第二章热力学第一定律习题物理化学天津大学第五版上课专用课件

习题2.1习题2.2习题2.3习题2.4习题2.5途径a途径bW=¡p(V2¡V1)=¡nR¢T=¡8:314JW=¡[p2V2(l)¡p1V1(g)]¼p1V1(g)=RT=8:314£(273:15+100)=3:102kJW=¡[p2V2(H2;O2;g)¡p1V1(H2O;l)]¼p2V2(g)=XnBRT=(1+0:5)£8:314£(273:15+25)=3:718kJ¢U=Qb+Wb=Qa+Wa=2:078¡4:157=¡2:079kJ)Wb=¢U¡Qb=¡2:079¡(¡0:692)=¡1:387kJ5molT0=25oCP=200kPaV05molT1=-28.57oCP1=100kPaV5molT2p=200kPaV绝热膨胀恒容加热5molT0=25oCP=200kPaV05molT2P=200kPaV恒压加热习题2.6习题2.7(1)(2)习题2.8习题2.9¢U=Qb+Wb=Qa+Wa=25:42¡5:57=19:85kJWb=¡p¢V=¡p³nRT1p1¡nRT0p0´=¡nRp³T1p1¡T0p0´=¡5£8:314£200£103£³244:58100£103¡298:25200£103´=¡7:940kJ)Wb=¢U¡Qb=19:85+7:940=27:79kJH=U+pV)¢H¡¢U=¢(pV))¢H¡¢U=¢(pV)=¢(nRT)=nR¢T=4£8:314£20=665:1J¢H=¢U+¢(pV)¼¢(pV)=V¢p¢H¼V¢p=18:02£10¡3997:04£100£103=1:8J¢H¼V¢p=18:02£10¡3997:04£(106¡100£103)=16:3J¢U=Q+W=nCV;m¢TW=0)¢U=Q=nCV;m¢T=5£1:5£8:314£50=3:118kJ¢H=nCp;m¢T=n(CV;m+R)=5£2:5£8:314£50=5:196kJ¢U=Q+W=nCV;m¢T=5£2:5£8:314£(¡50)=¡5:196kJW=¡p¢V=¡nR¢T=¡5£8:314£(¡50)=2:079kJ¢H=Q=¢U¡W=¡5:196¡2:079=¡7:275kJ习题2.10习题2.11习题2.12p0V0=p2V2)T0=T2)¢U=0;¢H=0W=¡pe¢V=¡200£103£(25¡50)£10¡3=5:00kJ¢U=Q+W=0)Q=¡W=¡5:00kJ¢U=nCV;m¢T=1£20:92£(97¡27)=1:464kJW=¡p1¢V=¡p1³RT2p2¡RT0p0´=2:497kJp1=p2T0T2=250:00£103£300:15370:15=202:7kPa¢U=Q+W)Q=¢U¡W=¡1:033kJ¢H=nCp;m¢T=n(CV;m+R)¢T=1£29:23£(97¡27)=2:046kJCp;m=RT2T1Cp;mdTT2¡T1=R800300(26:75+42:258£10¡3T¡14:25£10¡6T2)dT800¡300=(26:75T+1242:258£10¡3T2¡1314:25£10¡6T3)¯¯¯800300500=45:38J¢mol¡1¢K¡1¢H=nCp;m¢T=10344:01£45:38£500=515:6kJ习题2.15恒容绝热过程习题2.16把水煤气和水看作一个系统，则此系统与环境的热交换Q＝0习题2.17等压绝热过程¢U=QV=n(Ar;g)CV;m(Ar;g)¢T(Ar;g)+n(Cu;s)CV;m(Cu;s)¢T(Cu;s)=0¢T(Ar;g)=T;¢T(Cu;s)=T¡150)T=n(Cu;s)CV;m(Cu;s)T(Cu;s)n(Ar;g)CV;m(Ar;g)+n(Cu;s)CV;m(Cu;s)=74:23oC¢H=n(Ar;g)Cp;m(Ar;g)¢T(Ar;g)+n(Cu;s)Cp;m(Cu;s)¢T(Cu;s)=2:47kJCV;m(Ar;g)=Cp;m(Ar;g)¡R;CV;m(Cu;s)¼Cp;m(Cu;s)n(CO)=n(H2)=300£103M(CO)+M(H2)=9990molQ=n(CO;g)ZT2T1Cp;m(CO;g)dT+n(H2;g)ZT2T1Cp;m(H2;g)dT+m(H2O;l)Cp(H2O;l)¢T(H2O;l)=0m(H2O;l)=¡n(CO;g)RT2T1Cp;m(CO;g)dT+n(H2;g)RT2T1Cp;m(H2;g)dTCp(H2O;l)¢T(H2O;l)=2863:8kg¢U=X¢UB=XyBnCV;m(B)¢T=3£32R¢T+2£52R¢T=192R(T2¡T1)¢H=X¢HB=XyBnCp;m(B)¢T=¢U+¢(pV)=¢U+nR¢T=292R¢T=¡8:315kJW=¡p2¢V=¡p2(V2¡V1)=¡nRT2+p2p1nRT1=52RT1¡5RT2¢U=Q+W=W)T2=2429T1=331:0K¢U=Q+W=W=192R¢T=¡5:448kJ习题2.18恒压绝热过程习题2.20可逆相变习题2.21（1）（2）¢H=X¢HB=XnBCp;m(B)¢T(B)=0)nACp;m(A)¢T(A)+nBCp;m(B)¢T(B)=02£52R[T¡T1(A)]+5£72R[T¡T1(B)]=0)T=350:9KW=¢U=X¢UB=XnBCV;m(B)¢T(B)=¡369:2JQp=¢H=¡n¢vapHm=¡10318:02£40:668=¡2257kJW¼¡peV=nRT=172:2kJ¢U=W+Q=¡2085kJQp=¢H=n¢vapHm=10318:02£40:668=2257kJW¼¡peV=¡nRT=¡172:2kJ¢U=W+Q=2085kJW=0¢H=2257kJQV=¢U=¢H¡¢(pV)=2085kJ习题2.22习题2.23¢H=¢H1+¢H2=0=¡n¢vapHm+nCp;m(H2O;l)¢T(H2O;l)+mCp(Cu;s)¢T(Cu;s)=¡3:9118:02£40:668£103+3:9118:02£75:32(97:6¡100)+212£Cp(Cu;s)(97:6¡15)Cp(Cu;s)=501:7J¢kg¡1¢K¡1