature and Thermal Equilibrium • Temperature is operationally defined as the quantity measured by a thermometer. It is proportional to the average kinetic energy of atoms and molecules in a system. • Thermal equilibrium occurs when two bodies can freely exchange energy but no net energy is transferred between them. • The zeroth law of thermodynamics states that when two systems, A and B, are in thermal equilibrium with each other, and B is in thermal equilibrium with a third system C, then A is also in thermal equilibrium with C. 1.2 Thermometers and Temperature Scales • Three types of thermometers are alcohol, liquid crystal, and infrared radiation (pyrometer). • The three main temperature scales are Celsius, Fahrenheit, and Kelvin. Temperatures can be converted from one scale to another using temperature conversion equations. • The three phases of water (ice, liquid water, and water vapor) can coexist at a single pressure and temperature known as the triple point. 1.3 Thermal Expansion • Thermal expansion is the increase of the size (length, area, or volume) of a body due to a change in temperature, usually a rise. Thermal contraction is the decrease in size due to a change in temperature, usually a fall in temperature. • Thermal stress is created when thermal expansion or contraction is constrained. 1.4 Heat Transfer, Specific Heat, and Calorimetry • Heat and work are the two distinct methods of energy transfer. • Heat transfer to an object when its temperature changes is often approximated well by where m is the object’s mass and c is the specific heat of the substance. 1.5 Phase Changes • Most substances have three distinct phases (under ordinary conditions on Earth), and they depend on temperature and pressure. • Two phases coexist (i.e., they are in thermal equilibrium) at a set of pressures and temperatures. • Phase changes occur at fixed temperatures for a given substance at a given pressure, and these temperatures are called boiling, freezing (or melting), and sublimation points. 1.6 Mechanisms of Heat Transfer • Heat is transferred by three different methods: conduction, convection, and radiation. • Heat conduction is the transfer of heat between 1 • Chapter Review 49
two objects in direct contact with each other. • The rate of heat transfer P (energy per unit time) is proportional to the temperature difference via conduction through a slab of material with ends in contact with two objects at different temperatures and is proportional to the temperature difference and the contact area A, and inversely proportional to the distance d between the ends. • Convection is heat transfer by the macroscopic movement of mass. Convection can be natural or forced, and generally transfers thermal energy faster than conduction. • Radiation is heat transfer through the emission or absorption of electromagnetic waves. • The rate of radiative heat transfer is proportional to the emissivity e. For a perfect blackbody, , whereas a perfectly white, clear, or reflective body has , with real objects having values of e between 1 and 0. • The rate of heat emission depends on the surface area and the fourth power of the absolute temperature: where is the Stefan-Boltzmann constant and e is the emissivity of the body. The net rate of heat transfer from an object by radiation is where is the temperature of the object surrounded by an environment with uniform temperature and e is the emissivity of the object. Conceptual Questions 1.1 Temperature and Thermal Equilibrium 1 . What does it mean to say that two systems are in thermal equilibrium? 2 . Give an example in which A has some kind of non-thermal equilibrium relationship with B, and B has the same relationship with C, but A does not have that relationship with C. 1.2 Thermometers and Temperature Scales 3 . If a thermometer is allowed to come to equilibrium with the air, and a glass of water is not in equilibrium with the air, what will happen to the thermometer reading when it is placed in the water? 4 . Give an example of a physical property that varies with temperature and describe how it is used to measure temperature. 1.3 Thermal Expansion 5 . Pouring cold water into hot glass or ceramic cookware can easily break it. What causes the breaking? Explain why Pyrex®, a glass with a small coefficient of linear expansion, is less susceptible. 6 . One method of getting a tight fit, say of a metal peg in a hole in a metal block, is to manufacture the peg slightly larger than the hole. The peg is then inserted when at a different temperature than the block. Should the block be hotter or colder than the peg during insertion? Explain your answer. 7. Does it really help to run hot water over a tight metal lid on a glass jar before trying to open it? Explain your answer. 8. When a cold alcohol thermometer is placed in a hot liquid, the column of alcohol goes down slightly before going up. Explain why. 9. Calculate the length of a 1-meter rod of a material with thermal expansion coefficient when the temperature is raised from 300 K to 600 K. Taking your answer as the new initial length, find the length after the rod is cooled back down to 300 K. Is your answer 1 meter? Should it be? How can you account for the result you got? 10. Noting the large stresses that can be caused by thermal expansion, an amateur weapon inventor decides to use it to make a new kind of gun. He plans to jam a bullet against an aluminum rod inside a closed invar tube. When he heats the tube, the rod will expand more than the tube and a very strong force will build up. Then, by a method yet to be determined, he will open the tube in a split second and let the force of the rod launch the bullet at very high speed. What is he overlooking? 1.4 Heat Transfer, Specific Heat, and Calorimetry 11. How is heat transfer related to temperature? 12. Describe a situation in which heat transfer occurs. 13. When heat transfers into a system, is the energy 50 1 • Chapter Review Access for free at openstax.org
stored as heat? Explain briefly. 14. The brakes in a car increase in temperature by when bringing the car to rest from a speed v. How much greater would be if the car initially had twice the speed? You may assume the car stops fast enough that no heat transfers out of the brakes. 1.5 Phase Changes 15. A pressure cooker contains water and steam in equilibrium at a pressure greater than atmospheric pressure. How does this greater pressure increase cooking speed? 16. As shown below, which is the phase diagram for carbon dioxide, what is the vapor pressure of solid carbon dioxide (dry ice) at (Note that the axes in the figure are nonlinear and the graph is not to scale.) 17. Can carbon dioxide be liquefied at room temperature ( )? If so, how? If not, why not? (See the phase diagram in the preceding problem.) 18. What is the distinction between gas and vapor? 19. Heat transfer can cause temperature and phase changes. What else can cause these changes? 20. How does the latent heat of fusion of water help slow the decrease of air temperatures, perhaps preventing temperatures from falling significantly below in the vicinity of large bodies of water? 21. What is the temperature of ice right after it is formed by freezing water? 22. If you place ice into water in an insulated container, what will the net result be? Will there be less ice and more liquid water, or more ice and less liquid water, or will the amounts stay the same? 23. What effect does condensation on a glass of ice water have on the rate at which the ice melts? Will the condensation speed up the melting process or slow it down? 24. In Miami, Florida, which has a very humid climate and numerous bodies of water nearby, it is unusual for temperatures to rise above about ( ). In the desert climate of Phoenix, Arizona, however, temperatures rise above that almost every day in July and August. Explain how the type of heat transfer accounts for this difference and how it helps limit high temperatures in humid climates. 25. In winter, it is often warmer in San Francisco than in Sacramento, 150 km inland. In summer, it is nearly always hotter in Sacramento. Explain how the bodies of water surrounding San Francisco moderate its extreme temperatures. 26. Freeze-dried foods have been dehydrated in a vacuum. During the process, the food freezes and must be heated to facilitate dehydration. Explain both how the vacuum speeds up dehydration and why the food freezes as a result. 27. In a physics classroom demonstration, an instructor inflates a balloon by mouth and then cools it in liquid nitrogen. When cold, the shrunken balloon has a small amount of light blue liquid in it, as well as some snow-like crystals. As it warms up, the liquid boils, and part of the crystals sublime, with some crystals lingering for a while and then producing a liquid. Identify the blue liquid and the two solids in the cold balloon. Justify your identifications using data from Table 1.4. 1.6 Mechanisms of Heat Transfer 28. What are the main methods of heat transfer from the hot core of Earth to its surface? From Earth’s surface to outer space? 29. When our bodies get too warm, they respond by sweating and increasing blood circulation to the surface to transfer thermal energy away from the core. What effect will those processes have on a person in a hot tub? 1 • Chapter Review 51
30. Shown below is a cut-away drawing of a thermos bottle (also known as a Dewar flask), which is a device designed specifically to slow down all forms of heat transfer. Explain the functions of the various parts, such as the vacuum, the silvering of the walls, the thin-walled long glass neck, the rubber support, the air layer, and the stopper. 31. Some electric stoves have a flat ceramic surface with heating elements hidden beneath. A pot placed over a heating element will be heated, while the surface only a few centimeters away is safe to touch. Why is ceramic, with a conductivity less than that of a metal but greater than that of a good insulator, an ideal choice for the stove top? 32. Loose-fitting white clothing covering most of the body, shown below, is ideal for desert dwellers, both in the hot Sun and during cold evenings. Explain how such clothing is advantageous during both day and night. 33. One way to make a fireplace more energy- efficient is to have room air circulate around the outside of the fire box and back into the room. Detail the methods of heat transfer involved. 34. On cold, clear nights horses will sleep under the cover of large trees. How does this help them keep warm? 35. When watching a circus during the day in a large, dark-colored tent, you sense significant heat transfer from the tent. Explain why this occurs. 36. Satellites designed to observe the radiation from cold (3 K) dark space have sensors that are shaded from the Sun, Earth, and the Moon and are cooled to very low temperatures. Why must the sensors be at low temperature? 37. Why are thermometers that are used in weather stations shielded from the sunshine? What does a thermometer measure if it is shielded from the sunshine? What does it measure if it is not? 38. Putting a lid on a boiling pot greatly reduces the heat transfer necessary to keep it boiling. Explain why. 39. Your house will be empty for a while in cold weather, and you want to save energy and money. Should you turn the thermostat down to the lowest level that will protect the house from damage such as freezing pipes, or leave it at the normal temperature? (If you don’t like coming back to a cold house, imagine that a timer controls the heating system so the house will be 52 1 • Chapter Review Access for free at openstax.org
warm when you get back.) Explain your answer. 40. You pour coffee into an unlidded cup, intending to drink it 5 minutes later. You can add cream when you pour the cup or right before you drink it. (The cream is at the same temperature either way. Assume that the cream and coffee come into thermal equilibrium with each other very quickly.) Which way will give you hotter coffee? What feature of this question is different from the previous one? 41. Broiling is a method of cooking by radiation, which produces somewhat different results from cooking by conduction or convection. A gas flame or electric heating element produces a very high temperature close to the food and above it. Why is radiation the dominant heat- transfer method in this situation? 42. On a cold winter morning, why does the metal of a bike feel colder than the wood of a porch? Problems 1.2 Thermometers and Temperature Scales 43. While traveling outside the United States, you feel sick. A companion gets you a thermometer, which says your temperature is 39. What scale is that on? What is your Fahrenheit temperature? Should you seek medical help? 44. What are the following temperatures on the Kelvin scale? (a) an indoor temperature sometimes recommended for energy conservation in winter (b) one of the highest atmospheric temperatures ever recorded on Earth (Death Valley, California, 1913) (c) the temperature of the surface of the Sun 45. (a) Suppose a cold front blows into your locale and drops the temperature by 40.0 Fahrenheit degrees. How many degrees Celsius does the temperature decrease when it decreases by ? (b) Show that any change in temperature in Fahrenheit degrees is nine-fifths the change in Celsius degrees 46. An Associated Press article on climate change said, “Some of the ice shelf’s disappearance was probably during times when the planet was 36 degrees Fahrenheit (2 degrees Celsius) to 37 degrees Fahrenheit (3 degrees Celsius) warmer than it is today.” What mistake did the reporter make? 47. (a) At what temperature do the Fahrenheit and Celsius scales have the same numerical value? (b) At what temperature do the Fahrenheit and Kelvin scales have the same numerical value? 48. A person taking a reading of the temperature in a freezer in Celsius makes two mistakes: first omitting the negative sign and then thinking the temperature is Fahrenheit. That is, the person reads as . Oddly enough, the result is the correct Fahrenheit temperature. What is the original Celsius reading? Round your answer to three significant figures. 1.3 Thermal Expansion 49. The height of the Washington Monument is measured to be 170.00 m on a day when the temperature is What will its height be on a day when the temperature falls to ? Although the monument is made of limestone, assume that its coefficient of thermal expansion is the same as that of marble. Give your answer to five significant figures. 50. How much taller does the Eiffel Tower become at the end of a day when the temperature has increased by Its original height is 321 m and you can assume it is made of steel. 51. What is the change in length of a 3.00-cm-long column of mercury if its temperature changes from to , assuming the mercury is constrained to a cylinder but unconstrained in length? Your answer will show why thermometers contain bulbs at the bottom instead of simple columns of liquid. 52. How large an expansion gap should be left between steel railroad rails if they may reach a maximum temperature greater than when they were laid? Their original length is 10.0 m. 53. You are looking to buy a small piece of land in Hong Kong. The price is “only” $60,000 per square meter. The land title says the dimensions are . By how much would the total price change if you measured the parcel with a steel tape measure on a day when the temperature was above the temperature that the tape measure was designed for? The dimensions of the land do not change. 1 • Chapter Review 53
54. Global warming will produce rising sea levels partly due to melting ice caps and partly due to the expansion of water as average ocean temperatures rise. To get some idea of the size of this effect, calculate the change in length of a column of water 1.00 km high for a temperature increase of . Assume the column is not free to expand sideways. As a model of the ocean, that is a reasonable approximation, as only parts of the ocean very close to the surface can expand sideways onto land, and only to a limited degree. As another approximation, neglect the fact that ocean warming is not uniform with depth. 55. (a) Suppose a meter stick made of steel and one made of aluminum are the same length at . What is their difference in length at ? (b) Repeat the calculation for two 30.0-m-long surveyor’s tapes. 56. (a) If a 500-mL glass beaker is filled to the brim with ethyl alcohol at a temperature of , how much will overflow when the alcohol’s temperature reaches the room temperature of ? (b) How much less water would overflow under the same conditions? 57. Most cars have a coolant reservoir to catch radiator fluid that may overflow when the engine is hot. A radiator is made of copper and is filled to its 16.0-L capacity when at . What volume of radiator fluid will overflow when the radiator and fluid reach a temperature of given that the fluid’s volume coefficient of expansion is ? (Your answer will be a conservative estimate, as most car radiators have operating temperatures greater than ). 58. A physicist makes a cup of instant coffee and notices that, as the coffee cools, its level drops 3.00 mm in the glass cup. Show that this decrease cannot be due to thermal contraction by calculating the decrease in level if the of coffee is in a 7.00-cm-diameter cup and decreases in temperature from to . (Most of the drop in level is actually due to escaping bubbles of air.) 59. The density of water at is very nearly (it is actually ), whereas the density of ice at is Calculate the pressure necessary to keep ice from expanding when it freezes, neglecting the effect such a large pressure would have on the freezing temperature. (This problem gives you only an indication of how large the forces associated with freezing water might be.) 60. Show that by calculating the infinitesimal change in volume dV of a cube with sides of length L when the temperature changes by dT. 1.4 Heat Transfer, Specific Heat, and Calorimetry 61. On a hot day, the temperature of an 80,000-L swimming pool increases by . What is the net heat transfer during this heating? Ignore any complications, such as loss of water by evaporation. 62. To sterilize a 50.0-g glass baby bottle, we must raise its temperature from to . How much heat transfer is required? 63. The same heat transfer into identical masses of different substances produces different temperature changes. Calculate the final temperature when 1.00 kcal of heat transfers into 1.00 kg of the following, originally at : (a) water; (b) concrete; (c) steel; and (d) mercury. 64. Rubbing your hands together warms them by converting work into thermal energy. If a woman rubs her hands back and forth for a total of 20 rubs, at a distance of 7.50 cm per rub, and with an average frictional force of 40.0 N, what is the temperature increase? The mass of tissues warmed is only 0.100 kg, mostly in the palms and fingers. 65. A block of a pure material is heated from to by the addition of 4.35 kJ of energy. Calculate its specific heat and identify the substance of which it is most likely composed. 66. Suppose identical amounts of heat transfer into different masses of copper and water, causing identical changes in temperature. What is the ratio of the mass of copper to water? 67. (a) The number of kilocalories in food is determined by calorimetry techniques in which the food is burned and the amount of heat transfer is measured. How many kilocalories per gram are there in a 5.00-g peanut if the energy from burning it is transferred to 0.500 kg of water held in a 0.100-kg aluminum cup, causing a temperature increase? Assume the process takes place in an ideal calorimeter, in 54 1 • Chapter Review Access for free at openstax.org
other words a perfectly insulated container. (b) Compare your answer to the following labeling information found on a package of dry roasted peanuts: a serving of 33 g contains 200 calories. Comment on whether the values are consistent. 68. Following vigorous exercise, the body temperature of an 80.0 kg person is . At what rate in watts must the person transfer thermal energy to reduce the body temperature to in 30.0 min, assuming the body continues to produce energy at the rate of 150 W? 69. In a study of healthy young men1 , doing 20 push-ups in 1 minute burned an amount of energy per kg that for a 70.0-kg man corresponds to 8.06 calories (kcal). How much would a 70.0-kg man’s temperature rise if he did not lose any heat during that time? 70. A 1.28-kg sample of water at is in a calorimeter. You drop a piece of steel with a mass of 0.385 kg at into it. After the sizzling subsides, what is the final equilibrium temperature? (Make the reasonable assumptions that any steam produced condenses into liquid water during the process of equilibration and that the evaporation and condensation don’t affect the outcome, as we’ll see in the next section.) 71. Repeat the preceding problem, assuming the water is in a glass beaker with a mass of 0.200 kg, which in turn is in a calorimeter. The beaker is initially at the same temperature as the water. Before doing the problem, should the answer be higher or lower than the preceding answer? Comparing the mass and specific heat of the beaker to those of the water, do you think the beaker will make much difference? 1.5 Phase Changes 72. How much heat transfer (in kilocalories) is required to thaw a 0.450-kg package of frozen vegetables originally at if their heat of fusion is the same as that of water? 73. A bag containing ice is much more effective in absorbing energy than one containing the same amount of water. (a) How much heat transfer is necessary to raise the temperature of 0.800 kg of water from to ? (b) How much heat transfer is required to first melt 0.800 kg of ice and then raise its temperature to ? (c) Explain how your answer supports the contention that the ice is more effective. 74. (a) How much heat transfer is required to raise the temperature of a 0.750-kg aluminum pot containing 2.50 kg of water from to the boiling point and then boil away 0.750 kg of water? (b) How long does this take if the rate of heat transfer is 500 W? 75. Condensation on a glass of ice water causes the ice to melt faster than it would otherwise. If 8.00 g of vapor condense on a glass containing both water and 200 g of ice, how many grams of the ice will melt as a result? Assume no other heat transfer occurs. Use for water at as a better approximation than for water at .) 76. On a trip, you notice that a 3.50-kg bag of ice lasts an average of one day in your cooler. What is the average power in watts entering the ice if it starts at and completely melts to water in exactly one day? 77. On a certain dry sunny day, a swimming pool’s temperature would rise by if not for evaporation. What fraction of the water must evaporate to carry away precisely enough energy to keep the temperature constant? 78. (a) How much heat transfer is necessary to raise the temperature of a 0.200-kg piece of ice from to , including the energy needed for phase changes? (b) How much time is required for each stage, assuming a constant 20.0 kJ/s rate of heat transfer? (c) Make a graph of temperature versus time for this process. 79. In 1986, an enormous iceberg broke away from the Ross Ice Shelf in Antarctica. It was an approximately rectangular prism 160 km long, 40.0 km wide, and 250 m thick. (a) What is the mass of this iceberg, given that the density of ice is ? (b) How much heat transfer (in joules) is needed to melt it? (c) How many years would it take sunlight alone to melt ice this thick, if the ice absorbs an average of , 12.00 h per day? 80. How many grams of coffee must evaporate from 350 g of coffee in a 100-g glass cup to cool the coffee and the cup from to ? Assume the coffee has the same thermal 1 JW Vezina, “An examination of the differences between two methods of estimating energy expenditure in resistance training activities,” Journal of Strength and Conditioning Research, April 28, 2014, http://www.ncbi.nlm.nih.gov/pubmed/24402448 1 • Chapter Review 55
properties as water and that the average heat of vaporization is 2340 kJ/kg (560 kcal/g). Neglect heat losses through processes other than evaporation, as well as the change in mass of the coffee as it cools. Do the latter two assumptions cause your answer to be higher or lower than the true answer? 81. (a) It is difficult to extinguish a fire on a crude oil tanker, because each liter of crude oil releases of energy when burned. To illustrate this difficulty, calculate the number of liters of water that must be expended to absorb the energy released by burning 1.00 L of crude oil, if the water’s temperature rises from to , it boils, and the resulting steam’s temperature rises to at constant pressure. (b) Discuss additional complications caused by the fact that crude oil is less dense than water. 82. The energy released from condensation in thunderstorms can be very large. Calculate the energy released into the atmosphere for a small storm of radius 1 km, assuming that 1.0 cm of rain is precipitated uniformly over this area. 83. To help prevent frost damage, 4.00 kg of water at is sprayed onto a fruit tree. (a) How much heat transfer occurs as the water freezes? (b) How much would the temperature of the 200-kg tree decrease if this amount of heat transferred from the tree? Take the specific heat to be , and assume that no phase change occurs in the tree. 84. A 0.250-kg aluminum bowl holding of soup at is placed in a freezer. What is the final temperature if 388 kJ of energy is transferred from the bowl and soup, assuming the soup’s thermal properties are the same as that of water? 85. A 0.0500-kg ice cube at is placed in 0.400 kg of water in a very well- insulated container. What is the final temperature? 86. If you pour 0.0100 kg of water onto a 1.20-kg block of ice (which is initially at ), what is the final temperature? You may assume that the water cools so rapidly that effects of the surroundings are negligible. 87. Indigenous people sometimes cook in watertight baskets by placing hot rocks into water to bring it to a boil. What mass of granite must be placed in 4.00 kg of water to bring its temperature to , if 0.0250 kg of water escapes as vapor from the initial sizzle? You may neglect the effects of the surroundings. 88. What would the final temperature of the pan and water be in Example 1.7 if 0.260 kg of water were placed in the pan and 0.0100 kg of the water evaporated immediately, leaving the remainder to come to a common temperature with the pan? 1.6 Mechanisms of Heat Transfer 89. (a) Calculate the rate of heat conduction through house walls that are 13.0 cm thick and have an average thermal conductivity twice that of glass wool. Assume there are no windows or doors. The walls’ surface area is and their inside surface is at , while their outside surface is at . (b) How many 1-kW room heaters would be needed to balance the heat transfer due to conduction? 90. The rate of heat conduction out of a window on a winter day is rapid enough to chill the air next to it. To see just how rapidly the windows transfer heat by conduction, calculate the rate of conduction in watts through a window that is 0.634 cm thick (1/4 in.) if the temperatures of the inner and outer surfaces are and , respectively. (This rapid rate will not be maintained—the inner surface will cool, even to the point of frost formation.) 91. Calculate the rate of heat conduction out of the human body, assuming that the core internal temperature is , the skin temperature is , the thickness of the fatty tissues between the core and the skin averages 1.00 cm, and the surface area is . 92. Suppose you stand with one foot on ceramic flooring and one foot on a wool carpet, making contact over an area of with each foot. Both the ceramic and the carpet are 2.00 cm thick and are on their bottom sides. At what rate must heat transfer occur from each foot to keep the top of the ceramic and carpet at ? 93. A man consumes 3000 kcal of food in one day, converting most of it to thermal energy to maintain body temperature. If he loses half this energy by evaporating water (through breathing 56 1 • Chapter Review Access for free at openstax.org
and sweating), how many kilograms of water evaporate? 94. A firewalker runs across a bed of hot coals without sustaining burns. Calculate the heat transferred by conduction into the sole of one foot of a firewalker given that the bottom of the foot is a 3.00-mm-thick callus with a conductivity at the low end of the range for wood and its density is . The area of contact is the temperature of the coals is , and the time in contact is 1.00 s. Ignore the evaporative cooling of sweat. 95. (a) What is the rate of heat conduction through the 3.00-cm-thick fur of a large animal having a surface area? Assume that the animal’s skin temperature is , that the air temperature is , and that fur has the same thermal conductivity as air. (b) What food intake will the animal need in one day to replace this heat transfer? 96. A walrus transfers energy by conduction through its blubber at the rate of 150 W when immersed in water. The walrus’s internal core temperature is , and it has a surface area of . What is the average thickness of its blubber, which has the conductivity of fatty tissues without blood? 97. Compare the rate of heat conduction through a 13.0-cm-thick wall that has an area of and a thermal conductivity twice that of glass wool with the rate of heat conduction through a 0.750-cm-thick window that has an area of , assuming the same temperature difference across each. 98. Suppose a person is covered head to foot by wool clothing with average thickness of 2.00 cm and is transferring energy by conduction through the clothing at the rate of 50.0 W. What is the temperature difference across the clothing, given the surface area is ? 99. Some stove tops are smooth ceramic for easy cleaning. If the ceramic is 0.600 cm thick and heat conduction occurs through a 14 cm radius and at a rate of 2256 W, what is the temperature difference across it? Ceramic has the same thermal conductivity as glass and brick. 100. One easy way to reduce heating (and cooling) costs is to add extra insulation in the attic of a house. Suppose a single-story cubical house already had 15 cm of fiberglass insulation in the attic and in all the exterior surfaces. If you added an extra 8.0 cm of fiberglass to the attic, by what percentage would the heating cost of the house drop? Take the house to have dimensions 10 m by 15 m by 3.0 m. Ignore air infiltration and heat loss through windows and doors, and assume that the interior is uniformly at one temperature and the exterior is uniformly at another. 101. Many decisions are made on the basis of the payback period: the time it will take through savings to equal the capital cost of an investment. Acceptable payback times depend upon the business or philosophy one has. (For some industries, a payback period is as small as 2 years.) Suppose you wish to install the extra insulation in the preceding problem. If energy cost per million joules and the insulation was $4.00 per square meter, then calculate the simple payback time. Take the average for the 120-day heating season to be Additional Problems 102. In 1701, the Danish astronomer Ole Rømer proposed a temperature scale with two fixed points, freezing water at 7.5 degrees, and boiling water at 60.0 degrees. What is the boiling point of oxygen, 90.2 K, on the Rømer scale? 103. What is the percent error of thinking the melting point of tungsten is instead of the correct value of 3695 K? 104. An engineer wants to design a structure in which the difference in length between a steel beam and an aluminum beam remains at 0.500 m regardless of temperature, for ordinary temperatures. What must the lengths of the beams be? 105. How much stress is created in a steel beam if its temperature changes from to but it cannot expand? For steel, the Young’s modulus from Stress, Strain, and Elastic Modulus (http://openstax.org/ books/university-physics-volume-1/pages/ 12-3-stress-strain-and-elastic-modulus). (Ignore the change in area resulting from the expansion.) 106. A brass rod with a 1 • Chapter Review 57
diameter of 0.800 cm and a length of 1.20 m when the temperature is , is fixed at both ends. At what temperature is the force in it at 36,000 N? 107. A mercury thermometer still in use for meteorology has a bulb with a volume of and a tube for the mercury to expand into of inside diameter 0.130 mm. (a) Neglecting the thermal expansion of the glass, what is the spacing between marks apart? (b) If the thermometer is made of ordinary glass (not a good idea), what is the spacing? 108. Even when shut down after a period of normal use, a large commercial nuclear reactor transfers thermal energy at the rate of 150 MW by the radioactive decay of fission products. This heat transfer causes a rapid increase in temperature if the cooling system fails or and (a) Calculate the rate of temperature increase in degrees Celsius per second if the mass of the reactor core is and it has an average specific heat of . (b) How long would it take to obtain a temperature increase of , which could cause some metals holding the radioactive materials to melt? (The initial rate of temperature increase would be greater than that calculated here because the heat transfer is concentrated in a smaller mass. Later, however, the temperature increase would slow down because the 500,000-kg steel containment vessel would also begin to heat up.) 109. You leave a pastry in the refrigerator on a plate and ask your roommate to take it out before you get home so you can eat it at room temperature, the way you like it. Instead, your roommate plays video games for hours. When you return, you notice that the pastry is still cold, but the game console has become hot. Annoyed, and knowing that the pastry will not be good if it is microwaved, you warm up the pastry by unplugging the console and putting it in a clean trash bag (which acts as a perfect calorimeter) with the pastry on the plate. After a while, you find that the equilibrium temperature is a nice, warm . You know that the game console has a mass of 2.1 kg. Approximate it as having a uniform initial temperature of . The pastry has a mass of 0.16 kg and a specific heat of and is at a uniform initial temperature of . The plate is at the same temperature and has a mass of 0.24 kg and a specific heat of . What is the specific heat of the console? 110. Two solid spheres, A and B, made of the same material, are at temperatures of and , respectively. The spheres are placed in thermal contact in an ideal calorimeter, and they reach an equilibrium temperature of . Which is the bigger sphere? What is the ratio of their diameters? 111. In some countries, liquid nitrogen is used on dairy trucks instead of mechanical refrigerators. A 3.00-hour delivery trip requires 200 L of liquid nitrogen, which has a density of (a) Calculate the heat transfer necessary to evaporate this amount of liquid nitrogen and raise its temperature to . (Use and assume it is constant over the temperature range.) This value is the amount of cooling the liquid nitrogen supplies. (b) What is this heat transfer rate in kilowatt-hours? (c) Compare the amount of cooling obtained from melting an identical mass of ice with that from evaporating the liquid nitrogen. 112. Some gun fanciers make their own bullets, which involves melting lead and casting it into lead slugs. How much heat transfer is needed to raise the temperature and melt 0.500 kg of lead, starting from ? 113. A 0.800-kg iron cylinder at a temperature of is dropped into an insulated chest of 1.00 kg of ice at its melting point. What is the final temperature, and how much ice has melted? 114. Repeat the preceding problem with 2.00 kg of ice instead of 1.00 kg. 115. Repeat the preceding problem with 0.500 kg of ice, assuming that the ice is initially in a copper container of mass 1.50 kg in equilibrium with the ice. 116. A 30.0-g ice cube at its melting point is dropped into an aluminum calorimeter of mass 100.0 g in equilibrium at with 300.0 g of an unknown liquid. The final temperature is . What is the heat capacity of the liquid? 117. (a) Calculate the rate of heat conduction through a double-paned window that has a area and is made of two panes of 0.800-cm-thick glass separated by a 1.00-cm air gap. The inside 58 1 • Chapter Review Access for free at openstax.org
surface temperature is while that on the outside is (Hint: There are identical temperature drops across the two glass panes. First find these and then the temperature drop across the air gap. This problem ignores the increased heat transfer in the air gap due to convection.) (b) Calculate the rate of heat conduction through a 1.60-cm-thick window of the same area and with the same temperatures. Compare your answer with that for part (a). 118. (a) An exterior wall of a house is 3 m tall and 10 m wide. It consists of a layer of drywall with an R factor of 0.56, a layer 3.5 inches thick filled with fiberglass batts, and a layer of insulated siding with an R factor of 2.6. The wall is built so well that there are no leaks of air through it. When the inside of the wall is at and the outside is at , what is the rate of heat flow through the wall? (b) More realistically, the 3.5-inch space also contains 2-by-4 studs—wooden boards 1.5 inches by 3.5 inches oriented so that 3.5-inch dimension extends from the drywall to the siding. They are “on 16-inch centers,” that is, the centers of the studs are 16 inches apart. What is the heat current in this situation? Don’t worry about one stud more or less. 119. For the human body, what is the rate of heat transfer by conduction through the body’s tissue with the following conditions: the tissue thickness is 3.00 cm, the difference in temperature is , and the skin area is . How does this compare with the average heat transfer rate to the body resulting from an energy intake of about 2400 kcal per day? (No exercise is included.) 120. You have a Dewar flask (a laboratory vacuum flask) that has an open top and straight sides, as shown below. You fill it with water and put it into the freezer. It is effectively a perfect insulator, blocking all heat transfer, except on the top. After a time, ice forms on the surface of the water. The liquid water and the bottom surface of the ice, in contact with the liquid water, are at . The top surface of the ice is at the same temperature as the air in the freezer, Set the rate of heat flow through the ice equal to the rate of loss of heat of fusion as the water freezes. When the ice layer is 0.700 cm thick, find the rate in m/s at which the ice is thickening. 121. An infrared heater for a sauna has a surface area of and an emissivity of 0.84. What temperature must it run at if the required power is 360 W? Neglect the temperature of the environment. 122. (a) Determine the power of radiation from the Sun by noting that the intensity of the radiation at the distance of Earth is . Hint: That intensity will be found everywhere on a spherical surface with radius equal to that of Earth’s orbit. (b) Assuming that the Sun’s temperature is 5780 K and that its emissivity is 1, find its radius. Challenge Problems 123. A pendulum is made of a rod of length L and negligible mass, but capable of thermal expansion, and a weight of negligible size. (a) Show that when the temperature increases by dT, the period of the pendulum increases by a fraction . (b) A clock controlled by a brass pendulum keeps time correctly at . If the room temperature is , does the clock run faster or slower? What is its error in seconds per day? 124. At temperatures of a few hundred kelvins the specific heat capacity of copper approximately follows the empirical formula where and How much heat is needed to raise the temperature of a 2.00-kg piece of copper from to ? 125. In a calorimeter of negligible heat capacity, 200 1 • Chapter Review 59
g of steam at and 100 g of ice at are mixed. The pressure is maintained at 1 atm. What is the final temperature, and how much steam, ice, and water are present? 126. An astronaut performing an extra-vehicular activity (space walk) shaded from the Sun is wearing a spacesuit that can be approximated as perfectly white except for a patch in the form of the astronaut’s national flag. The patch has emissivity 0.300. The spacesuit under the patch is 0.500 cm thick, with a thermal conductivity , and its inner surface is at a temperature of . What is the temperature of the patch, and what is the rate of heat loss through it? Assume the patch is so thin that its outer surface is at the same temperature as the outer surface of the spacesuit under it. Also assume the temperature of outer space is 0 K. You will get an equation that is very hard to solve in closed form, so you can solve it numerically with a graphing calculator, with software, or even by trial and error with a calculator. 127. Find the growth of an ice layer as a function of time in a Dewar flask as seen in Exercise 1.120. Call the thickness of the ice layer L. (a) Derive an equation for dL/dt in terms of L , the temperature T above the ice, and the properties of ice (which you can leave in symbolic form instead of substituting the numbers). (b) Solve this differential equation assuming that at , you have If you have studied differential equations, you will know a technique for solving equations of this type: manipulate the equation to get dL/dt multiplied by a (very simple) function of L on one side, and integrate both sides with respect to time. Alternatively, you may be able to use your knowledge of the derivatives of various functions to guess the solution, which has a simple dependence on t. (c) Will the water eventually freeze to the bottom of the flask? 128. As the very first rudiment of climatology, estimate the temperature of Earth. Assume it is a perfect sphere and its temperature is uniform. Ignore the greenhouse effect. Thermal radiation from the Sun has an intensity (the “solar constant” S) of about at the radius of Earth’s orbit. (a) Assuming the Sun’s rays are parallel, what area must S be multiplied by to get the total radiation intercepted by Earth? It will be easiest to answer in terms of Earth’s radius, R. (b) Assume that Earth reflects about 30% of the solar energy it intercepts. In other words, Earth has an albedo with a value of . In terms of S, A, and R, what is the rate at which Earth absorbs energy from the Sun? (c) Find the temperature at which Earth radiates energy at the same rate. Assume that at the infrared wavelengths where it radiates, the emissivity e is 1. Does your result show that the greenhouse effect is important? (d) How does your answer depend on the the area of Earth? 129. Let’s stop ignoring the greenhouse effect and incorporate it into the previous problem in a very rough way. Assume the atmosphere is a single layer, a spherical shell around Earth, with an emissivity (chosen simply to give the right answer) at infrared wavelengths emitted by Earth and by the atmosphere. However, the atmosphere is transparent to the Sun’s radiation (that is, assume the radiation is at visible wavelengths with no infrared), so the Sun’s radiation reaches the surface. The greenhouse effect comes from the difference between the atmosphere’s transmission of visible light and its rather strong absorption of infrared. Note that the atmosphere’s radius is not significantly different from Earth’s, but since the atmosphere is a layer above Earth, it emits radiation both upward and downward, so it has twice Earth’s area. There are three radiative energy transfers in this problem: solar radiation absorbed by Earth’s surface; infrared radiation from the surface, which is absorbed by the atmosphere according to its emissivity; and infrared radiation from the atmosphere, half of which is absorbed by Earth and half of which goes out into space. Apply the method of the previous problem to get an equation for Earth’s surface and one for the atmosphere, and solve them for the two unknown temperatures, surface and atmosphere. a. In terms of Earth’s radius, the constant , and the unknown temperature of the surface, what is the power of the infrared radiation from the surface? b. What is the power of Earth’s radiation absorbed by the atmosphere? c. In terms of the unknown temperature of the atmosphere, what is the power radiated from the atmosphere? d. Write an equation that says the power of 60 1 • Chapter Review Access for free at openstax.org
the radiation the atmosphere absorbs from Earth equals the power of the radiation it emits. e. Half of the power radiated by the atmosphere hits Earth. Write an equation that says that the power Earth absorbs from the atmosphere and the Sun equals the power that it emits. f. Solve your two equations for the unknown temperature of Earth. 1 • Chapter Review 61
62 1 • Chapter Review Access for free at openstax.org
INTRODUCTION CHAPTER 2 The Kinetic Theory of Gases 2.1 Molecular Model of an Ideal Gas 2.2 Pressure, Temperature, and RMS Speed 2.3 Heat Capacity and Equipartition of Energy 2.4 Distribution of Molecular Speeds Gases are literally all around us—the air that we breathe is a mixture of gases. Other gases include those that make breads and cakes soft, those that make drinks fizzy, and those that burn to heat many homes. Engines and refrigerators depend on the behaviors of gases, as we will see in later chapters. As we discussed in the preceding chapter, the study of heat and temperature is part of an area of physics known as thermodynamics, in which we require a system to be macroscopic, that is, to consist of a huge number (such as ) of molecules. We begin by considering some macroscopic properties of gases: volume, pressure, and temperature. The simple model of a hypothetical “ideal gas” describes these properties of a gas very accurately under many conditions. We move from the ideal gas model to a more widely applicable approximation, called the Van der Waals model. To understand gases even better, we must also look at them on the microscopic scale of molecules. In gases, the molecules interact weakly, so the microscopic behavior of gases is relatively simple, and they serve as a good introduction to systems of many molecules. The molecular model of gases is called the kinetic theory of gases and is one of the classic examples of a molecular model that explains everyday behavior. FIGURE 2.1 A volcanic eruption releases tons of gas and dust into the atmosphere. Most of the gas is water vapor, but several other gases are common, including greenhouse gases such as carbon dioxide and acidic pollutants such as sulfur dioxide. However, the emission of volcanic gas is not all bad: Many geologists believe that in the earliest stages of Earth’s formation, volcanic emissions formed the early atmosphere. (credit: modification of work by “Boaworm”/Wikimedia Commons) CHAPTER OUTLINE
2.1 Molecular Model of an Ideal Gas LEARNING OBJECTIVES By the end of this section, you will be able to: • Apply the ideal gas law to situations involving the pressure, volume, temperature, and the number of molecules of a gas • Use the unit of moles in relation to numbers of molecules, and molecular and macroscopic masses • Explain the ideal gas law in terms of moles rather than numbers of molecules • Apply the van der Waals gas law to situations where the ideal gas law is inadequate In this section, we explore the thermal behavior of gases. Our word “gas” comes from the Flemish word meaning “chaos,” first used for vapors by the seventeenth-century chemist J. B. van Helmont. The term was more appropriate than he knew, because gases consist of molecules moving and colliding with each other at random. This randomness makes the connection between the microscopic and macroscopic domains simpler for gases than for liquids or solids. How do gases differ from solids and liquids? Under ordinary conditions, such as those of the air around us, the difference is that the molecules of gases are much farther apart than those of solids and liquids. Because the typical distances between molecules are large compared to the size of a molecule, as illustrated in Figure 2.2, the forces between them are considered negligible, except when they come into contact with each other during collisions. Also, at temperatures well above the boiling temperature, the motion of molecules is fast, and the gases expand rapidly to occupy all of the accessible volume. In contrast, in liquids and solids, molecules are closer together, and the behavior of molecules in liquids and solids is highly constrained by the molecules’ interactions with one another. The macroscopic properties of such substances depend strongly on the forces between the molecules, and since many molecules are interacting, the resulting “many-body problems” can be extremely complicated (see Condensed Matter Physics (http://openstax.org/books/university-physics-volume-3/pages/9-introduction)). FIGURE 2.2 Atoms and molecules in a gas are typically widely separated. Because the forces between them are quite weak at these distances, the properties of a gas depend more on the number of atoms per unit volume and on temperature than on the type of atom. The Gas Laws In the previous chapter, we saw one consequence of the large intermolecular spacing in gases: Gases are easily compressed. Table 1.2 shows that gases have larger coefficients of volume expansion than either solids or liquids. These large coefficients mean that gases expand and contract very rapidly with temperature changes. We also saw (in the section on thermal expansion) that most gases expand at the same rate or have the same coefficient of volume expansion, . This raises a question: Why do all gases act in nearly the same way, when all the various liquids and solids have widely varying expansion rates? To study how the pressure, temperature, and volume of a gas relate to one another, consider what happens when you pump air into a deflated car tire. The tire’s volume first increases in direct proportion to the amount of air injected, without much increase in the tire pressure. Once the tire has expanded to nearly its full size, the tire’s walls limit its volume expansion. If we continue to pump air into the tire, the pressure increases. When the car is driven and the tires flex, their temperature increases, and therefore the pressure increases even further (Figure 2.3). 64 2 • The Kinetic Theory of Gases Access for free at openstax.org
FIGURE 2.3 (a) When air is pumped into a deflated tire, its volume first increases without much increase in pressure. (b) When the tire is filled to a certain point, the tire walls resist further expansion, and the pressure increases with more air. (c) Once the tire is inflated, its pressure increases with temperature. Figure 2.4 shows data from the experiments of Robert Boyle (1627–1691), illustrating what is now called Boyle’s law: At constant temperature and number of molecules, the absolute pressure of a gas and its volume are inversely proportional. (Recall from Fluid Mechanics (http://openstax.org/books/university-physics-volume-1/pages/ 14-introduction) that the absolute pressure is the true pressure and the gauge pressure is the absolute pressure minus the ambient pressure, typically atmospheric pressure.) The graph in Figure 2.4 displays this relationship as an inverse proportionality of volume to pressure. FIGURE 2.4 Robert Boyle and his assistant found that volume and pressure are inversely proportional. Here their data are plotted as V versus 1/ p; the linearity of the graph shows the inverse proportionality. The number shown as the volume is actually the height in inches of air in a cylindrical glass tube. The actual volume was that height multiplied by the cross-sectional area of the tube, which Boyle did not publish. The data are from Boyle’s book A Defence of the Doctrine Touching the Spring and Weight of the Air…, p. 60.1 Figure 2.5 shows experimental data illustrating what is called Charles’s law, after Jacques Charles (1746–1823). Charles’s law states that at constant pressure and number of molecules, the volume of a gas is proportional to its absolute temperature. 1 http://bvpb.mcu.es/en/consulta/registro.cmd?id=406806 2.1 • Molecular Model of an Ideal Gas 65
FIGURE 2.5 Experimental data showing that at constant pressure, volume is approximately proportional to temperature. The best-fit line passes approximately through the origin.2 Similar is Amonton’s or Gay-Lussac’s law, which states that at constant volume and number of molecules, the pressure is proportional to the temperature. That law is the basis of the constant-volume gas thermometer, discussed in the previous chapter. (The histories of these laws and the appropriate credit for them are more complicated than can be discussed here.) It is known experimentally that for gases at low density (such that their molecules occupy a negligible fraction of the total volume) and at temperatures well above the boiling point, these proportionalities hold to a good approximation. Not surprisingly, with the other quantities held constant, either pressure or volume is proportional to the number of molecules. More surprisingly, when the proportionalities are combined into a single equation, the constant of proportionality is independent of the composition of the gas. The resulting equation for all gases applies in the limit of low density and high temperature; it’s the same for oxygen as for helium or uranium hexafluoride. A gas at that limit is called an ideal gas; it obeys the ideal gas law, which is also called the equation of state of an ideal gas. The constant is called the Boltzmann constant in honor of the Austrian physicist Ludwig Boltzmann (1844–1906) and has the value The ideal gas law describes the behavior of any real gas when its density is low enough or its temperature high enough that it is far from liquefaction. This encompasses many practical situations. In the next section, we’ll see why it’s independent of the type of gas. In many situations, the ideal gas law is applied to a sample of gas with a constant number of molecules; for instance, the gas may be in a sealed container. If N is constant, then solving for N shows that pV /T is constant. We IDEAL GAS LAW The ideal gas law states that where p is the absolute pressure of a gas, V is the volume it occupies, N is the number of molecules in the gas, and T is its absolute temperature. 2.1 2 http://chemed.chem.purdue.edu/genchem/history/charles.html 66 2 • The Kinetic Theory of Gases Access for free at openstax.org
can write that fact in a convenient form: where the subscripts 1 and 2 refer to any two states of the gas at different times. Again, the temperature must be expressed in kelvin and the pressure must be absolute pressure, which is the sum of gauge pressure and atmospheric pressure. EXAMPLE 2.1 Calculating Pressure Changes Due to Temperature Changes Suppose your bicycle tire is fully inflated, with an absolute pressure of (a gauge pressure of just under ) at a temperature of What is the pressure after its temperature has risen to on a hot day? Assume there are no appreciable leaks or changes in volume. Strategy The pressure in the tire is changing only because of changes in temperature. We know the initial pressure the initial temperature and the final temperature We must find the final pressure Since the number of molecules is constant, we can use the equation Since the volume is constant, and are the same and they divide out. Therefore, We can then rearrange this to solve for where the temperature must be in kelvin. Solution 1. Convert temperatures from degrees Celsius to kelvin 2. Substitute the known values into the equation, Significance The final temperature is about greater than the original temperature, so the final pressure is about greater as well. Note that absolute pressure (see Fluid Mechanics (http://openstax.org/books/university-physics-volume-1/ pages/14-introduction)) and absolute temperature (see Temperature and Heat) must be used in the ideal gas law. EXAMPLE 2.2 Calculating the Number of Molecules in a Cubic Meter of Gas How many molecules are in a typical object, such as gas in a tire or water in a glass? This calculation can give us an idea of how large N typically is. Let’s calculate the number of molecules in the air that a typical healthy young adult 2.2 2.1 • Molecular Model of an Ideal Gas 67
inhales in one breath, with a volume of 500 mL, at standard temperature and pressure (STP), which is defined as and atmospheric pressure. (Our young adult is apparently outside in winter.) Strategy Because pressure, volume, and temperature are all specified, we can use the ideal gas law, to find N. Solution 1. Identify the knowns. 2. Substitute the known values into the equation and solve for N. SignificanceN is huge, even in small volumes. For example, of a gas at STP contains molecules. Once again, note that our result for N is the same for all types of gases, including mixtures. As we observed in the chapter on fluid mechanics, pascals are , so Thus, our result for N is dimensionless, a pure number that could be obtained by counting (in principle) rather than measuring. As it is the number of molecules, we put “molecules” after the number, keeping in mind that it is an aid to communication rather than a unit. Moles and Avogadro’s Number It is often convenient to measure the amount of substance with a unit on a more human scale than molecules. The SI unit for this purpose was developed by the Italian scientist Amedeo Avogadro (1776–1856). (He worked from the hypothesis that equal volumes of gas at equal pressure and temperature contain equal numbers of molecules, independent of the type of gas. As mentioned above, this hypothesis has been confirmed when the ideal gas approximation applies.) A mole (abbreviated mol) is the base SI unit representing the number of elementary entities (atoms, molecules, particles, etc.) and was historically defined as the number of atoms in 12 grams of 12C. The number of molecules in one mole is called Avogadro’s number and the value of Avogadro’s number is now known to be We can now write , where n represents the number of moles of a substance. Avogadro’s number relates the mass of an amount of substance in grams to the number of protons and neutrons in an atom or molecule (12 for a carbon-12 atom), which roughly determine its mass. It’s natural to define a unit of mass such that the mass of an atom is approximately equal to its number of neutrons and protons. The unit of that kind accepted for use with the SI is the unified atomic mass unit (u), also called the dalton. Specifically, a carbon-12 atom has a mass of exactly 12 u, so that its molar mass M in grams per mole is numerically equal to the mass of one carbon-12 atom in u. That equality holds for any substance. In other words, is not only the conversion from numbers of molecules to moles, but it is also the conversion from u to grams: See Figure 2.6. FIGURE 2.6 How big is a mole? On a macroscopic level, Avogadro’s number of table tennis balls would cover Earth to a depth of about 40 km. 68 2 • The Kinetic Theory of Gases Access for free at openstax.org
Now letting stand for the mass of a sample of a substance, we have Letting m stand for the mass of a molecule, we have CHECK YOUR UNDERSTANDING 2.1 The recommended daily amount of vitamin or niacin, for women who are not pregnant or nursing, is 14 mg. Find the number of molecules of niacin in that amount. CHECK YOUR UNDERSTANDING 2.2 The density of air in a classroom and is . At what pressure is the density if the temperature is kept constant? The Ideal Gas Law Restated using Moles A very common expression of the ideal gas law uses the number of moles in a sample, n, rather than the number of molecules, N. We start from the ideal gas law, and multiply and divide the right-hand side of the equation by Avogadro’s number This gives us Note that is the number of moles. We define the universal gas constant as and obtain the ideal gas law in terms of moles. EXAMPLE 2.3 Density of Air at STP and in a Hot Air Balloon Calculate the density of dry air (a) under standard conditions and (b) in a hot air balloon at a temperature of . Dry air is approximately and . Strategy and Solution a. We are asked to find the density, or mass per cubic meter. We can begin by finding the molar mass. If we have a hundred molecules, of which 78 are nitrogen, 21 are oxygen, and 1 is argon, the average molecular mass is , or the mass of each constituent multiplied by its percentage. The same applies to the IDEAL GAS LAW (IN TERMS OF MOLES) In terms of number of moles n, the ideal gas law is written as In SI units, In other units, You can use whichever value of R is most convenient for a particular problem. 2.3 2.1 • Molecular Model of an Ideal Gas 69
molar mass, which therefore is Now we can find the number of moles per cubic meter. We use the ideal gas law in terms of moles, with , , , and . The most convenient choice for R in this case is because the known quantities are in SI units: Then, the mass of that air is Finally the density of air at STP is b. The air pressure inside the balloon is still 1 atm because the bottom of the balloon is open to the atmosphere. The calculation is the same except that we use a temperature of , which is 393 K. We can repeat the calculation in (a), or simply observe that the density is proportional to the number of moles, which is inversely proportional to the temperature. Then using the subscripts 1 for air at STP and 2 for the hot air, we have Significance Using the methods of Archimedes’ Principle and Buoyancy (http://openstax.org/books/university-physics-volume-1/ pages/14-4-archimedes-principle-and-buoyancy), we can find that the net force on of air at is or enough to lift about 867 kg. The mass density and molar density of air at STP, found above, are often useful numbers. From the molar density, we can easily determine another useful number, the volume of a mole of any ideal gas at STP, which is 22.4 L. CHECK YOUR UNDERSTANDING 2.3 Liquids and solids have densities on the order of 1000 times greater than gases. Explain how this implies that the distances between molecules in gases are on the order of 10 times greater than the size of their molecules. The ideal gas law is closely related to conservation of energy: The units on both sides of the equation are joules. The right-hand side of the ideal gas law equation is This term is roughly the total translational kinetic energy (which, when discussing gases, refers to the energy of translation of a molecule, not that of vibration of its atoms or rotation) of N molecules at an absolute temperature T, as we will see formally in the next section. The left-hand side of the ideal gas law equation is pV. As mentioned in the example on the number of molecules in an ideal gas, pressure multiplied by volume has units of energy. The energy of a gas can be changed when the gas does work as it increases in volume, something we explored in the preceding chapter, and the amount of work is related to the pressure. This is the process that occurs in gasoline or steam engines and turbines, as we’ll see in the next chapter. PROBLEM-SOLVING STRATEGY The Ideal Gas Law Step 1. Examine the situation to determine that an ideal gas is involved. Most gases are nearly ideal unless they are close to the boiling point or at pressures far above atmospheric pressure. Step 2. Make a list of what quantities are given or can be inferred from the problem as stated (identify the known quantities). 70 2 • The Kinetic Theory of Gases Access for free at openstax.org
Step 3. Identify exactly what needs to be determined in the problem (identify the unknown quantities). A written list is useful. Step 4. Determine whether the number of molecules or the number of moles is known or asked for to decide whether to use the ideal gas law as where N is the number of molecules, or where n is the number of moles. Step 5. Convert known values into proper SI units (K for temperature, Pa for pressure, for volume, molecules forN, and moles for n). If the units of the knowns are consistent with one of the non-SI values of R, you can leave them in those units. Be sure to use absolute temperature and absolute pressure. Step 6. Solve the ideal gas law for the quantity to be determined (the unknown quantity). You may need to take a ratio of final states to initial states to eliminate the unknown quantities that are kept fixed. Step 7. Substitute the known quantities, along with their units, into the appropriate equation and obtain numerical solutions complete with units. Step 8. Check the answer to see if it is reasonable: Does it make sense? The Van der Waals Equation of State We have repeatedly noted that the ideal gas law is an approximation. How can it be improved upon? The van der Waals equation of state (named after the Dutch physicist Johannes van der Waals, 1837−1923) improves it by taking into account two factors. First, the attractive forces between molecules, which are stronger at higher density and reduce the pressure, are taken into account by adding to the pressure a term equal to the square of the molar density multiplied by a positive coefficient a. Second, the volume of the molecules is represented by a positive constant b, which can be thought of as the volume of a mole of molecules. This is subtracted from the total volume to give the remaining volume that the molecules can move in. The constants a and b are determined experimentally for each gas. The resulting equation is In the limit of low density (small n), the a and b terms are negligible, and we have the ideal gas law, as we should for low density. On the other hand, if is small, meaning that the molecules are very close together, the pressure must be higher to give the same nRT, as we would expect in the situation of a highly compressed gas. However, the increase in pressure is less than that argument would suggest, because at high density the term is significant. Since it’s positive, it causes a lower pressure to give the same nRT. The van der Waals equation of state works well for most gases under a wide variety of conditions. As we’ll see in the next module, it even predicts the gas-liquid transition. pV Diagrams We can examine aspects of the behavior of a substance by plotting a pV diagram, which is a graph of pressure versus volume. When the substance behaves like an ideal gas, the ideal gas law describes the relationship between its pressure and volume. On a pV diagram, it’s common to plot an isotherm, which is a curve showing p as a function of V with the number of molecules and the temperature fixed. Then, for an ideal gas, For example, the volume of the gas decreases as the pressure increases. The resulting graph is a hyperbola. However, if we assume the van der Waals equation of state, the isotherms become more interesting, as shown in Figure 2.7. At high temperatures, the curves are approximately hyperbolas, representing approximately ideal behavior at various fixed temperatures. At lower temperatures, the curves look less and less like hyperbolas—that is, the gas is not behaving ideally. There is a critical temperature at which the curve has a point with zero slope. Below that temperature, the curves do not decrease monotonically; instead, they each have a “hump,” meaning that for a certain range of volume, increasing the volume increases the pressure. 2.4 2.1 • Molecular Model of an Ideal Gas 71
FIGURE 2.7 pV diagram for a Van der Waals gas at various temperatures. The red curves are calculated at temperatures above the critical temperature and the blue curves at temperatures below it. The blue curves have an oscillation in which volume (V) increases with increasing pressure (P), an impossible situation, so they must be corrected as in Figure 2.8. (credit: “Eman”/Wikimedia Commons) Such behavior would be completely unphysical. Instead, the curves are understood as describing a liquid-gas phase transition. The oscillating part of the curve is replaced by a horizontal line, showing that as the volume increases at constant temperature, the pressure stays constant. That behavior corresponds to boiling and condensation; when a substance is at its boiling temperature for a particular pressure, it can increase in volume as some of the liquid turns to gas, or decrease as some of the gas turns to liquid, without any change in temperature or pressure. Figure 2.8 shows similar isotherms that are more realistic than those based on the van der Waals equation. The steep parts of the curves to the left of the transition region show the liquid phase, which is almost incompressible—a slight decrease in volume requires a large increase in pressure. The flat parts show the liquid-gas transition; the blue regions that they define represent combinations of pressure and volume where liquid and gas can coexist. FIGURE 2.8 pV diagrams. (a) Each curve (isotherm) represents the relationship between p and V at a fixed temperature; the upper curves are at higher temperatures. The lower curves are not hyperbolas because the gas is no longer an ideal gas. (b) An expanded portion of the 72 2 • The Kinetic Theory of Gases Access for free at openstax.org
pV diagram for low temperatures, where the phase can change from a gas to a liquid. The term “vapor” refers to the gas phase when it exists at a temperature below the boiling temperature. The isotherms above do not go through the liquid-gas transition. Therefore, liquid cannot exist above that temperature, which is the critical temperature (described in the chapter on temperature and heat). At sufficiently low pressure above that temperature, the gas has the density of a liquid but will not condense; the gas is said to be supercritical. At higher pressure, it is solid. Carbon dioxide, for example, has no liquid phase at a temperature above . The critical pressure is the maximum pressure at which the liquid can exist. The point on the pV diagram at the critical pressure and temperature is the critical point (which you learned about in the chapter on temperature and heat). Table 2.1 lists representative critical temperatures and pressures. Substance Critical temperature Critical pressure K Pa atm Water 647.4 374.3 219.0 Sulfur dioxide 430.7 157.6 78.0 Ammonia 405.5 132.4 111.7 Carbon dioxide 304.2 31.1 73.2 Oxygen 154.8 –118.4 50.3 Nitrogen 126.2 –146.9 33.6 Hydrogen 33.3 –239.9 12.9 Helium 5.3 –267.9 2.27 TABLE 2.1 Critical Temperatures and Pressures for Various Substances 2.2 Pressure, Temperature, and RMS Speed LEARNING OBJECTIVES By the end of this section, you will be able to: • Explain the relations between microscopic and macroscopic quantities in a gas • Solve problems involving mixtures of gases • Solve problems involving the distance and time between a gas molecule’s collisions We have examined pressure and temperature based on their macroscopic definitions. Pressure is the force divided by the area on which the force is exerted, and temperature is measured with a thermometer. We can gain a better understanding of pressure and temperature from the kinetic theory of gases, the theory that relates the macroscopic properties of gases to the motion of the molecules they consist of. First, we make two assumptions about molecules in an ideal gas. 1. There is a very large number N of molecules, all identical and each having mass m. 2. The molecules obey Newton’s laws and are in continuous motion, which is random and isotropic, that is, the same in all directions. To derive the ideal gas law and the connection between microscopic quantities such as the energy of a typical molecule and macroscopic quantities such as temperature, we analyze a sample of an ideal gas in a rigid container, about which we make two further assumptions: 3. The molecules are much smaller than the average distance between them, so their total volume is much less than that of their container (which has volume V). In other words, we take the Van der Waals constant b, the 2.2 • Pressure, Temperature, and RMS Speed 73
volume of a mole of gas molecules, to be negligible compared to the volume of a mole of gas in the container. 4. The molecules make perfectly elastic collisions with the walls of the container and with each other. Other forces on them, including gravity and the attractions represented by the Van der Waals constant a, are negligible (as is necessary for the assumption of isotropy). The collisions between molecules do not appear in the derivation of the ideal gas law. They do not disturb the derivation either, since collisions between molecules moving with random velocities give new random velocities. Furthermore, if the velocities of gas molecules in a container are initially not random and isotropic, molecular collisions are what make them random and isotropic. We make still further assumptions that simplify the calculations but do not affect the result. First, we let the container be a rectangular box. Second, we begin by considering monatomic gases, those whose molecules consist of single atoms, such as helium. Then, we can assume that the atoms have no energy except their translational kinetic energy; for instance, they have neither rotational nor vibrational energy. (Later, we discuss the validity of this assumption for real monatomic gases and dispense with it to consider diatomic and polyatomic gases.) Figure 2.9 shows a collision of a gas molecule with the wall of a container, so that it exerts a force on the wall (by Newton’s third law). These collisions are the source of pressure in a gas. As the number of molecules increases, the number of collisions, and thus the pressure, increases. Similarly, if the average velocity of the molecules is higher, the gas pressure is higher. FIGURE 2.9 When a molecule collides with a rigid wall, the component of its momentum perpendicular to the wall is reversed. A force is thus exerted on the wall, creating pressure. In a sample of gas in a container, the randomness of the molecular motion causes the number of collisions of molecules with any part of the wall in a given time to fluctuate. However, because a huge number of molecules collide with the wall in a short time, the number of collisions on the scales of time and space we measure fluctuates by only a tiny, usually unobservable fraction from the average. We can compare this situation to that of a casino, where the outcomes of the bets are random and the casino’s takings fluctuate by the minute and the hour. However, over long times such as a year, the casino’s takings are very close to the averages expected from the odds. A tank of gas has enormously more molecules than a casino has bettors in a year, and the molecules make enormously more collisions in a second than a casino has bets. A calculation of the average force exerted by molecules on the walls of the box leads us to the ideal gas law and to the connection between temperature and molecular kinetic energy. (In fact, we will take two averages: one over time to get the average force exerted by one molecule with a given velocity, and then another average over molecules with different velocities.) This approach was developed by Daniel Bernoulli (1700–1782), who is best 74 2 • The Kinetic Theory of Gases Access for free at openstax.org
known in physics for his work on fluid flow (hydrodynamics). Remarkably, Bernoulli did this work before Dalton established the view of matter as consisting of atoms. Figure 2.10 shows a container full of gas and an expanded view of an elastic collision of a gas molecule with a wall of the container, broken down into components. We have assumed that a molecule is small compared with the separation of molecules in the gas, and that its interaction with other molecules can be ignored. Under these conditions, the ideal gas law is experimentally valid. Because we have also assumed the wall is rigid and the particles are points, the collision is elastic (by conservation of energy—there’s nowhere for a particle’s kinetic energy to go). Therefore, the molecule’s kinetic energy remains constant, and hence, its speed and the magnitude of its momentum remain constant as well. This assumption is not always valid, but the results in the rest of this module are also obtained in models that let the molecules exchange energy and momentum with the wall. FIGURE 2.10 Gas in a box exerts an outward pressure on its walls. A molecule colliding with a rigid wall has its velocity and momentum in the x-direction reversed. This direction is perpendicular to the wall. The components of its velocity momentum in the y- and z-directions are not changed, which means there is no force parallel to the wall. If the molecule’s velocity changes in the x-direction, its momentum changes from to Thus, its change in momentum is According to the impulse-momentum theorem given in the chapter on linear momentum and collisions, the force exerted on the ith molecule, where i labels the molecules from 1 to N, is given by (In this equation alone, p represents momentum, not pressure.) There is no force between the wall and the molecule except while the molecule is touching the wall. During the short time of the collision, the force between the molecule and wall is relatively large, but that is not the force we are looking for. We are looking for the average force, so we take to be the average time between collisions of the given molecule with this wall, which is the time in which we expect to find one collision. Let l represent the length of the box in the x-direction. Then is the time the molecule would take to go across the box and back, a distance 2l, at a speed of Thus and the expression for the force becomes This force is due to one molecule. To find the total force on the wall, F, we need to add the contributions of all N 2.2 • Pressure, Temperature, and RMS Speed 75
molecules: We now use the definition of the average, which we denote with a bar, to find the force: We want the force in terms of the speed v, rather than the x-component of the velocity. Note that the total velocity squared is the sum of the squares of its components, so that With the assumption of isotropy, the three averages on the right side are equal, so Substituting this into the expression for F gives The pressure is F/A, so we obtain where we used for the volume. This gives the important result Combining this equation with gives We can get the average kinetic energy of a molecule, , from the left-hand side of the equation by dividing outN and multiplying by 3/2. The equation is the average kinetic energy per molecule. Note in particular that nothing in this equation depends on the molecular mass (or any other property) of the gas, the pressure, or anything but the temperature. If samples of helium and xenon gas, with very different molecular masses, are at the same temperature, the molecules have the same average kinetic energy. The internal energy of a thermodynamic system is the sum of the mechanical energies of all of the molecules in it. We can now give an equation for the internal energy of a monatomic ideal gas. In such a gas, the molecules’ only energy is their translational kinetic energy. Therefore, denoting the internal energy by we simply have or 2.5 AVERAGE KINETIC ENERGY PER MOLECULE The average kinetic energy of a molecule is directly proportional to its absolute temperature: 2.6 76 2 • The Kinetic Theory of Gases Access for free at openstax.org
Often we would like to use this equation in terms of moles: We can solve for a typical speed of a molecule in an ideal gas in terms of temperature to determine what is known as the root-mean-square ( rms) speed of a molecule. The rms speed is not the average or the most likely speed of molecules, as we will see in Distribution of Molecular Speeds, but it provides an easily calculated estimate of the molecules’ speed that is related to their kinetic energy. Again we can write this equation in terms of the gas constant R and the molar mass M in kg/mol: We digress for a moment to answer a question that may have occurred to you: When we apply the model to atoms instead of theoretical point particles, does rotational kinetic energy change our results? To answer this question, we have to appeal to quantum mechanics. In quantum mechanics, rotational kinetic energy cannot take on just any value; it’s limited to a discrete set of values, and the smallest value is inversely proportional to the rotational inertia. The rotational inertia of an atom is tiny because almost all of its mass is in the nucleus, which typically has a radius less than . Thus the minimum rotational energy of an atom is much more than for any attainable temperature, and the energy available is not enough to make an atom rotate. We will return to this point when discussing diatomic and polyatomic gases in the next section. EXAMPLE 2.4 Calculating Kinetic Energy and Speed of a Gas Molecule (a) What is the average kinetic energy of a gas molecule at (room temperature)? (b) Find the rms speed of a nitrogen molecule at this temperature. Strategy (a) The known in the equation for the average kinetic energy is the temperature: Before substituting values into this equation, we must convert the given temperature into kelvin: We can find the rms speed of a nitrogen molecule by using the equation but we must first find the mass of a nitrogen molecule. Obtaining the molar mass of nitrogen from the periodic table, we find 2.7 RMS SPEED OF A MOLECULE The root-mean-square (rms) speed of a molecule, or the square root of the average of the square of the speed , is 2.8 2.9 2.2 • Pressure, Temperature, and RMS Speed 77
Solution a. The temperature alone is sufficient for us to find the average translational kinetic energy. Substituting the temperature into the translational kinetic energy equation gives b. Substituting this mass and the value for into the equation for yields Significance Note that the average kinetic energy of the molecule is independent of the type of molecule. The average translational kinetic energy depends only on absolute temperature. The kinetic energy is very small compared to macroscopic energies, so that we do not feel when an air molecule is hitting our skin. On the other hand, it is much greater than the typical difference in gravitational potential energy when a molecule moves from, say, the top to the bottom of a room, so our neglect of gravitation is justified in typical real-world situations. The rms speed of the nitrogen molecule is surprisingly large. These large molecular velocities do not yield macroscopic movement of air, since the molecules move in all directions with equal likelihood. The mean free path (the distance a molecule moves on average between collisions, discussed a bit later in this section) of molecules in air is very small, so the molecules move rapidly but do not get very far in a second. The high value for rms speed is reflected in the speed of sound, which is about 340 m/s at room temperature. The higher the rms speed of air molecules, the faster sound vibrations can be transferred through the air. The speed of sound increases with temperature and is greater in gases with small molecular masses, such as helium (see Figure 2.11). FIGURE 2.11 (a) In an ordinary gas, so many molecules move so fast that they collide billions of times every second. (b) Individual molecules do not move very far in a small amount of time, but disturbances like sound waves are transmitted at speeds related to the molecular speeds. EXAMPLE 2.5 Calculating Temperature: Escape Velocity of Helium Atoms To escape Earth’s gravity, an object near the top of the atmosphere (at an altitude of 100 km) must travel away from Earth at 11.1 km/s. This speed is called the escape velocity. At what temperature would helium atoms have an rms 78 2 • The Kinetic Theory of Gases Access for free at openstax.org
speed equal to the escape velocity? Strategy Identify the knowns and unknowns and determine which equations to use to solve the problem. Solution 1. Identify the knowns: v is the escape velocity, 11.1 km/s. 2. Identify the unknowns: We need to solve for temperature, T. We also need to solve for the mass m of the helium atom. 3. Determine which equations are needed. ◦ To get the mass m of the helium atom, we can use information from the periodic table: ◦ To solve for temperature T, we can rearrange to yield 4. Substitute the known values into the equations and solve for the unknowns, and Significance This temperature is much higher than atmospheric temperature, which is approximately 250 K at high elevation. Very few helium atoms are left in the atmosphere, but many were present when the atmosphere was formed, and more are always being created by radioactive decay (see the chapter on nuclear physics). The reason for the loss of helium atoms is that a small number of helium atoms have speeds higher than Earth’s escape velocity even at normal temperatures. The speed of a helium atom changes from one collision to the next, so that at any instant, there is a small but nonzero chance that the atom’s speed is greater than the escape velocity. The chance is high enough that over the lifetime of Earth, almost all the helium atoms that have been in the atmosphere have reached escape velocity at high altitudes and escaped from Earth’s gravitational pull. Heavier molecules, such as oxygen, nitrogen, and water, have smaller rms speeds, and so it is much less likely that any of them will have speeds greater than the escape velocity. In fact, the likelihood is so small that billions of years are required to lose significant amounts of heavier molecules from the atmosphere. Figure 2.12 shows the effect of a lack of an atmosphere on the Moon. Because the gravitational pull of the Moon is much weaker, it has lost almost its entire atmosphere. The atmospheres of Earth and other bodies are compared in this chapter’s exercises. 2.2 • Pressure, Temperature, and RMS Speed 79
FIGURE 2.12 This photograph of Apollo 17 Commander Eugene Cernan driving the lunar rover on the Moon in 1972 looks as though it was taken at night with a large spotlight. In fact, the light is coming from the Sun. Because the acceleration due to gravity on the Moon is so low (about 1/6 that of Earth), the Moon’s escape velocity is much smaller. As a result, gas molecules escape very easily from the Moon, leaving it with virtually no atmosphere. Even during the daytime, the sky is black because there is no gas to scatter sunlight. (credit: Harrison H. Schmitt/NASA) CHECK YOUR UNDERSTANDING 2.4 If you consider a very small object, such as a grain of pollen, in a gas, then the number of molecules striking its surface would also be relatively small. Would you expect the grain of pollen to experience any fluctuations in pressure due to statistical fluctuations in the number of gas molecules striking it in a given amount of time? Vapor Pressure, Partial Pressure, and Dalton’s Law If two or more gases are mixed, they will come to thermal equilibrium as a result of collisions between molecules; the process is analogous to heat conduction as described in the chapter on temperature and heat. As we have seen from kinetic theory, when the gases have the same temperature, their molecules have the same average kinetic energy. Thus, each gas obeys the ideal gas law separately and exerts the same pressure on the walls of a container that it would if it were alone. The pressure that a gas in a mixture would create if only it occupied the total volume available is called the gas’s partial pressure. Therefore, in a mixture of gases, the total pressure is the sum of partial pressures of the component gases, assuming ideal gas behavior and no chemical reactions between the components. This law is known as Dalton’s law of partial pressures, after the English scientist John Dalton (1766–1844) who proposed it. Dalton’s law is consistent with the fact that pressures add according to Pascal’s principle. In a mixture of ideal gases in thermal equilibrium, the number of molecules of each gas is proportional to its partial pressure. This result follows from applying the ideal gas law to each in the form Because the right- hand side is the same for any gas at a given temperature in a container of a given volume, the left-hand side is the same as well. • Partial pressure is the pressure a gas would create if it existed alone. • Dalton’s law states that the total pressure is the sum of the partial pressures of all of the gases present. • For any two gases (labeled 1 and 2) in equilibrium in a container, An important application of partial pressure is that, in chemistry, it functions as the concentration of a gas in determining the rate of a reaction. Here, we mention only that the partial pressure of oxygen in a person’s lungs is 80 2 • The Kinetic Theory of Gases Access for free at openstax.org
crucial to life and health. Breathing air that has a partial pressure of oxygen below 0.16 atm can impair coordination and judgment, particularly in people not acclimated to a high elevation. Lower partial pressures of have more serious effects; partial pressures below 0.06 atm can be quickly fatal, and permanent damage is likely even if the person is rescued. However, the sensation of needing to breathe, as when holding one’s breath, is caused much more by high concentrations of carbon dioxide in the blood than by low concentrations of oxygen. Thus, if a small room or closet is filled with air having a low concentration of oxygen, perhaps because a leaking cylinder of some compressed gas is stored there, a person will not feel any “choking” sensation and may go into convulsions or lose consciousness without noticing anything wrong. Safety engineers give considerable attention to this danger. Another important application of partial pressure is vapor pressure, which is the partial pressure of a vapor at which it is in equilibrium with the liquid (or solid, in the case of sublimation) phase of the same substance. At any temperature, the partial pressure of the water in the air cannot exceed the vapor pressure of the water at that temperature, because whenever the partial pressure reaches the vapor pressure, water condenses out of the air. Dew is an example of this condensation. The temperature at which condensation occurs for a sample of air is called the dew point. It is easily measured by slowly cooling a metal ball; the dew point is the temperature at which condensation first appears on the ball. The vapor pressures of water at some temperatures of interest for meteorology are given in Table 2.2. T Vapor Pressure (Pa) 0 610.5 3 757.9 5 872.3 8 1073 10 1228 13 1497 15 1705 18 2063 20 2338 23 2809 25 3167 30 4243 35 5623 40 7376 TABLE 2.2 Vapor Pressure of Water at Various Temperatures The relative humidity (R.H.) at a temperature T is defined by 2.2 • Pressure, Temperature, and RMS Speed 81
A relative humidity of means that the partial pressure of water is equal to the vapor pressure; in other words, the air is saturated with water. EXAMPLE 2.6 Calculating Relative Humidity What is the relative humidity when the air temperature is and the dew point is ? Strategy We simply look up the vapor pressure at the given temperature and that at the dew point and find the ratio. Solution Significance R.H. is important to our comfort. The value of is within the range of recommended for comfort indoors. As noted in the chapter on temperature and heat, the temperature seldom falls below the dew point, because when it reaches the dew point or frost point, water condenses and releases a relatively large amount of latent heat of vaporization. Mean Free Path and Mean Free Time We now consider collisions explicitly. The usual first step (which is all we’ll take) is to calculate the mean free path, the average distance a molecule travels between collisions with other molecules, and the mean free time , the average time between the collisions of a molecule. If we assume all the molecules are spheres with a radius r, then a molecule will collide with another if their centers are within a distance 2 r of each other. For a given particle, we say that the area of a circle with that radius, , is the “cross-section” for collisions. As the particle moves, it traces a cylinder with that cross-sectional area. The mean free path is the length such that the expected number of other molecules in a cylinder of length and cross-section is 1. If we temporarily ignore the motion of the molecules other than the one we’re looking at, the expected number is the number density of molecules, N/V, times the volume, and the volume is , so we have or Taking the motion of all the molecules into account makes the calculation much harder, but the only change is a factor of The result is In an ideal gas, we can substitute to obtain The mean free time is simply the mean free path divided by a typical speed, and the usual choice is the rms speed. Then 2.10 2.11 82 2 • The Kinetic Theory of Gases Access for free at openstax.org
EXAMPLE 2.7 Calculating Mean Free Time Find the mean free time for argon atoms at a temperature of and a pressure of 1.00 atm. Take the radius of an argon atom to be Solution 1. Identify the knowns and convert into SI units. We know the molar mass is 0.0399 kg/mol, the temperature is 273 K, the pressure is and the radius is 2. Find the rms speed: . 3. Substitute into the equation for the mean free time: Significance We can hardly compare this result with our intuition about gas molecules, but it gives us a picture of molecules colliding with extremely high frequency. CHECK YOUR UNDERSTANDING 2.5 Which has a longer mean free path, liquid water or water vapor in the air? 2.3 Heat Capacity and Equipartition of Energy LEARNING OBJECTIVES By the end of this section, you will be able to: • Solve problems involving heat transfer to and from ideal monatomic gases whose volumes are held constant • Solve similar problems for non-monatomic ideal gases based on the number of degrees of freedom of a molecule • Estimate the heat capacities of metals using a model based on degrees of freedom In the chapter on temperature and heat, we defined the specific heat capacity with the equation or . However, the properties of an ideal gas depend directly on the number of moles in a sample, so here we define specific heat capacity in terms of the number of moles, not the mass. Furthermore, when talking about solids and liquids, we ignored any changes in volume and pressure with changes in temperature—a good approximation for solids and liquids, but for gases, we have to make some condition on volume or pressure changes. Here, we focus on the heat capacity with the volume held constant. We can calculate it for an ideal gas. Heat Capacity of an Ideal Monatomic Gas at Constant Volume We define the molar heat capacity at constant volume as This is often expressed in the form 2.12 2.13 2.3 • Heat Capacity and Equipartition of Energy 83
If the volume does not change, there is no overall displacement, so no work is done, and the only change in internal energy is due to the heat flow (This statement is discussed further in the next chapter.) We use the equation to write and substitute for Q to find , which gives the following simple result for an ideal monatomic gas: It is independent of temperature, which justifies our use of finite differences instead of a derivative. This formula agrees well with experimental results. In the next chapter we discuss the molar specific heat at constant pressure which is always greater than EXAMPLE 2.8 Calculating Temperature A sample of 0.125 kg of xenon is contained in a rigid metal cylinder, big enough that the xenon can be modeled as an ideal gas, at a temperature of . The cylinder is moved outside on a hot summer day. As the xenon comes into equilibrium by reaching the temperature of its surroundings, 180 J of heat are conducted to it through the cylinder walls. What is the equilibrium temperature? Ignore the expansion of the metal cylinder. Solution 1. Identify the knowns: We know the initial temperature is , the heat Q is 180 J, and the mass m of the xenon is 0.125 kg. 2. Identify the unknown. We need the final temperature, so we’ll need . 3. Determine which equations are needed. Because xenon gas is monatomic, we can use Then we need the number of moles, 4. Substitute the known values into the equations and solve for the unknowns. The molar mass of xenon is 131.3 g, so we obtain Therefore, the final temperature is . The problem could equally well be solved in kelvin; as a kelvin is the same size as a degree Celsius of temperature change, you would get Significance The heating of an ideal or almost ideal gas at constant volume is important in car engines and many other practical systems. CHECK YOUR UNDERSTANDING 2.6 Suppose 2 moles of helium gas at 200 K are mixed with 2 moles of krypton gas at 400 K in a calorimeter. What is the final temperature? We would like to generalize our results to ideal gases with more than one atom per molecule. In such systems, the molecules can have other forms of energy beside translational kinetic energy, such as rotational kinetic energy and vibrational kinetic and potential energies. We will see that a simple rule lets us determine the average energies present in these forms and solve problems in much the same way as we have for monatomic gases. Degrees of Freedom In the previous section, we found that and , from which it follows that . 84 2 • The Kinetic Theory of Gases Access for free at openstax.org
The same equation holds for and for . Thus, we can look at our energy of as the sum of contributions of from each of the three dimensions of translational motion. Shifting to the gas as a whole, we see that the 3 in the formula also reflects those three dimensions. We define a degree of freedom as an independent possible motion of a molecule, such as each of the three dimensions of translation. Then, letting d represent the number of degrees of freedom, the molar heat capacity at constant volume of a monatomic ideal gas is where . The branch of physics called statistical mechanics tells us, and experiment confirms, that of any ideal gas is given by this equation, regardless of the number of degrees of freedom. This fact follows from a more general result, the equipartition theorem, which holds in classical (non-quantum) thermodynamics for systems in thermal equilibrium under technical conditions that are beyond our scope. Here, we mention only that in a system, the energy is shared among the degrees of freedom by collisions. This result is due to the Scottish physicist James Clerk Maxwell (1831−1879), whose name will appear several more times in this book. For example, consider a diatomic ideal gas (a good model for nitrogen, and oxygen, Such a gas has more degrees of freedom than a monatomic gas. In addition to the three degrees of freedom for translation, it has two degrees of freedom for rotation perpendicular to its axis. Furthermore, the molecule can vibrate along its axis. This motion is often modeled by imagining a spring connecting the two atoms, and we know from simple harmonic motion that such motion has both kinetic and potential energy. Each of these forms of energy corresponds to a degree of freedom, giving two more. We might expect that for a diatomic gas, we should use 7 as the number of degrees of freedom; classically, if the molecules of a gas had only translational kinetic energy, collisions between molecules would soon make them rotate and vibrate. However, as explained in the previous module, quantum mechanics controls which degrees of freedom are active. The result is shown in Figure 2.13. Both rotational and vibrational energies are limited to discrete values. For temperatures below about 60 K, the energies of hydrogen molecules are too low for a collision to bring the rotational state or vibrational state of a molecule from the lowest energy to the second lowest, so the only form of energy is translational kinetic energy, and or as in a monatomic gas. Above that temperature, the two rotational degrees of freedom begin to contribute, that is, some molecules are excited to the rotational state with the second-lowest energy. (This temperature is much lower than that where rotations of monatomic gases contribute, because diatomic molecules have much higher rotational inertias and hence much lower rotational energies.) From about room temperature (a bit less than 300 K) to about 600 K, the rotational degrees of freedom are fully active, but the vibrational ones are not, and . Then, finally, above about 3000 K, the vibrational degrees of freedom are fully active, and as the classical theory predicted. EQUIPARTITION THEOREM The energy of a thermodynamic system in equilibrium is partitioned equally among its degrees of freedom. Accordingly, the molar heat capacity of an ideal gas is proportional to its number of degrees of freedom, d: 2.14 2.3 • Heat Capacity and Equipartition of Energy 85
FIGURE 2.13 The molar heat capacity of hydrogen as a function of temperature (on a logarithmic scale). The three “steps” or “plateaus” show different numbers of degrees of freedom that the typical energies of molecules must achieve to activate. Translational kinetic energy corresponds to three degrees of freedom, rotational to another two, and vibrational to yet another two. Polyatomic molecules typically have one additional rotational degree of freedom at room temperature, since they have comparable moments of inertia around any axis. Thus, at room temperature, they have At high temperatures, the number of degrees of freedom increases by 2 for each vibrational mode. We usually assume that gases have the theoretical room-temperature values of d. As shown in Table 2.3, the results agree well with experiments for many monatomic and diatomic gases, but the agreement for triatomic gases is only fair. The differences arise from interactions that we have ignored between and within molecules. Gas at and 1 atm Ar 1.50 He 1.50 Ne 1.50 CO 2.50 2.47 2.50 2.53 2.8 3.48 3.13 3.66 TABLE 2.3 for Various Monatomic, Diatomic, and Triatomic Gases 86 2 • The Kinetic Theory of Gases Access for free at openstax.org
What about internal energy for diatomic and polyatomic gases? For such gases, is a function of temperature (Figure 2.13), so we do not have the kind of simple result we have for monatomic ideal gases. Molar Heat Capacity of Solid Elements The idea of equipartition leads to an estimate of the molar heat capacity of solid elements at ordinary temperatures. We can model the atoms of a solid as attached to neighboring atoms by springs (Figure 2.14). FIGURE 2.14 In a simple model of a solid element, each atom is attached to others by six springs, two for each possible motion: x, y, and z. Each of the three motions corresponds to two degrees of freedom, one for kinetic energy and one for potential energy. Thus Analogously to the discussion of vibration in the previous module, each atom has six degrees of freedom: one kinetic and one potential for each of the x-, y-, and z-directions. Accordingly, the molar specific heat of a metal should be 3R. This result, known as the Law of Dulong and Petit, works fairly well experimentally at room temperature. (For every element, it fails at low temperatures for quantum-mechanical reasons. Since quantum effects are particularly important for low-mass particles, the Law of Dulong and Petit already fails at room temperature for some light elements, such as beryllium and carbon. It also fails for some heavier elements for various reasons beyond what we can cover.) PROBLEM-SOLVING STRATEGY Heat Capacity and Equipartition The strategy for solving these problems is the same as the one in Phase Changes for the effects of heat transfer. The only new feature is that you should determine whether the case just presented—ideal gases at constant volume—applies to the problem. (For solid elements, looking up the specific heat capacity is generally better than estimating it from the Law of Dulong and Petit.) In the case of an ideal gas, determine the number d of degrees of freedom from the number of atoms in the gas molecule and use it to calculate (or use to solve for d). EXAMPLE 2.9 Calculating Temperature: Calorimetry with an Ideal Gas A 300-g piece of solid gallium (a metal used in semiconductor devices) at its melting point of only is in contact with 12.0 moles of air (assumed diatomic) at in an insulated container. When the air reaches equilibrium with the gallium, 202 g of the gallium have melted. Based on those data, what is the heat of fusion of gallium? Assume the volume of the air does not change and there are no other heat transfers. 2.3 • Heat Capacity and Equipartition of Energy 87
Strategy We’ll use the equation As some of the gallium doesn’t melt, we know the final temperature is still the melting point. Then the only is the heat lost as the air cools, where The only is the latent heat of fusion of the gallium, It is positive because heat flows into the gallium. Solution 1. Set up the equation: 2. Substitute the known values and solve: We solve to find that the heat of fusion of gallium is 80.2 kJ/kg. 2.4 Distribution of Molecular Speeds LEARNING OBJECTIVES By the end of this section, you will be able to: • Describe the distribution of molecular speeds in an ideal gas • Find the average and most probable molecular speeds in an ideal gas Particles in an ideal gas all travel at relatively high speeds, but they do not travel at the same speed. The rms speed is one kind of average, but many particles move faster and many move slower. The actual distribution of speeds has several interesting implications for other areas of physics, as we will see in later chapters. The Maxwell-Boltzmann Distribution The motion of molecules in a gas is random in magnitude and direction for individual molecules, but a gas of many molecules has a predictable distribution of molecular speeds. This predictable distribution of molecular speeds is known as the Maxwell-Boltzmann distribution, after its originators, who calculated it based on kinetic theory, and it has since been confirmed experimentally (Figure 2.15). To understand this figure, we must define a distribution function of molecular speeds, since with a finite number of molecules, the probability that a molecule will have exactly a given speed is 0. FIGURE 2.15 The Maxwell-Boltzmann distribution of molecular speeds in an ideal gas. The most likely speed is less than the rms speed . Although very high speeds are possible, only a tiny fraction of the molecules have speeds that are an order of magnitude greater than 88 2 • The Kinetic Theory of Gases Access for free at openstax.org
We define the distribution function by saying that the expected number of particles with speeds between and is given by [Since N is dimensionless, the unit of f( v) is seconds per meter.] We can write this equation conveniently in differential form: In this form, we can understand the equation as saying that the number of molecules with speeds between v and is the total number of molecules in the sample times f( v) times dv. That is, the probability that a molecule’s speed is between v and is f( v) dv. We can now quote Maxwell’s result, although the proof is beyond our scope. The factors before the are a normalization constant; they make sure that ∞ by making sure that∞ Let’s focus on the dependence on v. The factor of means that and for small v, the curve looks like a parabola. The factor of means that ∞ and the graph has an exponential tail, which indicates that a few molecules may move at several times the rms speed. The interaction of these factors gives the function the single-peaked shape shown in the figure. EXAMPLE 2.10 Calculating the Ratio of Numbers of Molecules Near Given Speeds In a sample of nitrogen with a molar mass of 28.0 g/mol) at a temperature of , find the ratio of the number of molecules with a speed very close to 300 m/s to the number with a speed very close to 100 m/s. Strategy Since we’re looking at a small range, we can approximate the number of molecules near 100 m/s as Then the ratio we want is All we have to do is take the ratio of the two f values. Solution 1. Identify the knowns and convert to SI units if necessary. 2. Substitute the values and solve. MAXWELL-BOLTZMANN DISTRIBUTION OF SPEEDS The distribution function for speeds of particles in an ideal gas at temperature T is 2.15 2.4 • Distribution of Molecular Speeds 89
Figure 2.16 shows that the curve is shifted to higher speeds at higher temperatures, with a broader range of speeds. FIGURE 2.16 The Maxwell-Boltzmann distribution is shifted to higher speeds and broadened at higher temperatures. INTERACTIVE With only a relatively small number of molecules, the distribution of speeds fluctuates around the Maxwell- Boltzmann distribution. However, you can view this simulation (https://openstax.org/l/21maxboltzdisim) to see the essential features that more massive molecules move slower and have a narrower distribution. Use the set-up “2 Gases, Random Speeds”. Note the display at the bottom comparing histograms of the speed distributions with the theoretical curves. We can use a probability distribution to calculate average values by multiplying the distribution function by the quantity to be averaged and integrating the product over all possible speeds. (This is analogous to calculating averages of discrete distributions, where you multiply each value by the number of times it occurs, add the results, and divide by the number of values. The integral is analogous to the first two steps, and the normalization is analogous to dividing by the number of values.) Thus the average velocity is Similarly, ∞ as in Pressure, Temperature, and RMS Speed. The most probable speed, also called the peak speed is the speed at the peak of the velocity distribution. (In statistics it would be called the mode.) It is less than the rms speed The most probable speed can be calculated by the more familiar method of setting the derivative of the ∞ 2.16 90 2 • The Kinetic Theory of Gases Access for free at openstax.org
distribution function, with respect to v, equal to 0. The result is which is less than In fact, the rms speed is greater than both the most probable speed and the average speed. The peak speed provides a sometimes more convenient way to write the Maxwell-Boltzmann distribution function: In the factor , it is easy to recognize the translational kinetic energy. Thus, that expression is equal to The distribution f(v) can be transformed into a kinetic energy distribution by requiring that Boltzmann showed that the resulting formula is much more generally applicable if we replace the kinetic energy of translation with the total mechanical energy E. Boltzmann’s result is The first part of this equation, with the negative exponential, is the usual way to write it. We give the second part only to remark that in the denominator is ubiquitous in quantum as well as classical statistical mechanics. PROBLEM-SOLVING STRATEGY Speed Distribution Step 1. Examine the situation to determine that it relates to the distribution of molecular speeds. Step 2. Make a list of what quantities are given or can be inferred from the problem as stated (identify the known quantities). Step 3. Identify exactly what needs to be determined in the problem (identify the unknown quantities). A written list is useful. Step 4. Convert known values into proper SI units (K for temperature, Pa for pressure, for volume, molecules forN, and moles for n). In many cases, though, using R and the molar mass will be more convenient than using and the molecular mass. Step 5. Determine whether you need the distribution function for velocity or the one for energy, and whether you are using a formula for one of the characteristic speeds (average, most probably, or rms), finding a ratio of values of the distribution function, or approximating an integral. Step 6. Solve the appropriate equation for the ideal gas law for the quantity to be determined (the unknown quantity). Note that if you are taking a ratio of values of the distribution function, the normalization factors divide out. Or if approximating an integral, use the method asked for in the problem. Step 7. Substitute the known quantities, along with their units, into the appropriate equation and obtain numerical solutions complete with units. We can now gain a qualitative understanding of a puzzle about the composition of Earth’s atmosphere. Hydrogen is by far the most common element in the universe, and helium is by far the second-most common. Moreover, helium is constantly produced on Earth by radioactive decay. Why are those elements so rare in our atmosphere? The answer is that gas molecules that reach speeds above Earth’s escape velocity, about 11 km/s, can escape from the atmosphere into space. Because of the lower mass of hydrogen and helium molecules, they move at higher speeds than other gas molecules, such as nitrogen and oxygen. Only a few exceed escape velocity, but far fewer heavier molecules do. Thus, over the billions of years that Earth has existed, far more hydrogen and helium molecules have 2.17 2.18 2.4 • Distribution of Molecular Speeds 91
escaped from the atmosphere than other molecules, and hardly any of either is now present. We can also now take another look at evaporative cooling, which we discussed in the chapter on temperature and heat. Liquids, like gases, have a distribution of molecular energies. The highest-energy molecules are those that can escape from the intermolecular attractions of the liquid. Thus, when some liquid evaporates, the molecules left behind have a lower average energy, and the liquid has a lower temperature. 92 2 • The Kinetic Theory of Gases Access for free at openstax.org
Chapter Review Key Terms Avogadro’s number the number of molecules in one mole of a substance; particles/mole Boltzmann constant a physical constant that relates energy to temperature and appears in the ideal gas law; critical temperature at which the isotherm has a point with zero slope Dalton’s law of partial pressures physical law that states that the total pressure of a gas is the sum of partial pressures of the component gases degree of freedom independent kind of motion possessing energy, such as the kinetic energy of motion in one of the three orthogonal spatial directions equipartition theorem theorem that the energy of a classical thermodynamic system is shared equally among its degrees of freedom ideal gas gas at the limit of low density and high temperature ideal gas law physical law that relates the pressure and volume of a gas, far from liquefaction, to the number of gas molecules or number of moles of gas and the temperature of the gas internal energy sum of the mechanical energies of all of the molecules in it kinetic theory of gases theory that derives the macroscopic properties of gases from the motion of the molecules they consist of Maxwell-Boltzmann distribution function that can be integrated to give the probability of finding ideal gas molecules with speeds in the range between the limits of integration mean free path average distance between collisions of a particle mean free time average time between collisions of a particle mole quantity of a substance whose mass (in grams) is equal to its molecular mass most probable speed speed near which the speeds of most molecules are found, the peak of the speed distribution function partial pressure pressure a gas would create if it occupied the total volume of space available peak speed same as “most probable speed” pV diagram graph of pressure vs. volume root-mean-square (rms) speed square root of the average of the square (of a quantity) supercritical condition of a fluid being at such a high temperature and pressure that the liquid phase cannot exist universal gas constant R, the constant that appears in the ideal gas law expressed in terms of moles, given by van der Waals equation of state equation, typically approximate, which relates the pressure and volume of a gas to the number of gas molecules or number of moles of gas and the temperature of the gas vapor pressure partial pressure of a vapor at which it is in equilibrium with the liquid (or solid, in the case of sublimation) phase of the same substance Key Equations Ideal gas law in terms of molecules Ideal gas law ratios if the amount of gas is constant Ideal gas law in terms of moles Van der Waals equation Pressure, volume, and molecular speed Root-mean-square speed Mean free path Mean free time The following two equations apply only to a monatomic ideal gas: Average kinetic energy of a molecule Internal energy 2 • Chapter Review 93
Heat in terms of molar heat capacity at constant volume Molar heat capacity at constant volume for an ideal gas with d degrees of freedom Maxwell–Boltzmann speed distribution Average velocity of a molecule Peak velocity of a molecule Summary 2.1 Molecular Model of an Ideal Gas • The ideal gas law relates the pressure and volume of a gas to the number of gas molecules and the temperature of the gas. • A mole of any substance has a number of molecules equal to the number of atoms in a 12-g sample of carbon-12. The number of molecules in a mole is called Avogadro’s number • A mole of any substance has a mass in grams numerically equal to its molecular mass in unified mass units, which can be determined from the periodic table of elements. The ideal gas law can also be written and solved in terms of the number of moles of gas: where n is the number of moles and R is the universal gas constant, • The ideal gas law is generally valid at temperatures well above the boiling temperature. • The van der Waals equation of state for gases is valid closer to the boiling point than the ideal gas law. • Above the critical temperature and pressure for a given substance, the liquid phase does not exist, and the sample is “supercritical.” 2.2 Pressure, Temperature, and RMS Speed • Kinetic theory is the atomic description of gases as well as liquids and solids. It models the properties of matter in terms of continuous random motion of molecules. • The ideal gas law can be expressed in terms of the mass of the gas’s molecules and the average of the molecular speed squared, instead of the temperature. • The temperature of gases is proportional to the average translational kinetic energy of molecules. Hence, the typical speed of gas molecules is proportional to the square root of the temperature and inversely proportional to the square root of the molecular mass. • In a mixture of gases, each gas exerts a pressure equal to the total pressure times the fraction of the mixture that the gas makes up. • The mean free path (the average distance between collisions) and the mean free time of gas molecules are proportional to the temperature and inversely proportional to the molar density and the molecules’ cross-sectional area. 2.3 Heat Capacity and Equipartition of Energy • Every degree of freedom of an ideal gas contributes per atom or molecule to its changes in internal energy. • Every degree of freedom contributes to its molar heat capacity at constant volume • Degrees of freedom do not contribute if the temperature is too low to excite the minimum energy of the degree of freedom as given by quantum mechanics. Therefore, at ordinary temperatures, for monatomic gases, for diatomic gases, and for polyatomic gases. 2.4 Distribution of Molecular Speeds • The motion of individual molecules in a gas is random in magnitude and direction. However, a gas of many molecules has a predictable distribution of molecular speeds, known as the Maxwell-Boltzmann distribution. • The average and most probable velocities of molecules having the Maxwell-Boltzmann speed distribution, as well as the rms velocity, can be calculated from the temperature and molecular mass. 94 2 • Chapter Review Access for free at openstax.org
Conceptual Questions 2.1 Molecular Model of an Ideal Gas 1 . Two molecules can react with one molecule to produce two molecules. How many moles of hydrogen molecules are needed to react with one mole of oxygen molecules? 2 . Under what circumstances would you expect a gas to behave significantly differently than predicted by the ideal gas law? 3 . A constant-volume gas thermometer contains a fixed amount of gas. What property of the gas is measured to indicate its temperature? 4 . Inflate a balloon at room temperature. Leave the inflated balloon in the refrigerator overnight. What happens to the balloon, and why? 5 . In the last chapter, free convection was explained as the result of buoyant forces on hot fluids. Explain the upward motion of air in flames based on the ideal gas law. 2.2 Pressure, Temperature, and RMS Speed 6 . How is momentum related to the pressure exerted by a gas? Explain on the molecular level, considering the behavior of molecules. 7 . If one kind of molecule has double the radius of another and eight times the mass, how do their mean free paths under the same conditions compare? How do their mean free times compare? 8 . What is the average velocity of the air molecules in the room where you are right now? 9 . Why do the atmospheres of Jupiter, Saturn, Uranus, and Neptune, which are much more massive and farther from the Sun than Earth is, contain large amounts of hydrogen and helium? 10. Statistical mechanics says that in a gas maintained at a constant temperature through thermal contact with a bigger system (a “reservoir”) at that temperature, the fluctuations in internal energy are typically a fraction of the internal energy. As a fraction of the total internal energy of a mole of gas, how big are the fluctuations in the internal energy? Are we justified in ignoring them? 11. Which is more dangerous, a closet where tanks of nitrogen are stored, or one where tanks of carbon dioxide are stored? 2.3 Heat Capacity and Equipartition of Energy 12. Experimentally it appears that many polyatomic molecules’ vibrational degrees of freedom can contribute to some extent to their energy at room temperature. Would you expect that fact to increase or decrease their heat capacity from the value R? Explain. 13. One might think that the internal energy of diatomic gases is given by For diatomic gases near room temperature, what additional considerations would cause the internal energy to be less than that? What would cause the internal energy to be more than that? Hint: Their internal energy includes the total energy added in raising the temperature from the boiling point (very low) to room temperature. 14. You mix 5 moles of at 300 K with 5 moles of He at 360 K in a perfectly insulated calorimeter. Is the final temperature higher or lower than 330 K? 2.4 Distribution of Molecular Speeds 15. One cylinder contains helium gas and another identical cylinder contains krypton gas at the same temperature. Mark each of these statements true, false, or impossible to determine from the given information. (a) The rms speeds of atoms in the two gases are the same. (b) The average kinetic energies of atoms in the two gases are the same. (c) The internal energies of 1 mole of gas in each cylinder are the same. (d) The pressures in the two cylinders are the same. 16. Repeat the previous question if one gas is still helium but the other is changed to fluorine, . 17. An ideal gas is at a temperature of 300 K. To double the average speed of its molecules, what does the temperature need to be changed to? Problems 2.1 Molecular Model of an Ideal Gas 18. The gauge pressure in your car tires is at a temperature of when you drive it onto a ship in Los Angeles to be sent to Alaska. What is their gauge pressure 2 • Chapter Review 95
on a night in Alaska when their temperature has dropped to ? Assume the tires have not gained or lost any air. 19. Suppose a gas-filled incandescent light bulb is manufactured so that the gas inside the bulb is at atmospheric pressure when the bulb has a temperature of . (a) Find the gauge pressure inside such a bulb when it is hot, assuming its average temperature is (an approximation) and neglecting any change in volume due to thermal expansion or gas leaks. (b) The actual final pressure for the light bulb will be less than calculated in part (a) because the glass bulb will expand. Is this effect significant? 20. People buying food in sealed bags at high elevations often notice that the bags are puffed up because the air inside has expanded. A bag of pretzels was packed at a pressure of 1.00 atm and a temperature of When opened at a summer picnic in Santa Fe, New Mexico, at a temperature of the volume of the air in the bag is 1.38 times its original volume. What is the pressure of the air? 21. How many moles are there in (a) 0.0500 g of gas (b) 10.0 g of gas (c) How many molecules are present in each case? 22. A cubic container of volume 2.00 L holds 0.500 mol of nitrogen gas at a temperature of What is the net force due to the nitrogen on one wall of the container? Compare that force to the sample’s weight. 23. Calculate the number of moles in the 2.00-L volume of air in the lungs of the average person. Note that the air is at (body temperature) and that the total volume in the lungs is several times the amount inhaled in a typical breath as given in Example 2.2. 24. An airplane passenger has of air in his stomach just before the plane takes off from a sea-level airport. What volume will the air have at cruising altitude if cabin pressure drops to ? 25. A company advertises that it delivers helium at a gauge pressure of in a cylinder of volume 43.8 L. How many balloons can be inflated to a volume of 4.00 L with that amount of helium? Assume the pressure inside the balloons is and the temperature in the cylinder and the balloons is . 26. According to http://hyperphysics.phy- astr.gsu.edu/hbase/solar/venusenv.html, the atmosphere of Venus is approximately and by volume. On the surface, where the temperature is about 750 K and the pressure is about 90 atm, what is the density of the atmosphere? 27. An expensive vacuum system can achieve a pressure as low as at How many molecules are there in a cubic centimeter at this pressure and temperature? 28. The number density N/V of gas molecules at a certain location in the space above our planet is about and the pressure is in this space. What is the temperature there? 29. A bicycle tire contains 2.00 L of gas at an absolute pressure of and a temperature of . What will its pressure be if you let out an amount of air that has a volume of at atmospheric pressure? Assume tire temperature and volume remain constant. 30. In a common demonstration, a bottle is heated and stoppered with a hard-boiled egg that’s a little bigger than the bottle’s neck. When the bottle is cooled, the pressure difference between inside and outside forces the egg into the bottle. Suppose the bottle has a volume of 0.500 L and the temperature inside it is raised to while the pressure remains constant at 1.00 atm because the bottle is open. (a) How many moles of air are inside? (b) Now the egg is put in place, sealing the bottle. What is the gauge pressure inside after the air cools back to the ambient temperature of but before the egg is forced into the bottle? 31. A high-pressure gas cylinder contains 50.0 L of toxic gas at a pressure of and a temperature of . The cylinder is cooled to dry ice temperature to reduce the leak rate and pressure so that it can be safely repaired. (a) What is the final pressure in the tank, assuming a negligible amount of gas leaks while being cooled and that there is no phase change? (b) What is the final pressure if one- tenth of the gas escapes? (c) To what temperature must the tank be cooled to reduce 96 2 • Chapter Review Access for free at openstax.org
the pressure to 1.00 atm (assuming the gas does not change phase and that there is no leakage during cooling)? (d) Does cooling the tank as in part (c) appear to be a practical solution? 32. Find the number of moles in 2.00 L of gas at and under of pressure. 33. Calculate the depth to which Avogadro’s number of table tennis balls would cover Earth. Each ball has a diameter of 3.75 cm. Assume the space between balls adds an extra to their volume and assume they are not crushed by their own weight. 34. (a) What is the gauge pressure in a car tire containing 3.60 mol of gas in a 30.0-L volume? (b) What will its gauge pressure be if you add 1.00 L of gas originally at atmospheric pressure and ? Assume the temperature remains at and the volume remains constant. 2.2 Pressure, Temperature, and RMS Speed In the problems in this section, assume all gases are ideal. 35. A person hits a tennis ball with a mass of 0.058 kg against a wall. The average component of the ball’s velocity perpendicular to the wall is 11 m/ s, and the ball hits the wall every 2.1 s on average, rebounding with the opposite perpendicular velocity component. (a) What is the average force exerted on the wall? (b) If the part of the wall the person hits has an area of what is the average pressure on that area? 36. A person is in a closed room (a racquetball court) with hitting a ball around at random without any pauses. The average kinetic energy of the ball is 2.30 J. (a) What is the average value of Does it matter which direction you take to be x? (b) Applying the methods of this chapter, find the average pressure on the walls? (c) Aside from the presence of only one “molecule” in this problem, what is the main assumption in Pressure, Temperature, and RMS Speed that does not apply here? 37. Five bicyclists are riding at the following speeds: 5.4 m/s, 5.7 m/s, 5.8 m/s, 6.0 m/s, and 6.5 m/s. (a) What is their average speed? (b) What is their rms speed? 38. Some incandescent light bulbs are filled with argon gas. What is for argon atoms near the filament, assuming their temperature is 2500 K? 39. Typical molecular speeds are large, even at low temperatures. What is for helium atoms at 5.00 K, less than one degree above helium’s liquefaction temperature? 40. What is the average kinetic energy in joules of hydrogen atoms on the surface of the Sun? (b) What is the average kinetic energy of helium atoms in a region of the solar corona where the temperature is ? 41. What is the ratio of the average translational kinetic energy of a nitrogen molecule at a temperature of 300 K to the gravitational potential energy of a nitrogen-molecule−Earth system at the ceiling of a 3-m-tall room with respect to the same system with the molecule at the floor? 42. What is the total translational kinetic energy of the air molecules in a room of volume if the pressure is (the room is at fairly high elevation) and the temperature is ? Is any item of data unnecessary for the solution? 43. The product of the pressure and volume of a sample of hydrogen gas at is 80.0 J. (a) How many moles of hydrogen are present? (b) What is the average translational kinetic energy of the hydrogen molecules? (c) What is the value of the product of pressure and volume at 44. What is the gauge pressure inside a tank of of compressed nitrogen with a volume of if the rms speed is 514 m/s? 45. If the rms speed of oxygen molecules inside a refrigerator of volume is 465 m/s, what is the partial pressure of the oxygen? There are 5.71 moles of oxygen in the refrigerator, and the molar mass of oxygen is 32.0 g/mol. 46. The escape velocity of any object from Earth is 11.1 km/s. At what temperature would oxygen molecules (molar mass is equal to 32.0 g/mol) have root-mean-square velocity equal to Earth’s escape velocity of 11.1 km/s? 47. The escape velocity from the Moon is much smaller than that from the Earth, only 2.38 km/s. At what temperature would hydrogen molecules 2 • Chapter Review 97
(molar mass is equal to 2.016 g/mol) have a root-mean-square velocity equal to the Moon’s escape velocity? 48. Nuclear fusion, the energy source of the Sun, hydrogen bombs, and fusion reactors, occurs much more readily when the average kinetic energy of the atoms is high—that is, at high temperatures. Suppose you want the atoms in your fusion experiment to have average kinetic energies of . What temperature is needed? 49. Suppose that the typical speed of carbon dioxide molecules (molar mass is 44.0 g/mol) in a flame is found to be 1350 m/s. What temperature does this indicate? 50. (a) Hydrogen molecules (molar mass is equal to 2.016 g/mol) have equal to 193 m/s. What is the temperature? (b) Much of the gas near the Sun is atomic hydrogen (H rather than Its temperature would have to be for the rms speed to equal the escape velocity from the Sun. What is that velocity? 51. There are two important isotopes of uranium, and ; these isotopes are nearly identical chemically but have different atomic masses. Only is very useful in nuclear reactors. Separating the isotopes is called uranium enrichment (and is often in the news as of this writing, because of concerns that some countries are enriching uranium with the goal of making nuclear weapons.) One of the techniques for enrichment, gas diffusion, is based on the different molecular speeds of uranium hexafluoride gas, . (a) The molar masses of and are 349.0 g/mol and 352.0 g/ mol, respectively. What is the ratio of their typical speeds ? (b) At what temperature would their typical speeds differ by 1.00 m/s? (c) Do your answers in this problem imply that this technique may be difficult? 52. The partial pressure of carbon dioxide in the lungs is about 470 Pa when the total pressure in the lungs is 1.0 atm. What percentage of the air molecules in the lungs is carbon dioxide? Compare your result to the percentage of carbon dioxide in the atmosphere, about 0.033%. 53. Dry air consists of approximately by mole, with trace amounts of other gases. A tank of compressed dry air has a volume of 1.76 cubic feet at a gauge pressure of 2200 pounds per square inch and a temperature of 293 K. How much oxygen does it contain in moles? 54. (a) Using data from the previous problem, find the mass of nitrogen, oxygen, and argon in 1 mol of dry air. The molar mass of is 28.0 g/mol, that of is 32.0 g/mol, and that of argon is 39.9 g/mol. (b) Dry air is mixed with pentane molar mass 72.2 g/mol), an important constituent of gasoline, in an air-fuel ratio of 15:1 by mass (roughly typical for car engines). Find the partial pressure of pentane in this mixture at an overall pressure of 1.00 atm. 55. (a) Given that air is oxygen, find the minimum atmospheric pressure that gives a relatively safe partial pressure of oxygen of 0.16 atm. (b) What is the minimum pressure that gives a partial pressure of oxygen above the quickly fatal level of 0.06 atm? (c) The air pressure at the summit of Mount Everest (8848 m) is 0.334 atm. Why have a few people climbed it without oxygen, while some who have tried, even though they had trained at high elevation, had to turn back? 56. (a) If the partial pressure of water vapor is 8.05 torr, what is the dew point? (b) On a warm day when the air temperature is and the dew point is , what are the partial pressure of the water in the air and the relative humidity? 2.3 Heat Capacity and Equipartition of Energy 57. To give a helium atom nonzero angular momentum requires about 21.2 eV of energy (that is, 21.2 eV is the difference between the energies of the lowest-energy or ground state and the lowest-energy state with angular momentum). The electron-volt or eV is defined as Find the temperature T where this amount of energy equals Does this explain why we can ignore the rotational energy of helium for most purposes? (The results for other monatomic gases, and for diatomic gases rotating around the axis connecting the two atoms, have comparable orders of magnitude.) 58. (a) How much heat must be added to raise the temperature of 1.5 mol of air from to at constant volume? Assume air is completely diatomic. (b) Repeat the problem for the same number of moles of xenon, Xe. 98 2 • Chapter Review Access for free at openstax.org
59. A sealed, rigid container of 0.560 mol of an unknown ideal gas at a temperature of is cooled to . In the process, 980 J of heat are removed from the gas. Is the gas monatomic, diatomic, or polyatomic? 60. A sample of neon gas (Ne, molar mass at a temperature of is put into a steel container of mass 47.2 g that’s at a temperature of . The final temperature is . (No heat is exchanged with the surroundings, and you can neglect any change in the volume of the container.) What is the mass of the sample of neon? 61. A steel container of mass 135 g contains 24.0 g of ammonia, , which has a molar mass of 17.0 g/mol. The container and gas are in equilibrium at . How much heat has to be removed to reach a temperature of ? Ignore the change in volume of the steel and treat the ammonia as an ideal gas with 6 degrees of freedom (3 translational and 3 rotational; at room temperature, the vibrational degree of freedom is inactive.) 62. A sealed room has a volume of . It’s filled with air, which may be assumed to be diatomic, at a temperature of and a pressure of A 1.00-kg block of ice at its melting point is placed in the room. Assume the walls of the room transfer no heat. What is the equilibrium temperature? 63. Heliox, a mixture of helium and oxygen, is sometimes given to hospital patients who have trouble breathing, because the low mass of helium makes it easier to breathe than air. Suppose helium at is mixed with oxygen at to make a mixture that is helium by mole. What is the final temperature? Ignore any heat flow to or from the surroundings, and assume the final volume is the sum of the initial volumes. 64. Professional divers sometimes use heliox, consisting of helium and oxygen by mole. Suppose a perfectly rigid scuba tank with a volume of 11 L contains heliox at an absolute pressure of at a temperature of . (a) How many moles of helium and how many moles of oxygen are in the tank? (b) The diver goes down to a point where the sea temperature is while using a negligible amount of the mixture. As the gas in the tank reaches this new temperature, how much heat is removed from it? 65. In car racing, one advantage of mixing liquid nitrous oxide with air is that the boiling of the “nitrous” absorbs latent heat of vaporization and thus cools the air and ultimately the fuel-air mixture, allowing more fuel-air mixture to go into each cylinder. As a very rough look at this process, suppose 1.0 mol of nitrous oxide gas at its boiling point, , is mixed with 4.0 mol of air (assumed diatomic) at . What is the final temperature of the mixture? Use the measured heat capacity of at , which is . (The primary advantage of nitrous oxide is that it consists of 1/3 oxygen, which is more than air contains, so it supplies more oxygen to burn the fuel. Another advantage is that its decomposition into nitrogen and oxygen releases energy in the cylinder.) 2.4 Distribution of Molecular Speeds 66. In a sample of hydrogen sulfide at a temperature of estimate the ratio of the number of molecules that have speeds very close to to the number that have speeds very close to 67. Using the approximation for small , estimate the fraction of nitrogen molecules at a temperature of that have speeds between 290 m/s and 291 m/s. 68. Using the method of the preceding problem, estimate the fraction of nitric oxide (NO) molecules at a temperature of 250 K that have energies between and . 69. By counting squares in the following figure, estimate the fraction of argon atoms at that have speeds between 600 m/s and 700 m/s. The curve is correctly normalized. The value of a square is its length as measured on the x-axis times its height as measured on the y-axis, with the units given on those axes. 2 • Chapter Review 99
70. Using a numerical integration method such as Simpson’s rule, find the fraction of molecules in a sample of oxygen gas at a temperature of 250 K that have speeds between 100 m/s and 150 m/s. The molar mass of oxygen is 32.0 g/ mol. A precision to two significant digits is enough. 71. Find (a) the most probable speed, (b) the average speed, and (c) the rms speed for nitrogen molecules at 295 K. 72. Repeat the preceding problem for nitrogen molecules at 2950 K. 73. At what temperature is the average speed of carbon dioxide molecules 510 m/s? 74. The most probable speed for molecules of a gas at 296 K is 263 m/s. What is the molar mass of the gas? (You might like to figure out what the gas is likely to be.) 75. a) At what temperature do oxygen molecules have the same average speed as helium atoms have at 300 K? b) What is the answer to the same question about most probable speeds? c) What is the answer to the same question about rms speeds? Additional Problems 76. In the deep space between galaxies, the density of molecules (which are mostly single atoms) can be as low as and the temperature is a frigid 2.7 K. What is the pressure? (b) What volume (in ) is occupied by 1 mol of gas? (c) If this volume is a cube, what is the length of its sides in kilometers? 77. (a) Find the density in SI units of air at a pressure of 1.00 atm and a temperature of , assuming that air is , (b) Find the density of the atmosphere on Venus, assuming that it’s , with a temperature of 737 K and a pressure of 92.0 atm. 78. The air inside a hot-air balloon has a temperature of 370 K and a pressure of 101.3 kPa, the same as that of the air outside. Using the composition of air as , find the density of the air inside the balloon. 79. When an air bubble rises from the bottom to the top of a freshwater lake, its volume increases by . If the temperatures at the bottom and the top of the lake are 4.0 and 10 , respectively, how deep is the lake? 80. (a) Use the ideal gas equation to estimate the temperature at which 1.00 kg of steam (molar mass ) at a pressure of occupies a volume of . (b) The van der Waals constants for water are and . Use the Van der Waals equation of state to estimate the temperature under the same conditions. (c) The actual temperature is 779 K. Which estimate is better? 81. One process for decaffeinating coffee uses carbon dioxide at a molar density of about and a temperature of about . (a) Is CO2 a solid, liquid, gas, or supercritical fluid under those conditions? (b) The van der Waals constants for carbon dioxide are and Using the van der Waals equation, estimate the pressure of at that temperature and density. 82. On a winter day when the air temperature is the relative humidity is . Outside air comes inside and is heated to a room temperature of . What is the relative humidity of the air inside the room. (Does this problem show why inside air is so dry in winter?) 83. On a warm day when the air temperature is , a metal can is slowly cooled by adding bits of ice to liquid water in it. Condensation first appears when the can reaches . What is the relative humidity of the air? 100 2 • Chapter Review Access for free at openstax.org
84. (a) People often think of humid air as “heavy.” Compare the densities of air with relative humidity and relative humidity when both are at 1 atm and . Assume that the dry air is an ideal gas composed of molecules with a molar mass of 29.0 g/mol and the moist air is the same gas mixed with water vapor. (b) As discussed in the chapter on the applications of Newton’s laws, the air resistance felt by projectiles such as baseballs and golf balls is approximately , where is the mass density of the air, A is the cross-sectional area of the projectile, and C is the projectile’s drag coefficient. For a fixed air pressure, describe qualitatively how the range of a projectile changes with the relative humidity. (c) When a thunderstorm is coming, usually the humidity is high and the air pressure is low. Do those conditions give an advantage or disadvantage to home-run hitters? 85. The mean free path for helium at a certain temperature and pressure is The radius of a helium atom can be taken as . What is the measure of the density of helium under those conditions (a) in molecules per cubic meter and (b) in moles per cubic meter? 86. The mean free path for methane at a temperature of 269 K and a pressure of is Find the effective radius r of the methane molecule. 87. In the chapter on fluid mechanics, Bernoulli’s equation for the flow of incompressible fluids was explained in terms of changes affecting a small volume dV of fluid. Such volumes are a fundamental idea in the study of the flow of compressible fluids such as gases as well. For the equations of hydrodynamics to apply, the mean free path must be much less than the linear size of such a volume, For air in the stratosphere at a temperature of 220 K and a pressure of 5.8 kPa, how big should a be for it to be 100 times the mean free path? Take the effective radius of air molecules to be which is roughly correct for . 88. Find the total number of collisions between molecules in 1.00 s in 1.00 L of nitrogen gas at standard temperature and pressure ( , 1.00 atm). Use as the effective radius of a nitrogen molecule. (The number of collisions per second is the reciprocal of the collision time.) Keep in mind that each collision involves two molecules, so if one molecule collides once in a certain period of time, the collision of the molecule it hit cannot be counted. 89. (a) Estimate the specific heat capacity of sodium from the Law of Dulong and Petit. The molar mass of sodium is 23.0 g/mol. (b) What is the percent error of your estimate from the known value, ? 90. A sealed, perfectly insulated container contains 0.630 mol of air at and an iron stirring bar of mass 40.0 g. The stirring bar is magnetically driven to a kinetic energy of 50.0 J and allowed to slow down by air resistance. What is the equilibrium temperature? 91. Find the ratio for hydrogen gas at a temperature of 77.0 K. 92. Unreasonable results. (a) Find the temperature of 0.360 kg of water, modeled as an ideal gas, at a pressure of if it has a volume of . (b) What is unreasonable about this answer? How could you get a better answer? 93. Unreasonable results. (a) Find the average speed of hydrogen sulfide, , molecules at a temperature of 250 K. Its molar mass is 31.4 g/ mol (b) The result isn’t very unreasonable, but why is it less reliable than those for, say, neon or nitrogen? Challenge Problems 94. An airtight dispenser for drinking water is in horizontal dimensions and 20 cm tall. It has a tap of negligible volume that opens at the level of the bottom of the dispenser. Initially, it contains water to a level 3.0 cm from the top and air at the ambient pressure, 1.00 atm, from there to the top. When the tap is opened, water will flow out until the gauge pressure at the bottom of the dispenser, and thus at the opening of the tap, is 0. What volume of water flows out? Assume the temperature is constant, the dispenser is perfectly rigid, and the water has a constant density of . 2 • Chapter Review 101
95. Eight bumper cars, each with a mass of 322 kg, are running in a room 21.0 m long and 13.0 m wide. They have no drivers, so they just bounce around on their own. The rms speed of the cars is 2.50 m/s. Repeating the arguments of Pressure, Temperature, and RMS Speed, find the average force per unit length (analogous to pressure) that the cars exert on the walls. 96. Verify that . 97. Verify the normalization equation ∞ In doing the integral, first make the substitution This “scaling” transformation gives you all features of the answer except for the integral, which is a dimensionless numerical factor. You’ll need the formula ∞ to find the numerical factor and verify the normalization. 98. Verify that Make the same scaling transformation as in the preceding problem. 99. Verify that . 102 2 • Chapter Review Access for free at openstax.org
INTRODUCTION CHAPTER 3 The First Law of Thermodynamics 3.1 Thermodynamic Systems 3.2 Work, Heat, and Internal Energy 3.3 First Law of Thermodynamics 3.4 Thermodynamic Processes 3.5 Heat Capacities of an Ideal Gas 3.6 Adiabatic Processes for an Ideal Gas Heat is the transfer of energy due to a temperature difference between two systems. Heat describes the process of converting from one form of energy into another. A car engine, for example, burns gasoline. Heat is produced when the burned fuel is chemically transformed into mostly and which are gases at the combustion temperature. These gases exert a force on a piston through a displacement, doing work and converting the piston’s kinetic energy into a variety of other forms—into the car’s kinetic energy; into electrical energy to run the spark plugs, radio, and lights; and back into stored energy in the car’s battery. Energy is conserved in all processes, including those associated with thermodynamic systems. The roles of heat transfer and internal energy change vary from process to process and affect how work is done by the system in that process. We will see that the first law of thermodynamics explains that a change in the internal energy of a system comes from changes in heat or work. Understanding the laws that govern thermodynamic processes and the relationship between the system and its surroundings is therefore paramount in gaining scientific knowledge of energy and energy consumption. FIGURE 3.1 A weak cold front of air pushes all the smog in northeastern China into a giant smog blanket over the Yellow Sea, as captured by NASA’s Terra satellite in 2012. To understand changes in weather and climate, such as the event shown here, you need a thorough knowledge of thermodynamics. (credit: modification of work by NASA) CHAPTER OUTLINE
3.1 Thermodynamic Systems LEARNING OBJECTIVES By the end of this section, you will be able to: • Define a thermodynamic system, its boundary, and its surroundings • Explain the roles of all the components involved in thermodynamics • Define thermal equilibrium and thermodynamic temperature • Link an equation of state to a system A thermodynamic system includes anything whose thermodynamic properties are of interest. It is embedded in its surroundings or environment; it can exchange heat with, and do work on, its environment through a boundary, which is the imagined wall that separates the system and the environment (Figure 3.2). In reality, the immediate surroundings of the system are interacting with it directly and therefore have a much stronger influence on its behavior and properties. For example, if we are studying a car engine, the burning gasoline inside the cylinder of the engine is the thermodynamic system; the piston, exhaust system, radiator, and air outside form the surroundings of the system. The boundary then consists of the inner surfaces of the cylinder and piston. FIGURE 3.2 (a) A system, which can include any relevant process or value, is self-contained in an area. The surroundings may also have relevant information; however, the surroundings are important to study only if the situation is an open system. (b) The burning gasoline in the cylinder of a car engine is an example of a thermodynamic system. Normally, a system must have some interactions with its surroundings. A system is called an isolated and closed system if it is completely separated from its environment—for example, a gas that is surrounded by immovable and thermally insulating walls. In reality, a closed system does not exist unless the entire universe is treated as the system, or it is used as a model for an actual system that has minimal interactions with its environment. Most systems are known as an open system, which can exchange energy and/or matter with its surroundings (Figure 3.3). 104 3 • The First Law of Thermodynamics Access for free at openstax.org
FIGURE 3.3 (a) This boiling tea kettle is an open thermodynamic system. It transfers heat and matter (steam) to its surroundings. (b) A pressure cooker is a good approximation to a closed system. A little steam escapes through the top valve to prevent explosion. (credit a: modification of work by Gina Hamilton; credit b: modification of work by Jane Whitney) When we examine a thermodynamic system, we ignore the difference in behavior from place to place inside the system for a given moment. In other words, we concentrate on the macroscopic properties of the system, which are the averages of the microscopic properties of all the molecules or entities in the system. Any thermodynamic system is therefore treated as a continuum that has the same behavior everywhere inside. We assume the system is in equilibrium. You could have, for example, a temperature gradient across the system. However, when we discuss a thermodynamic system in this chapter, we study those that have uniform properties throughout the system. Before we can carry out any study on a thermodynamic system, we need a fundamental characterization of the system. When we studied a mechanical system, we focused on the forces and torques on the system, and their balances dictated the mechanical equilibrium of the system. In a similar way, we should examine the heat transfer between a thermodynamic system and its environment or between the different parts of the system, and its balance should dictate the thermal equilibrium of the system. Intuitively, such a balance is reached if the temperature becomes the same for different objects or parts of the system in thermal contact, and the net heat transfer over time becomes zero. Thus, when we say two objects (a thermodynamic system and its environment, for example) are in thermal equilibrium, we mean that they are at the same temperature, as we discussed in Temperature and Heat. Let us consider three objects at temperatures and respectively. How do we know whether they are in thermal equilibrium? The governing principle here is the zeroth law of thermodynamics, as described in Temperature and Heat on temperature and heat:If object 1 is in thermal equilibrium with objects 2 and 3, respectively, then objects 2 and 3 must also be in thermal equilibrium. Mathematically, we can simply write the zeroth law of thermodynamics as This is the most fundamental way of defining temperature: Two objects must be at the same temperature thermodynamically if the net heat transfer between them is zero when they are put in thermal contact and have reached a thermal equilibrium. The zeroth law of thermodynamics is equally applicable to the different parts of a closed system and requires that the temperature everywhere inside the system be the same if the system has reached a thermal equilibrium. To simplify our discussion, we assume the system is uniform with only one type of material—for example, water in a tank. The measurable properties of the system at least include its volume, pressure, and temperature. The range of specific relevant variables depends upon the system. For example, for a stretched rubber band, the relevant variables would be length, tension, and temperature. The relationship between these three basic properties of the system is called the equation of state of the system and is written symbolically for a closed system as 3.1 3.1 • Thermodynamic Systems 105
where V, p, and T are the volume, pressure, and temperature of the system at a given condition. In principle, this equation of state exists for any thermodynamic system but is not always readily available. The forms of for many materials have been determined either experimentally or theoretically. In the preceding chapter, we saw an example of an equation of state for an ideal gas, We have so far introduced several physical properties that are relevant to the thermodynamics of a thermodynamic system, such as its volume, pressure, and temperature. We can separate these quantities into two generic categories. The quantity associated with an amount of matter is an extensive variable, such as the volume and the number of moles. The other properties of a system are intensive variables, such as the pressure and temperature. An extensive variable doubles its value if the amount of matter in the system doubles, provided all the intensive variables remain the same. For example, the volume or total energy of the system doubles if we double the amount of matter in the system while holding the temperature and pressure of the system unchanged. 3.2 Work, Heat, and Internal Energy LEARNING OBJECTIVES By the end of this section, you will be able to: • Describe the work done by a system, heat transfer between objects, and internal energy change of a system • Calculate the work, heat transfer, and internal energy change in a simple process We discussed the concepts of work and energy earlier in mechanics. Examples and related issues of heat transfer between different objects have also been discussed in the preceding chapters. Here, we want to expand these concepts to a thermodynamic system and its environment. Specifically, we elaborated on the concepts of heat and heat transfer in the previous two chapters. Here, we want to understand how work is done by or to a thermodynamic system; how heat is transferred between a system and its environment; and how the total energy of the system changes under the influence of the work done and heat transfer. Work Done by a System A force created from any source can do work by moving an object through a displacement. Then how does a thermodynamic system do work? Figure 3.4 shows a gas confined to a cylinder that has a movable piston at one end. If the gas expands against the piston, it exerts a force through a distance and does work on the piston. If the piston compresses the gas as it is moved inward, work is also done—in this case, on the gas. The work associated with such volume changes can be determined as follows: Let the gas pressure on the piston face be p. Then the force on the piston due to the gas is pA, where A is the area of the face. When the piston is pushed outward an infinitesimal distance dx, the magnitude of the work done by the gas is Since the change in volume of the gas is this becomes For a finite change in volume from we can integrate this equation from to find the net work: 3.2 3.3 3.4 106 3 • The First Law of Thermodynamics Access for free at openstax.org
FIGURE 3.4 The work done by a confined gas in moving a piston a distance dx is given by This integral is only meaningful for a quasi-static process, which means a process that takes place in infinitesimally small steps, keeping the system at thermal equilibrium. (We examine this idea in more detail later in this chapter.) Only then does a well-defined mathematical relationship (the equation of state) exist between the pressure and volume. This relationship can be plotted on a pV diagram of pressure versus volume, where the curve is the change of state. We can approximate such a process as one that occurs slowly, through a series of equilibrium states. The integral is interpreted graphically as the area under the pV curve (the shaded area of Figure 3.5). Work done by the gas is positive for expansion and negative for compression. FIGURE 3.5 When a gas expands slowly from the work done by the system is represented by the shaded area under the pV curve. Consider the two processes involving an ideal gas that are represented by paths AC and ABC in Figure 3.6. The first process is an isothermal expansion, with the volume of the gas changing its volume from . This isothermal process is represented by the curve between points A and C. The gas is kept at a constant temperature T by keeping it in thermal equilibrium with a heat reservoir at that temperature. From Equation 3.4 and the ideal gas law, FIGURE 3.6 The paths ABC, AC, and ADC represent three different quasi-static transitions between the equilibrium states A and C. The expansion is isothermal, so T remains constant over the entire process. Since n and R are also constant, the only 3.2 • Work, Heat, and Internal Energy 107
variable in the integrand is V, so the work done by an ideal gas in an isothermal process is Notice that if (expansion), W is positive, as expected. The straight lines from A to B and then from B to C represent a different process. Here, a gas at a pressure first expands isobarically (constant pressure) and quasi-statically from , after which it cools quasi-statically at the constant volume until its pressure drops to . From A to B, the pressure is constant at p, so the work over this part of the path is From B to C, there is no change in volume and therefore no work is done. The net work over the path ABC is then A comparison of the expressions for the work done by the gas in the two processes of Figure 3.6 shows that they are quite different. This illustrates a very important property of thermodynamic work: It is path dependent. We cannot determine the work done by a system as it goes from one equilibrium state to another unless we know its thermodynamic path. Different values of the work are associated with different paths. EXAMPLE 3.1 Isothermal Expansion of a van der Waals Gas Studies of a van der Waals gas require an adjustment to the ideal gas law that takes into consideration that gas molecules have a definite volume (see The Kinetic Theory of Gases). One mole of a van der Waals gas has an equation of state where a and b are two parameters for a specific gas. Suppose the gas expands isothermally and quasi-statically from volume to volume How much work is done by the gas during the expansion? Strategy Because the equation of state is given, we can use Equation 3.4 to express the pressure in terms of V and T. Furthermore, temperature T is a constant under the isothermal condition, so V becomes the only changing variable under the integral. Solution To evaluate this integral, we must express p as a function of V. From the given equation of state, the gas pressure is Because T is constant under the isothermal condition, the work done by 1 mol of a van der Waals gas in expanding from a volume to a volume is thus 108 3 • The First Law of Thermodynamics Access for free at openstax.org
Significance By taking into account the volume of molecules, the expression for work is much more complex. If, however, we set and we see that the expression for work matches exactly the work done by an isothermal process for one mole of an ideal gas. CHECK YOUR UNDERSTANDING 3.1 How much work is done by the gas, as given in Figure 3.6, when it expands quasi-statically along the path ADC? Internal Energy The internal energy of a thermodynamic system is, by definition, the sum of the mechanical energies of all the molecules or entities in the system. If the kinetic and potential energies of molecule i are and respectively, then the internal energy of the system is the average of the total mechanical energy of all the entities: where the summation is over all the molecules of the system, and the bars over K and U indicate average values. The kinetic energy of an individual molecule includes contributions due to its rotation and vibration, as well as its translational energy where is the molecule’s speed measured relative to the center of mass of the system. The potential energy is associated only with the interactions between molecule i and the other molecules of the system. In fact, neither the system’s location nor its motion is of any consequence as far as the internal energy is concerned. The internal energy of the system is not affected by moving it from the basement to the roof of a 100-story building or by placing it on a moving train. In an ideal monatomic gas, each molecule is a single atom. Consequently, there is no rotational or vibrational kinetic energy and . Furthermore, there are no interatomic interactions (collisions notwithstanding), so , which we set to zero. The internal energy is therefore due to translational kinetic energy only and From the discussion in the preceding chapter, we know that the average kinetic energy of a molecule in an ideal monatomic gas is where T is the Kelvin temperature of the gas. Consequently, the average mechanical energy per molecule of an ideal monatomic gas is also that is, The internal energy is just the number of molecules multiplied by the average mechanical energy per molecule. Thus for n moles of an ideal monatomic gas, Notice that the internal energy of a given quantity of an ideal monatomic gas depends on just the temperature and is completely independent of the pressure and volume of the gas. For other systems, the internal energy cannot be expressed so simply. However, an increase in internal energy can often be associated with an increase in temperature. 3.5 3.6 3.2 • Work, Heat, and Internal Energy 109
We know from the zeroth law of thermodynamics that when two systems are placed in thermal contact, they eventually reach thermal equilibrium, at which point they are at the same temperature. As an example, suppose we mix two monatomic ideal gases. Now, the energy per molecule of an ideal monatomic gas is proportional to its temperature. Thus, when the two gases are mixed, the molecules of the hotter gas must lose energy and the molecules of the colder gas must gain energy. This continues until thermal equilibrium is reached, at which point, the temperature, and therefore the average translational kinetic energy per molecule, is the same for both gases. The approach to equilibrium for real systems is somewhat more complicated than for an ideal monatomic gas. Nevertheless, we can still say that energy is exchanged between the systems until their temperatures are the same. 3.3 First Law of Thermodynamics LEARNING OBJECTIVES By the end of this section, you will be able to: • State the first law of thermodynamics and explain how it is applied • Explain how heat transfer, work done, and internal energy change are related in any thermodynamic process Now that we have seen how to calculate internal energy, heat, and work done for a thermodynamic system undergoing change during some process, we can see how these quantities interact to affect the amount of change that can occur. This interaction is given by the first law of thermodynamics. British scientist and novelist C. P. Snow (1905–1980) is credited with a joke about the four laws of thermodynamics. His humorous statement of the first law of thermodynamics is stated “you can’t win,” or in other words, you cannot get more energy out of a system than you put into it. We will see in this chapter how internal energy, heat, and work all play a role in the first law of thermodynamics. Suppose Q represents the heat exchanged between a system and the environment, and W is the work done by or on the system. The first law states that the change in internal energy of that system is given by . Since added heat increases the internal energy of a system, Q is positive when it is added to the system and negative when it is removed from the system. When a gas expands, it does work and its internal energy decreases. Thus, W is positive when work is done by the system and negative when work is done on the system. This sign convention is summarized in Table 3.1. The first law of thermodynamics is stated as follows: Thermodynamic Sign Conventions for Heat and Work Process Convention Heat added to system Heat removed from system TABLE 3.1 FIRST LAW OF THERMODYNAMICS Associated with every equilibrium state of a system is its internal energy The change in for any transition between two equilibrium states is where Q and W represent, respectively, the heat exchanged by the system and the work done by or on the system. 3.7 110 3 • The First Law of Thermodynamics Access for free at openstax.org
Thermodynamic Sign Conventions for Heat and Work Process Convention Work done by system Work done on system TABLE 3.1 The first law is a statement of energy conservation. It tells us that a system can exchange energy with its surroundings by the transmission of heat and by the performance of work. The net energy exchanged is then equal to the change in the total mechanical energy of the molecules of the system (i.e., the system’s internal energy). Thus, if a system is isolated, its internal energy must remain constant. Although Q and W both depend on the thermodynamic path taken between two equilibrium states, their difference does not. Figure 3.7 shows the pV diagram of a system that is making the transition from A to B repeatedly along different thermodynamic paths. Along path 1, the system absorbs heat and does work along path 2, it absorbs heat and does work and so on. The values of and may vary from path to path, but we have or That is, the change in the internal energy of the system between A and B is path independent. In the chapter on potential energy and the conservation of energy, we encountered another path-independent quantity: the change in potential energy between two arbitrary points in space. This change represents the negative of the work done by a conservative force between the two points. The potential energy is a function of spatial coordinates, whereas the internal energy is a function of thermodynamic variables. For example, we might write for the internal energy. Functions such as internal energy and potential energy are known as state functions because their values depend solely on the state of the system. FIGURE 3.7 Different thermodynamic paths taken by a system in going from state A to state B. For all transitions, the change in the internal energy of the system is the same. Often the first law is used in its differential form, which is Here is an infinitesimal change in internal energy when an infinitesimal amount of heat dQ is exchanged with the system and an infinitesimal amount of work dW is done by (positive in sign) or on (negative in sign) the system. 3.8 3.3 • First Law of Thermodynamics 111
EXAMPLE 3.2 Changes of State and the First Law During a thermodynamic process, a system moves from state A to state B, it is supplied with 400 J of heat and does 100 J of work. (a) For this transition, what is the system’s change in internal energy? (b) If the system then moves from state B back to state A, what is its change in internal energy? (c) If in moving from A to B along a different path, of work is done on the system, how much heat does it absorb? Strategy The first law of thermodynamics relates the internal energy change, work done by the system, and the heat transferred to the system in a simple equation. The internal energy is a function of state and is therefore fixed at any given point regardless of how the system reaches the state. Solution a. From the first law, the change in the system’s internal energy is b. Consider a closed path that passes through the states A and B. Internal energy is a state function, so is zero for a closed path. Thus and This yields c. The change in internal energy is the same for any path, so and the heat exchanged is The negative sign indicates that the system loses heat in this transition. Significance When a closed cycle is considered for the first law of thermodynamics, the change in internal energy around the whole path is equal to zero. If friction were to play a role in this example, less work would result from this heat added. Example 3.3 takes into consideration what happens if friction plays a role. If the system were an ideal gas, the work would be the integral of , and positive work done by the system would imply an expansion ( ). Changing the path from A to B could not change the sign of the work done, since negative work would imply that , contradicting the conclusion reached in part a. Not every system is a gas, however, and not all work changes the volume of the system. For example, the work in this case may be a combination of positive work done by a spring being compressed a fixed amount and a variable, path dependent, amount of negative work by friction. In the first path of part a, friction does less work than the spring, so the total work is positive, but in the second path in part c, more of the work is done by friction, so the total work is negative. Notice that in Example 3.2, we did not assume that the transitions were quasi-static. This is because the first law is not subject to such a restriction. It describes transitions between equilibrium states but is not concerned with the intermediate states. The system does not have to pass through only equilibrium states. For example, if a gas in a steel container at a well-defined temperature and pressure is made to explode by means of a spark, some of the gas may condense, different gas molecules may combine to form new compounds, and there may be all sorts of turbulence in the container—but eventually, the system will settle down to a new equilibrium state. This system is 112 3 • The First Law of Thermodynamics Access for free at openstax.org
clearly not in equilibrium during its transition; however, its behavior is still governed by the first law because the process starts and ends with the system in equilibrium states. EXAMPLE 3.3 Polishing a Fitting A machinist polishes a 0.50-kg copper fitting with a piece of emery cloth for 2.0 min. He moves the cloth across the fitting at a constant speed of 1.0 m/s by applying a force of 20 N, tangent to the surface of the fitting. (a) What is the total work done on the fitting by the machinist? (b) What is the increase in the internal energy of the fitting? Assume that the change in the internal energy of the cloth is negligible and that no heat is exchanged between the fitting and its environment. (c) What is the increase in the temperature of the fitting? Strategy The machinist’s force over a distance that can be calculated from the speed and time given is the work done on the system. The work, in turn, increases the internal energy of the system. This energy can be interpreted as the heat that raises the temperature of the system via its heat capacity. Be careful with the sign of each quantity. Solution a. The power created by a force on an object or the rate at which the machinist does frictional work on the fitting is . Thus, in an elapsed time (2.0 min), the work done on the fitting is b. By assumption, no heat is exchanged between the fitting and its environment, so the first law gives for the change in the internal energy of the fitting: c. Since is path independent, the effect of the of work is the same as if it were supplied at atmospheric pressure by a transfer of heat. Thus, and the increase in the temperature of the fitting is where we have used the value for the specific heat of copper, . Significance If heat were released, the change in internal energy would be less and cause less of a temperature change than what was calculated in the problem. CHECK YOUR UNDERSTANDING 3.2 The quantities below represent four different transitions between the same initial and final state. Fill in the blanks. Q (J) W (J) –80 –120 90 TABLE 3.2 3.3 • First Law of Thermodynamics 113
Q (J) W (J) 40 –40 TABLE 3.2 EXAMPLE 3.4 An Ideal Gas Making Transitions between Two States Consider the quasi-static expansions of an ideal gas between the equilibrium states A and C of Figure 3.6. If 515 J of heat are added to the gas as it traverses the path ABC, how much heat is required for the transition along ADC? Assume that and Strategy The difference in work done between process ABC and process ADC is the area enclosed by ABCD. Because the change of the internal energy (a function of state) is the same for both processes, the difference in work is thus the same as the difference in heat transferred to the system. Solution For path ABC, the heat added is and the work done by the gas is the area under the path on the pV diagram, which is Along ADC, the work done by the gas is again the area under the path: Then using the strategy we just described, we have which leads to Significance The work calculations in this problem are made simple since no work is done along AD and BC and along AB and DC; the pressure is constant over the volume change, so the work done is simply . An isothermal line could also have been used, as we have derived the work for an isothermal process as . EXAMPLE 3.5 Isothermal Expansion of an Ideal Gas Heat is added to 1 mol of an ideal monatomic gas confined to a cylinder with a movable piston at one end. The gas expands quasi-statically at a constant temperature of 300 K until its volume increases from V to 3V. (a) What is the change in internal energy of the gas? (b) How much work does the gas do? (c) How much heat is added to the gas? Strategy (a) Because the system is an ideal gas, the internal energy only changes when the temperature changes. (b) The heat added to the system is therefore purely used to do work that has been calculated in Work, Heat, and Internal 114 3 • The First Law of Thermodynamics Access for free at openstax.org
Energy. (c) Lastly, the first law of thermodynamics can be used to calculate the heat added to the gas. Solution a. We saw in the preceding section that the internal energy of an ideal monatomic gas is a function only of temperature. Since , for this process, b. The quasi-static isothermal expansion of an ideal gas was considered in the preceding section and was found to be c. With the results of parts (a) and (b), we can use the first law to determine the heat added: which leads to Significance An isothermal process has no change in the internal energy. Based on that, the first law of thermodynamics reduces to . CHECK YOUR UNDERSTANDING 3.3 Why was it necessary to state that the process of Example 3.5 is quasi-static? EXAMPLE 3.6 Vaporizing Water When 1.00 g of water at changes from the liquid to the gas phase at atmospheric pressure, its change in volume is (a) How much heat must be added to vaporize the water? (b) How much work is done by the water against the atmosphere in its expansion? (c) What is the change in the internal energy of the water? Strategy We can first figure out how much heat is needed from the latent heat of vaporization of the water. From the volume change, we can calculate the work done from because the pressure is constant. Then, the first law of thermodynamics provides us with the change in the internal energy. Solution a. With representing the latent heat of vaporization, the heat required to vaporize the water is b. Since the pressure on the system is constant at , the work done by the water as it is vaporized is c. From the first law, the thermal energy of the water during its vaporization changes by Significance We note that in part (c), we see a change in internal energy, yet there is no change in temperature. Ideal gases that are not undergoing phase changes have the internal energy proportional to temperature. Internal energy in general is the sum of all energy in the system. 3.3 • First Law of Thermodynamics 115
CHECK YOUR UNDERSTANDING 3.4 When 1.00 g of ammonia boils at atmospheric pressure and its volume changes from 1.47 to . Its heat of vaporization at this pressure is What is the change in the internal energy of the ammonia when it vaporizes? INTERACTIVE Engage the Phet simulation below to learn about the first law of thermodynamics. First, pump some heavy species molecules into the chamber. Then, play around by doing work (pushing the wall to the right where the person is located) to see how the internal energy changes (as seen by temperature). Then, look at how heat added changes the internal energy. Finally, you can set a parameter constant such as temperature and see what happens when you do work to keep the temperature constant ( Note: You might see a change in these variables initially if you are moving around quickly in the simulation, but ultimately, this value will return to its equilibrium value). Access multimedia content (https://openstax.org/l/211stlawthermo) 3.4 Thermodynamic Processes LEARNING OBJECTIVES By the end of this section, you will be able to: • Define a thermodynamic process • Distinguish between quasi-static and non-quasi-static processes • Calculate physical quantities, such as the heat transferred, work done, and internal energy change for isothermal, adiabatic, and cyclical thermodynamic processes In solving mechanics problems, we isolate the body under consideration, analyze the external forces acting on it, and then use Newton’s laws to predict its behavior. In thermodynamics, we take a similar approach. We start by identifying the part of the universe we wish to study; it is also known as our system. (We defined a system at the beginning of this chapter as anything whose properties are of interest to us; it can be a single atom or the entire Earth.) Once our system is selected, we determine how the environment, or surroundings, interact with the system. Finally, with the interaction understood, we study the thermal behavior of the system with the help of the laws of thermodynamics. The thermal behavior of a system is described in terms of thermodynamic variables. For an ideal gas, these variables are pressure, volume, temperature, and the number of molecules or moles of the gas. Different types of systems are generally characterized by different sets of variables. For example, the thermodynamic variables for a stretched rubber band are tension, length, temperature, and mass. The state of a system can change as a result of its interaction with the environment. The change in a system can be fast or slow and large or small. The manner in which a state of a system can change from an initial state to a final state is called a thermodynamic process. For analytical purposes in thermodynamics, it is helpful to divide up processes as either quasi-static or non-quasi-static, as we now explain. Quasi-static and Non-quasi-static Processes A quasi-static process refers to an idealized or imagined process where the change in state is made infinitesimally slowly so that at each instant, the system can be assumed to be at a thermodynamic equilibrium with itself and with the environment. For instance, imagine heating 1 kg of water from a temperature to at a constant pressure of 1 atmosphere. To heat the water very slowly, we may imagine placing the container with water in a large bath that can be slowly heated such that the temperature of the bath can rise infinitesimally slowly from to . If we put 1 kg of water at directly into a bath at , the temperature of the water will rise rapidly to in a non-quasi-static way. Quasi-static processes are done slowly enough that the system remains at thermodynamic equilibrium at each instant, despite the fact that the system changes over time. The thermodynamic equilibrium of the system is 116 3 • The First Law of Thermodynamics Access for free at openstax.org
necessary for the system to have well-defined values of macroscopic properties such as the temperature and the pressure of the system at each instant of the process. Therefore, quasi-static processes can be shown as well- defined paths in state space of the system. Since quasi-static processes cannot be completely realized for any finite change of the system, all processes in nature are non-quasi-static. Examples of quasi-static and non-quasi-static processes are shown in Figure 3.8. Despite the fact that all finite changes must occur essentially non-quasi-statically at some stage of the change, we can imagine performing infinitely many quasi-static process corresponding to every quasi-static process. Since quasi-static processes can be solved analytically, we mostly study quasi-static processes in this book. We have already seen that in a quasi-static process the work by a gas is given by ∫ pdV. FIGURE 3.8 Quasi-static and non-quasi-static processes between states A and B of a gas. In a quasi-static process, the path of the process between A and B can be drawn in a state diagram since all the states that the system goes through are known. In a non-quasi-static process, the states between A and B are not known, and hence no path can be drawn. It may follow the dashed line as shown in the figure or take a very different path. Isothermal Processes An isothermal process is a change in the state of the system at a constant temperature. This process is accomplished by keeping the system in thermal equilibrium with a large heat bath during the process. Recall that a heat bath is an idealized “infinitely” large system whose temperature does not change. In practice, the temperature of a finite bath is controlled by either adding or removing a finite amount of energy as the case may be. As an illustration of an isothermal process, consider a cylinder of gas with a movable piston immersed in a large water tank whose temperature is maintained constant. Since the piston is freely movable, the pressure inside is balanced by the pressure outside by some weights on the piston, as in Figure 3.9. FIGURE 3.9 Expanding a system at a constant temperature. Removing weights on the piston leads to an imbalance of forces on the piston, which causes the piston to move up. As the piston moves up, the temperature is lowered momentarily, which causes heat to flow from the heat bath to the system. The energy to move the piston eventually comes from the heat bath. As weights on the piston are removed, an imbalance of forces on the piston develops. The net nonzero force on the piston would cause the piston to accelerate, resulting in an increase in volume. The expansion of the gas cools the gas to a lower temperature, which makes it possible for the heat to enter from the heat bath into the system until the temperature of the gas is reset to the temperature of the heat bath. If weights are removed in infinitesimal steps, the pressure in the system decreases infinitesimally slowly. This way, an isothermal process can be 3.4 • Thermodynamic Processes 117
conducted quasi-statically. An isothermal line on a ( p, V) diagram is represented by a curved line from starting pointA to finishing point B, as seen in Figure 3.10. For an ideal gas, an isothermal process is hyperbolic, since for an ideal gas at constant temperature, . FIGURE 3.10 An isothermal expansion from a state labeled A to another state labeled B on a pV diagram. The curve represents the relation between pressure and volume in an ideal gas at constant temperature. An isothermal process studied in this chapter is quasi-statically performed, since to be isothermal throughout the change of volume, you must be able to state the temperature of the system at each step, which is possible only if the system is in thermal equilibrium continuously. The system must go out of equilibrium for the state to change, but for quasi-static processes, we imagine that the process is conducted in infinitesimal steps such that these departures from equilibrium can be made as brief and as small as we like. Other quasi-static processes of interest for gases are isobaric and isochoric processes. An isobaric process is a process where the pressure of the system does not change, whereas an isochoric process is a process where the volume of the system does not change. Adiabatic Processes In an adiabatic process, the system is insulated from its environment so that although the state of the system changes, no heat is allowed to enter or leave the system, as seen in Figure 3.11. An adiabatic process can be conducted either quasi-statically or non-quasi-statically. When a system expands adiabatically, it must do work against the outside world, and therefore its energy goes down, which is reflected in the lowering of the temperature of the system. An adiabatic expansion leads to a lowering of temperature, and an adiabatic compression leads to an increase of temperature. We discuss adiabatic expansion again in Adiabatic Processes for an ideal Gas. FIGURE 3.11 An insulated piston with a hot, compressed gas is released. The piston moves up, the volume expands, and the pressure and temperature decrease. The internal energy goes into work. If the expansion occurs within a time frame in which negligible heat can enter the system, then the process is called adiabatic. Ideally, during an adiabatic process no heat enters or exits the system. Cyclic Processes We say that a system goes through a cyclic process if the state of the system at the end is same as the state at the beginning. Therefore, state properties such as temperature, pressure, volume, and internal energy of the system do not change over a complete cycle: When the first law of thermodynamics is applied to a cyclic process, we obtain a simple relation between heat into 118 3 • The First Law of Thermodynamics Access for free at openstax.org
the system and the work done by the system over the cycle: Thermodynamic processes are also distinguished by whether or not they are reversible. A reversible process is one that can be made to retrace its path by differential changes in the environment. Such a process must therefore also be quasi-static. Note, however, that a quasi-static process is not necessarily reversible, since there may be dissipative forces involved. For example, if friction occurred between the piston and the walls of the cylinder containing the gas, the energy lost to friction would prevent us from reproducing the original states of the system. We considered several thermodynamic processes: 1. An isothermal process, during which the system’s temperature remains constant 2. An adiabatic process, during which no heat is transferred to or from the system 3. An isobaric process, during which the system’s pressure does not change 4. An isochoric process, during which the system’s volume does not change Many other processes also occur that do not fit into any of these four categories. INTERACTIVE View this site (https://openstax.org/l/21idegaspvdiag) to set up your own process in a pV diagram. See if you can calculate the values predicted by the simulation for heat, work, and change in internal energy. 3.5 Heat Capacities of an Ideal Gas LEARNING OBJECTIVES By the end of this section, you will be able to: • Define heat capacity of an ideal gas for a specific process • Calculate the specific heat of an ideal gas for either an isobaric or isochoric process • Explain the difference between the heat capacities of an ideal gas and a real gas • Estimate the change in specific heat of a gas over temperature ranges We learned about specific heat and molar heat capacity in Temperature and Heat; however, we have not considered a process in which heat is added. We do that in this section. First, we examine a process where the system has a constant volume, then contrast it with a system at constant pressure and show how their specific heats are related. Let’s start with looking at Figure 3.12, which shows two vessels A and B, each containing the same type of ideal gas at a temperature T and a volume V. The only difference between the two vessels is that the piston at the top of A is fixed, whereas the one at the top of B is free to move against a constant external pressure p. We now consider what happens when the temperature of the gas in each vessel is slowly increased to with the addition of heat. FIGURE 3.12 Two vessels are identical except that the piston at the top of A is fixed, whereas that atop B is free to move against a constant external pressure p. Since the piston of vessel A is fixed, the volume of the enclosed gas does not change. Consequently, the gas does no work, and we have from the first law 3.5 • Heat Capacities of an Ideal Gas 119
We represent the fact that the heat is exchanged at constant volume by writing where is the molar heat capacity at constant volume of the gas, and, as before, n is the number of moles of the gas. In addition, since for this particular process, We obtained this equation assuming the volume of the gas was fixed. However, internal energy is a state function that depends on only the temperature of an ideal gas. Therefore, gives the change in internal energy of an ideal gas for any process involving a temperature change dT. When the gas in vessel B is heated, it expands against the movable piston and does work In this case, the heat is added at constant pressure, and we write where is the molar heat capacity at constant pressure of the gas. Furthermore, since the ideal gas expands against a constant pressure, becomes Finally, inserting the expressions for dQ and pdV into the first law, we obtain We have found for both an isochoric and an isobaric process. Because the internal energy of an ideal gas depends only on the temperature, must be the same for both processes. Thus, and The derivation of Equation 3.10 was based only on the ideal gas law. Consequently, this relationship is approximately valid for all dilute gases, whether monatomic like He, diatomic like or polyatomic like In the preceding chapter, we found the molar heat capacity of an ideal gas under constant volume to be where d is the number of degrees of freedom of a molecule in the system. Table 3.3 shows the molar heat capacities of some dilute ideal gases at room temperature Table 2.3 lists the experimental values for some gases. The heat capacities of real gases are somewhat higher than those predicted by the expressions of and given in Equation 3.10. This indicates that vibrational motion in polyatomic molecules is significant, even at room temperature. Nevertheless, the difference in the molar heat capacities, is very close to R, even for the polyatomic gases. 3.9 3.10 120 3 • The First Law of Thermodynamics Access for free at openstax.org
Molar Heat Capacities of Dilute Ideal Gases at Room Temperature Type of Molecule (J/mol K) (J/mol K) Monatomic Diatomic Polyatomic TABLE 3.3 3.6 Adiabatic Processes for an Ideal Gas LEARNING OBJECTIVES By the end of this section, you will be able to: • Define adiabatic expansion of an ideal gas • Demonstrate the qualitative difference between adiabatic and isothermal expansions When an ideal gas is compressed adiabatically work is done on it and its temperature increases; in an adiabatic expansion, the gas does work and its temperature drops. Adiabatic compressions actually occur in the cylinders of a car, where the compressions of the gas-air mixture take place so quickly that there is no time for the mixture to exchange heat with its environment. Nevertheless, because work is done on the mixture during the compression, its temperature does rise significantly. In fact, the temperature increases can be so large that the mixture can explode without the addition of a spark. Such explosions, since they are not timed, make a car run poorly—it usually “knocks.” Because ignition temperature rises with the octane of gasoline, one way to overcome this problem is to use a higher-octane gasoline. Another interesting adiabatic process is the free expansion of a gas. Figure 3.13 shows a gas confined by a membrane to one side of a two-compartment, thermally insulated container. When the membrane is punctured, gas rushes into the empty side of the container, thereby expanding freely. Because the gas expands “against a vacuum” , it does no work, and because the vessel is thermally insulated, the expansion is adiabatic. With and in the first law, so for the free expansion. FIGURE 3.13 The gas in the left chamber expands freely into the right chamber when the membrane is punctured. If the gas is ideal, the internal energy depends only on the temperature. Therefore, when an ideal gas expands freely, its temperature does not change. A quasi-static, adiabatic expansion of an ideal gas is represented in Figure 3.14, which shows an insulated cylinder that contains an ideal gas. The gas is made to expand quasi-statically by removing one grain of sand at a time from the top of the piston. When the gas expands by dV, the change in its temperature is dT. The work done by the gas in the expansion is because the cylinder is insulated; and the change in the internal energy of the gas is, from Equation 3.9, Therefore, from the first law, 3.6 • Adiabatic Processes for an Ideal Gas 121
so FIGURE 3.14 When sand is removed from the piston one grain at a time, the gas expands adiabatically and quasi-statically in the insulated vessel. Also, for an ideal gas, so and We now have two equations for dT. Upon equating them, we find that Now, we divide this equation by npV and use . We are then left with which becomes where we define as the ratio of the molar heat capacities: 3.11 122 3 • The First Law of Thermodynamics Access for free at openstax.org
Thus, and Finally, using , we can write this in the form This equation is the condition that must be obeyed by an ideal gas in a quasi-static adiabatic process. For example, if an ideal gas makes a quasi-static adiabatic transition from a state with pressure and volume and to a state with and then it must be true that The adiabatic condition of Equation 3.12 can be written in terms of other pairs of thermodynamic variables by combining it with the ideal gas law. In doing this, we find that and A reversible adiabatic expansion of an ideal gas is represented on the pV diagram of Figure 3.15. The slope of the curve at any point is FIGURE 3.15 Quasi-static adiabatic and isothermal expansions of an ideal gas. The dashed curve shown on this pV diagram represents an isothermal expansion where T (and therefore pV) is constant. The slope of this curve is useful when we consider the second law of thermodynamics in the next chapter. This slope is Because the isothermal curve is not as steep as that for the adiabatic expansion. EXAMPLE 3.7 Compression of an Ideal Gas in an Automobile Engine Gasoline vapor is injected into the cylinder of an automobile engine when the piston is in its expanded position. The 3.12 3.13 3.14 3.6 • Adiabatic Processes for an Ideal Gas 123
temperature, pressure, and volume of the resulting gas-air mixture are , and , respectively. The mixture is then compressed adiabatically to a volume of . Note that in the actual operation of an automobile engine, the compression is not quasi-static, although we are making that assumption here. A second, important approximation is that although the vapor is a gasoline-air mixture of nitrogen, oxygen, carbon dioxide, water, and hydrocarbons, we treat it as an ideal, diatomic gas. (a) What are the pressure and temperature of the mixture after the compression? (b) How much work is done by the mixture during the compression? Strategy Because we are modeling the process as a quasi-static adiabatic compression of an ideal gas, we have and . The work needed can then be evaluated with . Solution a. For an adiabatic compression we have Since we are treating the vapor as an ideal, diatomic gas, we can use . So after the compression, the pressure of the mixture is From the ideal gas law, the temperature of the mixture after the compression is b. The work done by the mixture during the compression is With the adiabatic condition of Equation 3.12, we may write p as where The work is therefore Significance The negative sign on the work done indicates that the piston does work on the gas-air mixture. The engine would not work if the gas-air mixture did work on the piston. 124 3 • The First Law of Thermodynamics Access for free at openstax.org
Chapter Review Key Terms adiabatic process process during which no heat is transferred to or from the system boundary imagined walls that separate the system and its surroundings closed system system that is mechanically and thermally isolated from its environment cyclic process process in which the state of the system at the end is same as the state at the beginning environment outside of the system being studied equation of state describes properties of matter under given physical conditions equilibrium thermal balance established between two objects or parts within a system extensive variable variable that is proportional to the amount of matter in the system first law of thermodynamics the change in internal energy for any transition between two equilibrium states is intensive variable variable that is independent of the amount of matter in the system internal energy average of the total mechanical energy of all the molecules or entities in the system isobaric process process during which the system’s pressure does not change isochoric process process during which the system’s volume does not change isothermal process process during which the system’s temperature remains constant molar heat capacity at constant pressure quantifies the ratio of the amount of heat added removed to the temperature while measuring at constant pressure molar heat capacity at constant volume quantifies the ratio of the amount of heat added removed to the temperature while measuring at constant volume open system system that can exchange energy and/ or matter with its surroundings quasi-static process evolution of a system that goes so slowly that the system involved is always in thermodynamic equilibrium reversible process process that can be reverted to restore both the system and its environment back to their original states together surroundings environment that interacts with an open system thermodynamic process manner in which a state of a system can change from initial state to final state thermodynamic system object and focus of thermodynamic study Key Equations Equation of state for a closed system Net work for a finite change in volume Internal energy of a system (average total energy) Internal energy of a monatomic ideal gas First law of thermodynamics Molar heat capacity at constant pressure Ratio of molar heat capacities Condition for an ideal gas in a quasi-static adiabatic process Summary 3.1 Thermodynamic Systems • A thermodynamic system, its boundary, and its surroundings must be defined with all the roles of the components fully explained before we can analyze a situation. • Thermal equilibrium is reached with two objects if a third object is in thermal equilibrium with the other two separately. • A general equation of state for a closed system has the form with an ideal gas as an illustrative example. 3.2 Work, Heat, and Internal Energy • Positive (negative) work is done by a thermodynamic system when it expands (contracts) under an external pressure. 3 • Chapter Review 125
• Heat is the energy transferred between two objects (or two parts of a system) because of a temperature difference. • Internal energy of a thermodynamic system is its total mechanical energy. 3.3 First Law of Thermodynamics • The internal energy of a thermodynamic system is a function of state and thus is unique for every equilibrium state of the system. • The increase in the internal energy of the thermodynamic system is given by the heat added to the system less the work done by the system in any thermodynamics process. 3.4 Thermodynamic Processes • The thermal behavior of a system is described in terms of thermodynamic variables. For an ideal gas, these variables are pressure, volume, temperature, and number of molecules or moles of the gas. • For systems in thermodynamic equilibrium, the thermodynamic variables are related by an equation of state. • A heat reservoir is so large that when it exchanges heat with other systems, its temperature does not change. • A quasi-static process takes place so slowly that the system involved is always in thermodynamic equilibrium. • A reversible process is one that can be made to retrace its path and both the temperature and pressure are uniform throughout the system. • There are several types of thermodynamic processes, including (a) isothermal, where the system’s temperature is constant; (b) adiabatic, where no heat is exchanged by the system; (c) isobaric, where the system’s pressure is constant; and (d) isochoric, where the system’s volume is constant. • As a consequence of the first law of thermodymanics, here is a summary of the thermodymaic processes: (a) isothermal: (b) adiabatic: (c) isobaric: and (d) isochoric: 3.5 Heat Capacities of an Ideal Gas • For an ideal gas, the molar capacity at constant pressure is given by , where d is the number of degrees of freedom of each molecule/ entity in the system. • A real gas has a specific heat close to but a little bit higher than that of the corresponding ideal gas with 3.6 Adiabatic Processes for an Ideal Gas • A quasi-static adiabatic expansion of an ideal gas produces a steeper pV curve than that of the corresponding isotherm. • A realistic expansion can be adiabatic but rarely quasi-static. Conceptual Questions 3.1 Thermodynamic Systems 1 . Consider these scenarios and state whether work is done by the system on the environment (SE) or by the environment on the system (ES): (a) opening a carbonated beverage; (b) filling a flat tire; (c) a sealed empty gas can expands on a hot day, bowing out the walls. 3.2 Work, Heat, and Internal Energy 2 . Is it possible to determine whether a change in internal energy is caused by heat transferred, by work performed, or by a combination of the two? 3 . When a liquid is vaporized, its change in internal energy is not equal to the heat added. Why? 4 . Why does a bicycle pump feel warm as you inflate your tire? 5. Is it possible for the temperature of a system to remain constant when heat flows into or out of it? If so, give examples. 3.3 First Law of Thermodynamics 6. What does the first law of thermodynamics tell us about the energy of the universe? 7. Does adding heat to a system always increase its internal energy? 8. A great deal of effort, time, and money has been spent in the quest for a so-called perpetual- motion machine, which is defined as a hypothetical machine that operates or produces useful work indefinitely and/or a hypothetical machine that produces more work or energy than it consumes. Explain, in terms of the first law of thermodynamics, why or why not such a 126 3 • Chapter Review Access for free at openstax.org
machine is likely to be constructed. 3.4 Thermodynamic Processes 9 . When a gas expands isothermally, it does work. What is the source of energy needed to do this work? 10. If the pressure and volume of a system are given, is the temperature always uniquely determined? 11. It is unlikely that a process can be isothermal unless it is a very slow process. Explain why. Is the same true for isobaric and isochoric processes? Explain your answer. 3.5 Heat Capacities of an Ideal Gas 12. How can an object transfer heat if the object does not possess a discrete quantity of heat? 13. Most materials expand when heated. One notable exception is water between and which actually decreases in volume with the increase in temperature. Which is greater for water in this temperature region, or ? 14. Why are there two specific heats for gases and , yet only one given for solid? 3.6 Adiabatic Processes for an Ideal Gas 15. Is it possible for to be smaller than unity? 16. Would you expect to be larger for a gas or a solid? Explain. 17. There is no change in the internal energy of an ideal gas undergoing an isothermal process since the internal energy depends only on the temperature. Is it therefore correct to say that an isothermal process is the same as an adiabatic process for an ideal gas? Explain your answer. 18. Does a gas do any work when it expands adiabatically? If so, what is the source of the energy needed to do this work? Problems 3.1 Thermodynamic Systems 19. A gas follows on an isothermal curve, where p is the pressure, V is the volume, b is a constant, and c is a function of temperature. If c can be determined from the measurement of a single other thermodynamic parameter, it establishes a temperature scale for this gas. Show that a temperature scale under an isochoric process can be established with this gas, with pressure being the measured parameter, and is identical to that of an ideal gas. 20. A mole of gas has isobaric expansion coefficient and isochoric pressure- temperature coefficient . Show that the equation of state is consistent with these coefficients. 21. Find the isobaric expansion coefficient and the isothermal pressure-volume coefficient of a solid that has equation of state . 3.2 Work, Heat, and Internal Energy 22. A gas at a pressure of 2.00 atm undergoes a quasi-static isobaric expansion from 3.00 to 5.00 L. How much work is done by the gas? 23. It takes 500 J of work to compress quasi- statically 0.50 mol of an ideal gas to one-fifth its original volume. Calculate the temperature of the gas, assuming it remains constant during the compression. 24. It is found that, when a dilute gas expands quasi-statically from 0.50 to 4.0 L, it does 250 J of work. Assuming that the gas temperature remains constant at 300 K, how many moles of gas are present? 25. In a quasi-static isobaric expansion, 500 J of work are done by the gas. If the gas pressure is 0.80 atm, what is the fractional increase in the volume of the gas, assuming it was originally at 20.0 L? 26. When a gas undergoes a quasi-static isobaric change in volume from 10.0 to 2.0 L, 15 J of work from an external source are required. What is the pressure of the gas? 27. An ideal gas expands quasi-statically and isothermally from a state with pressure p and volume V to a state with volume 4V. Show that the work done by the gas in the expansion is pV(ln 4). 28. As shown below, calculate the work done by the gas in the quasi-static processes represented by the paths (a) AB; (b) ADB; (c) ACB; and (d) ADCB. 3 • Chapter Review 127
29. (a) Calculate the work done on the gas along the closed path shown below. The curved section between R and S is semicircular when plotted using the indicated unit. (b) If the process is carried out in the opposite direction, what is the work done on the gas? 30. An ideal gas expands quasi-statically to three times its original volume. Which process requires more work from the gas, an isothermal process or an isobaric one? Determine the ratio of the work done in these processes. 31. A dilute gas at a pressure of 2.0 atm and a volume of 4.0 L is taken through the following quasi-static steps: (a) an isobaric expansion to a volume of 10.0 L, (b) an isochoric change to a pressure of 0.50 atm, (c) an isobaric compression to a volume of 4.0 L, and (d) an isochoric change to a pressure of 2.0 atm. Show these steps on a pV diagram and determine from your graph the net work done by the gas. 32. What is the average mechanical energy of the atoms of an ideal monatomic gas at 300 K? 33. What is the internal energy of 6.00 mol of an ideal monatomic gas at ? 34. Calculate the internal energy of 15 mg of helium at a temperature of 35. Two monatomic ideal gases A and B are at the same temperature. If 1.0 g of gas A has the same internal energy as 0.10 g of gas B, what are (a) the ratio of the number of moles of each gas and (b) the ratio of the atomic masses of the two gases? 36. The van der Waals coefficients for oxygen are and . Use these values to draw a van der Waals isotherm of oxygen at 100 K. On the same graph, draw isotherms of one mole of an ideal gas. 37. Find the work done in the quasi-static processes shown below. The states are given as (p, V) values for the points in the pV plane: 1 (3 atm, 4 L), 2 (3 atm, 6 L), 3 (5 atm, 4 L), 4 (2 atm, 6 L), 5 (4 atm, 2 L), 6 (5 atm, 5 L), and 7 (2 atm, 5 L). 3.3 First Law of Thermodynamics 38. When a dilute gas expands quasi-statically from 0.50 to 4.0 L, it does 250 J of work. Assuming that the gas temperature remains constant at 300 K, (a) what is the change in the internal energy of the gas? (b) How much heat is absorbed by the gas in this process? 39. In an expansion of gas, 500 J of work are done by the gas. If the internal energy of the gas increased by 80 J in the expansion, how much heat does the gas absorb? 128 3 • Chapter Review Access for free at openstax.org
40. An ideal gas expands quasi-statically and isothermally from a state with pressure p and volume V to a state with volume 4V. How much heat is added to the expanding gas? 41. As shown below, if the heat absorbed by the gas along AB is 400 J, determine the quantities of heat absorbed along (a) ADB; (b) ACB; and (c) ADCB. 42. During the isobaric expansion from A to B represented below, 3,100 J of heat are added to the gas. What is the change in its internal energy? 43. (a) What is the change in internal energy for the process represented by the closed path shown below? (b) How much heat is exchanged? (c) If the path is traversed in the opposite direction, how much heat is exchanged? 44. When a gas expands along path AC shown below, it does 100 J of work and absorbs either 200 or 400 J of heat. (a) Suppose you are told that along path ABC, the gas absorbs either 200 or 400 J of heat. Which of these values is correct? (b) Give the correct answer from part (a), how much work is done by the gas along ABC? (c) Along CD, the internal energy of the gas decreases by 50 J. How much heat is exchanged by the gas along this path? 45. When a gas expands along AB (see below), it does 20 J of work and absorbs 30 J of heat. When the gas expands along AC, it does 40 J of work and absorbs 70 J of heat. (a) How much heat does the gas exchange along BC? (b) When the gas makes the transition from C to A along CDA, 60 J of work are done on it from C to D. How much heat does it exchange along CDA? 46. A dilute gas is stored in the left chamber of a container whose walls are perfectly insulating (see below), and the right chamber is evacuated. When the partition is removed, the gas expands and fills the entire container. Calculate the work done by the gas. Does the internal energy of the gas change in this process? 47. Ideal gases A and B are stored in the left and right chambers of an insulated container, as shown below. The partition is removed and the gases mix. 3 • Chapter Review 129
Is any work done in this process? If the temperatures of A and B are initially equal, what happens to their common temperature after they are mixed? 48. An ideal monatomic gas at a pressure of and a temperature of 300 K undergoes a quasi-static isobaric expansion from (a) What is the work done by the gas? (b) What is the temperature of the gas after the expansion? (c) How many moles of gas are there? (d) What is the change in internal energy of the gas? (e) How much heat is added to the gas? 49. Consider the process for steam in a cylinder shown below. Suppose the change in the internal energy in this process is 30 kJ. Find the heat entering the system. 50. The state of 30 moles of steam in a cylinder is changed in a cyclic manner from a-b-c-a, where the pressure and volume of the states are: a (30 atm, 20 L), b (50 atm, 20 L), and c (50 atm, 45 L). Assume each change takes place along the line connecting the initial and final states in the pV plane. (a) Display the cycle in the pV plane. (b) Find the net work done by the steam in one cycle. (c) Find the net amount of heat flow in the steam over the course of one cycle. 51. A monatomic ideal gas undergoes a quasi-static process that is described by the function , where the pressure and volume are measured in SI units, the starting state is , and the final state is . Assume the system consists of n moles of the gas in a container that can exchange heat with the environment and whose volume can change freely. (a) Evaluate the work done by the gas during the change in the state. (b) Find the change in internal energy of the gas. (c) Find the heat input to the gas during the change. (d) What are initial and final temperatures? 52. A metallic container of fixed volume of immersed in a large tank of temperature contains two compartments separated by a freely movable wall. Initially, the wall is kept in place by a stopper so that there are 0.02 mol of the nitrogen gas on one side and 0.03 mol of the oxygen gas on the other side, each occupying half the volume. When the stopper is removed, the wall moves and comes to a final position. The movement of the wall is controlled so that the wall moves in infinitesimal quasi-static steps. (a) Find the final volumes of the two sides assuming the ideal gas behavior for the two gases. (b) How much work does each gas do on the other? (c) What is the change in the internal energy of each gas? (d) Find the amount of heat that enters or leaves each gas. 53. A gas in a cylindrical closed container is adiabatically and quasi-statically expanded from a state A (3 MPa, 2 L) to a state B with volume of 6 L along the path (a) Plot the path in the pV plane. (b) Find the amount of work done by the gas and the change in the internal energy of the gas during the process. 3.4 Thermodynamic Processes 54. Two moles of a monatomic ideal gas at (5 MPa, 5 L) is expanded isothermally until the volume is doubled (step 1). Then it is cooled isochorically until the pressure is 1 MPa (step 2). The temperature drops in this process. The gas is now compressed isothermally until its volume is back to 5 L, but its pressure is now 2 MPa (step 3). Finally, the gas is heated isochorically to return to the initial state (step 4). (a) Draw the four processes in the pV plane. (b) Find the total work done by the gas. 55. Consider a transformation from point A to B in a two-step process. First, the pressure is lowered from 3 MPa at point A to a pressure of 1 MPa, while keeping the volume at 2 L by cooling the system. The state reached is labeled C. Then the system is heated at a constant pressure to reach 130 3 • Chapter Review Access for free at openstax.org
a volume of 6 L in the state B. (a) Find the amount of work done on the ACB path. (b) Find the amount of heat exchanged by the system when it goes from A to B on the ACB path. (c) Compare the change in the internal energy when the AB process occurs isothermally with the AB change through the two-step process on the ACB path. 56. Consider a cylinder with a movable piston containing n moles of an ideal gas. The entire apparatus is immersed in a constant temperature bath of temperature T kelvin. The piston is then pushed slowly so that the pressure of the gas changes quasi-statically from to at constant temperature T. Find the work done by the gas in terms of n, R, T, and 57. An ideal gas expands isothermally along AB and does 700 J of work (see below). (a) How much heat does the gas exchange along AB? (b) The gas then expands adiabatically along BC and does 400 J of work. When the gas returns to A along CA, it exhausts 100 J of heat to its surroundings. How much work is done on the gas along this path? 58. Consider the processes shown below for a monatomic gas. (a) Find the work done in each of the processes AB, BC, AD, and DC. (b) Find the internal energy change in processes AB and BC. (c) Find the internal energy difference between states C and A. (d) Find the total heat added in the ADC process. (e) From the information given, can you find the heat added in process AD? Why or why not? 59. Two moles of helium gas are placed in a cylindrical container with a piston. The gas is at room temperature and under a pressure of When the pressure from the outside is decreased while keeping the temperature the same as the room temperature, the volume of the gas doubles. (a) Find the work the external agent does on the gas in the process. (b) Find the heat exchanged by the gas and indicate whether the gas takes in or gives up heat. Assume ideal gas behavior. 60. An amount of n moles of a monatomic ideal gas in a conducting container with a movable piston is placed in a large thermal heat bath at temperature and the gas is allowed to come to equilibrium. After the equilibrium is reached, the pressure on the piston is lowered so that the gas expands at constant temperature. The process is continued quasi-statically until the final volume is 4/3 of the initial volume (a) Find the change in the internal energy of the gas. (b) Find the work done by the gas. (c) Find the heat exchanged by the gas, and indicate, whether the gas takes in or gives up heat. 3.5 Heat Capacities of an Ideal Gas 61. The temperature of an ideal monatomic gas rises by 8.0 K. What is the change in the internal energy of 1 mol of the gas at constant volume? 62. For a temperature increase of at constant volume, what is the heat absorbed by (a) 3.0 mol of a dilute monatomic gas; (b) 0.50 mol of a dilute diatomic gas; and (c) 15 mol of a dilute polyatomic gas? 63. If the gases of the preceding problem are initially at 300 K, what are their internal energies after they absorb the heat? 64. Consider 0.40 mol of dilute carbon dioxide at a pressure of 0.50 atm and a volume of 50 L. What 3 • Chapter Review 131
is the internal energy of the gas? 65. When 400 J of heat are slowly added to 10 mol of an ideal monatomic gas, its temperature rises by . What is the work done on the gas? 66. One mole of a dilute diatomic gas occupying a volume of 10.00 L expands against a constant pressure of 2.000 atm when it is slowly heated. If 400.0 J of heat are added in the process, what is its final volume? 3.6 Adiabatic Processes for an Ideal Gas 67. A monatomic ideal gas undergoes a quasi-static adiabatic expansion in which its volume is doubled. How is the pressure of the gas changed? 68. An ideal gas has a pressure of 0.50 atm and a volume of 10 L. It is compressed adiabatically and quasi-statically until its pressure is 3.0 atm and its volume is 2.8 L. Is the gas monatomic, diatomic, or polyatomic? 69. Pressure and volume measurements of a dilute gas undergoing a quasi-static adiabatic expansion are shown below. Plot ln p vs. ln V and determine for this gas from your graph. P (atm) V (L) 20.0 1.0 17.0 1.1 14.0 1.3 11.0 1.5 8.0 2.0 5.0 2.6 2.0 5.2 1.0 8.4 70. An ideal monatomic gas at 300 K expands adiabatically and reversibly to twice its volume. What is its final temperature? 71. An ideal diatomic gas at 80 K is slowly compressed adiabatically and reversibly to half its volume. What is its final temperature? 72. An ideal diatomic gas at 80 K is slowly compressed adiabatically to one-third its original volume. What is its final temperature? 73. Compare the charge in internal energy of an ideal gas for a quasi-static adiabatic expansion with that for a quasi-static isothermal expansion. What happens to the temperature of an ideal gas in an adiabatic expansion? 74. The temperature of n moles of an ideal gas changes from to in a quasi-static adiabatic transition. Show that the work done by the gas is given by 75. A dilute gas expands quasi-statically to three times its initial volume. Is the final gas pressure greater for an isothermal or an adiabatic expansion? Does your answer depend on whether the gas is monatomic, diatomic, or polyatomic? 76. (a) An ideal gas expands adiabatically from a volume of to . If the initial pressure and temperature were and 300 K, respectively, what are the final pressure and temperature of the gas? Use for the gas. (b) In an isothermal process, an ideal gas expands from a volume of to . If the initial pressure and temperature were and 300 K, respectively, what are the final pressure and temperature of the gas? 77. On an adiabatic process of an ideal gas pressure, volume and temperature change such that is constant with for monatomic gas such as helium and for diatomic gas such as hydrogen at room temperature. Use numerical values to plot two isotherms of 1 mol of helium gas using ideal gas law and two adiabatic processes mediating between them. Use for your plot. 78. Two moles of a monatomic ideal gas such as helium is compressed adiabatically and reversibly from a state (3 atm, 5 L) to a state with pressure 4 atm. (a) Find the volume and temperature of the final state. (b) Find the temperature of the initial state of the gas. (c) Find the work done by the gas in the process. (d) Find the change in internal energy of the gas in the process. 132 3 • Chapter Review Access for free at openstax.org
Additional Problems 79. Consider the process shown below. During steps AB and BC, 3600 J and 2400 J of heat, respectively, are added to the system. (a) Find the work done in each of the processes AB, BC, AD, and DC. (b) Find the internal energy change in processes AB and BC. (c) Find the internal energy difference between states C and A. (d) Find the total heat added in the ADC process. (e) From the information given, can you find the heat added in process AD? Why or why not? 80. A car tire contains of air at a pressure of (about 32 psi). How much more internal energy does this gas have than the same volume has at zero gauge pressure (which is equivalent to normal atmospheric pressure)? 81. A helium-filled rigid, sealed container has a gauge pressure of 0.200 atm and a volume of 10.0 L. How much greater is the internal energy of the helium in the container than it would be at zero gauge pressure? Assume the volume doesn't change and no gas leaks in or out. 82. Steam to drive an old-fashioned steam locomotive is supplied at a constant gauge pressure of (about 250 psi) to a piston with a 0.200-m radius. (a) By calculating , find the work done by the steam when the piston moves 0.800 m. Note that this is the net work output, since gauge pressure is used. (b) Now find the amount of work by calculating the force exerted times the distance traveled. Is the answer the same as in part (a)? 83. A hand-driven tire pump has a piston with a 2.50-cm diameter and a maximum stroke of 30.0 cm. (a) How much work do you do in one stroke if the average gauge pressure is (about 35 psi)? (b) What average force do you exert on the piston, neglecting friction and gravitational force? 84. Calculate the net work output of a heat engine following path ABCDA as shown below. 85. What is the net work output of a heat engine that follows path ABDA in the preceding problem with a straight line from B to D? Why is the work output less than for path ABCDA? 86. Five moles of a monatomic ideal gas in a cylinder at is expanded isothermally from a volume of 5 L to 10 L. (a) What is the change in internal energy? (b) How much work was done on the gas in the process? (c) How much heat was transferred to the gas? 87. Four moles of a monatomic ideal gas in a cylinder at is expanded at constant pressure equal to 1 atm until its volume doubles. (a) What is the change in internal energy? (b) How much work was done by the gas in the process? (c) How much heat was transferred to the gas? 88. Helium gas is cooled from to by expanding from 40 atm to 1 atm. If there is 1.4 mol of helium, (a) What is the final volume of helium? (b) What is the change in internal energy? 89. In an adiabatic process, oxygen gas in a container is compressed along a path that can be described by the following pressure in atm as a function of volume V, with : . The initial and final volumes during the process were 2 L and 1.5 L, respectively. Find the amount of work done on the gas. 90. A cylinder containing three moles of a monatomic ideal gas is heated at a constant pressure of 2 atm. The temperature of the gas changes from 300 K to 350 K as a result of the 3 • Chapter Review 133
expansion. Find work done (a) on the gas; and (b) by the gas. 91. A cylinder containing three moles of nitrogen gas is heated at a constant pressure of 2 atm. The temperature of the gas changes from 300 K to 350 K as a result of the expansion. Find work done (a) on the gas, and (b) by the gas by using van der Waals equation of state instead of ideal gas law. 92. Two moles of a monatomic ideal gas such as helium is compressed adiabatically and reversibly from a state (3 atm, 5 L) to a state with a pressure of 4 atm. (a) Find the volume and temperature of the final state. (b) Find the temperature of the initial state. (c) Find work done by the gas in the process. (d) Find the change in internal energy in the process. Assume and for the diatomic ideal gas in the conditions given. 93. An insulated vessel contains 1.5 moles of argon at 2 atm. The gas initially occupies a volume of 5 L. As a result of the adiabatic expansion the pressure of the gas is reduced to 1 atm. (a) Find the volume and temperature of the final state. (b) Find the temperature of the gas in the initial state. (c) Find the work done by the gas in the process. (d) Find the change in the internal energy of the gas in the process. Challenge Problems 94. One mole of an ideal monatomic gas occupies a volume of at a pressure of (a) What is the temperature of the gas? (b) The gas undergoes a quasi-static adiabatic compression until its volume is decreased to What is the new gas temperature? (c) How much work is done on the gas during the compression? (d) What is the change in the internal energy of the gas? 95. One mole of an ideal gas is initially in a chamber of volume and at a temperature of . (a) How much heat is absorbed by the gas when it slowly expands isothermally to twice its initial volume? (b) Suppose the gas is slowly transformed to the same final state by first decreasing the pressure at constant volume and then expanding it isobarically. What is the heat transferred for this case? (c) Calculate the heat transferred when the gas is transformed quasi- statically to the same final state by expanding it isobarically, then decreasing its pressure at constant volume. 96. A bullet of mass 10 g is traveling horizontally at 200 m/s when it strikes and embeds in a pendulum bob of mass 2.0 kg. (a) How much mechanical energy is dissipated in the collision? (b) Assuming that for the bob plus bullet is 3R, calculate the temperature increase of the system due to the collision. Take the molecular mass of the system to be 200 g/mol. 97. The insulated cylinder shown below is closed at both ends and contains an insulating piston that is free to move on frictionless bearings. The piston divides the chamber into two compartments containing gases A and B. Originally, each compartment has a volume of and contains a monatomic ideal gas at a temperature of and a pressure of 1.0 atm. (a) How many moles of gas are in each compartment? (b) Heat Q is slowly added to A so that it expands and B is compressed until the pressure of both gases is 3.0 atm. Use the fact that the compression of B is adiabatic to determine the final volume of both gases. (c) What are their final temperatures? (d) What is the value of Q? 98. In a diesel engine, the fuel is ignited without a spark plug. Instead, air in a cylinder is compressed adiabatically to a temperature above the ignition temperature of the fuel; at the point of maximum compression, the fuel is injected into the cylinder. Suppose that air at is taken into the cylinder at a volume and then compressed adiabatically and quasi- statically to a temperature of and a volume If what is the ratio (Note: In an operating diesel engine, the compression is not quasi-static.) 134 3 • Chapter Review Access for free at openstax.org
INTRODUCTION CHAPTER 4 The Second Law of Thermodynamics 4.1 Reversible and Irreversible Processes 4.2 Heat Engines 4.3 Refrigerators and Heat Pumps 4.4 Statements of the Second Law of Thermodynamics 4.5 The Carnot Cycle 4.6 Entropy 4.7 Entropy on a Microscopic Scale According to the first law of thermodynamics, the only processes that can occur are those that conserve energy. But this cannot be the only restriction imposed by nature, because many seemingly possible thermodynamic processes that would conserve energy do not occur. For example, when two bodies are in thermal contact, heat never flows from the colder body to the warmer one, even though this is not forbidden by the first law. So some other thermodynamic principles must be controlling the behavior of physical systems. One such principle is the second law of thermodynamics, which limits the use of energy within a source. Energy cannot arbitrarily pass from one object to another, just as we cannot transfer heat from a cold object to a hot one without doing any work. We cannot unmix cream from coffee without a chemical process that changes the physical characteristics of the system or its environment. We cannot use internal energy stored in the air to propel a car, or use the energy of the ocean to run a ship, without disturbing something around that object. In the chapter covering the first law of thermodynamics, we started our discussion with a joke by C. P. Snow stating that the first law means “you can’t win.” He paraphrased the second law as “you can’t break even, except on a very cold day.” Unless you are at zero kelvin, you cannot convert 100% of thermal energy into work. We start by discussing spontaneous processes and explain why some processes require work to occur even if energy would FIGURE 4.1 A xenon ion engine from the Jet Propulsion Laboratory shows the faint blue glow of charged atoms emitted from the engine. The ion propulsion engine is the first nonchemical propulsion to be used as the primary means of propelling a spacecraft. (credit: modification of work by NASA/JPL) CHAPTER OUTLINE
have been conserved. 4.1 Reversible and Irreversible Processes LEARNING OBJECTIVES By the end of this section, you will be able to: • Define reversible and irreversible processes • State the second law of thermodynamics via an irreversible process Consider an ideal gas that is held in half of a thermally insulated container by a wall in the middle of the container. The other half of the container is under vacuum with no molecules inside. Now, if we remove the wall in the middle quickly, the gas expands and fills up the entire container immediately, as shown in Figure 4.2. FIGURE 4.2 A gas expanding from half of a container to the entire container (a) before and (b) after the wall in the middle is removed. Because half of the container is under vacuum before the gas expands there, we do not expect any work to be done by the system—that is, —because no force from the vacuum is exerted on the gas during the expansion. If the container is thermally insulated from the rest of the environment, we do not expect any heat transfer to the system either, so . Then the first law of thermodynamics leads to the change of the internal energy of the system, For an ideal gas, if the internal energy doesn’t change, then the temperature stays the same. Thus, the equation of state of the ideal gas gives us the final pressure of the gas, where is the pressure of the gas before the expansion. The volume is doubled and the pressure is halved, but nothing else seems to have changed during the expansion. All of this discussion is based on what we have learned so far and makes sense. Here is what puzzles us: Can all the molecules go backward to the original half of the container in some future time? Our intuition tells us that this is going to be very unlikely, even though nothing we have learned so far prevents such an event from happening, regardless of how small the probability is. What we are really asking is whether the expansion into the vacuum half of the container is reversible. A reversible process is a process in which the system and environment can be restored to exactly the same initial states that they were in before the process occurred, if we go backward along the path of the process. The necessary condition for a reversible process is therefore the quasi-static requirement. Note that it is quite easy to restore a system to its original state; the hard part is to have its environment restored to its original state at the same time. For example, in the example of an ideal gas expanding into vacuum to twice its original volume, we can easily push it back with a piston and restore its temperature and pressure by removing some heat from the gas. The problem is that we cannot do it without changing something in its surroundings, such as dumping some heat there. A reversible process is truly an ideal process that rarely happens. We can make certain processes close to reversible and therefore use the consequences of the corresponding reversible processes as a starting point or reference. In reality, almost all processes are irreversible, and some properties of the environment are altered when the properties of the system are restored. The expansion of an ideal gas, as we have just outlined, is irreversible because the process is not even quasi-static, that is, not in an equilibrium state at any moment of the expansion. 136 4 • The Second Law of Thermodynamics Access for free at openstax.org
From the microscopic point of view, a particle described by Newton’s second law can go backward if we flip the direction of time. But this is not the case, in practical terms, in a macroscopic system with more than particles or molecules, where numerous collisions between these molecules tend to erase any trace of memory of the initial trajectory of each of the particles. For example, we can actually estimate the chance for all the particles in the expanded gas to go back to the original half of the container, but the current age of the universe is still not long enough for it to happen even once. An irreversible process is what we encounter in reality almost all the time. The system and its environment cannot be restored to their original states at the same time. Because this is what happens in nature, it is also called a natural process. The sign of an irreversible process comes from the finite gradient between the states occurring in the actual process. For example, when heat flows from one object to another, there is a finite temperature difference (gradient) between the two objects. More importantly, at any given moment of the process, the system most likely is not at equilibrium or in a well-defined state. This phenomenon is called irreversibility. Let us see another example of irreversibility in thermal processes. Consider two objects in thermal contact: one at temperature and the other at temperature , as shown in Figure 4.3. FIGURE 4.3 Spontaneous heat flow from an object at higher temperature to another at lower temperature We know from common personal experience that heat flows from a hotter object to a colder one. For example, when we hold a few pieces of ice in our hands, we feel cold because heat has left our hands into the ice. The opposite is true when we hold one end of a metal rod while keeping the other end over a fire. Based on all of the experiments that have been done on spontaneous heat transfer, the following statement summarizes the governing principle: This statement turns out to be one of several different ways of stating the second law of thermodynamics. The form of this statement is credited to German physicist Rudolf Clausius (1822−1888) and is referred to as the Clausius statement of the second law of thermodynamics. The word “spontaneously” here means no other effort has been made by a third party, or one that is neither the hotter nor colder object. We will introduce some other major statements of the second law and show that they imply each other. In fact, all the different statements of the second law of thermodynamics can be shown to be equivalent, and all lead to the irreversibility of spontaneous heat flow between macroscopic objects of a very large number of molecules or particles. Both isothermal and adiabatic processes sketched on a pV graph (discussed in The First Law of Thermodynamics) are reversible in principle because the system is always at an equilibrium state at any point of the processes and can go forward or backward along the given curves. Other idealized processes can be represented by pV curves; Table 4.1 summarizes the most common reversible processes. Process Constant Quantity and Resulting Fact Isobaric Constant pressure Isochoric Constant volume TABLE 4.1 Summary of Simple Thermodynamic Processes SECOND LAW OF THERMODYNAMICS (CLAUSIUS STATEMENT) Heat never flows spontaneously from a colder object to a hotter object. 4.1 • Reversible and Irreversible Processes 137
Process Constant Quantity and Resulting Fact Isothermal Constant temperature Adiabatic No heat transfer TABLE 4.1 Summary of Simple Thermodynamic Processes 4.2 Heat Engines LEARNING OBJECTIVES By the end of this section, you will be able to: • Describe the function and components of a heat engine • Explain the efficiency of an engine • Calculate the efficiency of an engine for a given cycle of an ideal gas A heat engine is a device used to extract heat from a source and then convert it into mechanical work that is used for all sorts of applications. For example, a steam engine on an old-style train can produce the work needed for driving the train. Several questions emerge from the construction and application of heat engines. For example, what is the maximum percentage of the heat extracted that can be used to do work? This turns out to be a question that can only be answered through the second law of thermodynamics. The second law of thermodynamics can be formally stated in several ways. One statement presented so far is about the direction of spontaneous heat flow, known as the Clausius statement. A couple of other statements are based on heat engines. Whenever we consider heat engines and associated devices such as refrigerators and heat pumps, we do not use the normal sign convention for heat and work. For convenience, we assume that the symbols and W represent only the amounts of heat transferred and work delivered, regardless what the givers or receivers are. Whether heat is entering or leaving a system and work is done to or by a system are indicated by proper signs in front of the symbols and by the directions of arrows in diagrams. It turns out that we need more than one heat source/sink to construct a heat engine. We will come back to this point later in the chapter, when we compare different statements of the second law of thermodynamics. For the moment, we assume that a heat engine is constructed between a heat source (high-temperature reservoir or hot reservoir) and a heat sink (low-temperature reservoir or cold reservoir), represented schematically in Figure 4.4. The engine absorbs heat from a heat source (hot reservoir) of Kelvin temperature uses some of that energy to produce useful work W, and then discards the remaining energy as heat into a heat sink (cold reservoir) of Kelvin temperature Power plants and internal combustion engines are examples of heat engines. Power plants use steam produced at high temperature to drive electric generators, while exhausting heat to the atmosphere or a nearby body of water in the role of the heat sink. In an internal combustion engine, a hot gas-air mixture is used to push a piston, and heat is exhausted to the nearby atmosphere in a similar manner. FIGURE 4.4 A schematic representation of a heat engine. Energy flows from the hot reservoir to the cold reservoir while doing work. Actual heat engines have many different designs. Examples include internal combustion engines, such as those used in most cars today, and external combustion engines, such as the steam engines used in old steam-engine trains. Figure 4.5 shows a photo of a nuclear power plant in operation. The atmosphere around the reactors acts as 138 4 • The Second Law of Thermodynamics Access for free at openstax.org
the cold reservoir, and the heat generated from the nuclear reaction provides the heat from the hot reservoir. FIGURE 4.5 The heat exhausted from a nuclear power plant goes to the cooling towers, where it is released into the atmosphere. Heat engines operate by carrying a working substance through a cycle. In a steam power plant, the working substance is water, which starts as a liquid, becomes vaporized, is then used to drive a turbine, and is finally condensed back into the liquid state. As is the case for all working substances in cyclic processes, once the water returns to its initial state, it repeats the same sequence. For now, we assume that the cycles of heat engines are reversible, so there is no energy loss to friction or other irreversible effects. Suppose that the engine of Figure 4.4 goes through one complete cycle and that and W represent the heats exchanged and the work done for that cycle. Since the initial and final states of the system are the same, for the cycle. We therefore have from the first law of thermodynamics, so that The most important measure of a heat engine is its efficiency (e), which is simply “what we get out” divided by “what we put in” during each cycle, as defined by With a heat engine working between two heat reservoirs, we get out W and put in so the efficiency of the engine is Here, we used Equation 4.1, in the final step of this expression for the efficiency. EXAMPLE 4.1 A Lawn Mower A lawn mower is rated to have an efficiency of and an average power of 3.00 kW. What are (a) the average work and (b) the minimum heat discharge into the air by the lawn mower in one minute of use? Strategy From the average power—that is, the rate of work production—we can figure out the work done in the given elapsed time. Then, from the efficiency given, we can figure out the minimum heat discharge with Solution a. The average work delivered by the lawn mower is b. The minimum heat discharged into the air is given by 4.1 4.2 4.2 • Heat Engines 139
which leads to Significance As the efficiency rises, the minimum heat discharged falls. This helps our environment and atmosphere by not having as much waste heat expelled. 4.3 Refrigerators and Heat Pumps LEARNING OBJECTIVES By the end of this section, you will be able to: • Describe a refrigerator and a heat pump and list their differences • Calculate the performance coefficients of simple refrigerators and heat pumps The cycles we used to describe the engine in the preceding section are all reversible, so each sequence of steps can just as easily be performed in the opposite direction. In this case, the engine is known as a refrigerator or a heat pump, depending on what is the focus: the heat removed from the cold reservoir or the heat dumped to the hot reservoir. Either a refrigerator or a heat pump is an engine running in reverse. For a refrigerator, the focus is on removing heat from a specific area. For a heat pump, the focus is on dumping heat to a specific area. We first consider a refrigerator (Figure 4.6). The purpose of this engine is to remove heat from the cold reservoir, which is the space inside the refrigerator for an actual household refrigerator or the space inside a building for an air-conditioning unit. FIGURE 4.6 A schematic representation of a refrigerator (or a heat pump). The arrow next to work (W) indicates work being put into the system. A refrigerator (or heat pump) absorbs heat from the cold reservoir at Kelvin temperature and discards heat to the hot reservoir at Kelvin temperature while work W is done on the engine’s working substance, as shown by the arrow pointing toward the system in the figure. A household refrigerator removes heat from the food within it while exhausting heat to the surrounding air. The required work, for which we pay in our electricity bill, is performed by the motor that moves a coolant through the coils. A schematic sketch of a household refrigerator is given in Figure 4.7. 140 4 • The Second Law of Thermodynamics Access for free at openstax.org
FIGURE 4.7 A schematic diagram of a household refrigerator. A coolant with a boiling temperature below the freezing point of water is sent through the cycle (clockwise in this diagram). The coolant extracts heat from the refrigerator at the evaporator, causing coolant to vaporize. It is then compressed and sent through the condenser, where it exhausts heat to the outside. The effectiveness or coefficient of performance of a refrigerator is measured by the heat removed from the cold reservoir divided by the work done by the working substance cycle by cycle: Note that we have used the condition of energy conservation, in the final step of this expression. The effectiveness or coefficient of performance of a heat pump is measured by the heat dumped to the hot reservoir divided by the work done to the engine on the working substance cycle by cycle: Once again, we use the energy conservation condition to obtain the final step of this expression. 4.4 Statements of the Second Law of Thermodynamics LEARNING OBJECTIVES By the end of this section, you will be able to: • Contrast the second law of thermodynamics statements according to Kelvin and Clausius formulations • Interpret the second of thermodynamics via irreversibility Earlier in this chapter, we introduced the Clausius statement of the second law of thermodynamics, which is based 4.3 4.4 4.4 • Statements of the Second Law of Thermodynamics 141
on the irreversibility of spontaneous heat flow. As we remarked then, the second law of thermodynamics can be stated in several different ways, and all of them can be shown to imply the others. In terms of heat engines, the second law of thermodynamics may be stated as follows: This is known as the Kelvin statement of the second law of thermodynamics. This statement describes an unattainable “perfect engine,” as represented schematically in Figure 4.8(a). Note that “without any other effect” is a very strong restriction. For example, an engine can absorb heat and turn it all into work, but not if it completes a cycle. Without completing a cycle, the substance in the engine is not in its original state and therefore an “other effect” has occurred. Another example is a chamber of gas that can absorb heat from a heat reservoir and do work isothermally against a piston as it expands. However, if the gas were returned to its initial state (that is, made to complete a cycle), it would have to be compressed and heat would have to be extracted from it. The Kelvin statement is a manifestation of a well-known engineering problem. Despite advancing technology, we are not able to build a heat engine that is efficient. The first law does not exclude the possibility of constructing a perfect engine, but the second law forbids it. FIGURE 4.8 (a) A “perfect heat engine” converts all input heat into work. (b) A “perfect refrigerator” transports heat from a cold reservoir to a hot reservoir without work input. Neither of these devices is achievable in reality. We can show that the Kelvin statement is equivalent to the Clausius statement if we view the two objects in the Clausius statement as a cold reservoir and a hot reservoir. Thus, the Clausius statement becomes: It is impossible to construct a refrigerator that transfers heat from a cold reservoir to a hot reservoir without aid from an external source. The Clausius statement is related to the everyday observation that heat never flows spontaneously from a cold object to a hot object. Heat transfer in the direction of increasing temperature always requires some energy input. A “perfect refrigerator,” shown in Figure 4.8(b), which works without such external aid, is impossible to construct. To prove the equivalence of the Kelvin and Clausius statements, we show that if one statement is false, it necessarily follows that the other statement is also false. Let us first assume that the Clausius statement is false, so that the perfect refrigerator of Figure 4.8(b) does exist. The refrigerator removes heat Q from a cold reservoir at a temperature and transfers all of it to a hot reservoir at a temperature Now consider a real heat engine working in the same temperature range. It extracts heat from the hot reservoir, does work W, and discards heat Q to the cold reservoir. From the first law, these quantities are related by . Suppose these two devices are combined as shown in Figure 4.9. The net heat removed from the hot reservoir is , no net heat transfer occurs to or from the cold reservoir, and work W is done on some external body. Since , the combination of a perfect refrigerator and a real heat engine is itself a perfect heat engine, thereby contradicting the Kelvin statement. Thus, if the Clausius statement is false, the Kelvin statement must also be false. SECOND LAW OF THERMODYNAMICS (KELVIN STATEMENT) It is impossible to convert the heat from a single source into work without any other effect. 142 4 • The Second Law of Thermodynamics Access for free at openstax.org
FIGURE 4.9 Combining a perfect refrigerator and a real heat engine yields a perfect heat engine because Using the second law of thermodynamics, we now prove two important properties of heat engines operating between two heat reservoirs. The first property is that any reversible engine operating between two reservoirs has a greater efficiency than any irreversible engine operating between the same two reservoirs. The second property to be demonstrated is that all reversible engines operating between the same two reservoirs have the same efficiency. To show this, we start with the two engines D and E of Figure 4.10(a), which are operating between two common heat reservoirs at temperatures First, we assume that D is a reversible engine and that E is a hypothetical irreversible engine that has a higher efficiency than D. If both engines perform the same amount of work per cycle, it follows from Equation 4.2 that . It then follows from the first law that FIGURE 4.10 (a) Two uncoupled engines D and E working between the same reservoirs. (b) The coupled engines, with D working in reverse. Suppose the cycle of D is reversed so that it operates as a refrigerator, and the two engines are coupled such that the work output of E is used to drive D, as shown in Figure 4.10(b). Since and the net result of each cycle is equivalent to a spontaneous transfer of heat from the cold reservoir to the hot reservoir, a process the second law does not allow. The original assumption must therefore be wrong, and it is impossible to construct an irreversible engine such that E is more efficient than the reversible engine D. Now it is quite easy to demonstrate that the efficiencies of all reversible engines operating between the same reservoirs are equal. Suppose that D and E are both reversible engines. If they are coupled as shown in Figure 4.10(b), the efficiency of E cannot be greater than the efficiency of D, or the second law would be violated. If both engines are then reversed, the same reasoning implies that the efficiency of D cannot be greater than the efficiency of E. Combining these results leads to the conclusion that all reversible engines working between the same two reservoirs have the same efficiency. CHECK YOUR UNDERSTANDING 4.1 What is the efficiency of a perfect heat engine? What is the coefficient of performance of a perfect refrigerator? 4.4 • Statements of the Second Law of Thermodynamics 143
CHECK YOUR UNDERSTANDING 4.2 Show that for the hypothetical engine of Figure 4.10(b). 4.5 The Carnot Cycle LEARNING OBJECTIVES • Describe the Carnot cycle with the roles of all four processes involved • Outline the Carnot principle and its implications • Demonstrate the equivalence of the Carnot principle and the second law of thermodynamics In the early 1820s, Sadi Carnot (1786−1832), a French engineer, became interested in improving the efficiencies of practical heat engines. In 1824, his studies led him to propose a hypothetical working cycle with the highest possible efficiency between the same two reservoirs, known now as the Carnot cycle. An engine operating in this cycle is called a Carnot engine. The Carnot cycle is of special importance for a variety of reasons. At a practical level, this cycle represents a reversible model for the steam power plant and the refrigerator or heat pump. Yet, it is also very important theoretically, for it plays a major role in the development of another important statement of the second law of thermodynamics. Finally, because only two reservoirs are involved in its operation, it can be used along with the second law of thermodynamics to define an absolute temperature scale that is truly independent of any substance used for temperature measurement. With an ideal gas as the working substance, the steps of the Carnot cycle, as represented by Figure 4.11, are as follows. 1. Isothermal expansion. The gas is placed in thermal contact with a heat reservoir at a temperature The gas absorbs heat from the heat reservoir and is allowed to expand isothermally, doing work Because the internal energy of an ideal gas is a function of the temperature only, the change of the internal energy is zero, that is, during this isothermal expansion. With the first law of thermodynamics, we find that the heat absorbed by the gas is FIGURE 4.11 The four processes of the Carnot cycle. The working substance is assumed to be an ideal gas whose thermodynamic path MNOP is represented in Figure 4.12. 144 4 • The Second Law of Thermodynamics Access for free at openstax.org
FIGURE 4.12 The total work done by the gas in the Carnot cycle is shown and given by the area enclosed by the loop MNOPM. 2. Adiabatic expansion. The gas is thermally isolated and allowed to expand further, doing work Because this expansion is adiabatic, the temperature of the gas falls—in this case, from From and the equation of state for an ideal gas, , we have so that 3. Isothermal compression. The gas is placed in thermal contact with a cold reservoir at temperature and compressed isothermally. During this process, work is done on the gas and it gives up heat to the cold reservoir. The reasoning used in step 1 now yields where is the heat dumped to the cold reservoir by the gas. 4. Adiabatic compression. The gas is thermally isolated and returned to its initial state by compression. In this process, work is done on the gas. Because the compression is adiabatic, the temperature of the gas rises—from in this particular case. The reasoning of step 2 now gives The total work done by the gas in the Carnot cycle is given by This work is equal to the area enclosed by the loop shown in the pV diagram of Figure 4.12. Because the initial and final states of the system are the same, the change of the internal energy of the gas in the cycle must be zero, that is, . The first law of thermodynamics then gives and To find the efficiency of this engine, we first divide When the adiabatic constant from step 2 is divided by that of step 4, we find Substituting this into the equation for we obtain 4.5 • The Carnot Cycle 145
Finally, with Equation 4.2, we find that the efficiency of this ideal gas Carnot engine is given by An engine does not necessarily have to follow a Carnot engine cycle. All engines, however, have the same net effect, namely the absorption of heat from a hot reservoir, the production of work, and the discarding of heat to a cold reservoir. This leads us to ask: Do all reversible cycles operating between the same two reservoirs have the same efficiency? The answer to this question comes from the second law of thermodynamics discussed earlier: All reversible engine cycles produce exactly the same efficiency. Also, as you might expect, all real engines operating between two reservoirs are less efficient than reversible engines operating between the same two reservoirs. This too is a consequence of the second law of thermodynamics shown earlier. The cycle of an ideal gas Carnot refrigerator is represented by the pV diagram of Figure 4.13. It is a Carnot engine operating in reverse. The refrigerator extracts heat from a cold-temperature reservoir at when the ideal gas expands isothermally. The gas is then compressed adiabatically until its temperature reaches after which an isothermal compression of the gas results in heat being discarded to a high-temperature reservoir at Finally, the cycle is completed by an adiabatic expansion of the gas, causing its temperature to drop to FIGURE 4.13 The work done on the gas in one cycle of the Carnot refrigerator is shown and given by the area enclosed by the loop MPONM. The work done on the ideal gas is equal to the area enclosed by the path of the pV diagram. From the first law, this work is given by An analysis just like the analysis done for the Carnot engine gives When combined with Equation 4.3, this yields for the coefficient of performance of the ideal-gas Carnot refrigerator. Similarly, we can work out the coefficient of performance for a Carnot heat pump as We have just found equations representing the efficiency of a Carnot engine and the coefficient of performance of a Carnot refrigerator or a Carnot heat pump, assuming an ideal gas for the working substance in both devices. However, these equations are more general than their derivations imply. We will soon show that they are both valid 4.5 4.6 4.7 146 4 • The Second Law of Thermodynamics Access for free at openstax.org
no matter what the working substance is. Carnot summarized his study of the Carnot engine and Carnot cycle into what is now known as Carnot’s principle: This principle can be viewed as another statement of the second law of thermodynamics and can be shown to be equivalent to the Kelvin statement and the Clausius statement. EXAMPLE 4.2 The Carnot Engine A Carnot engine has an efficiency of 0.60 and the temperature of its cold reservoir is 300 K. (a) What is the temperature of the hot reservoir? (b) If the engine does 300 J of work per cycle, how much heat is removed from the high-temperature reservoir per cycle? (c) How much heat is exhausted to the low-temperature reservoir per cycle? Strategy From the temperature dependence of the thermal efficiency of the Carnot engine, we can find the temperature of the hot reservoir. Then, from the definition of the efficiency, we can find the heat removed when the work done by the engine is given. Finally, energy conservation will lead to how much heat must be dumped to the cold reservoir. Solution a. From we have so that the temperature of the hot reservoir is b. By definition, the efficiency of the engine is , so that the heat removed from the high-temperature reservoir per cycle is c. From the first law, the heat exhausted to the low-temperature reservoir per cycle by the engine is Significance A Carnot engine has the maximum possible efficiency of converting heat into work between two reservoirs, but this does not necessarily mean it is efficient. As the difference in temperatures of the hot and cold reservoir increases, the efficiency of a Carnot engine increases. EXAMPLE 4.3 A Carnot Heat Pump Imagine a Carnot heat pump operates between an outside temperature of and an inside temperature of . What is the work needed if the heat delivered to the inside of the house is 30.0 kJ? Strategy Because the heat pump is assumed to be a Carnot pump, its performance coefficient is given by CARNOT’S PRINCIPLE No engine working between two reservoirs at constant temperatures can have a greater efficiency than a reversible engine. 4.5 • The Carnot Cycle 147
Thus, we can find the work W from the heat delivered Solution The work needed is obtained from Significance We note that this work depends not only on the heat delivered to the house but also on the temperatures outside and inside. The dependence on the temperature outside makes them impractical to use in areas where the temperature is much colder outside than room temperature. In terms of energy costs, the heat pump is a very economical means for heating buildings (Figure 4.14). Contrast this method with turning electrical energy directly into heat with resistive heating elements. In this case, one unit of electrical energy furnishes at most only one unit of heat. Unfortunately, heat pumps have problems that do limit their usefulness. They are quite expensive to purchase compared to resistive heating elements, and, as the performance coefficient for a Carnot heat pump shows, they become less effective as the outside temperature decreases. In fact, below about , the heat they furnish is less than the energy used to operate them. FIGURE 4.14 A photograph of a heat pump (large box) located outside a house. This heat pump is located in a warm climate area, like the southern United States, since it would be far too inefficient located in the northern half of the United States. (credit: modification of work by Peter Stevens) CHECK YOUR UNDERSTANDING 4.3 A Carnot engine operates between reservoirs at and . (a) What is the efficiency of the engine? (b) If the engine does 5.0 J of work per cycle, how much heat per cycle does it absorb from the high-temperature reservoir? (c) How much heat per cycle does it exhaust to the cold-temperature reservoir? (d) What temperatures at the cold reservoir would give the minimum and maximum efficiency? CHECK YOUR UNDERSTANDING 4.4 A Carnot refrigerator operates between two heat reservoirs whose temperatures are and . (a) What is the coefficient of performance of the refrigerator? (b) If 200 J of work are done on the working substance per cycle, how much heat per cycle is extracted from the cold reservoir? (c) How much heat per cycle is discarded to the hot reservoir? 148 4 • The Second Law of Thermodynamics Access for free at openstax.org
4.6 Entropy LEARNING OBJECTIVES By the end of this section you will be able to: • Describe the meaning of entropy • Calculate the change of entropy for some simple processes The second law of thermodynamics is best expressed in terms of a change in the thermodynamic variable known as entropy, which is represented by the symbol S. Entropy, like internal energy, is a state function. This means that when a system makes a transition from one state into another, the change in entropy is independent of path and depends only on the thermodynamic variables of the two states. We first consider for a system undergoing a reversible process at a constant temperature. In this case, the change in entropy of the system is given by where Q is the heat exchanged by the system kept at a temperature T (in kelvin). If the system absorbs heat—that is, with —the entropy of the system increases. As an example, suppose a gas is kept at a constant temperature of 300 K while it absorbs 10 J of heat in a reversible process. Then from Equation 4.8, the entropy change of the gas is Similarly, if the gas loses 5.0 J of heat; that is, , at temperature , we have the entropy change of the system given by EXAMPLE 4.4 Entropy Change of Melting Ice Heat is slowly added to a 50-g chunk of ice at until it completely melts into water at the same temperature. What is the entropy change of the ice? Strategy Because the process is slow, we can approximate it as a reversible process. The temperature is a constant, and we can therefore use Equation 4.8 in the calculation. Solution The ice is melted by the addition of heat: In this reversible process, the temperature of the ice-water mixture is fixed at or 273.15 K. Now from , the entropy change of the ice is when it melts to water at . Significance During a phase change, the temperature is constant, allowing us to use Equation 4.8 to solve this problem. The same equation could also be used if we changed from a liquid to a gas phase, since the temperature does not change during that process either. 4.8 4.6 • Entropy 149
The change in entropy of a system for an arbitrary, reversible transition for which the temperature is not necessarily constant is defined by modifying . Imagine a system making a transition from state A to B in small, discrete steps. The temperatures associated with these states are and respectively. During each step of the transition, the system exchanges heat reversibly at a temperature This can be accomplished experimentally by placing the system in thermal contact with a large number of heat reservoirs of varying temperatures , as illustrated in Figure 4.15. The change in entropy for each step is The net change in entropy of the system for the transition is We now take the limit as , and the number of steps approaches infinity. Then, replacing the summation by an integral, we obtain where the integral is taken between the initial state A and the final state B. This equation is valid only if the transition from A to B is reversible. FIGURE 4.15 The gas expands at constant pressure as its temperature is increased in small steps through the use of a series of heat reservoirs. As an example, let us determine the net entropy change of a reversible engine while it undergoes a single Carnot cycle. In the adiabatic steps 2 and 4 of the cycle shown in Figure 4.11, no heat exchange takes place, so In step 1, the engine absorbs heat at a temperature so its entropy change is Similarly, in step 3, The net entropy change of the engine in one cycle of operation is then However, we know that for a Carnot engine, so There is no net change in the entropy of the Carnot engine over a complete cycle. Although this result was obtained for a particular case, its validity can be shown to be far more general: There is no net change in the entropy of a system undergoing any complete reversible cyclic process. Mathematically, we write this statement as 4.9 4.10 150 4 • The Second Law of Thermodynamics Access for free at openstax.org
where represents the integral over a closed reversible path. We can use Equation 4.11 to show that the entropy change of a system undergoing a reversible process between two given states is path independent. An arbitrary, closed path for a reversible cycle that passes through the statesA and B is shown in Figure 4.16. From Equation 4.11, for this closed path. We may split this integral into two segments, one along I, which leads from A to B, the other along II, which leads from B to A. Then Since the process is reversible, FIGURE 4.16 The closed loop passing through states A and B represents a reversible cycle. Hence, the entropy change in going from A to B is the same for paths I and II. Since paths I and II are arbitrary, reversible paths, the entropy change in a transition between two equilibrium states is the same for all the reversible processes joining these states. Entropy, like internal energy, is therefore a state function. What happens if the process is irreversible? When the process is irreversible, we expect the entropy of a closed system, or the system and its environment (the universe), to increase. Therefore we can rewrite this expression as where S is the total entropy of the closed system or the entire universe, and the equal sign is for a reversible process. The fact is the entropy statement of the second law of thermodynamics: We can show that this statement is consistent with the Kelvin statement, the Clausius statement, and the Carnot principle. 4.11 4.12 SECOND LAW OF THERMODYNAMICS (ENTROPY STATEMENT) The entropy of a closed system and the entire universe never decreases. 4.6 • Entropy 151
EXAMPLE 4.5 Entropy Change of a System during an Isobaric Process Determine the entropy change of an object of mass m and specific heat c that is cooled rapidly (and irreversibly) at constant pressure from to Strategy The process is clearly stated as an irreversible process; therefore, we cannot simply calculate the entropy change from the actual process. However, because entropy of a system is a function of state, we can imagine a reversible process that starts from the same initial state and ends at the given final state. Then, the entropy change of the system is given by Equation 4.10, Solution To replace this rapid cooling with a process that proceeds reversibly, we imagine that the hot object is put into thermal contact with successively cooler heat reservoirs whose temperatures range from to Throughout the substitute transition, the object loses infinitesimal amounts of heat dQ, so we have From the definition of heat capacity, an infinitesimal exchange dQ for the object is related to its temperature changedT by Substituting this dQ into the expression for , we obtain the entropy change of the object as it is cooled at constant pressure from Note that here because In other words, the object has lost some entropy. But if we count whatever is used to remove the heat from the object, we would still end up with because the process is irreversible. Significance If the temperature changes during the heat flow, you must keep it inside the integral to solve for the change in entropy. If, however, the temperature is constant, you can simply calculate the entropy change as the heat flow divided by the temperature. EXAMPLE 4.6 Stirling Engine The steps of a Stirling engine are as follows. For this problem, we will use 0.0010 mol of a monatomic gas that starts at a temperature of and a volume of , which will be called point A. Then it goes through the following steps: 1. Step AB: isothermal expansion at from to 2. Step BC: isochoric cooling to 3. Step CD: isothermal compression at from to 4. Step DA: isochoric heating back to and (a) Draw the pV diagram for the Stirling engine with proper labels. 152 4 • The Second Law of Thermodynamics Access for free at openstax.org
(b) Fill in the following table. Step W (J) Q (J) (J/K) Step AB Step BC Step CD Step DA Complete cycle (c) How does the efficiency of the Stirling engine compare to the Carnot engine working within the same two heat reservoirs? Strategy Using the ideal gas law, calculate the pressure at each point so that they can be labeled on the pV diagram. Isothermal work is calculated using and an isochoric process has no work done. The heat flow is calculated from the first law of thermodynamics, where for monatomic gasses. Isothermal steps have a change in entropy of Q/T, whereas isochoric steps have The efficiency of a heat engine is calculated by using Solution a. The graph is shown below. b. The completed table is shown below. Step W (J) Q (J) (J/K) Step AB Isotherm 2.3 2.3 0.0057 Step BC Isochoric 0 –1.2 –0.0035 Step CD Isotherm –1.8 –1.8 –0.0057 4.6 • Entropy 153
Step W (J) Q (J) (J/K) Step DA Isochoric 0 1.2 0.0035 Complete cycle 0.5 0.5 ~ 0 c. The efficiency of the Stirling heat engine is If this were a Carnot engine operating between the same heat reservoirs, its efficiency would be Therefore, the Carnot engine would have a greater efficiency than the Stirling engine. Significance We have noted that all reversible engines operating between the same two reservoirs have the same efficiency. Since we found that the Stirling engine has a lower efficiency than the Carnot cycle, it must not be a reversible engine, and this is indeed the case. The isochoric steps are not reversible unless a regenerator is used to capture and reclaim the heat during these steps. With a perfect regenerator, the efficiency of the ideal Stirling engine would be the same as that of the Carnot engine. In the early days of steam engines, accidents would occur due to the high pressure of the steam in the boiler. Robert Stirling developed an engine in 1816 that did not use steam and therefore was safer. The Stirling engine was commonly used in the nineteenth century, but developments in steam and internal combustion engines have made it difficult to broaden the use of the Stirling engine. The Stirling engine uses compressed air as the working substance, which passes back and forth between two chambers with a porous plug, called the regenerator, which is made of material that does not conduct heat as well. In two of the steps, pistons in the two chambers move in phase. 4.7 Entropy on a Microscopic Scale LEARNING OBJECTIVES By the end of this section, you will be able to: • Interpret the meaning of entropy at a microscopic scale • Calculate a change in entropy for an irreversible process of a system and contrast with the change in entropy of the universe • Explain the third law of thermodynamics We have seen how entropy is related to heat exchange at a particular temperature. In this section, we consider entropy from a statistical viewpoint. Although the details of the argument are beyond the scope of this textbook, it turns out that entropy can be related to how disordered or randomized a system is—the more it is disordered, the higher is its entropy. For example, a new deck of cards is very ordered, as the cards are arranged numerically by suit. In shuffling this new deck, we randomize the arrangement of the cards and therefore increase its entropy (Figure 4.17). Thus, by picking one card off the top of the deck, there would be no indication of what the next selected card will be. 154 4 • The Second Law of Thermodynamics Access for free at openstax.org
FIGURE 4.17 The entropy of a new deck of cards goes up after the dealer shuffles them. (credit: “Rommel SK”/YouTube) The second law of thermodynamics requires that the entropy of the universe increase in any irreversible process. Thus, in terms of order, the second law may be stated as follows:In any irreversible process, the universe becomes more disordered. For example, the irreversible free expansion of an ideal gas, shown in Figure 4.2, results in a larger volume for the gas molecules to occupy. A larger volume means more possible arrangements for the same number of atoms, so disorder is also increased. As a result, the entropy of the gas has gone up. The gas in this case is a closed system, and the process is irreversible. Changes in phase also illustrate the connection between entropy and disorder. EXAMPLE 4.7 Entropy Change of the Universe Suppose we place 50 g of ice at in contact with a heat reservoir at . Heat spontaneously flows from the reservoir to the ice, which melts and eventually reaches a temperature of . Find the change in entropy of (a) the ice and (b) the universe. Strategy Because the entropy of a system is a function of its state, we can imagine two reversible processes for the ice: (1) ice is melted at and (2) melted ice (water) is warmed up from to under constant pressure. Then, we add the change in entropy of the reservoir when we calculate the change in entropy of the universe. Solution a. From Equation 4.10, the increase in entropy of the ice is b. During this transition, the reservoir gives the ice an amount of heat equal to This leads to a change (decrease) in entropy of the reservoir: The increase in entropy of the universe is therefore 4.7 • Entropy on a Microscopic Scale 155
Significance The entropy of the universe therefore is greater than zero since the ice gains more entropy than the reservoir loses. If we considered only the phase change of the ice into water and not the temperature increase, the entropy change of the ice and reservoir would be the same, resulting in the universe gaining no entropy. This process also results in a more disordered universe. The ice changes from a solid with molecules located at specific sites to a liquid whose molecules are much freer to move. The molecular arrangement has therefore become more randomized. Although the change in average kinetic energy of the molecules of the heat reservoir is negligible, there is nevertheless a significant decrease in the entropy of the reservoir because it has many more molecules than the melted ice cube. However, the reservoir’s decrease in entropy is still not as large as the increase in entropy of the ice. The increased disorder of the ice more than compensates for the increased order of the reservoir, and the entropy of the universe increases by 4.6 J/K. You might suspect that the growth of different forms of life might be a net ordering process and therefore a violation of the second law. After all, a single cell gathers molecules and eventually becomes a highly structured organism, such as a human being. However, this ordering process is more than compensated for by the disordering of the rest of the universe. The net result is an increase in entropy and an increase in the disorder of the universe. CHECK YOUR UNDERSTANDING 4.5 In Example 4.7, the spontaneous flow of heat from a hot object to a cold object results in a net increase in entropy of the universe. Discuss how this result can be related to an increase in disorder of the system. The second law of thermodynamics makes clear that the entropy of the universe never decreases during any thermodynamic process. For any other thermodynamic system, when the process is reversible, the change of the entropy is given by . But what happens if the temperature goes to zero, ? It turns out this is not a question that can be answered by the second law. A fundamental issue still remains: Is it possible to cool a system all the way down to zero kelvin? We understand that the system must be at its lowest energy state because lowering temperature reduces the kinetic energy of the constituents in the system. What happens to the entropy of a system at the absolute zero temperature? It turns out the absolute zero temperature is not reachable—at least, not though a finite number of cooling steps. This is a statement of the third law of thermodynamics, whose proof requires quantum mechanics that we do not present here. In actual experiments, physicists have continuously pushed that limit downward, with the lowest temperature achieved at about in a low-temperature lab at the Helsinki University of Technology in 2008. Like the second law of thermodynamics, the third law of thermodynamics can be stated in different ways. One of the common statements of the third law of thermodynamics is: The absolute zero temperature cannot be reached through any finite number of cooling steps. In other words, the temperature of any given physical system must be finite, that is, . This produces a very interesting question in physics: Do we know how a system would behave if it were at the absolute zero temperature? The reason a system is unable to reach 0 K is fundamental and requires quantum mechanics to fully understand its origin. But we can certainly ask what happens to the entropy of a system when we try to cool it down to 0 K. Because the amount of heat that can be removed from the system becomes vanishingly small, we expect that the change in entropy of the system along an isotherm approaches zero, that is, This can be viewed as another statement of the third law, with all the isotherms becoming isentropic, or into a reversible ideal adiabat. We can put this expression in words: A system becomes perfectly ordered when its temperature approaches absolute zero and its entropy approaches its absolute minimum. The third law of thermodynamics puts another limit on what can be done when we look for energy resources. If 4.13 156 4 • The Second Law of Thermodynamics Access for free at openstax.org
there could be a reservoir at the absolute zero temperature, we could have engines with efficiency of , which would, of course, violate the second law of thermodynamics. EXAMPLE 4.8 Entropy Change of an Ideal Gas in Free Expansion An ideal gas occupies a partitioned volume inside a box whose walls are thermally insulating, as shown in Figure 4.18(a). When the partition is removed, the gas expands and fills the entire volume of the box, as shown in part (b). What is the entropy change of the universe (the system plus its environment)? FIGURE 4.18 The adiabatic free expansion of an ideal gas from volume to volume . Strategy The adiabatic free expansion of an ideal gas is an irreversible process. There is no change in the internal energy (and hence temperature) of the gas in such an expansion because no work or heat transfer has happened. Since the entropy of the system is a state variable, the entropy change of the system during the irreversible process is the same as for any other path. A convenient alternative reversible path connecting the same two equilibrium states for which we can calculate the change in entropy is a slow (quasistatic), isothermal expansion from to . In this process, the gas could be expanding against a piston while in thermal contact with a heat reservoir, as in step 1 of the Carnot cycle. Solution Since the temperature is constant, the entropy change is given by where because Now, with the help of the ideal gas law, we have so the change in entropy of the gas is Because , is positive, and the entropy of the gas has gone up during the free expansion. Significance What about the environment? The walls of the container are thermally insulating, so no heat exchange takes place between the gas and its surroundings. The entropy of the environment is therefore constant during the expansion. The net entropy change of the universe is then simply the entropy change of the gas. Since this is positive, the entropy of the universe increases in the free expansion of the gas. 4.7 • Entropy on a Microscopic Scale 157
EXAMPLE 4.9 Entropy Change during Heat Transfer Heat flows from a steel object of mass 4.00 kg whose temperature is 400 K to an identical object at 300 K. Assuming that the objects are thermally isolated from the environment, what is the net entropy change of the universe after thermal equilibrium has been reached? Strategy Since the objects are identical, their common temperature at equilibrium is 350 K. To calculate the entropy changes associated with their transitions, we substitute the irreversible process of the heat transfer by two isobaric, reversible processes, one for each of the two objects. The entropy change for each object is then given by Solution Using , the specific heat of steel, we have for the hotter object Similarly, the entropy change of the cooler object is The net entropy change of the two objects during the heat transfer is then Significance The objects are thermally isolated from the environment, so its entropy must remain constant. Thus, the entropy of the universe also increases by 37 J/K. CHECK YOUR UNDERSTANDING 4.6 A quantity of heat Q is absorbed from a reservoir at a temperature by a cooler reservoir at a temperature What is the entropy change of the hot reservoir, the cold reservoir, and the universe? CHECK YOUR UNDERSTANDING 4.7 A 50-g copper piece at a temperature of is placed into a large insulated vat of water at . (a) What is the entropy change of the copper piece when it reaches thermal equilibrium with the water? (b) What is the entropy change of the water? (c) What is the entropy change of the universe? INTERACTIVE View this site (https://openstax.org/l/21reversereact) to learn about entropy and microstates. Start with a large barrier in the middle and 1000 molecules in only the left chamber. What is the total entropy of the system? Now remove the barrier and let the molecules travel from the left to the right hand side? What is the total entropy of the system now? Lastly, add heat and note what happens to the temperature. Did this increase entropy of the system? 158 4 • The Second Law of Thermodynamics Access for free at openstax.org
Chapter Review Key Terms Carnot cycle cycle that consists of two isotherms at the temperatures of two reservoirs and two adiabatic processes connecting the isotherms Carnot engine Carnot heat engine, refrigerator, or heat pump that operates on a Carnot cycle Carnot principle principle governing the efficiency or performance of a heat device operating on a Carnot cycle: any reversible heat device working between two reservoirs must have the same efficiency or performance coefficient, greater than that of an irreversible heat device operating between the same two reservoirs Clausius statement of the second law of thermodynamics heat never flows spontaneously from a colder object to a hotter object coefficient of performance measure of effectiveness of a refrigerator or heat pump cold reservoir sink of heat used by a heat engine disorder measure of order in a system; the greater the disorder is, the higher the entropy efficiency ( e) output work from the engine over the input heat to the engine from the hot reservoir entropy state function of the system that changes when heat is transferred between the system and the environment entropy statement of the second law of thermodynamics entropy of a closed system or the entire universe never decreases heat engine device that converts heat into work heat pump device that delivers heat to a hot reservoir hot reservoir source of heat used by a heat engine irreversibility phenomenon associated with a natural process irreversible process process in which neither the system nor its environment can be restored to their original states at the same time isentropic reversible adiabatic process where the process is frictionless and no heat is transferred Kelvin statement of the second law of thermodynamics it is impossible to convert the heat from a single source into work without any other effect perfect engine engine that can convert heat into work with efficiency perfect refrigerator (heat pump) refrigerator (heat pump) that can remove (dump) heat without any input of work refrigerator device that removes heat from a cold reservoir reversible process process in which both the system and the external environment theoretically can be returned to their original states third law of thermodynamics absolute zero temperature cannot be reached through any finite number of cooling steps Key Equations Result of energy conservation Efficiency of a heat engine Coefficient of performance of a refrigerator Coefficient of performance of a heat pump Resulting efficiency of a Carnot cycle Performance coefficient of a reversible refrigerator Performance coefficient of a reversible heat pump Entropy of a system undergoing a reversible process at a constant temperature Change of entropy of a system under a reversible process Entropy of a system undergoing any complete reversible cyclic process 4 • Chapter Review 159
Change of entropy of a closed system under an irreversible process Change in entropy of the system along an isotherm Summary 4.1 Reversible and Irreversible Processes • A reversible process is one in which both the system and its environment can return to exactly the states they were in by following the reverse path. • An irreversible process is one in which the system and its environment cannot return together to exactly the states that they were in. • The irreversibility of any natural process results from the second law of thermodynamics. 4.2 Heat Engines • The work done by a heat engine is the difference between the heat absorbed from the hot reservoir and the heat discharged to the cold reservoir, that is, • The ratio of the work done by the engine and the heat absorbed from the hot reservoir provides the efficiency of the engine, that is, 4.3 Refrigerators and Heat Pumps • A refrigerator or a heat pump is a heat engine run in reverse. • The focus of a refrigerator is on removing heat from the cold reservoir with a coefficient of performance • The focus of a heat pump is on dumping heat to the hot reservoir with a coefficient of performance 4.4 Statements of the Second Law of Thermodynamics • The Kelvin statement of the second law of thermodynamics: It is impossible to convert the heat from a single source into work without any other effect. • The Kelvin statement and Clausius statement of the second law of thermodynamics are equivalent. 4.5 The Carnot Cycle • The Carnot cycle is the most efficient engine for a reversible cycle designed between two reservoirs. • The Carnot principle is another way of stating the second law of thermodynamics. 4.6 Entropy • The change in entropy for a reversible process at constant temperature is equal to the heat divided by the temperature. The entropy change of a system under a reversible process is given by . • A system’s change in entropy between two states is independent of the reversible thermodynamic path taken by the system when it makes a transition between the states. 4.7 Entropy on a Microscopic Scale • Entropy can be related to how disordered a system is—the more it is disordered, the higher is its entropy. In any irreversible process, the universe becomes more disordered. • According to the third law of thermodynamics, absolute zero temperature is unreachable. Conceptual Questions 4.1 Reversible and Irreversible Processes 1 . State an example of a process that occurs in nature that is as close to reversible as it can be. 4.2 Heat Engines 2 . Explain in practical terms why efficiency is defined as 4.3 Refrigerators and Heat Pumps 3 . If the refrigerator door is left open, what happens to the temperature of the kitchen? 4. Is it possible for the efficiency of a reversible engine to be greater than 1.0? Is it possible for the coefficient of performance of a reversible refrigerator to be less than 1.0? 4.4 Statements of the Second Law of Thermodynamics 5. In the text, we showed that if the Clausius statement is false, the Kelvin statement must also be false. Now show the reverse, such that if the Kelvin statement is false, it follows that the 160 4 • Chapter Review Access for free at openstax.org
Clausius statement is false. 6 . Why don’t we operate ocean liners by extracting heat from the ocean or operate airplanes by extracting heat from the atmosphere? 7 . Discuss the practical advantages and disadvantages of heat pumps and electric heating. 8 . The energy output of a heat pump is greater than the energy used to operate the pump. Why doesn’t this statement violate the first law of thermodynamics? 9 . Speculate as to why nuclear power plants are less efficient than fossil-fuel plants based on temperature arguments. 10. An ideal gas goes from state to state when it is allowed to expand freely. Is it possible to represent the actual process on a pV diagram? Explain. 4.5 The Carnot Cycle 11. To increase the efficiency of a Carnot engine, should the temperature of the hot reservoir be raised or lowered? What about the cold reservoir? 12. How could you design a Carnot engine with efficiency? 13. What type of processes occur in a Carnot cycle? 4.6 Entropy 14. Does the entropy increase for a Carnot engine for each cycle? 15. Is it possible for a system to have an entropy change if it neither absorbs nor emits heat during a reversible transition? What happens if the process is irreversible? 4.7 Entropy on a Microscopic Scale 16. Are the entropy changes of the systems in the following processes positive or negative? (a) water vapor that condenses on a cold surface; (b) gas in a container that leaks into the surrounding atmosphere; (c) an ice cube that melts in a glass of lukewarm water; (d) the lukewarm water of part (c); (e) a real heat engine performing a cycle; (f) food cooled in a refrigerator. 17. Discuss the entropy changes in the systems of Exercise 4.16 in terms of disorder. Problems 4.1 Reversible and Irreversible Processes 18. A tank contains 111.0 g chlorine gas which is at temperature and absolute pressure The temperature of the air outside the tank is . The molar mass of is 70.9 g/mol. (a) What is the volume of the tank? (b) What is the internal energy of the gas? (c) What is the work done by the gas if the temperature and pressure inside the tank drop to and , respectively, due to a leak? 19. A mole of ideal monatomic gas at and 1.00 atm is warmed up to expand isobarically to triple its volume. How much heat is transferred during the process? 20. A mole of an ideal gas at pressure 4.00 atm and temperature 298 K expands isothermally to double its volume. What is the work done by the gas? 21. After a free expansion to quadruple its volume, a mole of ideal diatomic gas is compressed back to its original volume adiabatically and then cooled down to its original temperature. What is the minimum heat removed from the gas in the final step to restoring its state? 4.2 Heat Engines 22. An engine is found to have an efficiency of 0.40. If it does 200 J of work per cycle, what are the corresponding quantities of heat absorbed and discharged? 23. In performing 100.0 J of work, an engine discharges 50.0 J of heat. What is the efficiency of the engine? 24. An engine with an efficiency of 0.30 absorbs 500 J of heat per cycle. (a) How much work does it perform per cycle? (b) How much heat does it discharge per cycle? 25. It is found that an engine discharges 100.0 J while absorbing 125.0 J each cycle of operation. (a) What is the efficiency of the engine? (b) How much work does it perform per cycle? 26. The temperature of the cold reservoir of the engine is 300 K. It has an efficiency of 0.30 and absorbs 500 J of heat per cycle. (a) How much 4 • Chapter Review 161
work does it perform per cycle? (b) How much heat does it discharge per cycle? 27. An engine absorbs three times as much heat as it discharges. The work done by the engine per cycle is 50 J. Calculate (a) the efficiency of the engine, (b) the heat absorbed per cycle, and (c) the heat discharged per cycle. 28. A coal power plant consumes 100,000 kg of coal per hour and produces 500 MW of power. If the heat of combustion of coal is 30 MJ/kg, what is the efficiency of the power plant? 4.3 Refrigerators and Heat Pumps 29. A refrigerator has a coefficient of performance of 3.0. (a) If it requires 200 J of work per cycle, how much heat per cycle does it remove the cold reservoir? (b) How much heat per cycle is discarded to the hot reservoir? 30. During one cycle, a refrigerator removes 500 J from a cold reservoir and discharges 800 J to its hot reservoir. (a) What is its coefficient of performance? (b) How much work per cycle does it require to operate? 31. If a refrigerator discards 80 J of heat per cycle and its coefficient of performance is 6.0, what are (a) the quantity off heat it removes per cycle from a cold reservoir and (b) the amount of work per cycle required for its operation? 32. A refrigerator has a coefficient of performance of 3.0. (a) If it requires 200 J of work per cycle, how much heat per cycle does it remove the cold reservoir? (b) How much heat per cycle is discarded to the hot reservoir? 4.5 The Carnot Cycle 33. The temperature of the cold and hot reservoirs between which a Carnot heat pump operates are and , respectively. Which is its coefficient of performance? 34. Suppose a Carnot refrigerator operates between Calculate the amount of work required to extract 1.0 J of heat from the cold reservoir if (a) , ; (b) , (c) , ; and (d) , . 35. A Carnot engine operates between reservoirs at 600 and 300 K. If the engine absorbs 100 J per cycle at the hot reservoir, what is its work output per cycle? 36. A 500-W motor operates a Carnot refrigerator between and . (a) What is the amount of heat per second extracted from the inside of the refrigerator? (b) How much heat is exhausted to the outside air per second? 37. Sketch a Carnot cycle on a temperature-volume diagram. 38. A Carnot heat pump operates between and . How much heat is exhausted into the interior of a house for every 1.0 J of work done by the pump? 39. An engine operating between heat reservoirs at and extracts 1000 J per cycle from the hot reservoir. (a) What is the maximum possible work that engine can do per cycle? (b) For this maximum work, how much heat is exhausted to the cold reservoir per cycle? 40. Suppose a Carnot engine can be operated between two reservoirs as either a heat engine or a refrigerator. How is the coefficient of performance of the refrigerator related to the efficiency of the heat engine? 41. A Carnot engine is used to measure the temperature of a heat reservoir. The engine operates between the heat reservoir and a reservoir consisting of water at its triple point. (a) If 400 J per cycle are removed from the heat reservoir while 200 J per cycle are deposited in the triple-point reservoir, what is the temperature of the heat reservoir? (b) If 400 J per cycle are removed from the triple-point reservoir while 200 J per cycle are deposited in the heat reservoir, what is the temperature of the heat reservoir? 42. What is the minimum work required of a refrigerator if it is to extract 50 J per cycle from the inside of a freezer at and exhaust heat to the air at ? 4.6 Entropy 43. Two hundred joules of heat are removed from a heat reservoir at a temperature of 200 K. What is the entropy change of the reservoir? 44. In an isothermal reversible expansion at , an ideal gas does 20 J of work. What is the entropy change of the gas? 45. An ideal gas at 300 K is compressed isothermally to one-fifth its original volume. Determine the entropy change per mole of the 162 4 • Chapter Review Access for free at openstax.org
gas. 46. What is the entropy change of 10 g of steam at when it condenses to water at the same temperature? 47. A metal rod is used to conduct heat between two reservoirs at temperatures respectively. When an amount of heat Q flows through the rod from the hot to the cold reservoir, what is the net entropy change of the rod, the hot reservoir, the cold reservoir, and the universe? 48. For the Carnot cycle of Figure 4.12, what is the entropy change of the hot reservoir, the cold reservoir, and the universe? 49. A 5.0-kg piece of lead at a temperature of is placed in a lake whose temperature is . Determine the entropy change of (a) the lead piece, (b) the lake, and (c) the universe. 50. One mole of an ideal gas doubles its volume in a reversible isothermal expansion. (a) What is the change in entropy of the gas? (b) If 1500 J of heat are added in this process, what is the temperature of the gas? 51. An ideal monatomic gas is confined to a rigid container. When heat is added reversibly to the gas, its temperature changes from (a) How much heat is added? (b) What is the change in entropy of the gas? 52. (a) A 5.0-kg rock at a temperature of is dropped into a shallow lake also at from a height of . What is the resulting change in entropy of the universe? (b) If the temperature of the rock is when it is dropped, what is the change of entropy of the universe? Assume that air friction is negligible (not a good assumption) and that is the specific heat of the rock. 4.7 Entropy on a Microscopic Scale 53. A copper rod of cross-sectional area and length 5.0 m conducts heat from a heat reservoir at 373 K to one at 273 K. What is the time rate of change of the universe’s entropy for this process? 54. Fifty grams of water at is heated until it becomes vapor at . Calculate the change in entropy of the water in this process. 55. Fifty grams of water at are changed into vapor at . What is the change in entropy of the water in this process? 56. In an isochoric process, heat is added to 10 mol of monoatomic ideal gas whose temperature increases from 273 to 373 K. What is the entropy change of the gas? 57. Two hundred grams of water at is brought into contact with a heat reservoir at . After thermal equilibrium is reached, what is the temperature of the water? Of the reservoir? How much heat has been transferred in the process? What is the entropy change of the water? Of the reservoir? What is the entropy change of the universe? 58. Suppose that the temperature of the water in the previous problem is raised by first bringing it to thermal equilibrium with a reservoir at a temperature of and then with a reservoir at . Calculate the entropy changes of (a) each reservoir, (b) of the water, and (c) of the universe. 59. Two hundred grams of water at is brought into contact into thermal equilibrium successively with reservoirs at , , , and . (a) What is the entropy change of the water? (b) Of the reservoir? (c) What is the entropy change of the universe? 60. (a) Ten grams of starts as ice at . The ice absorbs heat from the air (just above ) until all of it melts. Calculate the entropy change of the , of the air, and of the universe. (b) Suppose that the air in part (a) is at rather than and that the ice absorbs heat until it becomes water at . Calculate the entropy change of the , of the air, and of the universe. (c) Is either of these processes reversible? 61. The Carnot cycle is represented by the temperature-entropy diagram shown below. (a) How much heat is absorbed per cycle at the high-temperature reservoir? (b) How much heat is exhausted per cycle at the low-temperature reservoir? (c) How much work is done per cycle by the engine? (d) What is the efficiency of the engine? 4 • Chapter Review 163
62. A Carnot engine operating between heat reservoirs at 500 and 300 K absorbs 1500 J per cycle at the high-temperature reservoir. (a) Represent the engine’s cycle on a temperature- entropy diagram. (b) How much work per cycle is done by the engine? 63. A monoatomic ideal gas (n moles) goes through a cyclic process shown below. Find the change in entropy of the gas in each step and the total entropy change over the entire cycle. 64. A Carnot engine has an efficiency of 0.60. When the temperature of its cold reservoir changes, the efficiency drops to 0.55. If initially , determine (a) the constant value of and (b) the final value of . 65. A Carnot engine performs 100 J of work while discharging 200 J of heat each cycle. After the temperature of the hot reservoir only is adjusted, it is found that the engine now does 130 J of work while discarding the same quantity of heat. (a) What are the initial and final efficiencies of the engine? (b) What is the fractional change in the temperature of the hot reservoir? 66. A Carnot refrigerator exhausts heat to the air, which is at a temperature of . How much power is used by the refrigerator if it freezes 1.5 g of water per second? Assume the water is at . Additional Problems 67. A 300-W heat pump operates between the ground, whose temperature is , and the interior of a house at . What is the maximum amount of heat per hour that the heat pump can supply to the house? 68. An engineer must design a refrigerator that does 300 J of work per cycle to extract 2100 J of heat per cycle from a freezer whose temperature is . What is the maximum air temperature for which this condition can be met? Is this a reasonable condition to impose on the design? 69. A Carnot engine employs 1.5 mol of nitrogen gas as a working substance, which is considered as an ideal diatomic gas with at the working temperatures of the engine. The Carnot cycle goes in the cycle ABCDA with AB being an isothermal expansion. The volume at points A and C of the cycle are and 0.15 m3, respectively. The engine operates between two thermal baths of temperature 500 K and 300 K. (a) Find the values of volume at B and D. (b) How much heat is absorbed by the gas in the AB isothermal expansion? (c) How much work is done by the gas in the AB isothermal expansion? (d) How much heat is given up by the gas in the CD isothermal compression? (e) How much work is done by the gas in the CD isothermal compression? (f) How much work is done by the gas in the BC adiabatic expansion? (g) How much work is done by the gas in the DA adiabatic compression? (h) Find the value of efficiency of the engine based on the net work and heat input. Compare this value to the efficiency of a Carnot engine based on the temperatures of the two baths. 70. A 5.0-kg wood block starts with an initial speed 164 4 • Chapter Review Access for free at openstax.org
of 8.0 m/s and slides across the floor until friction stops it. Estimate the resulting change in entropy of the universe. Assume that everything stays at a room temperature of . 71. A system consisting of 20.0 mol of a monoatomic ideal gas is cooled at constant pressure from a volume of 50.0 L to 10.0 L. The initial temperature was 300 K. What is the change in entropy of the gas? 72. A glass beaker of mass 400 g contains 500 g of water at . The beaker is heated reversibly so that the temperature of the beaker and water rise gradually to . Find the change in entropy of the beaker and water together. 73. A Carnot engine operates between and baths and produces 300 kJ of energy in each cycle. Find the change in entropy of the (a) hot bath and (b) cold bath, in each Carnot cycle? 74. An ideal gas at temperature T is stored in the left half of an insulating container of volume V using a partition of negligible volume (see below). What is the entropy change per mole of the gas in each of the following cases? (a) The partition is suddenly removed and the gas quickly fills the entire container. (b) A tiny hole is punctured in the partition and after a long period, the gas reaches an equilibrium state such that there is no net flow through the hole. (c) The partition is moved very slowly and adiabatically all the way to the right wall so that the gas finally fills the entire container. 75. A 0.50-kg piece of aluminum at is dropped into 1.0 kg of water at . After equilibrium is reached, what is the net entropy change of the system? 76. Suppose 20 g of ice at is added to 300 g of water at . What is the total change in entropy of the mixture after it reaches thermal equilibrium? 77. A heat engine operates between two temperatures such that the working substance of the engine absorbs 5000 J of heat from the high-temperature bath and discharges 3000 J to the low-temperature bath. The rest of the energy is converted into mechanical energy of the turbine. Find (a) the amount of work produced by the engine and (b) the efficiency of the engine. 78. A thermal engine produces 4 MJ of electrical energy while operating between two thermal baths of different temperatures. The working substance of the engine discharges 5 MJ of heat to the cold temperature bath. What is the efficiency of the engine? 79. A coal power plant consumes 100,000 kg of coal per hour and produces 500 MW of power. If the heat of combustion of coal is 30 MJ/kg, what is the efficiency of the power plant? 80. A Carnot engine operates in a Carnot cycle between a heat source at and a heat sink at Find the efficiency of the Carnot engine. 81. A Carnot engine working between two heat baths of temperatures 600 K and 273 K completes each cycle in 5 sec. In each cycle, the engine absorbs 10 kJ of heat. Find the power of the engine. 82. A Carnot cycle working between and is used to drive a refrigerator between and How much energy must the Carnot engine produce per second so that the refrigerator is able to discard 10 J of energy per second? Challenge Problems 83. (a) An infinitesimal amount of heat is added reversibly to a system. By combining the first and second laws, show that . (b) When heat is added to an ideal gas, its temperature and volume change from . Show that the entropy change of n moles of the gas is given by . 84. Using the result of the preceding problem, show that for an ideal gas undergoing an adiabatic process, is constant. 85. With the help of the two preceding problems, 4 • Chapter Review 165
show that between states 1 and 2 of n moles an ideal gas is given by . 86. A cylinder contains 500 g of helium at 120 atm and . The valve is leaky, and all the gas slowly escapes isothermally into the atmosphere. Use the results of the preceding problem to determine the resulting change in entropy of the universe. 87. A diatomic ideal gas is brought from an initial equilibrium state at and to a final stage with and Use the results of the previous problem to determine the entropy change per mole of the gas. 88. The gasoline internal combustion engine operates in a cycle consisting of six parts. Four of these parts involve, among other things, friction, heat exchange through finite temperature differences, and accelerations of the piston; it is irreversible. Nevertheless, it is represented by the ideal reversible Otto cycle, which is illustrated below. The working substance of the cycle is assumed to be air. The six steps of the Otto cycle are as follows: i. Isobaric intake stroke (OA). A mixture of gasoline and air is drawn into the combustion chamber at atmospheric pressure as the piston expands, increasing the volume of the cylinder from zero to . ii. Adiabatic compression stroke (AB). The temperature of the mixture rises as the piston compresses it adiabatically from a volume . iii. Ignition at constant volume ( BC). The mixture is ignited by a spark. The combustion happens so fast that there is essentially no motion of the piston. During this process, the added heat causes the pressure to increase from at the constant volume . iv. Adiabatic expansion ( CD). The heated mixture of gasoline and air expands against the piston, increasing the volume from . This is called the power stroke, as it is the part of the cycle that delivers most of the power to the crankshaft. v. Constant-volume exhaust ( DA). When the exhaust valve opens, some of the combustion products escape. There is almost no movement of the piston during this part of the cycle, so the volume remains constant at . Most of the available energy is lost here, as represented by the heat exhaust . vi. Isobaric compression (AO). The exhaust valve remains open, and the compression from to zero drives out the remaining combustion products. (a) Using ( i) ; (ii) ; and (iii) , , show that . (b) Use the fact that steps (ii) and (iv) are adiabatic to show that , where . The quantity r is called the compression ratio of the engine. (c) In practice, r is kept less than around 7. For larger values, the gasoline-air mixture is compressed to temperatures so high that it explodes before the finely timed spark is delivered. This preignition causes engine knock and loss of power. Show that for and (the value for air), , or an efficiency of Because of the many irreversible processes, an actual internal combustion engine has an efficiency much less than this ideal value. A typical efficiency for a tuned engine is about . 89. An ideal diesel cycle is shown below. This cycle consists of five strokes. In this case, only air is drawn into the chamber during the intake stroke OA. The air is then compressed adiabatically from state A to state B, raising its temperature high enough so that when fuel is added during 166 4 • Chapter Review Access for free at openstax.org
the power stroke BC, it ignites. After ignition ends at C, there is a further adiabatic power stroke CD. Finally, there is an exhaust at constant volume as the pressure drops from to , followed by a further exhaust when the piston compresses the chamber volume to zero. (a) Use , , and to show that . (b) Use the fact that and are adiabatic to show that . (c) Since there is no preignition (remember, the chamber does not contain any fuel during the compression), the compression ratio can be larger than that for a gasoline engine. Typically, . For these values and show that , or an efficiency of . Diesel engines actually operate at an efficiency of about compared with for gasoline engines. 90. Consider an ideal gas Joule cycle, also called the Brayton cycle, shown below. Find the formula for efficiency of the engine using this cycle in terms of , , and . 91. Derive a formula for the coefficient of performance of a refrigerator using an ideal gas as a working substance operating in the cycle shown below in terms of the properties of the three states labeled 1, 2, and 3. 92. Two moles of nitrogen gas, with for ideal diatomic gases, occupies a volume of in an insulated cylinder at temperature 300 K. The gas is adiabatically and reversibly compressed to a volume of 5 L. The piston of the cylinder is locked in its place, and the insulation around the cylinder is removed. The heat-conducting cylinder is then placed in a 300-K bath. Heat from the compressed gas leaves the gas, and the temperature of the gas becomes 300 K again. The gas is then slowly expanded at the fixed temperature 300 K until the volume of the gas becomes , thus making a complete cycle for the gas. For the entire cycle, calculate (a) the work done by the gas, (b) the heat into or out of the gas, (c) the change in the internal energy of the gas, and (d) the change in entropy of the gas. 93. A Carnot refrigerator, working between and is used to cool a bucket of water containing of water at to in 2 hours. Find the total amount of work needed. 4 • Chapter Review 167
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INTRODUCTION CHAPTER 5 Electric Charges and Fields 5.1 Electric Charge 5.2 Conductors, Insulators, and Charging by Induction 5.3 Coulomb's Law 5.4 Electric Field 5.5 Calculating Electric Fields of Charge Distributions 5.6 Electric Field Lines 5.7 Electric Dipoles Back when we were studying Newton’s laws, we identified several physical phenomena as forces. We did so based on the effect they had on a physical object: Specifically, they caused the object to accelerate. Later, when we studied impulse and momentum, we expanded this idea to identify a force as any physical phenomenon that changed the momentum of an object. In either case, the result is the same: We recognize a force by the effect that it has on an object. In Gravitation (http://openstax.org/books/university-physics-volume-1/pages/13-introduction), we examined the force of gravity, which acts on all objects with mass. In this chapter, we begin the study of the electric force, which acts on all objects with a property called charge. The electric force is much stronger than gravity (in most systems where both appear), but it can be a force of attraction or a force of repulsion, which leads to very different effects on objects. The electric force helps keep atoms together, so it is of fundamental importance in matter. But it also governs most everyday interactions we deal with, from chemical interactions to biological processes. FIGURE 5.1 Electric charges exist all around us. They can cause objects to be repelled from each other or to be attracted to each other. (credit: modification of work by Sean McGrath) CHAPTER OUTLINE
5.1 Electric Charge LEARNING OBJECTIVES By the end of this section, you will be able to: • Describe the concept of electric charge • Explain qualitatively the force electric charge creates You are certainly familiar with electronic devices that you activate with the click of a switch, from computers to cell phones to television. And you have certainly seen electricity in a flash of lightning during a heavy thunderstorm. But you have also most likely experienced electrical effects in other ways, maybe without realizing that an electric force was involved. Let’s take a look at some of these activities and see what we can learn from them about electric charges and forces. Discoveries You have probably experienced the phenomenon of static electricity: When you first take clothes out of a dryer, many (not all) of them tend to stick together; for some fabrics, they can be very difficult to separate. Another example occurs if you take a woolen sweater off quickly—you can feel (and hear) the static electricity pulling on your clothes, and perhaps even your hair. If you comb your hair on a dry day and then put the comb close to a thin stream of water coming out of a faucet, you will find that the water stream bends toward (is attracted to) the comb (Figure 5.2). FIGURE 5.2 An electrically charged comb attracts a stream of water from a distance. Note that the water is not touching the comb. (credit: Jane Whitney) Suppose you bring the comb close to some small strips of paper; the strips of paper are attracted to the comb and even cling to it (Figure 5.3). In the kitchen, quickly pull a length of plastic cling wrap off the roll; it will tend to cling to most any nonmetallic material (such as plastic, glass, or food). If you rub a balloon on a wall for a few seconds, it will stick to the wall. Probably the most annoying effect of static electricity is getting shocked by a doorknob (or a friend) after shuffling your feet on some types of carpeting. 170 5 • Electric Charges and Fields Access for free at openstax.org
FIGURE 5.3 After being used to comb hair, this comb attracts small strips of paper from a distance, without physical contact. Investigation of this behavior helped lead to the concept of the electric force. (credit: Jane Whitney) Many of these phenomena have been known for centuries. The ancient Greek philosopher Thales of Miletus (624–546 BCE) recorded that when amber (a hard, translucent, fossilized resin from extinct trees) was vigorously rubbed with a piece of fur, a force was created that caused the fur and the amber to be attracted to each other (Figure 5.4). Additionally, he found that the rubbed amber would not only attract the fur, and the fur attract the amber, but they both could affect other (nonmetallic) objects, even if not in contact with those objects (Figure 5.5). FIGURE 5.4 Borneo amber is mined in Sabah, Malaysia, from shale-sandstone-mudstone veins. When a piece of amber is rubbed with a piece of fur, the amber gains more electrons, giving it a net negative charge. At the same time, the fur, having lost electrons, becomes positively charged. (credit: “Sebakoamber”/Wikimedia Commons) FIGURE 5.5 When materials are rubbed together, charges can be separated, particularly if one material has a greater affinity for electrons than another. (a) Both the amber and cloth are originally neutral, with equal positive and negative charges. Only a tiny fraction of the 5.1 • Electric Charge 171
charges are involved, and only a few of them are shown here. (b) When rubbed together, some negative charge is transferred to the amber, leaving the cloth with a net positive charge. (c) When separated, the amber and cloth now have net charges, but the absolute value of the net positive and negative charges will be equal. The English physicist William Gilbert (1544–1603) also studied this attractive force, using various substances. He worked with amber, and, in addition, he experimented with rock crystal and various precious and semi-precious gemstones. He also experimented with several metals. He found that the metals never exhibited this force, whereas the minerals did. Moreover, although an electrified amber rod would attract a piece of fur, it would repel another electrified amber rod; similarly, two electrified pieces of fur would repel each other. This suggested there were two types of an electric property; this property eventually came to be called electric charge. The difference between the two types of electric charge is in the directions of the electric forces that each type of charge causes: These forces are repulsive when the same type of charge exists on two interacting objects and attractive when the charges are of opposite types. The SI unit of electric charge is the coulomb (C), after the French physicist Charles-Augustin de Coulomb (1736–1806). The most peculiar aspect of this new force is that it does not require physical contact between the two objects in order to cause an acceleration. This is an example of a so-called “long-range” force. (Or, as James Clerk Maxwell later phrased it, “action at a distance.”) With the exception of gravity, all other forces we have discussed so far act only when the two interacting objects actually touch. The American physicist and statesman Benjamin Franklin found that he could concentrate charge in a “Leyden jar,” which was essentially a glass jar with two sheets of metal foil, one inside and one outside, with the glass between them (Figure 5.6). This created a large electric force between the two foil sheets. FIGURE 5.6 A Leyden jar (an early version of what is now called a capacitor) allowed experimenters to store large amounts of electric charge. Benjamin Franklin used such a jar to demonstrate that lightning behaved exactly like the electricity he got from the equipment in his laboratory. Franklin pointed out that the observed behavior could be explained by supposing that one of the two types of charge remained motionless, while the other type of charge flowed from one piece of foil to the other. He further suggested that an excess of what he called this “electrical fluid” be called “positive electricity” and the deficiency of it be called “negative electricity.” His suggestion, with some minor modifications, is the model we use today. (With the 172 5 • Electric Charges and Fields Access for free at openstax.org
experiments that he was able to do, this was a pure guess; he had no way of actually determining the sign of the moving charge. Unfortunately, he guessed wrong; we now know that the charges that flow are the ones Franklin labeled negative, and the positive charges remain largely motionless. Fortunately, as we’ll see, it makes no practical or theoretical difference which choice we make, as long as we stay consistent with our choice.) Let’s list the specific observations that we have of this electric force: • The force acts without physical contact between the two objects. • The force can be either attractive or repulsive: If two interacting objects carry the same sign of charge, the force is repulsive; if the charges are of opposite sign, the force is attractive. These interactions are referred to as electrostatic repulsion and electrostatic attraction, respectively. • Not all objects are affected by this force. • The magnitude of the force decreases (rapidly) with increasing separation distance between the objects. To be more precise, we find experimentally that the magnitude of the force decreases as the square of the distance between the two interacting objects increases. Thus, for example, when the distance between two interacting objects is doubled, the force between them decreases to one fourth what it was in the original system. We can also observe that the surroundings of the charged objects affect the magnitude of the force. However, we will explore this issue in a later chapter. Properties of Electric Charge In addition to the existence of two types of charge, several other properties of charge have been discovered. • Charge is quantized. This means that electric charge comes in discrete amounts, and there is a smallest possible amount of charge that an object can have. In the SI system, this smallest amount is . No free particle can have less charge than this, and, therefore, the charge on any object—the charge on all objects—must be an integer multiple of this amount. All macroscopic, charged objects have charge because electrons have either been added or taken away from them, resulting in a net charge. • The magnitude of the charge is independent of the type. Phrased another way, the smallest possible positive charge (to four significant figures) is , and the smallest possible negative charge is ; these values are exactly equal. This is simply how the laws of physics in our universe turned out. • Charge is conserved. Charge be transferred from place to place, from one object to another, but the net charge (sum of all the charges) of the universe is constant. Frequently, we speak of two charges “canceling”; this is verbal shorthand. It means that if two objects that have equal and opposite charges are physically close to each other, then the (oppositely directed) forces they apply on some other charged object cancel, for a net force of zero. It is important that you understand that the charges on the objects by no means disappear, however.1 • Charge is conserved in closed systems. In principle, if a negative charge disappeared from your lab bench and reappeared on the Moon, conservation of charge would still hold. However, this never happens. If the total charge you have in your local system on your lab bench is changing, there will be a measurable flow of charge into or out of the system. Again, charges can and do move around, and their effects can and do cancel, but the net charge in your local environment (if closed) is conserved. The last two items are both referred to as the law of conservation of charge. The Source of Charges: The Structure of the Atom Once it became clear that all matter was composed of particles that came to be called atoms, it also quickly became clear that the constituents of the atom included both positively charged particles and negatively charged particles. The next question was, what are the physical properties of those electrically charged particles? 1 Net charge is always conserved and charge is neither created nor destroyed in all of the interactions we will be concerned with in this unit, but we will see that in particle physics individual particles can be created and destroyed. For example, if an electron (charge -e) encounters an anti-electron (a positron, charge +e), both will annihilate, leaving behind only a photon (zero charge). While individual charges, and in fact the particles themselves, disappear, and a new particle appears, the net charge (zero) in the system is still conserved. You can learn more about this in Introduction to Particle Physics (http://openstax.org/books/university-physics-volume-3/pages/ 11-introduction). 5.1 • Electric Charge 173
The negatively charged particle was the first one to be discovered. In 1897, the English physicist J. J. Thomson was studying what was then known as cathode rays. Some years before, the English physicist William Crookes had shown that these “rays” were negatively charged, but his experiments were unable to tell any more than that. (The fact that they carried a negative electric charge was strong evidence that these were not rays at all, but particles.) Thomson prepared a pure beam of these particles and sent them through crossed electric and magnetic fields, and adjusted the various field strengths until the net deflection of the beam was zero. With this experiment, he was able to determine the charge-to-mass ratio of the particle. This ratio showed that the mass of the particle was much smaller than that of any other previously known particle—1837 times smaller, in fact. Eventually, this particle came to be called the electron. Since the atom as a whole is electrically neutral, the next question was to determine how the positive and negative charges are distributed within the atom. Thomson himself imagined that his electrons were embedded within a sort of positively charged paste, smeared out throughout the volume of the atom. However, in 1908, the New Zealand physicist Ernest Rutherford showed that the positive charges of the atom existed within a tiny core—called a nucleus—that took up only a very tiny fraction of the overall volume of the atom, but held over 99% of the mass. (See Linear Momentum and Collisions (http://openstax.org/books/university-physics-volume-1/pages/ 9-introduction).) In addition, he showed that the negatively charged electrons perpetually orbited about this nucleus, forming a sort of electrically charged cloud that surrounds the nucleus (Figure 5.7). Rutherford concluded that the nucleus was constructed of small, massive particles that he named protons. FIGURE 5.7 This simplified model of a hydrogen atom shows a positively charged nucleus (consisting, in the case of hydrogen, of a single proton), surrounded by an electron “cloud.” The charge of the electron cloud is equal (and opposite in sign) to the charge of the nucleus, but the electron does not have a definite location in space; hence, its representation here is as a cloud. Normal macroscopic amounts of matter contain immense numbers of atoms and molecules, and, hence, even greater numbers of individual negative and positive charges. Since it was known that different atoms have different masses, and that ordinarily atoms are electrically neutral, it was natural to suppose that different atoms have different numbers of protons in their nucleus, with an equal number of negatively charged electrons orbiting about the positively charged nucleus, thus making the atoms overall electrically neutral. However, it was soon discovered that although the lightest atom, hydrogen, did indeed have a single proton as its nucleus, the next heaviest atom—helium—has twice the number of protons (two), but four times the mass of hydrogen. This mystery was resolved in 1932 by the English physicist James Chadwick, with the discovery of the neutron. The neutron is, essentially, an electrically neutral twin of the proton, with no electric charge, but (nearly) identical mass to the proton. The helium nucleus therefore has two neutrons along with its two protons. (Later experiments were to show that although the neutron is electrically neutral overall, it does have an internal charge structure. Furthermore, although the masses of the neutron and the proton are nearly equal, they aren’t exactly equal: The neutron’s mass is very slightly larger than the mass of the proton. That slight mass excess turned out to be of great importance. That, however, is a story that will have to wait until our study of modern physics in Nuclear Physics (http://openstax.org/ books/university-physics-volume-3/pages/10-introduction).) Thus, in 1932, the picture of the atom was of a small, massive nucleus constructed of a combination of protons and 174 5 • Electric Charges and Fields Access for free at openstax.org
neutrons, surrounded by a collection of electrons whose combined motion formed a sort of negatively charged “cloud” around the nucleus (Figure 5.8). In an electrically neutral atom, the total negative charge of the collection of electrons is equal to the total positive charge in the nucleus. The very low-mass electrons can be more or less easily removed or added to an atom, changing the net charge on the atom (though without changing its type). An atom that has had the charge altered in this way is called an ion. Positive ions have had electrons removed, whereas negative ions have had excess electrons added. We also use this term to describe molecules that are not electrically neutral. FIGURE 5.8 The nucleus of a carbon atom is composed of six protons and six neutrons. As in hydrogen, the surrounding six electrons do not have definite locations and so can be considered to be a sort of cloud surrounding the nucleus. The story of the atom does not stop there, however. In the latter part of the twentieth century, many more subatomic particles were discovered in the nucleus of the atom: pions, neutrinos, and quarks, among others. With the exception of the photon, none of these particles are directly relevant to the study of electromagnetism, so we defer further discussion of them until the chapter on particle physics (Particle Physics and Cosmology (http://openstax.org/books/university-physics-volume-3/pages/11-introduction)). A Note on Terminology As noted previously, electric charge is a property that an object can have. This is similar to how an object can have a property that we call mass, a property that we call density, a property that we call temperature, and so on. Technically, we should always say something like, “Suppose we have a particle that carries a charge of ” However, it is very common to say instead, “Suppose we have a charge.” Similarly, we often say something like, “Six charges are located at the vertices of a regular hexagon.” A charge is not a particle; rather, it is a property of a particle. Nevertheless, this terminology is extremely common (and is frequently used in this book, as it is everywhere else). So, keep in the back of your mind what we really mean when we refer to a “charge.” 5.2 Conductors, Insulators, and Charging by Induction LEARNING OBJECTIVES By the end of this section, you will be able to: • Explain what a conductor is • Explain what an insulator is • List the differences and similarities between conductors and insulators • Describe the process of charging by induction In the preceding section, we said that scientists were able to create electric charge only on nonmetallic materials and never on metals. To understand why this is the case, you have to understand more about the nature and structure of atoms. In this section, we discuss how and why electric charges do—or do not—move through materials (Figure 5.9). A more complete description is given in a later chapter. 5.2 • Conductors, Insulators, and Charging by Induction 175
FIGURE 5.9 This power adapter uses metal wires and connectors to conduct electricity from the wall socket to a laptop computer. The conducting wires allow electrons to move freely through the cables, which are shielded by rubber and plastic. These materials act as insulators that don’t allow electric charge to escape outward. (credit: modification of work by “Evan-Amos”/Wikimedia Commons) Conductors and Insulators As discussed in the previous section, electrons surround the tiny nucleus in the form of a (comparatively) vast cloud of negative charge. However, this cloud does have a definite structure to it. Let’s consider an atom of the most commonly used conductor, copper. For reasons that will become clear in Atomic Structure (http://openstax.org/books/university-physics-volume-3/ pages/8-introduction), there is an outermost electron that is only loosely bound to the atom’s nucleus. It can be easily dislodged; it then moves to a neighboring atom. In a large mass of copper atoms (such as a copper wire or a sheet of copper), these vast numbers of outermost electrons (one per atom) wander from atom to atom, and are the electrons that do the moving when electricity flows. These wandering, or “free,” electrons are called conduction electrons, and copper is therefore an excellent conductor (of electric charge). All conducting elements have a similar arrangement of their electrons, with one or two conduction electrons. This includes most metals. Insulators, in contrast, are made from materials that lack conduction electrons; charge flows only with great difficulty, if at all. Even if excess charge is added to an insulating material, it cannot move, remaining indefinitely in place. This is why insulating materials exhibit the electrical attraction and repulsion forces described earlier, whereas conductors do not; any excess charge placed on a conductor would instantly flow away (due to mutual repulsion from existing charges), leaving no excess charge around to create forces. Charge cannot flow along or through an insulator, so its electric forces remain for long periods of time. (Charge will dissipate from an insulator, given enough time.) As it happens, amber, fur, and most semi-precious gems are insulators, as are materials like wood, glass, and plastic. Charging by Induction Let’s examine in more detail what happens in a conductor when an electrically charged object is brought close to it. As mentioned, the conduction electrons in the conductor are able to move with nearly complete freedom. As a result, when a charged insulator (such as a positively charged glass rod) is brought close to the conductor, the (total) charge on the insulator exerts an electric force on the conduction electrons. Since the rod is positively charged, the conduction electrons (which themselves are negatively charged) are attracted, flowing toward the insulator to the near side of the conductor (Figure 5.10). Now, the conductor is still overall electrically neutral; the conduction electrons have changed position, but they are still in the conducting material. However, the conductor now has a charge distribution; the near end (the portion of the conductor closest to the insulator) now has more negative charge than positive charge, and the reverse is true of the end farthest from the insulator. The relocation of negative charges to the near side of the conductor results in an overall positive charge in the part of the conductor farthest from the insulator. We have thus created an electric charge distribution where one did not exist before. This process is referred to as inducing polarization—in this case, polarizing the conductor. The resulting separation of positive and negative charge is called polarization, and a material, or even a molecule, that exhibits polarization is said to be polarized. A similar situation occurs with a 176 5 • Electric Charges and Fields Access for free at openstax.org
negatively charged insulator, but the resulting polarization is in the opposite direction. FIGURE 5.10 Induced polarization. A positively charged glass rod is brought near the left side of the conducting sphere, attracting negative charge and leaving the other side of the sphere positively charged. Although the sphere is overall still electrically neutral, it now has a charge distribution, so it can exert an electric force on other nearby charges. Furthermore, the distribution is such that it will be attracted to the glass rod. The result is the formation of what is called an electric dipole, from a Latin phrase meaning “two ends.” The presence of electric charges on the insulator—and the electric forces they apply to the conduction electrons—creates, or “induces,” the dipole in the conductor. Neutral objects can be attracted to any charged object. The pieces of straw attracted to polished amber are neutral, for example. If you run a plastic comb through your hair, the charged comb can pick up neutral pieces of paper. Figure 5.11 shows how the polarization of atoms and molecules in neutral objects results in their attraction to a charged object. FIGURE 5.11 Both positive and negative objects attract a neutral object by polarizing its molecules. (a) A positive object brought near a neutral insulator polarizes its molecules. There is a slight shift in the distribution of the electrons orbiting the molecule, with unlike charges being brought nearer and like charges moved away. Since the electrostatic force decreases with distance, there is a net attraction. (b) A negative object produces the opposite polarization, but again attracts the neutral object. (c) The same effect occurs for a conductor; since the unlike charges are closer, there is a net attraction. When a charged rod is brought near a neutral substance, an insulator in this case, the distribution of charge in atoms and molecules is shifted slightly. Opposite charge is attracted nearer the external charged rod, while like charge is repelled. Since the electrostatic force decreases with distance, the repulsion of like charges is weaker than the attraction of unlike charges, and so there is a net attraction. Thus, a positively charged glass rod attracts neutral pieces of paper, as will a negatively charged rubber rod. Some molecules, like water, are polar molecules. Polar molecules have a natural or inherent separation of charge, although they are neutral overall. Polar molecules are particularly affected by other charged objects and show greater polarization effects than molecules with naturally uniform charge distributions. When the two ends of a dipole can be separated, this method of charging by induction may be used to create charged objects without transferring charge. In Figure 5.12, we see two neutral metal spheres in contact with one another but insulated from the rest of the world. A positively charged rod is brought near one of them, attracting negative charge to that side, leaving the other sphere positively charged. 5.2 • Conductors, Insulators, and Charging by Induction 177
FIGURE 5.12 Charging by induction. (a) Two uncharged or neutral metal spheres are in contact with each other but insulated from the rest of the world. (b) A positively charged glass rod is brought near the sphere on the left, attracting negative charge and leaving the other sphere positively charged. (c) The spheres are separated before the rod is removed, thus separating negative and positive charges. (d) The spheres retain net charges after the inducing rod is removed—without ever having been touched by a charged object. Another method of charging by induction is shown in Figure 5.13. The neutral metal sphere is polarized when a charged rod is brought near it. The sphere is then grounded, meaning that a conducting wire is run from the sphere to the ground. Since Earth is large and most of the ground is a good conductor, it can supply or accept excess charge easily. In this case, electrons are attracted to the sphere through a wire called the ground wire, because it supplies a conducting path to the ground. The ground connection is broken before the charged rod is removed, leaving the sphere with an excess charge opposite to that of the rod. Again, an opposite charge is achieved when charging by induction, and the charged rod loses none of its excess charge. 178 5 • Electric Charges and Fields Access for free at openstax.org
FIGURE 5.13 Charging by induction using a ground connection. (a) A positively charged rod is brought near a neutral metal sphere, polarizing it. (b) The sphere is grounded, allowing electrons to be attracted from Earth’s ample supply. (c) The ground connection is broken. (d) The positive rod is removed, leaving the sphere with an induced negative charge. 5.3 Coulomb's Law LEARNING OBJECTIVES By the end of this section, you will be able to: • Describe the electric force, both qualitatively and quantitatively • Calculate the force that charges exert on each other • Determine the direction of the electric force for different source charges • Correctly describe and apply the superposition principle for multiple source charges Experiments with electric charges have shown that if two objects each have electric charge, then they exert an electric force on each other. The magnitude of the force is linearly proportional to the net charge on each object and inversely proportional to the square of the distance between them. (Interestingly, the force does not depend on the mass of the objects.) The direction of the force vector is along the imaginary line joining the two objects and is dictated by the signs of the charges involved. Let • the vector force exerted by charge 1 on charge 2; • the vector displacement from to . The magnitude of the electric force on one of the charges is proportional to the magnitude of its own charge and the magnitude of the other charge, and is inversely proportional to the square of the distance between them: This proportionality becomes an equality with the introduction of a proportionality constant. For reasons that will become clear in a later chapter, the proportionality constant that we use is actually a collection of constants. (We discuss this constant shortly.) COULOMB’S LAW The magnitude of the electric force (or Coulomb force) between two electrically charged particles is equal to If the charges have the same sign, the force is in the same direction as showing a repelling force. If the charges 5.1 5.3 • Coulomb's Law 179
FIGURE 5.14 The electrostatic force between point charges and separated by a distance r is given by Coulomb’s law. Note that Newton’s third law (every force exerted creates an equal and opposite force) applies as usual—the force on is equal in magnitude and opposite in direction to the force it exerts on . (a) Like charges; (b) unlike charges. It is important to note that the electric force is not constant; it is a function of the separation distance between the two charges. If either the test charge or the source charge (or both) move, then changes, and so does the force. An immediate consequence of this is that direct application of Newton’s laws with this force can be mathematically difficult, depending on the specific problem at hand. It can (usually) be done, but we almost always look for easier methods of calculating whatever physical quantity we are interested in. (Conservation of energy is the most common choice.) Finally, the new constant in Coulomb’s law is called the permittivity of free space, or (better) the permittivity of vacuum. It has a very important physical meaning that we will discuss in a later chapter; for now, it is simply an empirical proportionality constant. Its numerical value (to three significant figures) turns out to be These units are required to give the force in Coulomb’s law the correct units of newtons. Note that in Coulomb’s law, the permittivity of vacuum is only part of the proportionality constant. For convenience, we often define a Coulomb’s constant: EXAMPLE 5.1 The Force on the Electron in Hydrogen A hydrogen atom consists of a single proton and a single electron. The proton has a charge of and the electron has . In the “ground state” of the atom, the electron orbits the proton at most probable distance of (Figure 5.15). Calculate the electric force on the electron due to the proton. FIGURE 5.15 A schematic depiction of a hydrogen atom, showing the force on the electron. This depiction is only to enable us to calculate the force; the hydrogen atom does not really look like this. Recall Figure 5.7. have different signs, the force is in the opposite direction of showing an attracting force. (Figure 5.14). 180 5 • Electric Charges and Fields Access for free at openstax.org
Strategy For the purposes of this example, we are treating the electron and proton as two point particles, each with an electric charge, and we are told the distance between them; we are asked to calculate the force on the electron. We thus use Coulomb’s law. Solution Our two charges and the distance between them are, The magnitude of the force on the electron is As for the direction, since the charges on the two particles are opposite, the force is attractive; the force on the electron points radially directly toward the proton, everywhere in the electron’s orbit. The force is thus expressed as Significance This is a three-dimensional system, so the electron (and therefore the force on it) can be anywhere in an imaginary spherical shell around the proton. In this “classical” model of the hydrogen atom, the electrostatic force on the electron points in the inward centripetal direction, thus maintaining the electron’s orbit. But note that the quantum mechanical model of hydrogen (discussed in Quantum Mechanics (http://openstax.org/books/university-physics- volume-3/pages/7-introduction)) is utterly different. CHECK YOUR UNDERSTANDING 5.1 What would be different if the electron also had a positive charge? Multiple Source Charges The analysis that we have done for two particles can be extended to an arbitrary number of particles; we simply repeat the analysis, two charges at a time. Specifically, we ask the question: Given N charges (which we refer to as source charge), what is the net electric force that they exert on some other point charge (which we call the test charge)? Note that we use these terms because we can think of the test charge being used to test the strength of the force provided by the source charges. Like all forces that we have seen up to now, the net electric force on our test charge is simply the vector sum of each individual electric force exerted on it by each of the individual source charges. Thus, we can calculate the net force on the test charge Q by calculating the force on it from each source charge, taken one at a time, and then adding all those forces together (as vectors). This ability to simply add up individual forces in this way is referred to as the principle of superposition, and is one of the more important features of the electric force. In mathematical form, this becomes In this expression, Q represents the charge of the particle that is experiencing the electric force , and is located at from the origin; the are the N source charges, and the vectors are the displacements from the 5.2 5.3 • Coulomb's Law 181
position of the ith charge to the position of Q. Each of the N unit vectors points directly from its associated source charge toward the test charge. All of this is depicted in Figure 5.16. Please note that there is no physical difference between Q and ; the difference in labels is merely to allow clear discussion, with Q being the charge we are determining the force on. FIGURE 5.16 The eight source charges each apply a force on the single test charge Q. Each force can be calculated independently of the other seven forces. This is the essence of the superposition principle. (Note that the force vector does not necessarily point in the same direction as the unit vector ; it may point in the opposite direction, . The signs of the source charge and test charge determine the direction of the force on the test charge.) There is a complication, however. Just as the source charges each exert a force on the test charge, so too (by Newton’s third law) does the test charge exert an equal and opposite force on each of the source charges. As a consequence, each source charge would change position. However, by Equation 5.2, the force on the test charge is a function of position; thus, as the positions of the source charges change, the net force on the test charge necessarily changes, which changes the force, which again changes the positions. Thus, the entire mathematical analysis quickly becomes intractable. Later, we will learn techniques for handling this situation, but for now, we make the simplifying assumption that the source charges are fixed in place somehow, so that their positions are constant in time. (The test charge is allowed to move.) With this restriction in place, the analysis of charges is known as electrostatics, where “statics” refers to the constant (that is, static) positions of the source charges and the force is referred to as an electrostatic force. EXAMPLE 5.2 The Net Force from Two Source Charges Three different, small charged objects are placed as shown in Figure 5.17. The charges and are fixed in place; is free to move. Given , , and , and that , what is the net force on the middle charge ? 182 5 • Electric Charges and Fields Access for free at openstax.org
FIGURE 5.17 Source charges and each apply a force on . Strategy We use Coulomb’s law again. The way the question is phrased indicates that is our test charge, so that and are source charges. The principle of superposition says that the force on from each of the other charges is unaffected by the presence of the other charge. Therefore, we write down the force on from each and add them together as vectors. Solution We have two source charges and a test charge distances and and we are asked to find a force. This calls for Coulomb’s law and superposition of forces. There are two forces: We can’t add these forces directly because they don’t point in the same direction: points only in the −x-direction, while points only in the + y-direction. The net force is obtained from applying the Pythagorean theorem to its x- and y-components: where and We find that at an angle of 5.3 • Coulomb's Law 183
that is, above the − x-axis, as shown in the diagram. Significance Notice that when we substituted the numerical values of the charges, we did not include the negative sign of either or . Recall that negative signs on vector quantities indicate a reversal of direction of the vector in question. But for electric forces, the direction of the force is determined by the types (signs) of both interacting charges; we determine the force directions by considering whether the signs of the two charges are the same or are opposite. If you also include negative signs from negative charges when you substitute numbers, you run the risk of mathematically reversing the direction of the force you are calculating. Thus, the safest thing to do is to calculate just the magnitude of the force, using the absolute values of the charges, and determine the directions physically. It’s also worth noting that the only new concept in this example is how to calculate the electric forces; everything else (getting the net force from its components, breaking the forces into their components, finding the direction of the net force) is the same as force problems you have done earlier. CHECK YOUR UNDERSTANDING 5.2 What would be different if were negative? 5.4 Electric Field LEARNING OBJECTIVES By the end of this section, you will be able to: • Explain the purpose of the electric field concept • Describe the properties of the electric field • Calculate the field of a collection of source charges of either sign As we showed in the preceding section, the net electric force on a test charge is the vector sum of all the electric forces acting on it, from all of the various source charges, located at their various positions. But what if we use a different test charge, one with a different magnitude, or sign, or both? Or suppose we have a dozen different test charges we wish to try at the same location? We would have to calculate the sum of the forces from scratch. Fortunately, it is possible to define a quantity, called the electric field, which is independent of the test charge. It only depends on the configuration of the source charges, and once found, allows us to calculate the force on any test charge. Defining a Field Suppose we have N source charges , applying N electrostatic forces on a test charge Q, at displacements from the source charges. The net force on Q is (see Equation 5.2) We can rewrite this as where 5.3 184 5 • Electric Charges and Fields Access for free at openstax.org
or, more compactly, This expression is called the electric field at position of the N source charges. Here, P is the location of the point in space where you are calculating the field and the vectors are the displacements from each source to the test point (Figure 5.18). Note that we have to impose a coordinate system to solve actual problems. FIGURE 5.18 Each of these five source charges creates its own electric field at every point in space; shown here are the field vectors at an arbitrary point P. Like the electric force, the net electric field obeys the superposition principle. Notice that the calculation of the electric field makes no reference to the test charge. Thus, the physically useful approach is to calculate the electric field and then use it to calculate the force on some test charge later, if needed. Different test charges experience different forces Equation 5.3, but it is the same electric field Equation 5.4. That being said, recall that there is no fundamental difference between a test charge and a source charge; these are merely convenient labels for the system of interest. Any charge produces an electric field; however, just as Earth’s orbit is not affected by Earth’s own gravity, a charge is not subject to a force due to the electric field it generates. Charges are only subject to forces from the electric fields of other charges. In this respect, the electric field of a point charge is similar to the gravitational field of Earth; once we have calculated the gravitational field at some point in space, we can use it any time we want to calculate the resulting force on any mass we choose to place at that point. In fact, this is exactly what we do when we say the gravitational field of Earth (near Earth’s surface) has a value of and then we calculate the resulting force (i.e., weight) on different masses. Also, the general expression for calculating at arbitrary distances from the center of Earth (i.e., not just near Earth’s surface) is very similar to the expression for : , where G is a proportionality constant, playing the same role for as does for . The value of is calculated once and is then used in an endless number of problems. To push the analogy further, notice the units of the electric field: From , the units of E are newtons per coulomb, N/C, that is, the electric field applies a force on each unit charge. Now notice the units of g: From , the units of g are newtons per kilogram, N/kg, that is, the gravitational field applies a force on each unit mass. We could say that the gravitational field of Earth, near Earth’s surface, has a value of 9.81 N/kg. 5.4 5.4 • Electric Field 185
The Meaning of “Field” Recall from your studies of gravity that the word “field” in this context has a precise meaning. A field, in physics, is a physical quantity whose value depends on (is a function of) position, relative to the source of the field. In the case of the electric field, Equation 5.4 shows that the value of (both the magnitude and the direction) depends on where in space the point P is located, with measured from the locations of the source charges . In addition, since the electric field is a vector quantity, the electric field is referred to as a vector field. (The gravitational field is also a vector field.) In contrast, a field that has only a magnitude at every point is a scalar field. The temperature in a room is an example of a scalar field. It is a field because the temperature, in general, is different at different locations in the room, and it is a scalar field because temperature is a scalar quantity. Also, as you did with the gravitational field of an object with mass, you should picture the electric field of a charge- bearing object (the source charge) as a continuous, immaterial substance that surrounds the source charge, filling all of space—in principle, to ∞ in all directions. The field exists at every physical point in space. To put it another way, the electric charge on an object alters the space around the charged object in such a way that all other electrically charged objects in space experience an electric force as a result of being in that field. The electric field, then, is the mechanism by which the electric properties of the source charge are transmitted to and through the rest of the universe. (Again, the range of the electric force is infinite.) We will see in subsequent chapters that the speed at which electrical phenomena travel is the same as the speed of light. There is a deep connection between the electric field and light. Superposition Yet another experimental fact about the field is that it obeys the superposition principle. In this context, that means that we can (in principle) calculate the total electric field of many source charges by calculating the electric field of only at position P, then calculate the field of at P, while—and this is the crucial idea—ignoring the field of, and indeed even the existence of, We can repeat this process, calculating the field of each individual source charge, independently of the existence of any of the other charges. The total electric field, then, is the vector sum of all these fields. That, in essence, is what Equation 5.4 says. In the next section, we describe how to determine the shape of an electric field of a source charge distribution and how to sketch it. The Direction of the Field Equation 5.4 enables us to determine the magnitude of the electric field, but we need the direction also. We use the convention that the direction of any electric field vector is the same as the direction of the electric force vector that the field would apply to a positive test charge placed in that field. Such a charge would be repelled by positive source charges (the force on it would point away from the positive source charge) but attracted to negative charges (the force points toward the negative source). INTERACTIVE Add charges to the Electric Field of Dreams (https://openstax.org/l/21elefiedream) and see how they react to the electric field. Turn on a background electric field and adjust the direction and magnitude. DIRECTION OF THE ELECTRIC FIELD By convention, all electric fields point away from positive source charges and point toward negative source charges. 186 5 • Electric Charges and Fields Access for free at openstax.org
EXAMPLE 5.3 The E-Field of an Atom In an ionized helium atom, the most probable distance between the nucleus and the electron is . What is the electric field due to the nucleus at the location of the electron? Strategy Note that although the electron is mentioned, it is not used in any calculation. The problem asks for an electric field, not a force; hence, there is only one charge involved, and the problem specifically asks for the field due to the nucleus. Thus, the electron is a red herring; only its distance matters. Also, since the distance between the two protons in the nucleus is much, much smaller than the distance of the electron from the nucleus, we can treat the two protons as a single charge +2e (Figure 5.19). FIGURE 5.19 A schematic representation of a helium atom. Again, helium physically looks nothing like this, but this sort of diagram is helpful for calculating the electric field of the nucleus. Solution The electric field is calculated by Since there is only one source charge (the nucleus), this expression simplifies to Here (since there are two protons) and r is given; substituting gives The direction of is radially away from the nucleus in all directions. Why? Because a positive test charge placed in this field would accelerate radially away from the nucleus (since it is also positively charged), and again, the convention is that the direction of the electric field vector is defined in terms of the direction of the force it would apply to positive test charges. 5.4 • Electric Field 187
EXAMPLE 5.4 The E-Field above Two Equal Charges (a) Find the electric field (magnitude and direction) a distance z above the midpoint between two equal charges that are a distance d apart (Figure 5.20). Check that your result is consistent with what you’d expect when . (b) The same as part (a), only this time make the right-hand charge instead of . FIGURE 5.20 Finding the field of two identical source charges at the point P. Due to the symmetry, the net field at P is entirely vertical. (Notice that this is not true away from the midline between the charges.) Strategy We add the two fields as vectors, per Equation 5.4. Notice that the system (and therefore the field) is symmetrical about the vertical axis; as a result, the horizontal components of the field vectors cancel. This simplifies the math. Also, we take care to express our final answer in terms of only quantities that are given in the original statement of the problem: q, z, d, and constants Solution a. By symmetry, the horizontal ( x)-components of cancel (Figure 5.21); . FIGURE 5.21 Note that the horizontal components of the electric fields from the two charges cancel each other out, while the 188 5 • Electric Charges and Fields Access for free at openstax.org