CramX Logo
RegisterLog in
Solution Manual For Thermal Physics, 2nd Edition - Document preview page 1

Solution Manual For Thermal Physics, 2nd Edition - Page 1

Document preview content for Solution Manual For Thermal Physics, 2nd Edition

Solution Manual For Thermal Physics, 2nd Edition

Solve your textbook questions faster and more accurately with Solution Manual For Thermal Physics, 2nd Edition, a reliable guide for every chapter.

Michael Davis
Contributor
4.3
0
12 months ago
Preview (16 of 151 Pages)
100%
Log in to unlock
Page 1 of 16
Solution Manual For Thermal Physics, 2nd Edition - Page 1 preview imageCHAPTER22.1)ENTROPYANDTEMPERATURE(a)o=logg=logC+(3N/2)logU.(80/3U)y=1/1=(3N/2)(1/U),henceU=3Nt/2.(®)(2%0/00%)N=(38/2)(-1/U%)<0.2.2)PARAMAGNETISMStateswithspinexcess2shavetheenergyU=-2smB,hences=-U/2mB.Insertioninto(40)yields(41).From(41)obtain(3a/3U)yB=1/t==U/m?B2N.SolveforU:U=<U>==n2B2N/t=-MB.HenceM/Nm=mB/t.2.3)QUANTUMHARMONICOSCILLATOR(a)From(1.55):0=logg=log(N+n-1)!=logn!-log(N-1)!=log(N+n)!-logn!-logN!ReplacingN-1byNisequivalenttoneglectingaterm~log((N+n)/N},whichforlargeNissmallcomparedtothefactorialterms.WiththeStirlingapproximation:o=(N+n)log(N+n)-nlogn-NlogN=(N+n)log[(N+n)/N]-nlog(n/N)=N[(1+n/N)log(l+n/N)-(n/N)log(n/N)].(b)Setn=U/MwandNfw=Uy(N):0(UN)=N[(1+U/U))log(1+U/U)=(U/Uy)log(u/ug)].-3-[2.3]—StudyXY
Page 2 of 16
Solution Manual For Thermal Physics, 2nd Edition - Page 2 preview imageDownloadedfromStudyXY.com®+StudyXYSdYe.o>\|iF’prE\3SStudyAnythingThisContentHasbeenPostedOnStudyXY.comassupplementarylearningmaterial.StudyXYdoesnotendroseanyuniversity,collegeorpublisher.Allmaterialspostedareundertheliabilityofthecontributors.wv8)www.studyxy.com
Page 3 of 16
Solution Manual For Thermal Physics, 2nd Edition - Page 3 preview imageWiththeabbreviationx=U/U,:1/t=(30/30)=(1/04)(30/3%)=(N/Uy)[log(1+x)-1logx]=(1/Mw)log(1+Uy/U).solvingforU/NyieldsU/N=Juw/[exp(Mw/t1)-1].Alternatemethod:Forlargen:d(logn!)/dn=logn!-log(n-1)!=logn.Withthis,1/t=(1/Mw)(3a/3n)y=(1/Hw)[log(N+n)-logn]=(1/Mw)log(1+N/n),whichistheequivalenttotheresultabove.2.4)THEMEANINGOFNEVER(a)Theprobabilityacorrectkeywillbestruckis1/44.Theprobabilityasequenceof10°keyswillbecorrectis(1/44)100,000=107164,345(b)Onasingletypewriterthenumberofkeysthatcanbestruckin10%8sat10keys/sis101°.TheHamletsequencemaystartwithanyoftheseexceptthelast10%-1.Thusthereare10%-10%+1=101%possiblestarts.TheprobabilityofacorrectHamletsequenceononetypewriteris10195707164,345_|-164,326For1010monkeys:1010x7~164,326.107164,316[2.3]-4-|—StudyXY
Page 4 of 16
Solution Manual For Thermal Physics, 2nd Edition - Page 4 preview imageComment.OneUCSBstudentsuggestedthatthemonkeysbepermittedatleastasmanytypographicalerrorsasinthelecturenotesfromwhichthemaintextwasprepared.Itisnotdifficulttoshowthateven1000errorswouldhaveincreasedtheprobabilitybylessthanafactor(43x10%)1000M106633toavaluethatisstillnegligible.2.5)ADDITIVITYOFENTROPYFORTWOSPINSYSTEMS(a)From(17),withN=N,=102%,6=101%;=2-919;=(979)ax®XP(-467/N1)=(979,)1,*exp(-4),(9195)/(979,pax=0-0183.Comparethiswith107174for6=10%2.(b)From(1.35):a(N,s)=(2/n)%x2Nxexp(-282/M).Thismaybeappliedbothtothetwoindividualsystems(N=Ny=Ny),andtothecombinedsystem(N=N+Ny).Forthetwoindividualsystemsintheirmostprobableconfigurationwehavefrom(14),§;=§,.Hence(919)dma=190481%=(27m)x2°Lxexp(-a8,2/m;)9192)max=[90,5010=1P(=45)Np)Forthecombinedsystem,s=28):29,(Npi5))g,(Ny5-51)=g(2N,281)S12N.1a2=@/mip)%x2txexp(-a3,2M)_5-[2.5]|—StudyXY
Page 5 of 16
Solution Manual For Thermal Physics, 2nd Edition - Page 5 preview imageThetworesultsdifferbythefactor9(2M,28,)/[9(8),81%=(N)/20)%=ax10t0(c)Thetrueentropyoftheinteractingcombinedsystemiso=logg(2N),25))=2M)log2~as,2/m,-log(nN,)x1.3863x10%%=1018=25.9=1.3862x10%2Theerrormadeintheentropybyestimatingtheentropyas109(9;9,)pay=09%0,islog(4x10'%)=24.4.Thisisafractionalerrorof1.76x1072L.2.6)INTEGRATEDDEVIATIONForNy=Ny.(17)becomes2919;=(919)paySXP(-467/Np)®.me2919%(919DayJSxwi-262/mya010,©=(919)payN/0)%[exp(-thatXwherex=26/N%.Wecountbothpositiveandnegative§with[8]>6).Then®®(18156)=[exp(-t?)at/[exp(-t?)atx[Jx1(NM52=(2/n%)erfc(x)=bonyFaexp(-46,/Ny)=0.28xexp(-400"=exp(-403.6)=107273|[2.5]-6-|—StudyXY
Page 6 of 16
Solution Manual For Thermal Physics, 2nd Edition - Page 6 preview imageCHAPTER33.1)FREEENERGYFORTWOSTATESYSTEM(a)Byinserting(13)into(55):F=-1tlog2=-tlog[l+exp(-c/1)].(b)Insertedinto(49),o=-(3F/31)=log[l+exp(-e/t)]+(¢/1)/[1+exp(e/1)],(81)U=F+to=¢/[l+exp(e/1)],thesameas(14)obtaineddirectlyfromZz.Comment.Ast»0andhenceexp(-£/t)»0,thelogarithmin(S1)maybeexpanded:log[l+exp(-e/t)]+exp(-£/1).Theno»(1l+e/t)exp(-¢£/1)»0.Theexponentialfactorgoestozerofasterthananyinversepowerof1goestoinfinity:BothoandallderivativesofovanishwhentT+»0,asshowninFig.11.Thehigh-temperaturelimitofoisobtainedbylettingexp(-£/t)>1,in(Sl):o>log2+e/2t»log2.3.2)MAGNETICSUSCEPTIBILITY(a)Forasinglemagnet,with&£=mB:zy=exp(mB/t)+exp(-mB/t)=2cosh(mB/t),(s1)<m>=[mexp(+mB/1)-mexp(-mB/1)1/2;=mtanh(mB/t).ThemagnetizationMisobtainedbymultiplyingbytheparticleconcentrationn:M=n<m>=nmtanh(mB/t).Forweakfields,mB<<t:—2.M=nm°B/t(s2)_7-[3.2]|—StudyXY
Page 7 of 16
Solution Manual For Thermal Physics, 2nd Edition - Page 7 preview imageForstrongfields,mB>>1:M=nm.ThesetwolimitsareseeninFig.3.12.(b)Inserting(S1)into(55)andmultiplyingbyn:F=-ntlog2q=-ntlog[2cosh(mB/t)].Toexpresscosh(mB/t)asafunctionofx=M/mm=tanh(mB/1)=tanhyweusetherelation1/cosh?y=(cosh?y-sinh®y)/cosh?y=1-tanh?y=1-x2.Wenextwritelog(2coshy]=-%log(1/4cosh®y)=-%log[(1-x%)/4].Withthis:F=+(n/2)logl(1-x")/4](c)Thesusceptibilityisdefinedasx=dM/dB.InthelimitmB<<t,from(S2):xX=nm?/t.3.3)FREEENERGYOFANHARMONICOSCILLATORThepartitionfunctionisageometricseries:z=poyexp(-sjw/t)=1/[1-exp(-Hw/1)].(a)Insertedinto(55):F=-1logZ=tlog[l-exp(-fuw/t)].(87)[3.2]-8-
Page 8 of 16
Solution Manual For Thermal Physics, 2nd Edition - Page 8 preview imageAthightemperatures,Mw/t<<1,sothatl-exp(-Mw/1)=Yw/t.Hencefrom(S1):F=tlog(fw/t).(s2)(b)Theexpression(88)followsdirectlybyinserting(87)into(49),o=-(3F/3T1).Comment.Thelow-temperature(1<<Mw)behavioroftheharmonicoscillatoristhesameasforthetwostatesystemwith¢=Jw,asisapparentfromcomparingFigs.3.13and3.14withFigs.3.11and3.4:Onlythetwoloweststatesmatter.Thehigh-temperaturebehavior(tv>>Mw)isquitedifferent,becausethenumberofacces-siblestatesisnotlimitedto2.Inthislimit,from(s2):0==(3F/31t)»1+log(t/Hw).Ifthisisinsertedinto(17a):Cy=1(da/31)>»1,infundamentalunits.3.4)ENERGYFLUCTUATIONSNotefirstthat<(e-<e>)2>=<6?—2g<e>+<e5%>=eZ—<g>|NextwritethepartitionfunctionasafunctionofBp=1/t:2=2exp(-Be,).2Then=9-[3.4]|—StudyXY
Page 9 of 16
Solution Manual For Thermal Physics, 2nd Edition - Page 9 preview image==1-=-1ladzU=<g>=zeexp(-Bey)=-75w?=156?exp(-pe,)=+=ozz45%2Z452Fromthese,withdg/dt=-1/12,2(3)Ze?(&)a8(3),at),3p),dr3p“4(5%)spfz1(=fdp\zapZgg2TZ\@=<e?=<>?Comment.Manipulationsinvolvingtemperaturediffer-entialsofthepartitionfunctionsarefrequentlysimpli-fiedbyusingB=1/1ratherthan1asindependentvari-able.3.5)OVERHAUSEREFFECTLetU,betheenergyofthereservoirwhentheenergyofthesystemiszero.Then,whenthesystemhastheenergye,thereservoirhas,byoursupposition,theenergyUg-&+ag=Uy-(1-x)e.TheprobabilityP(e)tofindthesysteminaparticularstatewithenergyegisthenproportionaltothenumberofstatesofthereservoirwiththeenergyUy-(1-a)eg,Peg)=gplUg-(1-a)e],withthesameproportionalityfactorforallstates.Hence,insteadof(2)and(3):Bley)_9g[Ug-(1-a)e,}_exp{op[Uy-(1-a)e,1}(51)B(e,)=9p(Uy-(1-a)e,]=expfoglUg-(1-a)e,11°IftheentropyofthereservoirisexpandedaboutU=Uysasin(7):[3.4]-10-|—StudyXY
Page 10 of 16
Solution Manual For Thermal Physics, 2nd Edition - Page 10 preview image0plUg=(1-a)e]=0,(Uy)=(305/8U)(1-a)e+...Ifthisisinsertedinto(S1),oneobtains,insteadof(9),P(e)expl-(l-a)e,/1]P(ey)expl-(1l-a)e,/t]whichisequivalentto(91).3.6)ROTATIONOFDIATOMICMOLECULE(a)Thereare2j+1statesatenergyj(i+l)c,,henceEN=Texploeg/)=z(23+1)expl[-i(i+l)eg/t]wherethesumoverallstateshasbeenconvertedintoasumoverallenergylevels.(b)Thesummaybeviewedasasumovertheareasofrectangleswiththewidthaj=1andwiththeheight£5=(23+1)exp[-j(i+l)eg/T],asshownbelow.Alsoshownisthecurvef(j),withjtreatedasacontinuousratherthandiscretevariable.Notethatthiscurvegoesthroughthemiddleoftheupperedgeofeachrectangle.£.,£03)3JAT]eg/T=0.1/ANN/|ANN/n<J:a3012345S11[3.6]|\—StudyXY
Page 11 of 16
Solution Manual For Thermal Physics, 2nd Edition - Page 11 preview imageWhen©>>£qrtheslopeofthecontinuouscurvef(j)variesonlyverylittlewithineachrectangle,andthesumcanbeapproximatedbytheintegral©z=/£(3)a3=kNotethatthelowerintegrationlimitisnotzero;thisisimportant.Substitutenewvariable:x=JG+e/T,dx=(2+1)(eo/1)dj:wz=[exp(-x)ax=I[1-exp(-x,)1(s1)EyEr0*owherex,=-e,/41isthevalueofxcorrespondingtoj=-1/2.Whent>>eg,X,<<1,andtheexponentialmaybeexpanded:exp(-x4)=z1-XoFoeIfthisisinsertedinto(S1),Z=t/ey+1/4+cee4(s2)whereallomittedtermsdecreaseatleastasrapidlyas1/twhen1>@.Amorecarefultreatmentofthesum(seebelow)yieldstheslightlydifferentresultZ2=t/e,+1/3+Lo..(s3)Thisistheformwewilluseinwhatfollows.(€)2=1+3exp(-2¢0/1)+...(54)wheretheomittedtermsvanishmorerapidlythanthesecondtermwhent>0.(d)Insertzinto(12),U(1)=t2(92/91)/2:[3.6]-12-|—StudyXY
Page 12 of 16
Solution Manual For Thermal Physics, 2nd Edition - Page 12 preview image—2—Treg:U(r)=©£02=T/(1+ey/31)HTx(l-ey/31)=1-£o/3.(85)cv)=1.TeyUT)=begexp(-2¢0/T)/2=beyexp(-2¢,/1)<<6eB(s6)cr)=(12¢4/19)exp(-2e0/T)Notethat,asinProblems3.1and3.3,theexponentialfactorgoestozeromuchfasterthan1/t°goestoin-finity,when1>0.(e)Seedrawingsonnextpage.Theapproximation(S6)dipsbelow(S5),suggestingthatU(t)approachestheasymptote($5)frombelow.ThismeansthatC=3aU/dt>1abovesometemperature,thatis,theheatcapacitygoesthroughamaximum.Anaccuratecalculationconfirmsthisprediction.TheinstructorshouldcheckthatthestudentdrawingsshowcorrectlythevanishingslopeofbothU(r)andC(t)ast+0,ratherthanexhibitinganon-descriptbehavior.DiscussionandElaborations.Unlessforewarned,manystudentsarelikelytoreplacethesumbyanintegralwithalowerintegrationlimitofzero.Asthegraphsof£5and£(j)show,thisisinvalid.Thispitfallmaybetakenasthepointofdepartureforaclassroomdiscussiononseveralmathematicalpointsconcerningthesummationofseries,startingwithawarningagainstthepurelyformalreplacementofsumsbyintegralswithoutagraphicalvisualizationofthedifferencebetweenthetwo.Thedifferencebetween(S2)and(S3)arisesfromthenon-zerosecondderivativeofthecontinuousfunction£(j),whichintegratestoafiniteerrorin(S2).Theformaltoolforamoresystematicaccuratereplacementof-13-[3.61|—Study
Page 13 of 16
Solution Manual For Thermal Physics, 2nd Edition - Page 13 preview imageuU=rpd£9ydJSJVipyYVaI—U=1-5/3//bs£92eqAllderivativesarezeroCyLeITT.py2¢eyAllderivativesarezero[3.61“14-+StudyXy
Page 14 of 16
Solution Manual For Thermal Physics, 2nd Edition - Page 14 preview imagesumsbyintegralsistheEuler-McLaurinsummingformula.TheSupplementaryMaterialtothisChaptergivesasimplederivationthatdoesnotdrawonthepropertiesoftheEulerpolynomials,whicharecommonlyutilizedintheproofsfoundinmosttextbooks.IftheEuler-McLaurinexpansionisappliedtoourparti-tionfunction,onecanobtainanexpansionofZbypowersofn=£o/Tr1,1,02Z=g*t3+igrom)(57)whereO(n?)standsforomittedtermsofordern2orhigh-er.Toobtainallcontributionsofordern,theEuler-McLaurinexpansionmustbecarrieduptotheterm(1/720)£3)(0).From(S7)oneobtains,bystandardmethods,_22U(r)==&5d(logz)/dn=1=24/3=£,°/45T+O(1/17)or)=1+(1/85)(e/0)7+01/3)Thedetailsarelefttothereader.TheseresultsdoindeedindicatethatU(1)approachesitsasymptotefrombelowandthatC(1)approachesitslimitingvaluefromabove.Asanalternativetotheseanalyticaltreatments,itisnotdifficulttosumthepartitionfunctionnumericallyonaprogrammablecalculator.ThesummationisgreatlyspeededupbycalculatingtheBoltzmannfactorsrecursive-ly:Bj=exp[-j(i*l)eg/1]=ByjxC5C5=exp[-2jey/t]=C51xByBy=Cy=1.-15-[3.61|—StudyXY
Page 15 of 16
Solution Manual For Thermal Physics, 2nd Edition - Page 15 preview imageInthiswayonlyoneexponentialmustbeevaluatedforeach1;theremainingexponentialsintheseriesfollowbysimplemultiplications,atalargesavingincalculationtime.If2issummednumerically,theenergyismostconve-nientlyobtainedfromthefirstequalityin(3.12),byalsosummingu=1&_exp(-e_/1)TzzseXP(eg/T-Inthiswaythedifferentiationofnumericaldataisavoided.similarly,theheatcapacitycanbeobtainedwithoutdifferentiationfrom=_1(12--v7c=52[izexp(-¢/1)v?].3.7)ZIPPERPROBLEM(a)Astateinwhichslinksareopencanberealizedinonlyoneway.ThusthepartitionfunctionisZ=1+exp(-¢/t)+exp(-2¢/t)+...+exp(~-Ne/1).NN+1=3x®=BEwherex=exp(-e/1)(93)2xs=0(b)TheaveragenumberofopenlinksisN=1Ss_da<s>=2Ysx=x3xlogz.(s1)s=0N+1If¢>>tr,thenx<<1,andwemayneglectthetermxin(93)toobtain<>=-xL1og(1-x)=X==1/[exp(e/1)-1]dxI-x:ThisisoftheformofthePlanckdistribution.[3.61-16-|—StudyXY
Page 16 of 16
Solution Manual For Thermal Physics, 2nd Edition - Page 16 preview imageExtension.Ourassumptionthateachlinkhasonlyoneopenstateisanunrealisticassumption,whichneglectsthatthetwohalvesofanopenlinkmayhavemanydiffer-entorientationsrelativetoeachother.Itisinstruc-tivetogeneralizetheproblembyassumingthateachopenlinkhasGopenstateswiththeenergy&.Thechangehasfar-reachingconsequences.AstateofthezipperwithsopenlinksisthenG°-folddegenerate,andthepartitionfunctionnowbecomesZ=1+Gexp(-£/1)+Glexp(-2e/1)+...NN+1coo+Mexp(anesr)=3xS=IEZ-xs=0whereXx=Gexp(-g/t),whichdiffersfromtheearlierformonlybythefactorG>1inthedefinitionofx.Becauseofthisfactor,valuesx>1arenowpossible.Thishasdrasticconse-quencesifthetotalnumberoflinksisverylarge,N>>1.InthiscasetheopeningofthezipperapproachesthebehaviorofanabruptphasetransitionatthesharptransitiontemperatureTy=€/logGwhichis‘thetemperatureforwhichx=1.Fortempera-turesverylittlebelowt,,onlyaverysmallfractionofthelinksareopen,fortemperaturesverylittleaboveTyalmostalllinksareopen.ThelargerN,thenarrowerthetemperatureintervaloverwhichtheopeningtakesplace.Wegivehereonlythekeypointsinthederivationofthisresult.Itisnotdifficulttoshowthattheexpression(S1)for<s>canbewrittenas_17-[3.71|—Study
Preview Mode

This document has 151 pages. Sign in to access the full document!

Study Now!

X-Copilot AI
Unlimited Access
Secure Payment
Instant Access
24/7 Support
Document Chat

Document Details

Subject
Physics
Textbook
Thermal Physics by Charles Kittel, ...

Related Documents

View all
Quick Guides

Physics – Waves and Sound

Quick Guides

Physics – Thermodynamics

Quick Guides

Physics – Modern Physics

Quick Guides

Physics – Light

Quick Guides

Physics – Electricity and Magnetism

Quick Guides

Physics – Classical Mechanics

Class Notes

Exploring Light Intensity Through Polarizers

Class Notes

IPL Lab Report

Solved Assignments

Distance Time Velocity Gizmo Handout-2

Solved Assignments

Energy Skate Park Activity