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Physical Chemistry
Physical Chemistry
Robert G. Mortimer
In this third edition, core applications have been added along with more recent developments in the theories of chemical reaction kinetics and molecular quantum mechanics, as well as in the experimental study of extremely rapid chemical reactions. * Fully revised concise edition covering recent developments in the field * Clear and comprehensive text ideal for undergraduate and graduate course study * Encourages readers to apply theory in practical situations
Categories:
Chemistry\\Physical Chemistry
Year:
2008
Edition:
3rd ed
Publisher:
Academic Press/Elsevier
Language:
english
Pages:
1405
ISBN 10:
0123706173
ISBN 13:
9780080878591
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Physical Chemistry Third Edition Physical Chemistry Third Edition Robert G. Mortimer Professor Emeritus Rhodes College Memphis, Tennessee AMSTERDAM • BOSTON • HEIDELBERG • LONDON NEW YORK • OXFORD • PARIS • SAN DIEGO SAN FRANCISCO • SINGAPORE • SYDNEY • TOKYO Academic Press is an imprint of Elsevier Cover Design: Eric DeCicco Cover Image: © iStockphoto Elsevier Academic Press 30 Corporate Drive, Suite 400, Burlington, MA 01803, USA 525 B Street, Suite 1900, San Diego, CA 921014495, USA 84 Theobald’s Road, London WC1X 8RR, UK ∞ This book is printed on acidfree paper. Copyright © 2008, Elsevier Inc. All rights reserved. No part of this publication may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopy, recording, or any information storage and retrieval system, without permission in writing from the publisher. Permissions may be sought directly from Elsevier’s Science & Technology Rights Department in Oxford, UK: phone: (+44) 1865 843830, fax: (+44) 1865 853333, Email: permissions@elsevier.co.uk. You may also complete your request online via the Elsevier homepage (http://elsevier.com), by selecting “Customer Support” and then “Obtaining Permissions.” Library of Congress CataloginPublishing Data Mortimer, Robert G. Physical chemistry / Robert G. Mortimer. – 3rd ed. p. cm. Includes bibliographical references and index. ISBN 9780123706171 (hardcover : alk. paper) 1. Chemistry, Physical and theoretical. I. Title. QD453.2.M67 2008 541–dc22 2008007675 British Library Cataloguing in Publication Data A catalogue record for this book is available from the British Library ISBN13: 9780123706171 For information on all Elsevier Academic Press publications visit our Web site at www.books.elsevier.com Printed in Canada 08 09 10 9 8 7 6 5 4 3 2 1 To my wife, Ann, and to my late father, William E. Mortimer, who was responsible for my taking my ﬁrst chemistry course Contents Periodic Table Inside front cover List of Numerical Tables in Appendix A Inside front cover Information Tables Inside back cover Preface xv Acknowledgments Part 1 Chapter 1 xvii Thermodynamics and the Macroscopic Description of Physical Systems 1 The Behavior of Gases and Liquids 3 1.1 Introduction 4 1.2 Systems and States in Physical Chemistry 12 1.3 Real Gases 21 1.4 The Coexistence of Phases and the Critical Point 27 Chapter 2 Work, Heat, and Energy: The First Law of Thermodynamics 39 2.1 Work and the State of a System 40 2.2 Heat 51 2.3 Internal Energy: The First Law of Thermodynamics 55 2.4 Calculation of Amounts of Heat and Energy Changes 60 2.5 Enthalpy 74 2.6 Calculation of Enthalpy Changes of Processes without Chemical Reactions 81 2.7 Calculation of Enthalpy Changes of a Class of Chemical Reactions 86 2.8 Calculation of Energy Changes of Chemical Reactions 94 Chapter 3 The Second and Third Laws of Thermodynamics: Entropy 105 3.1 The Second Law of Thermodynamics and the Carnot Heat Engine 106 vii viii Contents 3.2 3.3 3.4 3.5 The Mathematical Statement of the Second Law: Entropy 114 The Calculation of Entropy Changes 121 Statistical Entropy 133 The Third Law of Thermodynamics and Absolute Entropies 139 Chapter 4 The Thermodynamics of Real Systems 151 4.1 Criteria for Spontaneous Processes and for Equilibrium: The Gibbs and Helmholtz Energies 152 4.2 Fundamental Relations for Closed Simple Systems 158 4.3 Additional Useful Thermodynamic Identities 167 4.4 Gibbs Energy Calculations 175 4.5 Multicomponent Systems 182 4.6 Euler’s Theorem and the Gibbs–Duhem Relation 188 Chapter 5 Phase Equilibrium 199 5.1 The Fundamental Fact of Phase Equilibrium 200 5.2 The Gibbs Phase Rule 202 5.3 Phase Equilibria in OneComponent Systems 205 5.4 The Gibbs Energy and Phase Transitions 215 5.5 Surfaces in OneComponent Systems 222 5.6 Surfaces in Multicomponent Systems 230 Chapter 6 The Thermodynamics of Solutions 237 6.1 Ideal Solutions 238 6.2 Henry’s Law and Dilute Nonelectrolyte Solutions 6.3 Activity and Activity Coefﬁcients 258 6.4 The Activities of Nonvolatile Solutes 267 6.5 Thermodynamic Functions of Nonideal Solutions 6.6 Phase Diagrams of Nonideal Mixtures 282 6.7 Colligative Properties 292 248 275 Chapter 7 Chemical Equilibrium 303 7.1 Gibbs Energy Changes and the Equilibrium Constant 304 7.2 Reactions Involving Gases and Pure Solids or Liquids 310 7.3 Chemical Equilibrium in Solutions 315 7.4 Equilibria in Solutions of Strong Electrolytes 328 7.5 Buffer Solutions 331 7.6 The Temperature Dependence of Chemical Equilibrium. The Principle of Le Châtelier 335 7.7 Chemical Equilibrium and Biological Systems 343 Chapter 8 The Thermodynamics of Electrochemical Systems 351 8.1 The Chemical Potential and the Electric Potential 352 8.2 Electrochemical Cells 354 8.3 HalfCell Potentials and Cell Potentials 361 8.4 The Determination of Activities and Activity Coefﬁcients of Electrolytes 371 8.5 Thermodynamic Information from Electrochemistry 374 ix Contents Part 2 Chapter 9 Dynamics 381 Gas Kinetic Theory: The Molecular Theory of Dilute Gases at Equilibrium 383 9.1 Macroscopic and Microscopic States of Macroscopic Systems 384 9.2 A Model System to Represent a Dilute Gas 386 9.3 The Velocity Probability Distribution 394 9.4 The Distribution of Molecular Speeds 405 9.5 The Pressure of a Dilute Gas 411 9.6 Effusion and Wall Collisions 416 9.7 The Model System with Potential Energy 418 9.8 The HardSphere Gas 422 9.9 The Molecular Structure of Liquids 434 Chapter 10 Transport Processes 441 10.1 The Macroscopic Description of Nonequilibrium States 442 10.2 Transport Processes 444 10.3 The Gas Kinetic Theory of Transport Processes in HardSphere Gases 460 10.4 Transport Processes in Liquids 467 10.5 Electrical Conduction in Electrolyte Solutions 475 Chapter 11 Chapter 12 Chapter 13 The Rates of Chemical Reactions 485 11.1 The Macroscopic Description of Chemical Reaction Rates 486 11.2 Forward Reactions with One Reactant 488 11.3 Forward Reactions with More Than One Reactant 11.4 Inclusion of a Reverse Reaction. Chemical Equilibrium 507 11.5 A Simple Reaction Mechanism: Two Consecutive Steps 510 11.6 Competing Reactions 513 11.7 The Experimental Study of Fast Reactions 515 Chemical Reaction Mechanisms I: Rate Laws and Mechanisms 523 12.1 Reaction Mechanisms and Elementary Processes in Gases 524 12.2 Elementary Processes in Liquid Solutions 527 12.3 The Temperature Dependence of Rate Constants 12.4 Reaction Mechanisms and Rate Laws 540 12.5 Chain Reactions 556 499 533 Chemical Reaction Mechanisms II: Catalysis and Miscellaneous Topics 565 13.1 Catalysis 566 13.2 Competing Mechanisms and the Principle of Detailed Balance 583 13.3 Autocatalysis and Oscillatory Chemical Reactions 585 13.4 The Reaction Kinetics of Polymer Formation 589 x Contents 13.5 13.6 Part 3 Nonequilibrium Electrochemistry 595 Experimental Molecular Study of Chemical Reaction Mechanisms 608 The Molecular Nature of Matter 617 Chapter 14 Classical Mechanics and the Old Quantum Theory 14.1 Introduction 620 14.2 Classical Mechanics 621 14.3 Classical Waves 629 14.4 The Old Quantum Theory 640 Chapter 15 The Principles of Quantum Mechanics. I. De Broglie Waves and the Schrödinger Equation 653 15.1 De Broglie Waves 654 15.2 The Schrödinger Equation 657 15.3 The Particle in a Box and the Free Particle 663 15.4 The Quantum Harmonic Oscillator 674 Chapter 16 The Principles of Quantum Mechanics. II. The Postulates of Quantum Mechanics 683 16.1 The First Two Postulates of Quantum Mechanics 684 16.2 The Third Postulate. Mathematical Operators and Mechanical Variables 684 16.3 The Operator Corresponding to a Given Variable 688 16.4 Postulate 4 and Expectation Values 696 16.5 The Uncertainty Principle of Heisenberg 711 16.6 Postulate 5. Measurements and the Determination of the State of a System 717 Chapter 17 The Electronic States of Atoms. I. The Hydrogen Atom 725 17.1 The Hydrogen Atom and the Central Force System 726 17.2 The Relative Schrödinger Equation. Angular Momentum 729 17.3 The Radial Factor in the Hydrogen Atom Wave Function. The Energy Levels of the Hydrogen Atom 736 17.4 The Orbitals of the HydrogenLike Atom 741 17.5 Expectation Values in the Hydrogen Atom 749 17.6 The TimeDependent Wave Functions of the HydrogenAtom 17.7 The Intrinsic Angular Momentum of the Electron. “Spin” 755 Chapter 18 619 The Electronic States ofAtoms. II. The ZeroOrderApproximation for Multielectron Atoms 763 18.1 The HeliumLike Atom 764 18.2 The Indistinguishability of Electrons and the Pauli Exclusion Principle 766 18.3 The Ground State of the Helium Atom in Zero Order 768 18.4 Excited States of the Helium Atom 772 18.5 Angular Momentum in the Helium Atom 774 753 xi Contents 18.6 The Lithium Atom 781 18.7 Atoms with More Than Three Electrons 784 Chapter 19 The Electronic States of Atoms. III. HigherOrder Approximations 789 19.1 The Variation Method and Its Application to the Helium Atom 790 19.2 The SelfConsistent Field Method 796 19.3 The Perturbation Method and Its Application to the Ground State of the Helium Atom 799 19.4 Excited States of the HeliumAtom. Degenerate Perturbation Theory 803 19.5 The Density Functional Method 805 19.6 Atoms with More Than Two Electrons 806 Chapter 20 The Electronic States of Diatomic Molecules 823 20.1 The Born–Oppenheimer Approximation and the Hydrogen Molecule Ion 824 20.2 LCAOMOs.Approximate Molecular Orbitals ThatAre Linear Combinations of Atomic Orbitals 833 20.3 Homonuclear Diatomic Molecules 838 20.4 Heteronuclear Diatomic Molecules 851 Chapter 21 The Electronic Structure of Polyatomic Molecules 867 868 21.1 The BeH2 Molecule and the sp Hybrid Orbitals 871 21.2 The BH3 Molecule and the sp2 Hybrid Orbitals 21.3 The CH4 , NH3 , and H2 O Molecules and the sp3 Hybrid Orbitals 873 21.4 Molecules with Multiple Bonds 878 21.5 The ValenceBond Description of Polyatomic Molecules 21.6 Delocalized Bonding 885 21.7 The FreeElectron Molecular Orbital Method 892 21.8 Applications of Symmetry to Molecular Orbitals 894 21.9 Groups of Symmetry Operators 896 21.10 More Advanced Treatments of Molecular Electronic Structure. Computational Chemistry 904 881 Chapter 22 Translational, Rotational, and Vibrational States of Atoms and Molecules 915 22.1 The Translational States of Atoms 916 22.2 The Nonelectronic States of Diatomic Molecules 919 22.3 Nuclear Spins and Wave Function Symmetry 930 22.4 The Rotation and Vibration of Polyatomic Molecules 933 22.5 The Equilibrium Populations of Molecular States 942 Chapter 23 Optical Spectroscopy and Photochemistry 949 23.1 Emission/Absorption Spectroscopy and Energy Levels 23.2 The Spectra of Atoms 959 23.3 Rotational and Vibrational Spectra of Diatomic Molecules 961 23.4 Electronic Spectra of Diatomic Molecules 972 950 xii Contents 23.5 23.6 23.7 23.8 Chapter 24 Part 4 Spectra of Polyatomic Molecules 975 Fluorescence, Phosphorescence, and Photochemistry Raman Spectroscopy 985 Other Types of Spectroscopy 991 979 Magnetic Resonance Spectroscopy 1001 24.1 Magnetic Fields and Magnetic Dipoles 1002 24.2 Electronic and Nuclear Magnetic Dipoles 1006 24.3 Electron Spin Resonance Spectroscopy 1010 24.4 Nuclear Magnetic Resonance Spectroscopy 1014 24.5 Fourier Transform NMR Spectroscopy 1024 The Reconciliation of the Macroscopic and Molecular Theories of Matter 1037 Chapter 25 Equilibrium Statistical Mechanics I. The Probability Distribution for Molecular States 1039 25.1 The Quantum Statistical Mechanics of a Simple Model System 1040 25.2 The Probability Distribution for a Dilute Gas 1047 25.3 The Probability Distribution and the Molecular Partition Function 1055 25.4 The Calculation of Molecular Partition Functions 1064 Chapter 26 Equilibrium Statistical Mechanics. II. Statistical Thermodynamics 1081 26.1 The Statistical Thermodynamics of a Dilute Gas 1082 26.2 Working Equations for the Thermodynamic Functions of a Dilute Gas 1089 26.3 Chemical Equilibrium in Dilute Gases 1101 26.4 The Activated Complex Theory of Bimolecular Chemical Reaction Rates in Dilute Gases 1106 26.5 Miscellaneous Topics in Statistical Thermodynamics 1116 Chapter 27 Equilibrium Statistical Mechanics. III. Ensembles 1121 27.1 The Canonical Ensemble 1122 27.2 Thermodynamic Functions in the Canonical Ensemble 1128 27.3 The Dilute Gas in the Canonical Ensemble 1130 27.4 Classical Statistical Mechanics 1133 27.5 Thermodynamic Functions in the Classical Canonical Ensemble 1141 27.6 The Classical Statistical Mechanics of Dense Gases and Liquids 1147 Chapter 28 The Structure of Solids, Liquids, and Polymers 28.1 The Structure of Solids 1154 28.2 Crystal Vibrations 1162 28.3 The Electronic Structure of Crystalline Solids 28.4 Electrical Resistance in Solids 1179 1153 1171 xiii Contents 28.5 The Structure of Liquids 1184 28.6 Approximate Theories of Transport Processes in Liquids 1188 28.7 Polymer Conformation 1194 28.8 Polymers in Solution 1198 28.9 Rubber Elasticity 1200 28.10 Nanomaterials 1205 Appendices 1209 A. Tables of Numerical Data 1209 B. Some Useful Mathematics 1235 C. A Short Table of Integrals 1257 D. Some Derivations of Formulas and Methods 1261 E. Classical Mechanics 1267 F. Some Mathematics Used in Quantum Mechanics 1275 G. The Perturbation Method 1283 H. The Hückel Method 1289 I. Matrix Representations of Groups 1293 J. Symbols Used in This Book 1303 K. Answers to Numerical Exercises and OddNumbered Numerical Problems 1309 Index 1351 Preface This is the third edition of a physical chemistry textbook designed for a twosemester undergraduate physical chemistry course. The physical chemistry course is often the ﬁrst opportunity that a student has to synthesize descriptive, theoretical, and mathematical knowledge about chemistry into a coherent whole. To facilitate this synthesis, the book is constructed about the idea of deﬁning a system, studying the states in which it might be found, and analyzing the processes by which it can change its state. The book is divided into four parts. The ﬁrst part focuses on the macroscopic properties of physical systems. It begins with the descriptive study of gases and liquids, and proceeds to the study of thermodynamics, which is a comprehensive macroscopic theory of the behavior of material systems. The second part focuses on dynamics, including gas kinetic theory, transport processes, and chemical reaction kinetics. The third part presents quantum mechanics and spectroscopy. The fourth part presents the relationship between molecular and macroscopic properties of systems through the study of statistical mechanics. This theory is applied to the structure of condensed phases. The book is designed so that the ﬁrst three parts can be studied in any order, while the fourth part is designed to be a capstone in which the other parts are integrated into a cohesive whole. In addition to the standard tables of integrals and numerical values of various properties, the book contains several appendices that expand on discussions in the body of the text, such as more detailed discussions of perturbation theory, group theory, and several mathematical topics. Each chapter begins with a statement of the principal facts and ideas that are presented in the chapter. There is a summary at the end of each chapter to assist in synthesizing the material of each chapter into a coherent whole. There are also marginal notes throughout the chapters that present biographical information and some comments. Each chapter contains examples that illustrate various kinds of calculations, as well as exercises placed within the chapter. Both these exercises and the problems at the end of each section are designed to provide practice in applying techniques and insights obtained through study of the chapter. Answers to all of the numerical exercises and to the oddnumbered numerical problems are placed in Appendix K. A solutions manual, with complete solutions to all exercises and all oddnumbered problems, is available from the publisher. An instructor’s manual with solutions to the evennumbered problems is available online to instructors. The instructor can choose whether to allow students to have access to the solutions manual, but can assign evennumbered problems when he or she wants the students to work problems without access to solutions. xv xvi Preface The author encourages students and instructors to comment on any part of the book; please send comments and suggestions to the author’s attention. Robert G. Mortimer 2769 Mercury St. Bartlett, TN 38134, USA Acknowledgments The writing of the ﬁrst edition of this book was begun during a sabbatical leave from Rhodes College, and continued during summer grants from the Faculty Development Committee of Rhodes College. It is a pleasure to acknowledge this support. It has been my pleasure to have studied with many dedicated and proﬁcient teachers, and I acknowledge their inﬂuence, example, and inspiration. I am also grateful for the privilege of working with students, whose efforts to understand the workings of the physical universe make teaching the most desirable of all professions. I have beneﬁted from the expert advice of many reviewers. These include: Jonas Goldsmith Jason D. Hofstein Daniel Lawson Jennifer Mihalick Cynthia M. Woodbridge Bryn Mawr College Sienna College University of Michigan–Dearborn University of Wisconsin–Oshkosh Hillsdale College and the reviewers of the previous editions.All of these reviewers gave sound advice, and some of them went beyond the call of duty in searching out errors and unclarities and in suggesting remedies. The errors that remain are my responsibility, not theirs. I wish to thank the editorial staff of Elsevier/Academic Press for their guidance and help during a rather long and complicated project, and also wish to thank Erica Ellison, who was a valuable consultant. I thank my wife, Ann, for her patience, love, and support during this project. xvii 1 Thermodynamics and the Macroscopic Description of Physical Systems 1 The Behavior of Gases and Liquids PRINCIPAL FACTS AND IDEAS 1. The principal goal of physical chemistry is to understand the properties and behavior of material systems and to apply this understanding in useful ways. 2. The state of a system is speciﬁed by giving the values of a certain number of independent variables (state variables). 3. In an equilibrium onephase ﬂuid system of one substance, three macroscopic variables such as temperature, volume, and amount of substance can be independent variables and can be used to specify the macroscopic equilibrium state of the system. At least one of the variables used to specify the state of the system must be proportional to the size of the system (be extensive). Other macroscopic variables are mathematical functions of the independent variables. 4. The intensive state, which includes only intensive variables (variables that are independent of the size of the system), is speciﬁed by only two variables in the case of an equilibrium onephase ﬂuid system of one substance. 5. Nonideal gases and liquids are described mathematically by various equations of state. 6. The coexistence of phases can be described mathematically. 7. The liquid–gas coexistence curve terminates at the critical point, beyond which there is no distinction between liquid and gas phases. 8. The law of corresponding states asserts that in terms of reduced variables, all substances obey the same equation of state. 3 4 1 The Behavior of Gases and Liquids 1.1 Antoine Laurent Lavoisier, 1743–1794, was a great French chemist who was called the “father of modern chemistry” because of his discovery of the law of conservation of mass. He was beheaded during the French Revolution because of his involvement in his fatherinlaw’s ﬁrm, which was employed by the royal government to collect taxes. It is said that he arranged with a friend to observe his head to see how long he could blink his eyes after his head was severed. He blinked for 15 seconds. Joseph Proust, 1754–1826, was a French chemist who was the ﬁrst to isolate sugar from grapes. John Dalton, 1766–1844, was an English schoolmaster and chemist. After he became a famous chemist, he continued to teach at what we would now call the elementary school level. Galileo Galilei, 1564–1642, was a great Italian mathematician and physicist. He refuted the assertion of Aristotle that a heavier object should fall faster than a lighter one and is said to have dropped two balls of different masses from the leaning tower of Pisa to demonstrate that they fell at the same rate. He supported the hypothesis of Copernicus that the earth revolves around the sun and was convicted of heresy in 1633 by the Roman Catholic Church for this belief. He spent the rest of his life under house arrest. Introduction This book is a textbook for a standard twosemester physical chemistry course at the undergraduate level. Physical chemistry involves both physics and chemistry. Physics has been deﬁned as the study of the properties of matter that are shared by all substances, whereas chemistry has been deﬁned as the study of the properties of individual substances. Chemistry grew out of the ancient occult art of alchemy, which involved among other things the attempted transmutation of cheaper materials into gold. Chemistry began as a completely experimental science. Substances were named and studied without reference to their molecular structures. Sulfuric acid was called “oil of vitriol,” and chemists memorized the fact that when copper was treated with oil of vitriol a solution of “blue vitriol” (now known as copper(II) sulfate) resulted. In the late 18th century, Lavoisier established the law of conservation of mass in chemical reactions, and Proust established the law of deﬁnite proportion. In order to explain these laws, Dalton proposed his atomic theory in 1803, as well as announcing the law of multiple proportions. With this theory, chemistry could evolve into a molecular science, with properties of substances tied to their molecular structures. Systems We call any object that we wish to study our system. A large system containing many atoms or molecules is called a macroscopic system, and a system consisting of a single atom or molecule is called a microscopic system. We consider two principal types of properties of systems. Macroscopic properties such as temperature and pressure apply only to a macroscopic system and are properties of the whole system. They can be observed and studied without reference to the molecular nature of matter. Microscopic properties such as kinetic energy and momentum are mechanical in nature. They apply to either macroscopic or microscopic systems. The study of macroscopic properties involves thermodynamics, which is the major topic of this volume, along with gas kinetic theory, transport processes, and reaction kinetics. Quantum mechanics, spectroscopy, and statistical mechanics are molecular topics and are discussed in Parts 3 and 4 of this textbook. Mathematics in Physical Chemistry The study of any physical chemistry topics requires mathematics. Galileo once wrote, “The book of nature is written in the language of mathematics.” We will use mathematics in two different ways. First, we will use it to describe the behavior of systems without explaining the origin of the behavior. Second, we will use it to develop theories that explain why certain behaviors occur. This chapter is an example of the ﬁrst usage, and the next chapter is an example of the second usage. Much of the mathematical education that physical chemistry students have received has focused on mathematical theory rather than on practical applications. A student who was unable to apply an elementary calculus technique once said to the author, “I know that was in the calculus course, but nobody told me that I would ever have to use it.” Mathematical theory is not always important in physical chemistry, but you 5 1.1 Introduction need to be able to apply mathematical methods. There are several books that cover the application of mathematics to problems in physical chemistry.1 Arithmetic is the principal branch of numerical mathematics. It involves carrying out operations such as addition, subtraction, multiplication, and division on actual numbers. Geometry, algebra, and calculus are parts of symbolic mathematics, in which symbols that represent numerical quantities and operations are manipulated without doing the numerical operations. Both kinds of mathematics are applied in physical chemistry. Mathematical Functions A mathematical function involves two kinds of variables: An independent variable is one to which we can assign a value. A mathematical function is a rule that delivers the value of a dependent variable when values are assigned to the independent variable or variables. A function can be represented by a formula, a graph, a table, a mathematical series, and so on. Consider the ideal gas law: PV nRT (1.11) In this equation P represents the pressure of the gas, V represents its volume, n represents the amount of substance in moles, T represents the absolute temperature, and R stands for the ideal gas constant. The ideal gas law does a good but not perfect job of representing the equilibrium behavior of real gases under ordinary conditions. It is more nearly obeyed if the pressure of the gas is made smaller. A gas that is at a sufﬁciently low pressure that it obeys the ideal gas law to an adequate approximation is called a dilute gas. An ideal gas is deﬁned to obey this equation for all pressures and temperatures. An ideal gas does not exist in the real world, and we call it a model system. A model system is an imaginary system designed to resemble some real system. A model system is useful only if its behavior mimics that of a real system to a useful degree and if it can be more easily analyzed than the real system. We can solve the ideal gas law for V by symbolically dividing by P: V nRT P (1.12) The righthand side of Eq. (1.12) is a formula that represents a mathematical function. The variables T , P, and n are independent variables, and V is the dependent variable. If you have the numerical values of T , P, and n, you can now carry out the indicated arithmetic operations to ﬁnd the value of V . We can also solve Eq. (1.11) for P by symbolically dividing by V : P nRT V (1.13) We have now reassigned V to be one of the independent variables and P to be the dependent variable. This illustrates a general fact: If you have an equation containing 1 Robert G. Mortimer, Mathematics for Physical Chemistry, 3rd ed., Academic Press, San Diego, CA, U.S.A., 2005; James R. Barrante, Applied Mathematics for Physical Chemistry, 3rd ed., Pearson Prentice Hall, Upper Saddle River, NJ, 2004; Donald A. McQuarrie, Mathematical Methods for Scientists and Engineers, University Science Books, 2003. 6 1 The Behavior of Gases and Liquids several variables, you can manipulate the equation symbolically to turn any one of them into the dependent variable. The ideal gas law might not be accurate enough for some gases under some conditions. If so, we can ﬁnd some other function that will give the value of the pressure to greater accuracy. It is an experimental fact that the pressure of a gas or liquid of one substance at equilibrium is given by a function that depends on only three independent variables. We represent such a function by P P(T , V , n) (1.14) A mathematician would write P f (T , V , n) for the functional relation in Eq. (1.14), using the letter P for the variable and the letter f for the function. Chemists have too many variables to use two letters for each variable, so we use the same letter for the variable and the function. A functional relation that relates P, V , T , and n for a gas or a liquid at equilibrium is called an equation of state and is said to represent the volumetric behavior of the gas or liquid. We will introduce several equations of state later in this chapter. E X A M P L E 1.1 Assume that the volume of a liquid is a linearly decreasing function of P, is a linearly increasing function of T , and is proportional to n. Write a formula expressing this functional relationship. Solution Let V0 represent the volume at some reference temperature T0 , some reference pressure P0 , and some reference amount of substance n0 . V V0 n [1 − k(P − P0 ) + a(T − T0 )] nVm0 [1 − k(P − P0 ) + a(T − T0 )] n0 where k and a are constants and where Vm represents the molar volume, equal to V /n, and Vm0 represents V0 /n0 . A twodimensional graph can represent a function of one independent variable. You plot the value of the independent variable on the horizontal axis and represent the value of the dependent variable by the height of a curve in the graph. To make a twodimensional graph that represents the ideal gas law, we must keep two of the three independent variables ﬁxed. Figure 1.1a shows a set of graphical curves that represent the dependence of P on V for an ideal gas for n 1.000 mol and for several ﬁxed values of T . A threedimensional graph can represent a function of two independent variables. Figure 1.1b shows a perspective view of a graphical surface in three dimensions that represents the dependence of P on V and T for an ideal gas with a ﬁxed value of n (1.000 mol). Just as the height of a curve in Figure 1.1a gives the value of P for a particular value of V , the height of the surface in Figure 1.1b gives the value of P for a particular value of T and a particular value of V . Such graphs are not very useful for numerical purposes, but help in visualizing the general behavior of a function of two independent variables. 7 P 1.1 Introduction 5 3 105 5 3 105 3 3 105 P/Nm22 T 5 1000 K T 5 500 K 273 3 3 105 T 5 373 K T 5 273 K 0 1 3 105 373 500 T 1000 0.05 (a) 0.1 0.1 m 0.05 Vm/m3 mol21 V 0 (b) Figure 1.1 (a) The pressure of an ideal gas as a function of V at constant n and various constant values of T. (b) The pressure of an ideal gas as a function of V and T at constant n. A function can also be represented by a table of values. For a function of one independent variable, a set of values of the independent variable is placed in one column. The value of the dependent variable corresponding to each value of the independent variable is placed in another column on the same line. A mathematician would say that we have a set of ordered pairs of numbers. Prior to the advent of electronic calculators, such tables were used to represent logarithms and trigonometric functions. Such a table provides values only for a ﬁnite number of values of the independent variable, but interpolation between these values can be used to obtain other values. Units of Measurement The values of most physical variables consist of two parts, a number and a unit of measurement. Various units of measurement exist. For example, the same distance could be expressed as 1.000 mile, 1609 meters, 1.609 kilometer, 5280 feet, 63360 inches, 1760 yards, 106.7 rods, 8.000 furlongs, and so on. A given mass could be expressed as 1.000 kilogram, 1000 grams, 2.205 pounds, 0.1575 stone, 195.3 ounces, and so on. There are sets of units that are consistent with each other. For example, pounds are used with feet, kilograms are used with meters, and grams are used with centimeters. Here is an important fact: To carry out any numerical calculation correctly you must express all variables with consistent units. If any quantities are expressed in inconsistent units, you will almost certainly get the wrong answer. In September 1999, a space probe optimistically named the “Mars Climate Orbiter” crashed into the surface of Mars instead of orbiting that planet. The problem turned out to be that some engineers had used “English” units such as feet and pounds, while physicists working on the same project had used metric units such as meters and kilograms. Their failure to convert units correctly caused the loss of a space vehicle that cost many millions of U.S. dollars. In another instance, when a Canadian airline converted from English units to metric units, a ground crew that was accustomed to English units incorrectly calculated how much fuel in kilograms to put into an airliner for a certain ﬂight. The airplane ran out of 8 The newton is named for Sir Isaac Newton, 1642–1727, the great English mathematician and physicist who invented classical mechanics and who was one of the inventors of calculus. The pascal is named for Blaise Pascal, 1623–1662, a famous French philosopher, theologian, and mathematician. The joule is named for James Prescott Joule, 1818–1889, a great English physicist who pioneered in the thermodynamic study of work, heat, and energy in a laboratory that he constructed in his family’s brewery. 1 The Behavior of Gases and Liquids fuel before reaching its destination. Fortunately, the pilot was able to glide to a former military air ﬁeld and make a “deadstick” landing on an unused runway. Some people who were having a picnic on the runway were fortunately able to get out of the way in time. There was even a movie made about this incident. The ofﬁcial set of units that physicists and chemists use is the International System of Units, or SI units. The letters SI stand for Systeme Internationale, the French name for the set of units. In this system there are seven base units. The unit of length is the meter (m). The unit of mass is the kilogram (kg). The unit of time is the second (s). The unit of temperature is the kelvin (K). The unit of electric current is the ampere (A). The unit of luminous intensity is the candela (cd). The unit for the amount of a substance is the mole (mol). The SI units are called MKS (meterkilogramsecond) units. Prior to 1961, most chemists and some physicists used cgs (centimetergramsecond) units, but we now use SI units to avoid confusion. In addition to the seven base units, there are a number of derived units. The newton (N) is the SI unit of force: 1 N 1 kg m s−2 (deﬁnition) (1.15) The pascal (Pa) is the SI unit of pressure (force per unit area): 1 Pa 1 N m−2 1 kg m−1 s−2 (deﬁnition) (1.16) We have enclosed these deﬁning equations in boxes, and will enclose the most important equations in boxes throughout the rest of the book. A force exerted through a distance is equivalent to an amount of work, which is a form of energy. The SI unit of energy is the joule (J): 1 J 1 N m 1 kg m2 s−2 (deﬁnition) (1.17) Multiples and submultiples of SI units are indicated by preﬁxes, such as “milli” for 1/1000, “centi” for 1/100, “deci” for 1/10, “kilo” for 1000, and so on. These preﬁxes are listed inside the cover of this book. We do not use double preﬁxes such as millikilogram for the gram or microkilogram for the milligram. We will also use some nonSI units. The calorie (cal), which was originally deﬁned as the amount of heat required to raise the temperature of 1 gram of water by 1◦ C, is now deﬁned by: 1 cal 4.184 J (exactly, by deﬁnition) (1.18) We will use several nonSI units of pressure; the atmosphere (atm), the torr, and the bar. 1 atm 101325 Pa (exactly, by deﬁnition) (1.19) 760 torr 1 atm (exactly, by deﬁnition) (1.110) 1 bar 100000 Pa (exactly, by deﬁnition) (1.111) The angstrom (Å, equal to 10−10 m or 10−8 cm) has been a favorite unit of length among chemists, because it is roughly equal to a typical atomic radius. Picometers are nearly as convenient, with 100 pm equal to 1 Å. Chemists are also reluctant to abandon the liter (L), which is the same as 0.001 m3 or 1 dm3 (cubic decimeter). 9 1.1 Introduction The Mole and Avogadro’s Constant Lorenzo Romano Amadeo Carlo Avogadro, 1776–1856, was an Italian lawyer and professor of natural philosophy. He was the ﬁrst to postulate that equal volumes of gases under the same conditions contained the same number of molecules. The formula unit of a substance is the smallest amount of a substance that retains the identity of that substance. It can be an atom, a molecule, or an electrically neutral set of ions. A mole of any substance is an amount with the same number of formula units as the number of atoms contained in exactly 0.012 kg of the 12 C (carbon12) isotope. The atomic mass unit (amu or u) is deﬁned such that one atom of 12 C has a mass of exactly 12 amu. Therefore the mass of a mole of any substance expressed in grams is numerically equal to the mass of a formula unit expressed in atomic mass units. The number of formula units, N, in a sample of any substance is proportional to the amount of substance measured in moles, denoted by n: N NAv n Josef Loschmidt, 1821–1895, was an Austrian chemist who made various contributions, including being the ﬁrst to propose using two line segments to represent a double bond and three line segments to represent a triple bond. The proportionality constant NAv is called Avogadro’s constant in some countries and Loschmidt’s constant in others. It is known from experiment to have the value NAv 6.02214 × 1023 mol−1 (1.113) The ideal gas equation can be written in terms of the number of molecules, N: V Boltzmann’s constant is named for Ludwig Boltzmann, 1844–1906, an Austrian physicist who was one of the inventors of gas kinetic theory and statistical mechanics. (1.112) nNAv kB T NkB T nRT P P p (1.114) The ideal gas constant R is known from experiment to have the value 8.3145 J K −1 mol−1 . In common nonSI units, it is equal to 0.082058 L atm K −1 mol−1 . The constant kB is called Boltzmann’s constant: kB R 8.3145 J K −1 mol−1 1.3807 × 10−23 J K−1 NAv 6.02214 × 1023 mol−1 (1.115) Problem Solving Techniques If you have a home repair or automotive repair to do, the work will go better if you have the necessary tools at hand when you start the job. The same thing is true for physical chemistry problems. You should analyze the problem and make sure that you know what formulas and techniques are needed and make sure that you have them at hand. Think of your supply of formulas and techniques as your tools, and try to keep your toolbox organized. One of the most important things in problem solving is that you must use consistent units in any numerical calculation. The conversion to consistent units is conveniently done by the factorlabel method, which is a straightforward use of proportionality factors. It is illustrated in the following example, and you can review this method in almost any general chemistry textbook. E X A M P L E 1.2 Find the pressure in Pa and in atm of 20.00 g of neon gas (assumed to be ideal) at a temperature of 0.00◦ C and a volume of 22.400 L. 10 1 The Behavior of Gases and Liquids Solution The Celsius temperature differs from the absolute temperature by 273.15 K, but the Celsius degree is the same size as the kelvin. T 273.15 K + 0.00◦ C 273.15 K We convert amount of gas to moles and the volume to m3 using the factorlabel method: 1 mol n (20.00 g) 0.9911 mol 20.179 g 1 m3 V (22.400 L) 0.022400 m3 1000 L We now carry out the numerical calculation: −1 mol−1 (273.15 K) (0.9911 mol) 8.314 J K nRT P V 0.022400 m3 1.005 × 105 J m−3 1.005 × 105 N m−2 1.005 × 105 Pa You can see how the symbolic formula is used as a template for setting up the numerical calculation. The unit conversions can also be included in a single calculation: 20.00 g 8.314 J K −1 mol−1 (273.15 K) 1 mol 1000 L P (22.400 L) 20.179 g 1 m3 1.005 × 105 J m−3 1.005 × 105 N m2 1.005 × 105 Pa The pressure can be expressed in atmospheres by another conversion: 1 atm P (1.005 × 105 Pa) 0.9919 atm 101325 Pa A calculator displayed 100,486.28725 Pa for the pressure in the previous example. The answer was then rounded to four digits to display only signiﬁcant digits. In carrying out operations with a calculator, it is advisable to carry insigniﬁcant digits in intermediate steps in order to avoid roundoff error and then to round off insigniﬁcant digits in the ﬁnal answer. You can review signiﬁcant digits in any elementary chemistry textbook. The main idea is that if the calculation produces digits that are probably incorrect, they are insigniﬁcant digits and should be rounded away. An important rule is that in a set of multiplications and divisions, the result generally has as many signiﬁcant digits as the factor or divisor with the fewest signiﬁcant digits. Another important technique in problem solving is to ﬁgure out roughly how large your answer should be and what its units should be. For example, the author had a student under time pressure in an examination come up with an answer of roughly 1030 cm for a molecular dimension. A moment’s thought should have revealed that this distance is greater than the size of the known universe and cannot be correct. Many common mistakes produce an answer that either has the wrong units or is obviously too large or too small, and you can spot these errors if you look for them. You should always write the units on every factor or divisor when setting up a numerical calculation so that you will be more likely to spot an error in units. 11 1.1 Introduction E X A M P L E 1.3 The speed of sound in dry air at a density of 1.293 g L−1 and a temperature of 0◦ C is 331.45 m s−1 . Convert this speed to miles per hour. Solution (331.45 m s−1 ) 1 in 0.0254 m 1 ft 12 in 1 mile 5280 ft 3600 s 1 hour 741.43 miles hour−1 Note that the conversion ratios do not limit the number of signiﬁcant digits because they are deﬁned to be exact values. Exercise 1.1 a. Express the value of the ideal gas constant in cm3 bar K −1 mol−1 . Report only signiﬁcant digits. b. Find the volume of 2.000 mol of helium (assume ideal) at a temperature of 298.15 K and a pressure of 0.500 atm. c. Find the pressure of a sample of 2.000 mol of helium (assume ideal) at a volume of 20.00 L and a temperature of 500.0 K. Express your answer in terms of Pa, bar, atm, and torr. PROBLEMS Section 1.1: Introduction 1.1 Express the speed of light in furlongs per fortnight. A furlong is 1/8 mile, and a fortnight is 14 days. 1.2 In the “cgs” system, lengths are measured in centimeters, masses are measured in grams, and time is measured in seconds. The cgs unit of energy is the erg and the cgs unit of force is the dyne. a. Find the conversion factor between ergs and joules. b. Find the conversion factor between dynes and newtons. c. Find the acceleration due to gravity at the earth’s surface in cgs units. 1.3 In one English system of units, lengths are measured in feet, masses are measured in pounds, abbreviated lb (1 lb = 0.4536 kg), and time is measured in seconds. The absolute temperature scale is the Rankine scale, such that 1.8◦ R corresponds to 1◦ C and to 1 K. a. Find the acceleration due to gravity at the earth’s surface in English units. b. If the pound is a unit of mass, then the unit of force is called the poundal. Calculate the value of the ideal gas constant in ft poundals (◦ R)−1 mol−1 . c. In another English system of units, the pound is a unit of force, equal to the gravitational force at the earth’s surface, and the unit of mass is the slug. Find the acceleration due to gravity at the earth’s surface in this set of units. 1.4 A lightyear is the distance traveled by light in one year. a. Express the lightyear in meters and in kilometers. b. Express the lightyear in miles. c. If the size of the known universe is estimated to be 20 billion lightyears (2 × 1010 lightyears) estimate the size of the known universe in miles. d. If the closest star other than the sun is at a distance of 4 lightyears, express this distance in kilometers and in miles. e. The mean distance of the earth from the sun is 149,599,000 km. Express this distance in lightyears. 12 1 The Behavior of Gases and Liquids 1.5 The parsec is a distance used in astronomy, deﬁned to be a distance from the sun such that “the heliocentric parallax is 1 second of arc.” This means that the direction of observation of an object from the sun differs from the direction of observation from the earth by one second of arc. a. Find the value of 1 parsec in kilometers. Do this by constructing a right triangle with one side equal to 1 parsec and the other side equal to 1.49599 × 108 km, the distance from the earth to the sun. Make the angle opposite the short side equal to 1 second of arc. b. Find the value of 1 parsec in lightyears. c. Express the distance from the earth to the sun in parsec. 1.6 Making rough estimates of quantities is sometimes a useful skill. a. Estimate the number of grains of sand on all of the beaches of all continents on the earth, excluding islands. Do this by making suitable estimates of: (1) the average width of a beach; (2) the average depth of sand on a beach; (3) the length of the coastlines of all of the continents; (4) the average size of a grain of sand. b. Express your estimate in terms of moles of grains of sand, where a mole of grains of sand is 6.02214 × 1023 grains of sand. 1.7 Estimate the number of piano tuners in Chicago (or any other large city of your choice). Do this by estimating: (1) the number of houses, apartments, and other buildings in the city; (2) the fraction of buildings containing a piano; (3) the average frequency of tuning; (4) how many pianos a piano tuner can tune in 1 week. 1.8 Estimate the volume of the oceans of the earth in liters. Use the fact that the oceans cover about 71% of the earth’s area 1.2 and estimate the average depth of the oceans. The greatest depth of the ocean is about 7 miles, slightly greater than the altitude of the highest mountain on the earth. 1.9 Find the volume of CO2 gas produced from 100.0 g of CaCO3 if the CO2 is at a pressure of 746 torr and a temperature of 301.0 K. Assume the gas to be ideal. 1.10 According to Dalton’s law of partial pressures, the pressure of a mixture of ideal gases is the sum of the partial pressures of the gases. The partial pressure of a gas is deﬁned to be the pressure that would be exerted if that gas were alone in the volume occupied by the gas mixture. a. A sample of oxygen gas is collected over water at 25◦ C at a total pressure of 748.5 torr, with a partial pressure of water vapor equal to 23.8 torr. If the volume of the collected gas is equal to 454 mL, ﬁnd the mass of the oxygen. Assume the gas to be ideal. b. If the oxygen were produced by the decomposition of KClO3 , ﬁnd the mass of KClO3 . 1.11 The relative humidity is deﬁned as the ratio of the partial pressure of water vapor to the pressure of water vapor at equilibrium with the liquid at the same temperature. The equilibrium pressure of water vapor at 25◦ C is 23.756 torr. If the relative humidity is 49%, estimate the amount of water vapor in moles contained in a room that is 8.0 m by 8.0 m and 3.0 m in height. Calculate the mass of the water. 1.12 Assume that the atmosphere is at equilibrium at 25◦ C with a relative humidity of 100% and assume that the barometric pressure at sea level is 1.00 atm. Estimate the total rainfall depth that could occur if all of this moisture is removed from the air above a certain area of the earth. Systems and States in Physical Chemistry Figure 1.2 depicts a typical macroscopic system, a sample of a single gaseous substance that is contained in a cylinder with a movable piston. The cylinder is immersed in a constanttemperature bath that can regulate the temperature of the system. The volume of the system can be adjusted by moving the piston. There is a valve between the cylinder and a hose that leads to the atmosphere or to a tank of gas. When the valve is closed so that no matter can pass into or out of the system, the system is called a closed system. When the valve is open so that matter can be added to or removed from the system, it is called an open system. If the system were insulated from the rest of the universe so that no heat could pass into or out of the system, it would be called an adiabatic system and any process that it undergoes would be called an adiabatic 13 1.2 Systems and States in Physical Chemistry External force exerted here Constanttemperature bath Hose Valve Part of surroundings Figure 1.2 Piston System Cylinder A Typical Fluid System Contained in a Cylinder with Variable Volume. process. If the system were completely separated from the rest of the universe so that no heat, work, or matter could be transferred to or from the system, it would be called an isolated system. The portion of the universe that is outside of the system is called the surroundings. We must specify exactly what parts of the universe are included in the system. In this case we deﬁne the system to consist only of the gas. The cylinder, piston, and constanttemperature bath are parts of the surroundings. The State of a System Specifying the state of a system means describing the condition of the system by giving the values of a sufﬁcient set of numerical variables. We have already asserted that for an equilibrium onephase liquid or gaseous system of one substance, the pressure is a function of three independent variables. We now assert as an experimental fact that for any equilibrium onephase ﬂuid system (gas or liquid system) of one substance, there are only three macroscopic independent variables, at least one of which must be proportional to the size of the system. All other equilibrium macroscopic variables are dependent variables, with values given as functions of the independent variables. We say that three independent variables specify the equilibrium macroscopic state of a gas or liquid of one substance. We can generally choose which three independent variables to use so long as one is proportional to the size of the system. For ﬂuid system of one substance, we could choose T , V , and n to specify the equilibrium state. We could also choose T , P, and n, or we could choose T , P, and V . All other equilibrium macroscopic variables must be dependent variables that are functions of the variables chosen to specify the state of the system. We call both the independent variables and the dependent variables state functions or state variables. There are two principal classes of macroscopic variables. Extensive variables are proportional to the size of the system if P and T are constant, whereas intensive variables are independent of the size of the system if P and T are constant. For example, V , n, and m 14 1 The Behavior of Gases and Liquids (the mass of the system) are extensive variables, whereas P and T are intensive variables. The quotient of two extensive variables is an intensive variable. The density ρ is deﬁned as m/V , and the molar volume Vm is deﬁned to equal V /n. These are intensive variables. One test to determine whether a variable is extensive or intensive is to imagine combining two identical systems, keeping P and T ﬁxed. Any variable that has twice the value for the combined system as for one of the original systems is extensive, and any variable that has the same value is intensive. In later chapters we will deﬁne a number of extensive thermodynamic variables, such as the internal energy U, the enthalpy H, the entropy S, and the Gibbs energy G. We are sometimes faced with systems that are not at equilibrium, and the description of their states is more complicated. However, there are some nonequilibrium states that we can treat as though they were equilibrium states. For example, if liquid water at atmospheric pressure is carefully cooled below 0◦ C in a smooth container it can remain in the liquid form for a relatively long time. The water is said to be in a metastable state. At ordinary pressures, carbon in the form of diamond is in a metastable state, because it spontaneously tends to convert to graphite (although very slowly). Differential Calculus and State Variables Because a dependent variable depends on one or more independent variables, a change in an independent variable produces a corresponding change in the dependent variable. If f is a differentiable function of a single independent variable x, f f (x) (1.21) then an inﬁnitesimal change in x given by dx (the differential of x) produces a change in f given by df df dx dx (1.22) where df/dx represents the derivative of f with respect to x and where df represents the differential of the dependent variable f . The derivative df/dx gives the rate of change of f with respect to x and is deﬁned by df f (x + h) − f (x) lim dx h→0 h (1.23) if the limit exists. If the derivative exists, the function is said to be differentiable. There are standard formulas for the derivatives of many functions. For example, if f a sin(bx), where a and b represent constants, then df ab cos(x) dx (1.24) If a function depends on several independent variables, each independent variable makes a contribution like that in Eq. (1.22). If f is a differentiable function of x, y, and z, and if inﬁnitesimal changes dx, dy, and dz are imposed, then the differential df is given by 15 1.2 Systems and States in Physical Chemistry df ∂f ∂x dx + y,z ∂f ∂y dy + x,z ∂f ∂z dz (1.25) x,y where (∂f /∂x) y,z , (∂f /∂y)x,z , and (∂f /∂z)x,y are partial derivatives. If the function is represented by a formula, a partial derivative with respect to one independent variable is obtained by the ordinary procedures of differentiation, treating the other independent variables as though they were constants. The independent variables that are held constant are usually speciﬁed by subscripts. We assume that our macroscopic equilibrium state functions are differentiable except possibly at isolated points. For an equilibrium gas or liquid system of one phase and one substance, we can write ∂V ∂V ∂V dV dT + dP + dn (1.26) ∂T P,n ∂P T ,n ∂n T ,P This equation represents the value of an inﬁnitesimal change in volume that is produced when we impose arbitrary inﬁnitesimal changes dT , dP, and dn on the system, making sure that the system is at equilibrium after we make the changes. For an ideal gas dV nRT RT nR dT − 2 dP + dn P P P (ideal gas) (1.27) A mathematical identity is an equation that is valid for all values of the variables contained in the equation. There are several useful identities involving partial derivatives. Some of these are stated in Appendix B. An important identity is the cycle rule, which involves three variables such that each can be expressed as a differentiable function of the other two: ∂z ∂x y ∂x ∂y z ∂y ∂z −1 (cycle rule) (1.28) x If there is a fourth variable, it is held ﬁxed in all three of the partial derivatives. Some people are surprised by the occurrence of the negative sign in this equation. See Appendix B for further discussion. E X A M P L E 1.4 Take z xy and show that the three partial derivatives conform to the cycle rule. Solution ∂z ∂x y ∂x ∂y z ∂y ∂z x ∂x ∂y ∂z ∂x y ∂y z ∂z x y z − 2 y 1 x z z 1 −1 −y 2 − xy y x 16 1 The Behavior of Gases and Liquids Exercise 1.2 Take z ax ln(y/b), where a and b are constants. a. Find the partial derivatives (∂z/∂x)y , (∂x/∂y)z , and (∂y/∂z)x . b. Show that the derivatives of part a conform to the cycle rule. A second partial derivative is a partial derivative of a partial derivative. Both derivatives can be with respect to the same variable: 2 ∂ f ∂ ∂f (1.29) 2 ∂x ∂x y ∂x y y The following is called a mixed second partial derivative: ∂2 f ∂ ∂f ∂y∂x ∂y ∂x y (1.210) x The reciprocity relation is named for its discoverer, Leonhard Euler, 1707–1783, a great Swiss mathematician who spent most of his career in St. Petersburg, Russia, and who is considered by some to be the father of mathematical analysis. The base of natural logarithms is denoted by e after his initial. Euler’s reciprocity relation states that the two mixed second partial derivatives of a differentiable function must be equal to each other: ∂2 f ∂2 f ∂y∂x ∂x∂y (1.211) Exercise 1.3 Show that the three pairs of mixed partial derivatives that can be obtained from the derivatives in Eq. (1.27) conform to Euler’s reciprocity relation. Processes Aprocess is an occurrence that changes the state of a system. Every macroscopic process has a driving force that causes it to proceed. For example, a temperature difference is the driving force that causes a ﬂow of heat. A larger value of the driving force corresponds to a larger rate of the process. A zero rate must correspond to zero value of the driving force. A reversible process is one that can at any time be reversed in direction by an inﬁnitesimal change in the driving force. The driving force of a reversible process must be inﬁnitesimal in magnitude since it must reverse its sign with an inﬁnitesimal change. A reversible process must therefore occur inﬁnitely slowly, and the system has time to relax to equilibrium at each stage of the process. The system passes through a sequence of equilibrium states during a reversible process. A reversible process is sometimes called a quasiequilibrium process or a quasistatic process. There can be no truly reversible processes in the real universe, but we can often make calculations for them and apply the results to real processes, either exactly or approximately. An approximate version of Eq. (1.26) can be written for a ﬁnite reversible process corresponding to increments ∆P, ∆T , and ∆n: ∂V ∂V ∂V ∆V ≈ ∆T + ∆P + ∆n (1.212) ∂T P,n ∂P T ,n ∂n T ,P 17 1.2 Systems and States in Physical Chemistry where ≈ means “is approximately equal to” and where we use the common notation ∆V V (ﬁnal) − V (initial) (1.213) and so on. Calculations made with Eq. (1.212) will usually be more nearly correct if the ﬁnite increments ∆T , ∆P, and ∆n are small, and less nearly correct if the increments are large. Variables Related to Partial Derivatives The isothermal compressibility κT is deﬁned by 1 ∂V (deﬁnition of the κT − isothermal compressibility) V ∂P (1.214) The factor 1/V is included so that the compressibility is an intensive variable. The fact that T and n are ﬁxed in the differentiation means that measurements of the isothermal compressibility are made on a closed system at constant temperature. It is found experimentally that the compressibility of any system is positive. That is, every system decreases its volume when the pressure on it is increased. The coefﬁcient of thermal expansion is deﬁned by 1 ∂V α V ∂P P,n (deﬁnition of the coefﬁcient of thermal expansion) (1.215) The coefﬁcient of thermal expansion is an intensive quantity and is usually positive. That is, if the temperature is raised the volume usually increases. There are a few systems with negative values of the coefﬁcient of thermal expansion. For example, liquid water has a negative value of α between 0◦ C and 3.98◦ C. In this range of temperature the volume of a sample of water decreases if the temperature is raised. Values of the isothermal compressibility for a few pure liquids at several temperatures and at two different pressures are given in Table A.1 of Appendix A. The values of the coefﬁcient of thermal expansion for several substances are listed in Table A.2. Each value applies only to a single temperature and a single pressure, but the dependence on temperature and pressure is not large for typical liquids, and these values can usually be used over fairly wide ranges of temperature and pressure. For a closed system (constant n) Eq. (1.212) can be written ∆V ≈ V α∆T − V κT ∆P (1.216) E X A M P L E 1.5 The isothermal compressibility of liquid water at 298.15 K and 1.000 atm is equal to 4.57 × 10−5 bar−1 4.57 × 10−10 Pa−1 . Find the fractional change in the volume of a sample of water if its pressure is changed from 1.000 bar to 50.000 bar at a constant temperature of 298.15 K. Solution The compressibility is relatively small in magnitude so we can use Eq. (1.216): ∆V ≈ −V κT ∆P (1.217) 18 1 The Behavior of Gases and Liquids The fractional change is ∆V ≈ −κT ∆P −(4.57 × 10−5 bar−1 )(49.00 bar) −2.24 × 10−3 V E X A M P L E 1.6 For liquid water at 298.15 K and 1.000 atm, α 2.07 × 10−4 K −1 . Find the fractional change in the volume of a sample of water at 1.000 atm if its temperature is changed from 298.15 K to 303.15 K. Solution To a good approximation, ∆V ≈ V α∆T The fractional change in volume is ∆V ≈ α∆T (2.07 × 10−4 K−1 )(5.000 K) 1.04 × 10−3 V Exercise 1.4 a. Find expressions for the isothermal compressibility and coefﬁcient of thermal expansion for an ideal gas. b. Find the value of the isothermal compressibility in atm−1 , in bar −1 , and in Pa−1 for an ideal gas at 298.15 K and 1.000 atm. Find the ratio of this value to that of liquid water at the same temperature and pressure, using the value from Table A.1. c. Find the value of the coefﬁcient of thermal expansion of an ideal gas at 20◦ C and 1.000 atm. Find the ratio of this value to that of liquid water at the same temperature and pressure, using the value from Table A.2. In addition to the coefﬁcient of thermal expansion there is a quantity called the coefﬁcient of linear thermal expansion, deﬁned by 1 ∂L αL L ∂T P (deﬁnition of the coefﬁcient of linear thermal expansion) (1.218) where L is the length of the object. This coefﬁcient is usually used for solids, whereas the coefﬁcient of thermal expansion in Eq. (1.215) is used for gases and liquids. Unfortunately, the subscript L is sometimes omitted on the symbol for the coefﬁcient of linear thermal expansion, and the name “coefﬁcient of thermal expansion” is also sometimes used for it. Because the units of both coefﬁcients are the same (reciprocal temperature) there is opportunity for confusion between them. We can show that the linear coefﬁcient of thermal expansion is equal to onethird of the coefﬁcient of thermal expansion. Subject a cubical object of length L to an inﬁnitesimal change in temperature, dT . The new length of the object is ∂L L(T + dT ) L(T ) + dT L(T )(1 + αLdT ) (1.219) ∂T P 19 1.2 Systems and States in Physical Chemistry The volume of the object is equal to L3 , so V (T + dT ) L(T )3 1 + αLdT 3 L(T )3 1 + 3αLdT + 3(αLdT )2 + (αLdT )3 (1.220) Since dT is small, the last two terms are insigniﬁcant compared with the term that is proportional to dT . V (T + dT ) L(T )3 (1 + 3αLdT ) The volume at temperature T + dT is given by ∂V V (T + dT ) V (T ) + dT V (T )(1 + αdT ) ∂T (1.221) (1.222) Comparison of Eq. (1.222) with Eq. (1.221) shows that α 3αL (1.223) This relationship holds for objects that are not necessarily shaped like a cube. E X A M P L E 1.7 The linear coefﬁcient of expansion of borosilicate glass, such as Pyrex or Kimax , is equal to 3.2 × 10−6 K −1 . If a volumetric ﬂask contains 2.000000 L at 20.0◦ C, ﬁnd its volume at 25.0◦ C. Solution V (25◦ C) V (20◦ C)(1 + 3αL(5.0◦ C)) (2.000000 L)(1 + 3(3.2 × 10−6 )(5.0◦ C)) 2.000096 L Exercise 1.5 Find the volume of the volumetric ﬂask in Example 1.7 at 100.0◦ C. Moderate changes in temperature and pressure produce fairly small changes in the volume of a liquid, as in the examples just presented. The volumes of most solids are even more nearly constant. We therefore recommend the following practice: For ordinary calculations, assume that liquids and solids have ﬁxed volumes. For more precise calculations, calculate changes in volume proportional to changes in pressure or temperature as in Examples 1.5 and 1.6. Exercise 1.6 The compressibility of acetone at 20◦ C is 12.39 × 10−10 Pa−1 , and its density is 0.7899 g cm−3 at 20◦ C and 1.000 bar. a. Find the molar volume of acetone at 20◦ C and a pressure of 1.000 bar. b. Find the molar volume of acetone at 20◦ C and a pressure of 100.0 bar. 20 1 The Behavior of Gases and Liquids PROBLEMS Section 1.2: Systems and States in Physical Chemistry 1.13 Show that the three partial derivatives obtained from PV nRT with n ﬁxed conform to the cycle rule, Eq. (B15) of Appendix B. 1.14 For 1.000 mol of an ideal gas at 298.15 K and 1.000 bar, ﬁnd the numerical value of each of the three partial derivatives in the previous problem and show numerically that they conform to the cycle rule. 1.15 Finish the equation for an ideal gas and evaluate the partial derivatives for V 22.4 L, T 273.15 K , and n 1.000 mol. ∂P dV + ? dP ∂V T ,n 1.16 Take z aye x/b , where a and b are constants. a. Find the partial derivatives (∂z/∂x)y , (∂x/∂y)z , and (∂y/∂z)x . b. Show that the derivatives of part a conform to the cycle rule, Eq. (B15) of Appendix B. 1.17 a. Find the fractional change in the volume of a sample of liquid water if its temperature is changed from 20.00◦ C to 30.00◦ C and its pressure is changed from 1.000 bar to 26.000 bar. b. Estimate the percent change in volume of a sample of benzene if it is heated from 0◦ C to 45◦ C at 1.000 atm. c. Estimate the percent change in volume of a sample of benzene if it is pressurized at 55◦ C from 1.000 atm to 50.0 atm. 1.18 a. Estimate the percent change in the volume of a sample of carbon tetrachloride if it is pressurized from 1.000 atm to 10.000 atm at 25◦ C. b. Estimate the percent change in the volume of a sample of carbon tetrachloride if its temperature is changed from 20◦ C to 40◦ C. 1.19 Find the change in volume of 100.0 cm3 of liquid carbon tetrachloride if its temperature is changed from 20.00◦ C to 25.00◦ C and its pressure is changed from 1.000 atm to 10.000 atm. 1.20 Let f (u) sin(au2 ) and u x2 + y2 , where a is a constant. Using the chain rule, ﬁnd (∂f /∂x)y and (∂f /∂y)x . (See Appendix B.) 1.21 Show that for any system, α κT ∂P ∂T V ,n 1.22 The coefﬁcient of linear expansion of borosilicate glass is equal to 3.2 ×10−6 K −1 . a. Calculate the pressure of a sample of helium (assumed ideal) in a borosilicate glass vessel at 150◦ C if its pressure at 0◦ C is equal to 1.000 atm. Compare with the value of the pressure calculated assuming that the volume of the vessel is constant. b. Repeat the calculation of part a using the virial equation of state truncated at the B2 term. The value of B2 for helium is 11.8 cm3 mol−1 at 0◦ C and 11.0 cm3 mol−1 at 150◦ C. 1.23 Assuming that the coefﬁcient of thermal expansion of gasoline is roughly equal to that of benzene, estimate the fraction of your gasoline expense that could be saved by purchasing gasoline in the morning instead of in the afternoon, assuming a temperature difference of 5◦ C. 1.24 The volume of a sample of a liquid at constant pressure can be represented by Vm (tC ) Vm (0◦ C)(1 + α tC + β tC2 + γ tC3 ) where α , β , and γ are constants and tC is the Celsius temperature. a. Find an expression for the coefﬁcient of thermal expansion as a function of tC . b. Evaluate the coefﬁcient of thermal expansion of benzene at 20.00◦ C, using α 1.17626 × 10−3 (◦ C)−1 , β 1.27776 × 10−6 (◦ C)−2 , and γ 0.80648 × 10−8 (◦ C)−3 . Compare your value with the value in Table A.2. 1.25 The coefﬁcient of thermal expansion of ethanol equals 1.12 × 10−3 K −1 at 20◦ C and 1.000 atm. The density at 20◦ C is equal to 0.7893 g cm−3 . a. Find the volume of 1.000 mol of ethanol at 10.00◦ C and 1.000 atm. b. Find the volume of 1.000 mol of ethanol at 30.00◦ C and 1.000 atm. 21 1.3 Real Gases 1.3 The van der Waals equation of state is named for Johannes Diderik van der Waals,1837–1923, a Dutch physicist who received the 1910 Nobel Prize in physics for his work on equations of state. Real Gases Most gases obey the ideal gas law to a good approximation when near room temperature and at a moderate pressure. At higher pressures one might need a better description. Several equations of state have been devised for this purpose. The van der Waals equation of state is P+ an2 (V − nb) nRT V2 (1.31) The symbols a and b represent constant parameters that have different values for different substances. Table A.3 in Appendix A gives values of van der Waals parameters for several substances. We solve the van der Waals equation for P and note that P is actually a function of only two intensive variables, the temperature T and the molar volume Vm , deﬁned to equal V /n. P nRT an2 a RT − 2 − V − nb Vm − b Vm2 V (1.32) This dependence illustrates the fact that intensive variables such as pressure cannot depend on extensive variables and that the intensive state of a gas or liquid of one substance is speciﬁed by only two intensive variables. E X A M P L E 1.8 Use the van der Waals equation to calculate the pressure of nitrogen gas at 273.15 K and a molar volume of 22.414 L mol−1 . Compare with the pressure of an ideal gas at the same temperature and molar volume. Solution P 8.134 J K−1 mol−1 (273.15 K) 0.022414 m3 mol−1 − 0.0000391 m3 mol−1 0.1408 Pa m3 mol−1 2 0.022414 m3 mol−1 − 1.0122 × 105 Pa 0.9990 atm For the ideal gas P RT Vm 8.134 J K−1 mol−1 (273.15 K) 0.022414 m3 mol−1 1.0132 × 105 Pa 1.0000 atm Exercise 1.7 a. Show that in the limit that Vm becomes large, the van der Waals equation becomes identical to the ideal gas law. b. Find the pressure of 1.000 mol of nitrogen at a volume of 24.466 L and a temperature of 298.15 K using the van der Waals equation of state. Find the pressure of an ideal gas under the same conditions. 22 1 The Behavior of Gases and Liquids c. Find the pressure of 1.000 mol of nitrogen at a volume of 1.000 L and a temperature of 298.15 K using the van der Waals equation of state. Find the pressure of an ideal gas under the same conditions. Another common equation of state is the virial equation of state: B3 B4 B2 PVm + 2 + 3 + ··· 1+ RT Vm Vm Vm (1.33) which is a power series in the independent variable 1/Vm . The B coefﬁcients are called virial coefﬁcients. The ﬁrst virial coefﬁcient, B1 , is equal to unity. The other virial coefﬁcients must depend on temperature in order to provide an adequate representation. Table A.4 gives values of the second virial coefﬁcient for several gases at several temperatures. An equation of state that is a power series in P is called the pressure virial equation of state: PVm RT + A2 P + A3 P 2 + A4 P 3 + · · · (1.34) The coefﬁcients A2 , A3 , etc., are called pressure virial coefﬁcients and also must depend on the temperature. It can be shown that A2 and B2 are equal. E X A M P L E 1.9 Show that A2 B2 . Solution We solve Eq. (1.33) for P and substituting this expression for each P in Eq. (1.34). P RT B2 RT B3 RT + + + ··· 2 3 Vm Vm Vm We substitute this expression into the lefthand side of Eq. (1.34). PVm RT + RT B2 RT B3 + + ··· 2 Vm Vm We substitute this expression into the second term on the righthand side of Eq. (1.34). PVm RT + A2 RT B2 RT B3 RT + + + ··· 2 3 Vm Vm Vm If two power series in the same variable are equal to each other for all values of the variable, the coefﬁcients of the terms of the same power of the variable must be equal to each other. We equate the coefﬁcients of the 1/Vm terms and obtain the desired result: A2 B2 Exercise 1.8 Show that A3 1 B3 − B22 . RT 23 1.3 Real Gases Table 1.1 displays several additional equations of state, and values of parameters for several gases are found in Table A.3. The parameters for a given gas do not necessarily have the same values in different equations even if the same letters are used. The accuracy of several of the equations of state has been evaluated.2 The Redlich–Kwong equation of state seemed to perform better than the other twoparameter equations, with the van der Waals equation coming in second best. The Gibbons–Laughton modiﬁcation of the Redlich–Kwong equation (with four parameters) is more accurate than the twoparameter equations. Table 1.1 Some Equations of State The letters a and b stand for constant parameters that have different values for different substances. These parameters do not necessarily have the values for the same substance in different equations of state. The Berthelot Equation of State a (Vm − b) RT P+ TVm2 The Dieterici Equation of State Pea/Vm RT (Vm − b) RT The Redlich–Kwong Equation of State P RT a − Vm − b T 1/2 Vm (Vm + b) The Soave Modiﬁcation of the Redlich–Kwong Equation of State P RT aα(T ) − Vm − b Vm (Vm + b) where α(T ) {1 + m 1 − (T /Tc )1/2 }2 , where m is a constant parameter and where Tc is the critical temperature. See the article by Soave for values of the parameter m. The Gibbons–Laughton Modiﬁcation of the Redlich–KwongSoave Equation of State The equation is the same as the Soave modiﬁcation, but α(T ) is given by α(T ) 1 + X(T /Tc ) − 1 + Y (T /Tc )1/2 − 1 where X and Y are constant parameters. See the article by Gibbons and Laughton for values of these parameters. Other equations of state can be found in the book by Hirschfelder, Curtiss, and Bird, including the Beattie–Bridgeman equation, with ﬁve parameters, and the Benedict– Webb–Rubin equation, with eight parameters. 2 J. B. Ott, J. R. Goates, and H. T. Hall, Jr., J. Chem. Educ., 48, 515 (1971); M. W. Kemp, R. E. Thompson, and D. J. Zigrang, J. Chem. Educ., 52, 802 (1975). 24 1 The Behavior of Gases and Liquids Graphical Representation of Volumetric Data for Gases The compression factor, denoted by Z, is sometimes used to describe the behavior of real gases: Z PVm RT (1.35) Some authors call Z the compressibility factor. We avoid this name because it might be confused with the compressibility. The compression factor equals unity for an ideal gas. Figure 1.3 shows a graph of the compression factor of nitrogen gas as a function of pressure at several temperatures. At low temperatures, the value of Z is less than unity for moderate pressures, but rises above unity for larger pressures. At higher temperatures, the value of Z is greater than unity for all pressures. Attractions between the molecules tend to reduce the value of Z and repulsions between the molecules tend to increase the value of Z. Attractions are more important at lower temperatures and smaller pressures, and repulsions are more important at higher temperatures and higher pressures. The temperature at which the curve has zero slope at zero pressure is called the Boyle temperature. This is the temperature at which the gas most nearly approaches ideality for small pressures. For a van der Waals gas, the compression factor is given by Z PVm Vm a ay 1 − − RT Vm − b RTVm 1 − by RT (1.36) where we let y 1/Vm . Since a and b are both positive for all gases, the ﬁrst term on the righthand side of Eq. (1.36) gives a positive contribution to Z, and the second term gives a negative contribution. The parameter b describes the effect of repulsive 2.0 Z 200 K 250 K 1900 K 1.0 0.8 150 K 0.6 0.4 0.3 0.2 0.1 5 10 20 50 100 200 P/bar 500 Figure 1.3 The Compression Factor of Nitrogen as a Function of Pressure at Several Temperatures. 25 1.3 Real Gases intermolecular forces and the parameter a describes the effect of attractive intermolecular forces. For higher temperatures the second term is relatively unimportant, and the compression factor will exceed unity for all values of y. For temperatures below the Boyle temperature the second term becomes relatively more important, and a value of Z less than unity will occur if y is not too large. E X A M P L E 1.10 a. Find an expression for the Boyle temperature of a van der Waals gas. b. Find the value of the Boyle temperature of nitrogen gas as predicted by the van der Waals equation. Solution a. Since y is proportional to P for small values of P, we seek the temperature at which ∂Z b a a 0 − b− 2 ∂y y0 RT RT (1 − by) y0 where the subscript y 0 indicates the value of y at which the derivative is evaluated. The Boyle temperature is TBoyle a/Rb b. For nitrogen, 0.1408 Pa m2 mol−1 433 K 8.134 J K−1 mol−1 3.913 × 10−5 m3 mol−1 TBoyle Exercise 1.9 a. Find an expression for the Boyle temperature of a gas obeying the Dieterici equation of state. b. Find the value of the Boyle temperature of nitrogen according to the Dieterici equation of state. c. Find the expression for the molar volume at which Z 1 for the van der Waals gas for a given temperature below the Boyle temperature. Hint: Find the nonzero value of y in Eq. (1.36) that makes Z 1. d. Find the value of the molar volume and the pressure at which Z 1 for nitrogen at 273.15 K, according to the van der Waals equation. PROBLEMS Section 1.3: Real Gases 1.26 For the van der Waals equation of state, obtain formulas for the partial derivatives (∂P/∂T )V ,n , (∂P/∂V )T ,n , and (∂P/∂n)T ,V . 1.27 For the virial equation of state, a. Find the expressions for (∂P/∂V )T ,n and (∂P/∂T )V ,n . b. Show that (∂2 P/∂V ∂T )n (∂2 P/∂T ∂V )n . 26 1 The Behavior of Gases and Liquids 1.28 Evaluate each of the partial derivatives in Problem 1.26 for carbon dioxide at 298.15 K and 10.000 bar. 1.29 a. Derive an expression for the isothermal compressibility of a gas obeying the van der Waals equation of state. Hint: Use the reciprocal identity, Eq. (B8). b. Evaluate the isothermal compressibility of carbon dioxide gas at a temperature of 298.15 K and a molar volume of 0.01000 m3 mol−1 . Compare with the value obtained from the ideal gas law. 1.30 Write the expressions giving the compression factor Z as a function of temperature and molar volume for the van der Waals, Dieterici, and Redlich–Kwong equations of state. 1.31 a. For the van der Waals equation of state at temperatures below the Boyle temperature, ﬁnd an expression for a value of the pressure other than P 0 for which PVm RT . Vm 0.1497 L mol−1 is 1.1336. Find the values of Z predicted by the van der Waals, Dieterici, and Redlich–Kwong equations of state for these conditions. Calculate the percent error for each. 1.38 The parameters for the van der Waals equation of state for a mixture of gases can be approximated by use of the mixing rules: a a1 x12 + a12 x1 x2 + a2 x22 b b1 x12 + b12 x1 x2 + b2 x22 where x1 and x2 are the mole fractions of the two substances and where a1 , b1 , a2 , and b2 are the van der Waals parameters of the two substances. The quantities a12 and b12 are deﬁned by a12 (a1 a2 )1/2 and b. Find the value of this pressure for nitrogen gas at 298.15 K. 1.32 a. By differentiation, ﬁnd an expression for the isothermal compressibility of a gas obeying the Dieterici equation of state. b. Find the value of the isothermal compressibility of nitrogen gas at 298.15 K and Vm 24.4 L. Compare with that of an ideal gas. 1.33 a. By differentiation, ﬁnd an expression for the coefﬁcient of thermal expansion of a gas obeying the van der Waals equation of state. b. Find the value of the coefﬁcient of thermal expansion of nitrogen gas at 298.15 K and Vm 24.4 L mol−1 . 1.34 By differentiation, ﬁnd an expression for the coefﬁcient of thermal expansion of a gas obeying the Dieterici equation of state. 1.35 Manipulate the Dieterici equation of state into the virial form. Use the identity e−x 1 − x + x3 xn x2 − + · · · + (−1)n + ··· 2! 3! n! where n! n(n − 1)(n − 2)(n − 3) . . . (3)(2)(1). Write expressions for the second, third, and fourth virial coefﬁcients. 1.36 Write an expression for the isothermal compressibility of a nonideal gas obeying the Redlich–Kwong equation of state. 1.37 The experimental value of the compression factor Z PVm /RT for hydrogen gas at T 273.15 K and b12 1/3 b1 1/3 + b2 3 3 a. Using these mixing rules and the van der Waals equation of state, ﬁnd the pressure of a mixture of 0.79 mol of N2 and 0.21 mol of O2 at 298.15 K and at a mean molar volume (deﬁned as V /ntotal ) of 0.00350 m3 mol−1 . Compare your answer with the pressure of an ideal gas under the same conditions. b. Using the van der Waals equation of state, ﬁnd the pressure of pure N2 at 298.15 K and at a molar volume of 0.00350 m3 mol−1 . c. Using the van der Waals equation of state, ﬁnd the pressure of pure O2 at 298.15 K and at a molar volume of 0.00350 m3 mol−1 . 1.39 Find the value of the isothermal compressibility of carbon dioxide gas at 298.15 K and a molar volume of 24.4 L mol−1 , a. According to the ideal gas law. b. According to the truncated virial equation of state B2 PVm 1+ RT Vm For carbon dioxide at 298.15 K, B2 −12.5 × 10−5 m3 mol−1 . 1.40 Considering P to be a function of T , V , and n, obtain the expression for dP for a gas obeying the van der Waals equation of state. 27 1.4 The Coexistence of Phases and the Critical Point The Coexistence of Phases and the Critical Point Transitions from a gaseous state to a liquid state or from a liquid state to a solid state, and so forth, are called phase transitions and the samples of matter in the different states are called phases. Such transitions can take place abruptly. If a gas is initially at a temperature slightly above its condensation temperature, a small decrease in the temperature can produce a liquid phase that coexists with the gas phases, and a further small decrease in the temperature can cause the system to become a single liquid phase. This remarkable behavior is an exception to the general rule that in nature small changes produce small effects and large changes produce large effects. It is an experimental fact that for any pure substance the pressure at which two phases can coexist at equilibrium is a smooth function of the temperature. Equivalently, the temperature is a smooth function of the pressure. Figure 1.4 shows schematic curves representing these functions for a typical substance. The curves are called coexistence curves and the ﬁgure is called a phase diagram. The three curves shown are the solid– gas (sublimation) curve at the bottom of the ﬁgure, the liquid–gas (vaporization) curve at the upper right, and the solid–liquid (fusion, melting, or freezing) curve at the upper left. The three curves meet at a point called the triple point. This point corresponds to (3Step process to liquefy gas without phase transition) ce c urve Expansion at constant temperature Liquid Critical point Compression at constant temperature Vapor (gas) Li qu id va p or c oe x is te n Solid–liquid Solid coexistenc e cu rve Cooling at constant pressure P 1.4 1 atm Normal freezing temperature Triple point Normal boiling temperature T Solid–vapor coexistence curve Figure 1.4 The Coexistence Curves for a Typical Pure Substance (Schematic). 28 1 The Behavior of Gases and Liquids the unique value of the pressure and the unique value of the temperature at which all three phases can coexist. The equilibrium temperature for coexistence of the liquid and solid at a pressure equal to 1 atmosphere is called the normal melting temperature or normal freezing temperature. The equilibrium temperature for coexistence of the liquid and gas phases at a pressure equal to 1 atmosphere is called the normal boiling temperature. These temperatures are marked on Figure 1.4. If the triple point is at a higher pressure than 1 atmosphere the substance does not have a normal freezing temperature or a normal boiling temperature, but has a normal sublimation temperature at which the solid and gas coexist at a pressure equal to 1 atmosphere. The triple point of carbon dioxide occurs at a pressure of 5.112 atm and a temperature of 216.55 K (−56.60◦ C) and its normal sublimation temperature is equal to 194.6 K (−78.5◦ C). Equilibrium liquid carbon dioxide can be observed only at pressures greater than 5.112 atm. At lower pressures the solid sublimes directly into the vapor phase. The Critical Point There is a remarkable feature that is shown in Figure 1.4. The liquid–vapor coexistence curve terminates at a point that is called the critical point. The temperature, molar volume, and pressure at this point are called the critical temperature, denoted by Tc , the critical molar volume, denoted by Vmc , and the critical pressure, denoted by Pc . These three quantities are called critical constants. Table A.5 in the appendix gives values of the critical constants for several substances. At temperatures higher than the critical temperature and pressures higher than the critical pressure there is no transition between liquid and gas phases. It is possible to heat a gas to a temperature higher than the critical temperature, then to compress it until its density is as large as that of a liquid, and then to cool it until it is a liquid without ever having passed through a phase transition. A path representing this kind of process is drawn in Figure 1.4. Fluids at supercritical temperatures are often referred to as gases, but it is better to refer to them as supercritical ﬂuids. Some industrial extractions, such as the decaffeination of coffee, are carried out with supercritical ﬂuids such as carbon dioxide.3 Supercritical carbon dioxide is also used as a solvent in some HPLC applications.4 Using a chiral stationary phase, enantiomers can be separated. The liquid–solid coexistence curve apparently does not terminate at a critical point. Nobody has found such a termination, and it seems reasonable that the presence of a lattice structure in the solid, which makes it qualitatively different from the liquid, makes the existence of such a point impossible. Figure 1.5 schematically shows the pressure of a ﬂuid as a function of molar volume for several ﬁxed temperatures, with one curve for each ﬁxed temperature. These constanttemperature curves are called isotherms. For temperatures above the critical temperature there is only one ﬂuid phase, and the isotherms are smooth curves. The liquid branch is nearly vertical since the liquid is almost incompressible while the gas branch of the curve is similar to the curve for an ideal gas. For subcritical temperatures, the isotherm consists of two smooth curves (branches) and a horizontal line segment, which is called a tie line. A tie line connects the two points representing the molar volumes of the coexisting liquid and gas phases. As subcritical temperatures closer and closer to the critical temperature are chosen the tie lines become shorter and shorter 3 Chem. Eng. Sci., 36(11), 1769(1981); Env. Sci. Technol., 20(4), 319 (1986); Chemtech., 21(4), 250 (1991), Anal. Chem., 66(12), 106R (1994). 4A.M. Thayer, Chem. Eng. News, 83, 49 (September 5, 2005). 29 1.4 The Coexistence of Phases and the Critical Point Pc Liquid branches Supercritical isotherms (no distinction between liquid and vapor) Critical isotherm T 6 > Tc Tie line P T 5 > Tc T 4 > Tc Tc Tie line T2 <Tc T1 <Tc Vapor branch 0 Figure 1.5 Vm,c Vm Isotherms for a Typical Pure Substance (Schematic). until they shrink to zero length at the critical point. No two isotherms can intersect, so the isotherm that passes through the critical point must have a horizontal tangent line at the critical point. This point on the critical isotherm is an inﬂection point, with a zero value of (∂P/∂Vm )T and a zero value of (∂2 P/∂Vm2 )T . At the critical point, a ﬂuid exhibits some unusual properties such as strong scattering of light, inﬁnite heat capacity, and inﬁnite compressibility. If a sample of a pure ﬂuid is conﬁned in a rigid closed container such that the average molar volume is equal to that of the critical state and if the temperature is raised through the critical temperature, the meniscus between the liquid and gas phases becomes diffuse and then disappears at the critical temperature. Figure 1.6 shows photographs illustrating this behavior in carbon dioxide.5 The system contains three balls that are slightly different in density, with densities close to the critical density of carbon dioxide. Figure 1.7 depicts a perspective view of a threedimensional graph with a surface representing the pressure of a ﬂuid as a function of temperature and molar volume. The isotherms in Figure 1.5 are produced by passing planes of constant temperature through the surface of this graph. Several isotherms are drawn on the surface in Figure 1.7. The liquid–gas tie lines are seen in the tongueshaped region. When the threedimensional graph is viewed in a direction perpendicular to the T –P plane each liquid–gas tie line is seen as a point. The set of all such points makes up the gas–liquid coexistence curve seen in Figure 1.4. 5 J. V. Sengers and A. L. Sengers, Chem. Eng. News., 46, 54 (June 10, 1968). This ﬁgure can be seen on the Web at http://sfu.ca/chemcai/critical.html, courtesy of Dr. Steven Lower of Simon Fraser University. 30 1 The Behavior of Gases and Liquids (a) (b) (c) (d) Figure 1.6 Liquid–Gas Equilibrium near the Critical Point. (a) At a temperature slightly above the critical temperature. The density of the ﬂuid depends slightly on height, due to gravity. (b) At the critical temperature, and showing the scattering of light known as critical opalescence. (c) and (d) At subcritical temperatures, showing a deﬁnite meniscus. From J. V. Sengers and A. L. Sengers, Chem. Eng. News, June 10, 1968, p. 104. Used by permission of the copyright holder. P T Tc isotherm 0 Tie lines V m Figure 1.7 Surface Giving Pressure as a Function of Molar Volume and Temperature for a Typical Pure Substance in the Liquid–Vapor Region (Schematic). 31 1.4 The Coexistence of Phases and the Critical Point Because the entire ﬂuid (liquid and gas) surface in Figure 1.7 is connected, a completely successful equation of state should represent the entire surface. The equations of state that we have discussed yield surfaces that resemble the true surface in the liquid region as well as in the gas region, although they do not represent the tie lines. In Chapter 5 we will discuss a technique for constructing the tie lines for a particular equation of state. The modiﬁed Redlich–Kwong–Soave equation of Gibbons and Laughton seems to be fairly accurate in representing both the liquid and the gas, and the van der Waals equation is often used to give qualitative information. For any equation of state, we can obtain equations that locate the critical point. E X A M P L E 1.11 Derive formulas for the critical temperature and critical molar volume for a gas obeying the van der Waals equation of state. Solution We seek the point at which (∂P/∂Vm )T 0 (1.41) 2) 0 (∂2 P/∂Vm T (1.42) The ﬁrst derivative of Eq. (1.32) with respect to Vm is ∂P RT 2a − + 3 ∂Vm T (Vm − b)2 Vm (1.43) and the second derivative is ∂2 P 2 ∂Vm − T 2RT (Vm − b)3 + 6a 4 Vm (1.44) Setting the righthand side of each of these two equations equal to zero gives us two simultaneous algebraic equations, which are solved to give the values of the critical temperature Tc and the critical molar volume Vmc : Tc 8a , 27Rb Vmc 3b (1.45) Exercise 1.10 Solve the simultaneous equations to verify Eq. (1.45). One way to proceed is as follows: Obtain Eq. (I) by setting the righthand side of Eq. (1.43) equal to zero, and Eq. (II) by setting the righthand side of Eq. (1.44) equal to zero. Solve Eq. (I) for T and substitute this expression into Eq. (II). When the values of Tc and Vmc are substituted into the van der Waals equation of state the value of the critical pressure for a van der Waals gas is obtained: Pc a 27b2 (1.46) 32 1 The Behavior of Gases and Liquids For a van der Waals gas, the compression factor at the critical point is Zc Pc Vmc 3 0.375 RTc 8 (1.47) Exercise 1.11 Verify Eqs. (1.46) and (1.47). Equations (1.45) and (1.46) can be solved for a and b: 2 Pc a 3Vmc b 27R2 Tc2 9R Vmc Tc 8 64Pc Vmc RTc 3 8Pc (1.48) (1.49) There are two or three formulas for each parameter. Since no substance exactly ﬁts the equation different values can result from the different formulas. The best values of a and b are obtained by using Pc and Tc as independent variables. The values of the parameters for any twoparameter or threeparameter equation of state can be obtained from critical constants. Exercise 1.12 a. Show that for the Dieterici equation of state, Vmc 2b, Tc a , 4bR Pc a −2 e 4b2 (1.410) b. Show that for the Dieterici equation of state, Zc 2e−2 0.27067 c. Obtain the formulas giving the Dieterici parameters a and b as functions of Pc and Tc . Find the values of a and b for nitrogen and compare with the values in Table A.3. The parameters a and b in the Redlich–Kwong equation of state can be obtained from the relations 1/3 5/2 2 − 1 RTc R2 Tc , b (1.411) a 1/3 3Pc 9 2 − 1 Pc The value of the compression factor at the critical point according to the Redlich– Kwong equation of state is 1/3. Exercise 1.13 Find the values of a and b in the Redlich–Kwong equation of state for nitrogen. Figure 1.8 shows schematically a more complete view of the threedimensional graph of Figure 1.7, including the solid–liquid and solid–gas phase transitions. There 33 1.4 The Coexistence of Phases and the Critical Point Solid–liquid tie line P Liquid–gas tie line Triple point tie lines Solid–gas tie line 0 Tc V Tt T m Figure 1.8 Surface Giving Pressure as a Function of Molar Volume and Temperature Showing All Three Phases (Schematic). are three sets of tie lines, corresponding to the three curves in Figure 1.4. At the triple point, all three tie lines come together in a single tie line connecting three phases. As shown in Figures 1.7 and 1.8, the pressure of a onephase system of one substance is a function of only two intensive variables, T and Vm . Any intensive variable in a onecomponent ﬂuid system is also a function of two intensive variables. The intensive state of an equilibrium system is the state of the system so far as only intensive variables are concerned, and is speciﬁed by two independent intensive variables if an equilibrium system contains a single substance and a single ﬂuid phase. The size of the system is not speciﬁed. For a onephase ﬂuid (liquid or gas) system of c substances, c + 1 intensive variables specify the intensive state of the system. The Law of Corresponding States The van der Waals equation predicts that the value of the compression factor at the critical point is equal to 0.375 for all substances. There is even a greater degree of generality, expressed by an empirical law called the law of corresponding states:6 All substances obey the same equation of state in terms of reduced variables. The reduced variables are dimensionless variables deﬁned as follows: The reduced volume is the ratio of the molar volume to the critical molar volume: Vm Vr (1.412) Vmc The reduced pressure is the ratio of the pressure to the critical pressure: Pr P Pc 6 Hirschfelder, Curtiss, and Bird, op. cit., p. 235 [see Table 2.1]. (1.413) 34 1 The Behavior of Gases and Liquids The reduced temperature is the ratio of the temperature to the critical temperature: Tr T Tc (1.414) Using the deﬁnitions in Eqs. (1.412), (1.413), and (1.414) and the relations in Eqs. (1.45) and (1.46) we obtain for a ﬂuid obeying the van der Waals equation of state: P aPr , 27b2 Vm 3bVr , T 8aTr 27Rb When these relations are substituted into Eq. (1.31), the result is 3 1 8Tr Vr − Pr + 2 3 3 Vr (1.415) Exercise 1.14 Carry out the algebraic steps to obtain Eq. (1.415). In Eq. (1.415), the parameters a and b have canceled out. The van der Waals equation of state thus conforms to the law of corresponding states. The same equation of state without adjustable parameters applies to every substance that obeys the van der Waals equation of state if the reduced variables are used instead of P, Vm , and T . The other twoparameter equations of state also conform to the law of corresponding states. Figure 1.9 is a graph of the experimentally measured compression factor of a number of polar and nonpolar ﬂuids as a function of reduced pressure at a number of reduced Methane Isopentane 1.0 Ethylene nHeptane Ethane Nitrogen Propane Carbon dioxide nButane Water Tr ⫽ 2.00 0.9 1.50 0.8 1.30 Z⫽ PVm RT 0.7 0.6 1.20 0.5 1.10 0.4 1.00 0.3 0.2 0.1 0 0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 Reduced pressure, Pr 5.0 5.5 6.0 6.5 7.0 Figure 1.9 The Compression Factor as a Function of Reduced Pressure and Reduced Temperature for a Number of Gases. From G.J. Su, Ind. Eng. Chem., 38, 803 (1946). Used by permission of the copyright holder. 35 1.4 The Coexistence of Phases and the Critical Point temperatures.7 The agreement of the data for different substances with the law of corresponding states is generally better than the agreement of the data with any simple equation of state. Exercise 1.15 Show that the Dieterici equation of state conforms to the law of corresponding states by expressing it in terms of the reduced variables. PROBLEMS Section 1.4: The Coexistence of Phases and the Critical Point 1.41 a. Use the van der Waals equation of state in terms of reduced variables, Eq. (1.415), to calculate the pressure of 1.000 mol of CO2 in a volume of 1.000 L at 100.0◦ C. The critical constants are in Table A.5 in Appendix A. Since the critical compression factor of carbon dioxide does not conform to the van der Waals value, Zc 0.375, you must replace the experimental th (0.375)RT /P . critical molar volume by Vmc c c b. Repeat the calculation using the ordinary form of the van der Waals equation of state. 1.42 a. Find the formulas for the parameters a and b in the Soave and Gibbons–Laughton modiﬁcations of the Redlich–Kwong equation of state in terms of the critical constants. Show that information about the extra parameters is not needed. b. Find the values of the parameters a and b for nitrogen. 1.43 The critical temperature of xenon is 289.73 K, and its critical pressure is 5.840 MPa (5.840 × 106 Pa). a. Find the values of the van der Waals constants a and b for xenon. b. Find the value of the compression factor, Z, for xenon at a reduced temperature of 1.35 and a reduced pressure of 1.75. 1.44 a. Evaluate the parameters in the Dieterici equation of state for argon from critical point data. b. Find the Boyle temperature of argon according to the Dieterici equation of state. Summary of the Chapter A system is deﬁned as the material object that one is studying at a speciﬁc time. The state of a system is the circumstance in which it is found, expressed by numerical values of a sufﬁcient set of variables. A macroscopic system has two important kinds of states: the macroscopic state, which concerns only variables pertaining to the system as a whole, and the microscopic state, which pertains to the mechanical variables of individual molecules. The equilibrium macroscopic state of a onephase ﬂuid (liquid or gas) system of one component is speciﬁed by the values of three independent state variables. All other macroscopic state variables are dependent variables, with values given by mathematical functions of the independent variables. The volumetric (PVT) behavior of gases under ordinary pressures is described approximately by the ideal gas law. For higher pressures, several more accurate equations of state were introduced. A calculation practice was introduced: for ordinary calculations: Gases are treated as though they were ideal. The volumes of solids and liquids are computed with the compressibility and the coefﬁcient of thermal expansion. For ordinary calculations they are treated as though they had constant volume. 7 G.J. Su, Ind. Eng. Chem., 38, 803 (1946). 36 1 The Behavior of Gases and Liquids When two phases of a single substance are at equilibrium, the pressure is a function only of the temperature. A phase diagram for a pure substance contains three curves representing this dependence for the solid–liquid, solid–gas, and liquid–gas equilibria. These three curves meet at a point called the triple point. The liquid–vapor coexistence curve terminates at the critical point. Above the critical temperature, no gas–liquid phase transition occurs and there is only one ﬂuid phase. The law of corresponding states was introduced, according to which all substances obey the same equation of state in terms of reduced variables ADDITIONAL PROBLEMS 1.45 Assume that when Julius Caesar exhaled for the last time he exhaled 1.0 L of air. a. Estimate the number of nitrogen molecules that were exhaled in Julius Caesar’s last breath. The mole fraction of nitrogen in air is approximately 0.78. (The mole fraction of a substance is its amount in moles divided by the total amount of all substances.) b. Estimate the number of nitrogen molecules out of those in part a that are now present in the physical chemistry classroom.