Table Of Content

PHYSICAL CHEMISTRY Thermodynamics, Structure, and Change Tenth Edition Peter Atkins | Julio de Paula FUNDAMENTAL CONSTANTS Constant Symbol Value Power of 10 Units Speed of light c 2.997 924 58* 108 m s−1 Elementary charge e 1.602 176 565 10−19 C Planck’s constant h 6.626 069 57 10−34 J s ħ = h/2π 1.054 571 726 10−34 J s Boltzmann’s constant k 1.380 6488 10−23 J K−1 Avogadro’s constant N 6.022 141 29 1023 mol−1 A Gas constant R = N k 8.314 4621 J K−1 mol−1 A Faraday’s constant F = N e 9.648 533 65 104 C mol−1 A Mass Electron m 9.109 382 91 10−31 kg e Proton m 1.672 621 777 10−27 kg p Neutron m 1.674 927 351 10−27 kg n Atomic mass constant m 1.660 538 921 10−27 kg u Vacuum permeability μ 4π* 10−7 J s2 C−2 m−1 0 Vacuum permittivity ε = 1/μc2 8.854 187 817 10−12 J−1 C2 m−1 0 0 4πε 1.112 650 056 10−10 J−1 C2 m−1 0 Bohr magneton μ = eħ/2m 9.274 009 68 10−24 J T−1 B e Nuclear magneton μ = eħ/2m 5.050 783 53 10−27 J T−1 N p Proton magnetic moment μ 1.410 606 743 10−26 J T−1 p g-Value of electron g 2.002 319 304 e Magnetogyric ratio Electron γ = –ge/2m –1.001 159 652 1010 C kg−1 e e e Proton γ = 2μ/ħ 2.675 222 004 108 C kg−1 p p Bohr radius a = 4πεħ2/e2m 5.291 772 109 10−11 m 0 0 e Rydberg constant R(cid:31)∞=mee4/8h3cε02 1.097 373 157 105 cm−1 hcR(cid:31)∞/e 13.605 692 53 eV Fine-structure constant α = μe2c/2h 7.297 352 5698 10−3 0 α−1 1.370 359 990 74 102 Second radiation constant c = hc/k 1.438 777 0 10−2 m K 2 Stefan–Boltzmann constant σ = 2π5k4/15h3c2 5.670 373 10−8 W m−2 K−4 Standard acceleration of free fall g 9.806 65* m s−2 Gravitational constant G 6.673 84 10−11 N m2 kg−2 * Exact value. For current values of the constants, see the National Institute of Standards and Technology (NIST) website. PH YS I C A L C H E M I S T RY Thermodynamics, Structure, and Change Tenth edition Peter Atkins Fellow of Lincoln College, University of Oxford, Oxford, UK Julio de Paula Professor of Chemistry, Lewis & Clark College, Portland, Oregon, USA W. H. Freeman and Company New York Publisher: Jessica Fiorillo Associate Director of Marketing: Debbie Clare Associate Editor: Heidi Bamatter Media Acquisitions Editor: Dave Quinn Marketing Assistant: Samantha Zimbler Library of Congress Control Number: 2013939968 Physical Chemistry: Thermodynamics, Structure, and Change, Tenth Edition © 2014, 2010, 2006, and 2002 Peter Atkins and Julio de Paula All rights reserved ISBN-13: 978-1-4292-9019-7 ISBN-10: 1-4292-9019-6 Published in Great Britain by Oxford University Press This edition has been authorized by Oxford University Press for sales in the United States and Canada only and not export therefrom. First printing W. H. Freeman and Company 41 Madison Avenue New York, NY 10010 www.whfreeman.com PREFACE This new edition is the product of a thorough revision of Topic are useful distillations of the most important concepts content and its presentation. Our goal is to make the book and equations that appear in the exposition. even more accessible to students and useful to instructors by We continue to develop strategies to make mathematics, enhancing its flexibility. We hope that both categories of user which is so central to the development of physical chemistry, will perceive and enjoy the renewed vitality of the text and the accessible to students. In addition to associating Mathematical presentation of this demanding but engaging subject. background sections with appropriate chapters, we give more The text is still divided into three parts, but each chapter is help with the development of equations: we motivate them, now presented as a series of short and more readily mastered justify them, and comment on the steps taken to derive them. Topics. This new structure allows the instructor to tailor the text We also added a new feature: The chemist’s toolkit, which offers within the time constraints of the course as omissions will be quick and immediate help on a concept from mathematics or easier to make, emphases satisfied more readily, and the trajec- physics. tory through the subject modified more easily. For instance, This edition has more worked Examples, which require it is now easier to approach the material either from a ‘quan- students to organize their thoughts about how to proceed tum first’ or a ‘thermodynamics first’ perspective because it with complex calculations, and more Brief illustrations, is no longer necessary to take a linear path through chapters. which show how to use an equation or deploy a concept in Instead, students and instructors can match the choice of a straightforward way. Both have Self-tests to enable students Topics to their learning objectives. We have been very care- to assess their grasp of the material. We have structured the ful not to presuppose or impose a particular sequence, except end-of-chapter Discussion questions, Exercises, and Problems where it is demanded by common sense. to match the grouping of the Topics, but have added Topic- We open with a Foundations chapter, which reviews basic and Chapter-crossing Integrated activities to show that sev- concepts of chemistry and physics used through the text. Part eral Topics are often necessary to solve a single problem. The 1 now carries the title Thermodynamics. New to this edition is Resource section has been restructured and augmented by the coverage of ternary phase diagrams, which are important in addition of a list of integrals that are used (and referred to) applications of physical chemistry to engineering and mater- throughout the text. ials science. Part 2 (Structure) continues to cover quantum the- We are, of course, alert to the development of electronic ory, atomic and molecular structure, spectroscopy, molecular resources and have made a special effort in this edition to assemblies, and statistical thermodynamics. Part 3 (Change) encourage the use of web-based tools, which are identified in has lost a chapter dedicated to catalysis, but not the material. the Using the book section that follows this preface. Important Enzyme-catalysed reactions are now in Chapter 20, and hetero- among these tools are Impact sections, which provide examples geneous catalysis is now part of a new Chapter 22 focused on of how the material in the chapters is applied in such diverse surface structure and processes. areas as biochemistry, medicine, environmental science, and As always, we have paid special attention to helping students materials science. navigate and master this material. Each chapter opens with a Overall, we have taken this opportunity to refresh the text brief summary of its Topics. Then each Topic begins with three thoroughly, making it even more flexible, helpful, and up to questions: ‘Why do you need to know this material?’, ‘What is date. As ever, we hope that you will contact us with your sug- the key idea?’, and ‘What do you need to know already?’. The gestions for its continued improvement. answers to the third question point to other Topics that we con- sider appropriate to have studied or at least to refer to as back- PWA, Oxford ground to the current Topic. The Checklists at the end of each JdeP, Portland Contents The result of a measurement is a physical quantity that is certain other units, a decision has been taken to revise this reported as a numerical multiple of a unit: definA.i1t ioAnto, bmust it has not yet, in 2014, been implemented). Th2 e freezing( ap) oTihnet noufc lweaart emr o(dtheel melting point of ice) at 1 atm2 is physicalquantity=numericalvalue×unit then fou(bn) dT heex ppeerriiomdeicn ttaalblyle to lie 0.01 K below the triple poi2nt, It follows that units may be treated like algebraic quanti- so the fr(ce) eIzoinnsg point of water is 273.15 K. The Kelvin scale3 is ties and may be multiplied, divided, and cancelled. Thus, the unsuAi.2t aMbleo lfeocru elveesr yday measurements of temperature, and i3t is Atoms USING THE BOOK A.1 expression (physical quantity)/unit is the numerical value (a common(a) toLe uwsies stthreu cCtuerlesisu s scale, which is defined in terms3 of dimensionless quantity) of the measurement in the specified the Kelvin Bsrciaelf eil lausstration A.1: Octet expansion 4 units. For instance, the mass m of an object could be reported (b) VSEPR theory 4 Z as m = 2.5 kg or m/kg = 2.5. See Table A.1 in the Resource sec- θ/°C =BTrie/fK ill−us2t7ra3t.io1n5 A.2: MolecuDlaerfi nshitaiopne s Celsius scale (A4.4) tion for a list of units. Although it is good practice to use only (c) Polar bonds 4 nucleon number SI units, there will be occasions where accepted practice is Thus, the frBerieezf iinllugs ptroatiinotn o Af .3w: aNtoenr piso 0la °r Cm aonledc uitles sb woiitlhin g point (at For the tenth edsioti doene polyf rPohoytesdi ctahal tC phhyesmiciasl tqruya:n Thtitieers maroe dexypnraesmseidc su,s ing the 1v aatrmie)t iys fopofou lnalerd ab toronn dbisne g10 f0e °aCt u(mreosr ea plrreecaidseyly p 9r9e.9s7e4n °tC, )w. Ne 4ohtea ve singu-mber), A Structure, and Cothhaern, ngoen w-SeI uhnaitvs.e B yta iniltoerrneadt iotnhael ctoenxvte netvioenn, amll pohryes ical nificthaantA ti.n3ly tB heuinsl kthe mxatna Ttct eiendrv atrhiaeb lmy daetnhoetems tahteic ths esrumpopdoynrat mbiyc (aadb5sdoi-ng new closely to the neqeudasn otift isetsu adree nretpsr.e Fseinrstetd, tbhye o mbliaqtueer (isalol pwinitgh) isnym ebaoclhs; all Chelumtei)s tt’esm (at)p oPeorrolakptueirtre tib eaosn oxdfe btshu, alakt nmtedamt tcpehre reactkurleiss tosn o tfh ke eCye lcsoiuns csec5paltes at the units are roman (upright). are denotedBr θie f( itlhluesttara)t.ion A.4: Volume units 5 ber are the isotopes chapter has been Urenoitrsg maanyi bzee dm oidnitfioe dd biys car perteefi xt othpaitc dse tnoo teims ap farcotvore of a end of each topic. (b) The perfect gas equation 6 accessibility, claproiwtye,r oafn 1d0. Aflmexoinbgi tlhitey m. oSset ccoomndm, onin S I apdredfiixteios nar et toh ose A note onE gxaomodp lper Aa.c1t:i Uces inNgo tthee t pheartf ewcet gwarsi eteq uTa =ti o0n, not T = 0 K7. (a) listed in Table A.2 in the Resource section. Examples of the use General statements in science should be expressed without Checklist of concepts 7 of these prefixes are: reference to a specific set of units. Moreover, because T (unlike According to the Checklist of equations 8 θ) is absolute, the lowest point is 0 regardless of the scale used 1 nm = 10−9 m 1 ps = 10−12 s 1 µmol = 10−6 mol to express higher temperatures (such as the Kelvin scale). each of charge –e ( Organizing the information Similarly, we write m = 0, not m = 0 kg and l = 0, not l = 0 m. Powers of units apply to the prefix as well as the unit they mod- ify. For example, 1 cm3 = 1 (cm)3, and (10−2 m)3 = 10−6 m3. Note that 1 cm3 does not mean 1 c(m3) . When carrying out numeri- ➤ Innovative new structure (b) The perfect gas equation cal calculations, it is usually safest to write out the numerical ➤➤Why do you need to know this material? Each chaptevra lThuheae orsef aabnre eo esbensve ernvra eSboIl ebr giansa esn cuiinezniettsdi,fi wc ihnnioctthoa t aiorsenh l(ioasstr entd .n itnnon Tp ×ai bc10lsen, ) A..3 The Bpercoapuesreti ecsh tehmati sdteryfi nies tahbeo sutat tem oaft ate sry satnemd tahree ncohta inng geesn - aarcete arrirzaendg beyd t ihne eral independent of one another. The most important example making the inte txhte Rmesooruerc er seeactdioanb. lAe llf ootrh esr tpuhdyesincatls quaanndti times omraey be that it can undergo, both physically and chemically, the consists of n2 of a relation between them is provided by the idealized fluid flexible for inesxtprruescsteod rass. cEomacbhin atotiponics oof ptheensse bwasiteh u nai tcs o(smee mTaebnlet A.4 properties of matter underlie the entire discussion in this into n subshells known as a perfect gas (also, commonly, an ‘ideal gas’): in the Resource section). Molar concentration (more formally, book. on why it is important, a statement of the key idea, and a but very rarely, amount of substance concentration) for exam- brief summary of the background needed to understand p➤V➤=WnhRaTt is the key idea? Perfect gas equation (A.5) ple, which is an amount of substance divided by the volume it the topic. occupies, can be expressed using the derived units of mol dm−3 The bulk properties of matter are related to the identities Here R is the gas constant, a universal constant (in the sense (b) as a combination of the base units for amount of substance and arrangements of atoms and molecules in a sample. of being independent of the chemical identity of the gas) with and length. A number of these derived combinations of units have special names and symbols and we highlight them as the ➤va➤luWe h8a.3t 1d4o5 Jy Kou−1 mneoel−d1. toTh krnoouwgh aolurte athdiys? text, equations ➤ Notes otheny agrisoe.od practice applTichaisb leT oonplicy tore pvieerwfesc t mgaasteesr i(aal ndc oomthmero nidlye alcizoevde rseyds teimn s) are labelled, as here, with a number in blue. introductory chemistry. Our Notes on good practice will help you avoid making A note on good practice Although the term ‘ideal gas’ is To specify the state of a sample fully it is also necessary to common mistakes. They encourage conformity to the almost universally used in place of ‘perfect gas’, there are give its temperature, T. The temperature is formally a prop- reasons for preferring the latter term. In an ideal system internationearlt y lathnagt udaetgeerm ionfe s sicni ewnhciceh bdiyr ecsteiotnt inenge rgoyu wt iltl hfleo w as Ththee pinrteesreancttaiotinosn boeft wpheeynsi cmalo lcehceumleiss tirny ai nm tihxitsu treex ta ries baalls ethde o n conventionsh eaant dw hperno ctwedo usraemsp aleds oapret epdla bceyd tihne c oInnttaecrtn tahtriooungah l ther- tshaem eex. pIenr iam penertfaelclyt vgearsi fineodt ofanclty tahraet tmhea ttienrt ecroacntsiiosntss oafl la ttohme s. table are called Union of Pumrael layn cdon Adupcptilnige dw aCllhs:e emneirsgtyr flyo (wIsU frPoAmC t)h.e sample with the same but they are in fact zero. Few, though, make this useful higher temperature to the sample with the lower temperature. distinction. The symbol T is used to denote the thermodynamic temperature which is an absolute scale with T = 0 as the lowest point. Equation A.5, the perfect gas equation, is a summary of Temperatures above T = 0 are then most commonly expressed three empirical conclusions, namely Boyle’s law (p ∝ 1/V at by using the Kelvin scale, in which the gradations of tempera- constant temperature and amount), Charles’s law (p ∝ T at con- ture are expressed as multiples of the unit 1 kelvin (1 K). The stant volume and amount), and Avogadro’s principle (V ∝ n at Kelvin scale is currently defined by setting the triple poi0n1t_ Aotkfin s_Ch0c0oAn.insdtda n 2t temperature and pressure). ➤ Resource section RESOURCE SECTION The comprehensive Resource section at the end of the book contains a table of integrals, data tables, a summary of con- ventions about units, and character tables. Short extracts of t0h1_eAstkein st_Cahb00lAe.sin dod ft 6en appear in the topics themselves, prin- 8/22/2013 12:57:41 PM cipally to give an idea of the typical values of the physical quantities we are introducing. Contents 1 Common integrals 964 2 Units 965 3 Data 966 4 Character tables 996 stant volume by using the relation C − C = R.) Self-test 3A.11 p,m V,m Answer From eqn 3A.16 the entropy change in the isothermal expansion from V to V is Using the book vii i f ➤ Checklist of concepts Checklist of concepts A Checklist of key concepts is provided at the end of each topic so that you can tick off those concepts which you feel ☐ 1. The entropy acts as a signpost of spontaneous change. ☐ 6. The you have mastered. ☐ 2. Entropy change is defined in terms of heat transactions 118 3 The Second and Third Laws (the Clausius definition). ☐ 3. The Boltzmann formula defines absolute entro- pies in terms of the number of ways of achieving a ☐ 7. 2. Then to show that the result is true whatever the working q coTnfiguration. h =− h (3A.7) substance. ☐ q4. ThTe Carnot cycle is used to prove that entropy is a state ☐ 8. c c Presenti3n. Fgina ltlyh, toe sh mow tahatt thhee remsulta is ttriuce sfor any cycle. function. Su☐b st5it.u tThione eoffif tchieisn creyl aotfi oan h einatto e nthgein per iesc tehdein bga seiqsu oaft itohne dgeivfiens i- zero on tthioe nr iogfh tt,h we hthicehr mis owdhyant awme iwc atenmtepde troa pturorev es.cale and one ☐ 9. ➤ Justifi(ac) aTthieo nCasrnot cycle realization, the Kelvin scale. Justification 3A.1 Heating accompanying reversible A Carnot cycle, which is named after the French engineer Sadi MathematicCaalr ndoet,v ceolnospismts eonf tfo iusr arenv eirnsitbrlien sstaicge ps (aFritg .o 3fA p.7h):ysical adiabatic expansion chemistry, and to achieve full understanding you need This Justification is based on two features of the cycle. One fea- 1. Reversible isothermal expansion from A to B at T ; the to see how a particular expression is obtained and if ahny ture is that the two temperatures T and T in eqn 3A.7 lie on entropy change is q /T , where q is the energy supplied h c assumptions hatov eth be eseysnte mm aads hee. aThht freho mJu tshtei fihochat stoiounrcse .are set off the same adiabat in Fig. 3A.7. The second feature is that the energy transferred as heat during the two isothermal stages from the text2 .t Roe vleerts iyboleu a daidabjuatsitc etxhpea nlesivoenl froofm d Be ttao iCl . tNoo m eneeergt1 y7 _Atkins_Ch03aAr.eindd 124 your current neleeadvse sa tnhed s mystaekme a ist heeaast,i esor tthoe rcehvaniegwe i nm eantterorpiayl i.s zero. In the course of this expansion, the temperature q =nRT lnVB q =nRT lnVD falls from Th to Tc, the temperature of the cold sink. h h VA c c VC 3. Reversible isothermal compression from C to D at T. c We now show that the two volume ratios are related in a very Energy is released as heat to the cold sink; the change in simple way. From the relation between temperature and volume nenetgraotpivye o.f the system is qc/Tc; in this expression qc is f6or reFvoeursnibdlaet iaodniasbatic processes (VTc = constant, Topic 2D): 4. Reversible adiabatic compression from D to A. No energy VAThc=VDTcc VCTcc=VBThc ➤ Chemiste’nste trso thoe lskysittesm as heat, so the change in entropy is zero. The temperature rises from T to T . MuTlhtiep lcihcaetmioisnt ’os ft othoelk fiitr sAt. 1o f thQesuea enxtpitrieesss aionnds ubny itthse second c h gives New to the tenth edition, the Chemist’s toolkits are succinct The result of a measurement is a physical quantity that is The total change in entropy around the cycle is the sum of the reminders cohfa ntghees imn eaatchhe omf tahteiscea fol ucro sntecpes:pts and techniques reVpAoVrCteTdhc Tacsc =a VnuDVmBeTrhciTcaccl multiple of a unit: that you will need in order to understand a particular whichp, hoyns cicaanlcqeulalanttiiotny o=fn thuem teermicpaelrvaatluuree×s, usnimitplifies to derivation bein(cid:31)∫gd Sd=esTqchhr+ibTqeccd in the main text. ItiteVV fsDo al=nlodVVw Ams atyh abte umnuitlsti pmliaeyd ,b dei vtrideaedte, da nlidk cea anlcgeelblerda.i cTh quusa,n tthie- C B expression (physical quantity)/unit is the numerical value (a However, we show in the following Justification that for a Wditihm tehniss iroenlalteisosn q eusatanbtiltisyh) eodf , twhee cmaena wsurriteement in the specified perfect gas units. For instance, the mass m of an object could be reported V V V asq cm= =n 2R.T5c klng VorD m=/nkRgT =c 2ln.5V. ASe=e −TnaRbTlec Aln.1V Bin the Resource sec- θ/°C =T/ ➤ Mathematical backgrounds tion for a list ofC units. AlthoBugh it is good Apr actice to use only anSdI tuhnerietsfo, rtehere will be occasions where accepted practice is There are six MathematicAal background sections dispersed Masot dheeepmly roaotteidc tahla t bphayscickalg qruoanutitniesd ar e1 e xpDreisffsede urseinng tiation and integration tmolohrcadratoehturee gdtmoh a aotbt uteihct aaeb ltPressure, pe hlnceed o tn ootc efme xtphtat.e ssA t cTh4edthDrihaa ebappaythtt eyyrc1so iotcuoIvsa I oeswlntor hcht eehheireicmednrhmm itd3itos ei tAstru damynii.a ldob EBaseatttrh csr2hetea leonmvndaea n iiintns. Tpothhcwiadescyseo u oquplpswpii irtnuorsnoqcoqi hotUw taitashcefetehhci nlndseqtr= edtr et,r inahdo iistdr n−nrt soecnuua3 ie on i ifweRmTrgA ne nrmRse1hTan no .a-stT07baohomcSh y..ciflruel Is enerFlnAeba stn t ouv c Ao( en rm(mVtooni rVe ah.mm(ll 2Bipourd cvBett e/opsnrlpei h/asVn s.eddoVrg ierusB Anidi .rti Agetbfihhfity)ikhnh)affje oec)eientt=d,e ant)ec R s tr. tm−tdtsb oeeie,oom y sn ToTrbatuon nhthsacnyauri,t ae a tdcp rnoithctec rriioboie)eo eotr tl mmfin asnaeiieqntasx matci u ldhotteta eiicoh sancoiq osntsa(andce n st qln. nS li v dohseEIttie egpe niipxnpsanacitao trlntehon imesigtogtnveifioertn)pgeis xia q.al,bvs t eeautayeiessi ol em fv (loaaanse hwpfc rb . e( etihtoanhoah Thyltrtre eshse ta; ioi uo hsectoaf ssyae flaee ll tahrdedax td ixennn to=htinesxd tn eθ−x1t of these prefixes are: C d Volume, V MB11. 1n mD =i 1ff0−e9 rmentia1t ipos n= 1: 0d−1e2 fisniti1o µnmsol = 10−6 mol dθxeax=aeax Figure 3A.7 The basic structure of a Carnot cycle. In Step 1, DBiffrieerfe inlltuiasttrioatnio ins c3oAn.3c ernTehde wCiatrhn tohte c sylcolpees of functions, such d there is isothermal reversible expansion at the temperature asTh thPiefeoy Cw.r aFaetorrensr oooefftx cuachmnyaciptnlselg e aec,p a1opn cfly mab t e3ov =a rter h1ige a(a cbprmlrdee)e fiw3d,x ia atanhssd atw i (mer1el0lep −a.2r s Them tshe)ee3n =ftuoa 1nrt0miito− t6na hml eody3fe. mtfiNhnoeoid t-e- dxsinax c Th. Step 2 is a reversible adiabatic expansion in which the tiocnhta honaf gtt eh1s ec mtdae3k rdiinvogae tspi nvloae,ct dem fi/enda xna, n 1o fca (acm tfuu3)an .l c Witdiohenean lfi (zcxea)dr ir syeninggi noeu,t wnuhmereer i- temperature falls from Th to Tc. In Step 3 there is an isothermal hecaat li sc aclocnuvlaetritoends i,n itt ois w uosruka.l (lyH soawfeesvte tro, owthrietre coyuctl eths ea rneu cmloesreirc al (bd) lnax=1 froevlleorwsiebdle b cyo amnp ardeisasbioanti ca tr eTvc,e arnsidb lteh caot misoptrheesrsmioanl, swtehpic ihs aacpdvdcpaoxfrlThrou=dxe eδl iwoxirmm→efi t 0aaahtnfri e (oto xhnsbee+ss v eCδterδoxnavx )ra rS−nbeIloa febtl( iaxecns)ny escg culiienneD,n iee1ttfiss0in.,fi) 0iw tc iIJo hnn noi o cfat hnae tFn aiieroersnernt gg d l(iyieans rstiiev esn ad rwt.n iuviinnent n h nT d×ia(nr b1Mag0lweB ni 1)Ann..1 .)3 dx x restores the system to its initial state. in the Resource section. All other physical quantities may be As shown in Fig. MB1.1, the derivative can be interpreted as the expressed as combinations of these base units (see Table A.4 slope of the tangent to the graph of f(x). A positive first deriva- frokmno wd nt oa s ∂a in the Resource section). Molar concentration (more formally, tive biuntd viceartye sr atrhealyt ,t ahme fouunnct toiof nsu sblostpaensc eu pcownacrednst r(aatsi oxn i)n fcorre eaxseasm),- pV=nRT andp al en, ewghaticivhe i fis rasnt admeroivuantti voef isnudbisctaatnecse t hdeiv oidpepdo bsiyt et.h Iet viso sluommee -it timeosc ccuopniveesn, cieannt b teo edxepnroestese tdh ue sfiinrsgt tdheer diveartiivveed a usn fi ′t(sx o).f Thmoel dsemc-−3 onda sd ear civoamtibvien, adti2of/nd xo2f, tohfe ab afsuen uctnioitns fiosr tahme oduenritv oatfi svue bosft atnhcee Here R is the This equation has the same form as the original, but the coeffi- cients a and b, which differ from gas to gas, have disappeared. It follows that if the isotherms are plotted in terms of the reduced viii Using the book variables (as we did in fact in Fig. 1C.8 without drawing attention to the fact), then the same curves are obtained whatever the gas. This is precisely the content of the principle of corresponding states, so the van der Waals equation is compatible ➤ Annotated equations and with it. equation labels Looking for too mIInnutteecgghrraa sllAAig..22nificance in this apparent triumph isww m==is−−tannkRRenTT, ∫∫beVVcffadduVVse oth==er eq−−uanntRRioTTnsll nnofVV sfftate also accommodate VVii VV VVii We have annotated many equations to help you follow how PPeerrffeecctt ggaass,, WWoorrkk ooff they are developed. An annotation can take you across the rreevveerrssiibbllee,, ((22AA..99)) Checklist of conceptsiissootthheerrmmaall eexxppaannssiioonn equals sign: it is a reminder of the substitution used, an approximation made, the terms that have been assumed ☐ 1. The extent of deviations from perfect behaviour is sum- ☐ 5. The constant, the integral used, and so on. An annotation can marized by introducing the compression factor. also be a reminder of the significance of an individual ☐ 2. The virial equation is an empirical extension of the per- one (a term in an expression. We sometimes color a collection of fect gas equation that summarizes the behaviour of real other (b gases over a range of conditions. ☐ 6. numbers or symbols to show how they carry from one line ☐ 3. The isotherms of a real gas introduce the concept of to the next. Many of the equations are labelled to highlight vapour pressure and critical behaviour. their significance. ☐ 4. A gas can be liquefied by pressure alone only if its tem- ☐ 7. perature is at or below its critical temperature. ➤ Checklists of equations 52 1 The properties of gases Checklist of equations You don’t have to memorize every equation in the text. A checklist faotr tahlle g aesnesd thoaft eaarec hd etsocrpibice ds ubym thme avrainz edse rt Whea amls oeqsut ation Prope1rty Equation Comment important enqeaura tthieo ncrsit icaanl dp oitnht.e W ce osneed firtoiomn Tsa bulen 1dCe.r2 twhaht iaclthh ough Compression factor 2.0 Z=Vm/Vm(cid:31) Definition they apply. Z <3=0.375, it is approximately constant (at 0.3) and the dis- Z0.8 crcepa8ncy is reasonably small. Vctor, irial equation of state pVm=RT(1+B/Vm+C/Vm3+(cid:30)) B, C a0.6 von fan der Waals equat1i.o2n of state p = nRT/(V –N nitbr)o –g ae(nn/V)2 a Brief illustration 1C.4 Criteria for perfect gas behaviour essi0.4 Methane b For benzene a = 18.57 atm dm6 mol−2 (1.882 Pa m6 mol−2) and Rmpreduced vari1a.b0les Xr = Xm/Xc Propane X = p, V, or b = 0.1193 dm3 mol−1 (1.193 × 10−4 m3 mol−1); its normal boil- Co0.2 ing point is 353 K. Treated as a perfect gas at T = 400 K and Ethene p = 1.0 atm, benzene vapour has a molar volume of V = RT/p = 0 m 0 1 2 3 4 5 6 7 33 dm mol−1, so the criterion Vm ≫ b for perfect gas behaviour Reduced pressure, p/pc is satisfied. It follows that a/V2≈0.017atm, which is 1.7 per m cent of 1.0 atm. Therefore, we can expect benzene vapour to Figure 1C.9 The compression factors of four of the gases deviate only slightly from perfect gas behaviour at this tem- shown in Fig. 1C.3 plotted using reduced variables. The curves Setting puerpatu rae anndd pr esssourel.ving problems are labelled with the reduced temperature Tr = T/Tc. The use of reduced variables organizes the data on to single curves. Self-test 1C.5 Can argon gas be treated as a perfect gas at 400 K and 3.0 atm? ➤ Brief illustrations Answer: Yes Brief illustration 1C.5 Corresponding states A Brief illustration shows you how to use equations or The critical constants of argon and carbon dioxide are given in concepts that have just been introduced in the text. They Table 1C.2. Suppose argon is at 23 atm and 200 K, its reduced help you to(c ) leTahrne phroinwc itpol eu osef cdoartrae, spmoanndipiunlgat set autneists pressure and temperature are then correctly, anAdn ibmepcoortmanet gfeanmerialli aterc hwniiqtuhe itnh esc imencaeg nfoirt ucodmeps aroinf g the 23atm 200K p = =0.48 T = =1.33 properties. Thproepye ratriees aolfl oabcjceoctms ips aton ciehdoo bsye aa rSeelaltfe-dte fsutn qduamesetnitoanl p rop- r 48.0atm r 150.7K which you cearnty uofs eth teo s ammoe nkiintod ra nydo tuor s eptr uopg ar ereslsa.t ive scale on that basis. For carbon dioxide to be in a corresponding state, its pressure We have seen that the critical constants are characteristic prop- and temperature would need to be erties of gases, so it may be that a scale can be set up by using them as yardsticks. We therefore introduce the dimens0i7o_nAtlkeinsss_ Ch01C.indd 5p3=0.48×(72.9atm)=35atm T=1.33×304.2K=405K reduced variables of a gas by dividing the actual variable by the corresponding critical constant: Self-test 1C.6 What would be the corresponding state of ammonia? V p T Answer: 53 atm, 539 K V = m p = T = Definition Reduced variables (1C.8) r V r p r T c c c If the reduced pressure of a gas is given, we can easily calcu- The van der Waals equation sheds some light on the princi- late its actual pressure by using p = pp, and likewise for the ple. First, we express eqn 1C.5b in terms of the reduced vari- r c volume and temperature. van der Waals, who first tried this ables, which gives procedure, hoped that gases confined to the same reduced volume, Vr, at the same reduced temperature, Tr, would exert the p p = RTrTc − a same reduced pressure, pr. The hope was largely fulfilled (Fig. r c VrVc−b Vr2Vc2 1C.9). The illustration shows the dependence of the compression factor on the reduced pressure for a variety of gases at Then we express the critical constants in terms of a and b by various reduced temperatures. The success of the procedure using eqn 1C.8: is strikingly clear: compare this graph with Fig. 1C.3, where of a gas are different in the initial and final states. Because S is a T to T state function, we are free to choose the most convenient path i f from the initial state to the final state, such as reUvesrinsigb lteh iseo tbhoeor-k ix ∆ (Step 2) mal expansion to the final volume, followed by reversible heating at constant volume to the final temperature. Then the total entropy change is the sum of the two contributions. changes, is ➤ Worked examples Example 3A.2 Calculating the entropy change for a ∆S nRln composite process Worked Examples are more detailed illustrations of the Calculate the entropy change when argon at 25 °C and 1.00 application of the material, which require you to assemble bar in a container of volume 0.500 dm3 is allowed to expand to and develop concepts and equations. We provide a sug- 1.000 dm3 and is simultaneously heated to 100 °C. and obtain gested method for solving the problem and then implement Method As remarked in the text, use reversible isothermal ∆S=piViln it to reach the answer. Worked examples are also accompa- expansion to the final volume, followed by reversible heat- Ti nied by Self-test questions. ing at constant volume to the final temperature. The entropy change in the first step is given by eqn 3A.16 and that of the second step, provided C is independent of temperature, by V (1.0 CeqnH 3AA.20 (PwitTh CEV iRn pl a3ce o f Cp). In each case we need to ∆S= know n, the amount of gas molecules, and can calculate it from the perfect gas equation and the data for the initial state =+0.173 from n = pV/RT. The molar heat capacity at constant volume i i i Aisssu gmive ethna tb ayll tghasee se aqrue ippearfretcti tainodn t hthate doarteam ref ears t o3 2R9.8 (.1Th5 Ke u enqleussi potahretriw-ise stated. 2 tion theorem is reliable for monatomic gases: for others and in general use experimental data like that in Tables 2C.1 and errors. ➤ Discussion questions 2C.2 of the Resource section, converting to the value at con- TstOantP voIlCum 3e bAy u siEngn thter oreplatyion C − C = R.) Self-test 3A.11 p,m V,m Discussion questions appear at the end of every chapter, Answer From eqn 3A.16 the entropy change in the isothermal where they are organized by topic. These questions are Dexispcaunssiosnio frno mq uVei tsot Viof inss designed to encourage you to reflect on the material you 3A.1 The evolution of life requires the organization of a very large number 3A.3 have just read, and to view it conceptually. of molecules into biological cells. Does the formation of living organisms violate the Second Law of thermodynamics? State your conclusion clearly 3A.4 and present detailed arguments to support it. ➤ Exercises and Problems 3A.2 Discuss the significance of the terms ‘dispersal’ and ‘disorder’ in the Why? Cconhteext cof kthel iSsecton od Lfa wc.oncepts Exercises and Problems are also provided at the end of every chapter, and organized by topic. They prompt you to test ☐Ex e1.r cThisee esntropy acts as a signpost of spontaneous change. ☐ 6. The your understanding of the topics in that chapter. Exercises ☐ 2. Entropy change is defined in terms of heat transactions 3A.1(a) During a hypothetical process, the entropy of a system increases by 3A.8(b) Calculate Δ are designed as relatively straightforward numerical tests 125 J K−1( wthheil eC tlhaeu esniturosp dy eofif tnhiet siuornro).undings decreases by 125 J K−1. Is the ☐pr oc3e.s s Tsphonet aBneoolutsz?mann formula defines absolute entro- 25 °C and 1.50 whereas the problems are more challenging. The Exercises 3A.1(b) Dpuireisn gi an h tyeprotmhest icoafl ptrhoece sns,u thme benetrro opyf owf aa syyss teomf iancchreiaesevsi bnyg a ☐of ΔS7?. come in related pairs, with final numerical answers avail- 105 J K−1c wohnifileg thuer eanttiroonpy. of the surroundings decreases by 95 J K−1. Is the 3A.9(a) Calculate Δ process spontaneous? able on the Book Companion Site for the ‘a’ questions. ☐ 4. The Carnot cycle is used to prove that entropy is a state ☐50 8. Final numerical answers to the odd-numbered problems 3A.2(a) Af ucenrctatiino inde.al heat engine uses water at the triple point as the hot source and an organic liquid as the cold sink. It withdraws 10.00 kJ of heat are also available on the Book Companion Site. ☐fro m5 t.h eTh hoet seoffiurcceie anndc yg eonfe raa these 3a.t0 0e nkJg oifn weo irsk .t Wheh abta iss itsh eo tfe mthpee rdateufiren oif- 3A.9(b) Calculate Δ the orgatniioc nliq oufid t?he thermodynamic temperature scale and one ☐10 0 9. 3A.2(b) Ar ecaerltiaziant iidoenal, htehaet eKngeilnvei nus secs awlaet.er at the triple point as the hot ➤ Integrated activities source and an organic liquid as the cold sink. It withdraws 2.71 kJ of heat from the hot source and generates 0.71 kJ of work. What is the temperature of the 3A.10(a) organic liquid? At the end of most chapters, you will find questions that 3A.3(a) Calculate the change in entropy when 100 kJ of energy is transferred gas of mass 14 cross several topics and chapters, and are designed to help reversibly and isothermally as heat to a large block of copper at (a) 0 °C, (b) 50 °C. you use your knowledge creatively in a variety of ways. 3A.3(b) Calculate the change in entropy when 250 kJ of energy is transferred 3A.10(b) Some of the questions refer to the Living Graphs on the reversibly and isothermally as heat to a large block of lead at (a) 20 °C, (b) 100 °C. Book Companion Site, which you will find helpful for 17_Atkins_Ch033AA.i.n4d(da ) W124hich of F2(g) and I2(g) is likely to have the higher standard molar to 4.60 dm3 entropy at 298 K? answering them. 3A.4(b) Which of H2O(g) and CO2(g) is likely to have the higher standard expansion. molar entropy at 298 K? ➤ Solutions manuals 3A.5(a) Calculate the change in entropy when 15 g of carbon dioxide gas is 3A.11(a) allowed to expand from 1.0 dm3 to 3.0 dm3 at 300 K. Two solutions manuals have been written by Charles Th3Ae.5 I(bn) sCtarlcuulcatteo trh’es c Shaonlgue tinio enntrso pMy wahnenu 4a.0l0 p g roof nvitirdogeesn fisu allllo wseodl utot ions surroundings. Trapp, Marshall Cady, and Carmen Giunta to accompany to thexep an‘bd ’f roemx 5e0r0c cims3e tso 7a50n cdm 3 taot 3 0t0h Ke. even-numbered problems 3A.11(b) this book. (avai3lAa.6b(al)e P rteodi cdt tohew ennthloalapyd o ff vraopomriz atthioen oBf boeonzken Ce foromm iptsa nnorimoanl Site for boiling point, 80.1 °C. The Student Solutions Manual (ISBN 1-4641-2449-3) regis3tAe.6r(be)d Pr eaddicot tphet eenrtsh aolpfy tohf vea pboroizoatkio no onf lcyyc)l.ohexane from its normal surroundings. provides full solutions to the ‘a’ exercises and to the odd- boiling point, 80.7 °C. 3A.12(a) numbered problems. 3A.7(a) Calculate the molar entropy of a constant-volume sample of neon at −10.0 500 K given that it is 146.22 J K−1 mol−1 at 298 K. of 1 3A.7(b) Calculate the molar entropy of a constant-volume sample of argon at 75.291 J K−1 mol−1 250 K given that it is 154.84 J K−1 mol−1 at 298 K. 3A.12(b) −12.0 3A.8(a) Calculate ΔS (for the system) when the state of 3.00 mol of perfect gas atoms, for which Cp,m = 25R, is changed from 25 °C and 1.00 atm to 125 °C and 1 5.00 atm. How do you rationalize the sign of ΔS?