物理化学-03-05Maxwell关系式

0上次课主要内容2.A及G的计算ΔA＝ΔU－Δ(TS)ΔG＝ΔH－Δ(TS)＝ΔU+Δ(pV)－Δ(TS)＝ΔA+Δ(pV)3.热力学基本方程dU=TdS–pdVdH=TdS+VdpdA=-SdT–pdVdG=-SdT+Vdp1.Helmholtz函数与Gibbsb函数TSUAdefpVATSpVUTSHG＝＋＝defBGGmfBmr1TSVAVUpTSpGpVHpVSHSUTpVTGTAS4.八个重要的关系式dU=TdS–pdVdH=TdS+VdpdA=-SdT–pdVdG=-SdT+Vdp24.麦克斯韦关系式yxxNyMVpSTUdddpVSTHdddVpTSAdddpVTSGdddVSSpVTpSSVpTVTTpVSpTTVpSNdyMdxdyyZdxxZdZxyxyZyxZ2235.热力学函数关系式的推导和证明TCTSVVTdTCTdUTQdSVVr对于恒容过程TCTSpp例14VpSTUdddpTpTpVSTVUVTTdVVUdTCdVVUdTTUdUTvTVdVpTpTdTCdUVVdpTVTVdTCdHpp同理例2VTfU,5例3的变化关系：或随VpSpTTVpSpTVV1定义等压膨胀系数21ppVdpS代入上式积分得：pnRTVVnRTpVp12212121VVlnnRpplnnRdppnRVdpSpppp6)(dmαG)(dmβGT→T+dTp→p+dp1.Clapeyron方程T,p)(GBαm)(GBβm平衡)(G)(GBααmmd)(G)(GBββmmd平衡)(G)(Gmmα)(G)(G)(G)(Gmmmmddαα§3.9克拉佩龙方程(Clapeyronequation)7d)(d)()(dmmmpVTSGαααd)(d)()(dmmmpVTSGβββpVTSpVTSd)(d)(d)(d)(mmmmββααmmmmmmmm)()()()(ddVTHVSVVSSTp相变相变相变相变αβαβmmddVTHTp相变相变)(dmαG)(dmβG82mvapmmvap)g(ddRTpHTVHTpCRTHPapmvapln2.Clausius-Clapeyron方程将Clapeyron方程用于气—液及气—固平衡pRTgVlVgVVmmmm)()()(2mvapddlnRTHTp12mvap1211TTRHppln9*克—克方程的应用条件：1.因ΔVm≈Vg,在靠近临界点时不能用。2.Vm(g)=RT/p适用于理想气体，高压下不能用。3.因将ΔHm当常数，只有在温度变化不大时才适用。CKTBAPap/ln安托万(Antoine)方程是对克—克方程最简单的改进103.外压对液体蒸气压的影响蒸气液体加压)()(gGlGmm)()(),(),(gdGldGGgdpldpmmm及变化引起蒸气压力改变恒温下液体压力改变)()()()(gdGgGldGlGmmmm)()(gdGldGmm即VdpVdpSdTdGm恒温下)()()()(gdpgVldplVmm则)()()()(gVlVldpgdpmm11)()()()(gVlVldpgdpmm若蒸气为理想气体RTlVldpgpdgpRTgVmm)()()(ln)()()]()([)()()(ln1212lplpRTlVgpgpm12例酚的精制采取减压蒸馏方法。已知酚的正常沸点为181.9℃，如真空度为86.7kPa，酚的沸点应为多少？已知酚的蒸为，外压为100.0kPa。13molJ101.48解：kPa31011.p*T1，p1T2，p2△HK1455K227391811.)..(TkPa313kPa78601002.)..(p*13mvapmolJ10148.HT2=?14551K13145810148310131323./T..)..ln(T2=392.4K→119.2℃13例：已知纯A液体在360K的饱和蒸气压为81.06kPa,在此条件下,A(l)的摩尔气化热ΔvapHm=40kJ·mol-1.Cp,m(l)=75kJ·mol-1K-1Cp,m(g)=(30+10-2T/K)J·mol-1K-1Smθ(g,380K)=174.35J·K-1·mol-1假定A(g)为理想气体,忽略温度的变化对A(l)体积的影响.试计算下列始、末状态之间的ΔUm、ΔHm、ΔSm、ΔGm及ΔAmA(l,310K,81.06kPa)→A(g,380K,50.6625kPa)14解:A(l)T1=310Kp1=81.06kPaA(l)T2=360Kp2=p1A(g)T3=T2p3=p2A(g)T4=380Kp4=50.6625kPaΔH1=Cp,m(l)(T2－T1)=[75×(360－310)]J·mol-1=3750J·mol-1ΔH2=ΔvapHm=40kJ·mol-11380263123674103043molJdTmolJ)K/T(dT)g(CHKKTTm,pΔHm=ΔH1+ΔH2+ΔH3=44.42J·mol-1Δ(pV)=p4V4－p1V1≈p4V4=nRT4=3.159×103J·mol-1ΔU=ΔH－Δ(pV)=41.265kJ·mol-1ΔS1=Cp,m(l)ln(T2/T1)=11.215J·mol-1K-1ΔS2=ΔvapHm/T2=111.111J·mol-1K-11231543433TTm,ppplnRdTTCS1111433423473051030KmolJ.KmolJpplnRTTTTlnΔSm=ΔS1+ΔS2+ΔS3=128.06J·mol-1K-1S4=Smθ(g,380K)+ΔS(pθ→p4)=Smθ(g,380K)+Rln(pθ/p4)=180.00J·mol-1K-1S1=S4-ΔSm=51.91J·mol-1K-1Δ(TSm)=T4S4－T1S1=52.299J·mol-1ΔGm=ΔHm-Δ(TSm)=-7.875kJ·mol-1ΔAm=ΔUm-Δ(TSm)=-11.034kJ·mol-116销钉p(环)=101.325kPaA2mol400K1013.25kPa1013.25kPa400K2molB绝热活塞固定导热隔板如图所示，一带活塞(无质量，绝热)的绝热汽缸中放有一固定导热良好刚性隔板，此隔板将气缸分成左，右两室并分别放入A,B两种理想气体。开始时活塞用销钉固定，当销钉去掉后，活塞移动。当系统达平衡时，求此过程的W、H和S。A(g)与B(g)的Cv,m=3R/2。解：TpA2molT101.325kPa2molB分析：（1）过程绝热。（2）A气体体积不变。（3）A气体和B气体处于热平衡。A2mol400K1013.25kPa1013.25kPa400K2molB17WWQUUUBAJ.J)(.)KT(RUW564489400310314864006)pRTnpRTn(p)KT()B(Cn)KT()A(CnBexBexm,VBm,VA11400400(K)3108024006TRRT)KT(R)KJ.J..ln.ln.pplnRnTTln)CC(npplnRnTTlnCnTTlnCnSSSBm,pm,VBBm,pBm,VABA-12112211212(33213251012510133148240031031488(J)6.7482)400310(254,KKRTnCHHHmpBA