物理化学计算题1

一、1mol理想气体由1013.25kPa，5dm3，609.4K反抗恒外压101.325kPa膨胀至40dm3，压力等于外压，求此过程的W、Q、△U、△H、△S、△A、△G。已知CV,m=12.5J·mol－1·K－1；始态熵值S1=200J·K－1。解：KKTVpVpT5.4874.6095401011112223332111,2111,21101.325104053.546103.546112.5487.5609.41.524112.58.314487.5609.42.5371.5243.5462.022ambVmpmWpVVPamJkJUnCTTmolJmolKKkJHnCTTmolJmolKKkJQUWkJkJkJ1111122,111121487.51112.58.314ln8.314ln14.67609.41014.67200214.67pmTpSnClnnRlnmolJmolKJmolKJKTpSSSJKJK322112.53710487.5214.5609.420014.77GHTSTSJkJ二、始态为T1=300K，p1=200kPa的某双原子理想气体1mol，经先绝热可逆膨胀使压力降到100kPa，再恒压加热到T2=300K，p2=100kPa的终态，求整个过程的Q，△U，△H，△S，△G。解：JKJKSTHGKJmolRlnppnRlnSHU1729762.53000762.51210;01121JKRmolTTnCHQQKKTppTmp15681.2463002711.246300223,224.14.1111212四、甲醇（CH3OH）在101.325kPa下的沸点（正常沸点）为64.65℃，在此条件下的摩尔蒸发焓ΔvapHm=35.32kJ·mol-1。求在上述温度压力条件下，1kg液态甲醇全部成为甲醇蒸气时的Q、W、ΔU、ΔH，ΔS及ΔG。视甲醇蒸气为理想气体。解：根据题意可知，此蒸发过程是在相变温度及其平衡压力下进行的可逆相变化过程，若视甲醇蒸气为理想气体，且忽略液态甲醇的体积，则有kJkJHMmHnHQmvapmvapp30.1102)32.35321000(kJJRTgnlVgVpW65.87)]15.27365.64(3145.8321000[)()]()([kJkJWQU65.1014)65.8730.1102(1263.3)15.27365.64/(30.1102/KkJKkJTQSr0G等温恒外压δQr=01mol理想气体T1=300Kp1=200kPa1mol理想气体T2=P2=100kPa1mol理想气体T3=300KP3=200kPa三、1mol，25℃、pӨ的过冷水蒸气在pӨ下变为同温同压下的水，求过程的W，ΔU，ΔH，ΔS，ΔG。已知25℃时水的饱和蒸汽压力为3167Pa；25℃时水的凝聚焓1mH43.83kJmol。解：kJHHHkJmolkJmolHnHHVap83.43;083.4383.431;0211相变相变121lnppnRTG0相G（水的可逆相变过程）02G(纯凝聚相物质的G随压力变化很小)21GGGG相J855800ln12ppnRT118.17515.298855843830KJKJTGHS1298.152479glgWpVpVVpVnRTmolRKJ43830247941351gUHpVHpVHnRTJJ六、苯的正常沸点353K下的△vapHm=30.77kJ·mol－1，今将353K及101.325kPa下的1mol苯液体向真空蒸发为同温同压的苯蒸气（设为理想气体）。（1）试求算在此过程中苯吸收的热Q与做的功W；（2）求苯的摩尔气化熵△vapSm及摩尔气化吉布斯函数△vapGm；（3）求环境的熵变△Samb；（4）应用有关原理判断上述过程是否为可逆过程；（5）298K时苯的饱和蒸汽压是多大。解：（1）向真空蒸发，p=0,所以W=0；130773077vapmHnH(.)kJ.kJ3077183143532783UHpVHp[V(g)V(l)]Hn(g)RT.kJ.J.kJ（2）1130773538717vapmvapmSH/T.kJmol/K.JK△vapGm=0（3）求环境的熵变△Samb11278335378874msysmSQ/TU/T.kJmol/K.JK（4）0mvapmmSSS，所以上述过程不是可逆过程；（5）298K时苯的饱和蒸汽压是多大。212111ln()vapmHppRTT211307711ln()1013258314298353p.kJ/mol.kPa.JmolKKKp2=14.63kPa△H相变，△S相变△H2，△S2△H1，△S11mol,H2O(g)25℃,pθ1mol,H2O(l)25℃,pθ1mol,H2O(g)25℃,3167Pa1mol,H2O(l)25℃,3167PaGΔ1ΔG2ΔG相变GΔ五、设在273.15K和1.0×106Pa压力下，取10.0dm3单原子理想气体，用下列几种不同方式膨胀到末态压力为1.0×105Pa:(1)恒温可逆膨胀；(2)绝热可逆膨胀；(3)在外压恒定为1.000×105Pa下绝热膨胀。试计算上述各过程的Q、W、△U、△H、△S。(1).解：该过程的始末状态如下：KT15.2731KTT15.27312MPap000.11等温可逆膨胀Pap5210000.13100.10dmV?2Vmoln403.4moln403.4根据理想气体性质，在无化学变化、无相变化的等温过程中，011HU，根据热一律211lnppnRTWQPaPaKmolKJmol561110000.110000.1ln2.273314.8403.4=23.03kJ211lnppnRTQSrPaPamolKJmol561110000.110000.1ln314.8403.4=13.84KJ(2)绝热可逆过程始末态如下：理想气体理想气体moln403.4moln403.4KT15.27310rQ2T=Pap6110000.1Pap5210000.1首先利用过程方程求出末态温度2T，因为只要2T确定了，则Q、W、△U、△H、△S便可求出。根据题给数据11,47.12molKJCmv，则1111,,314.847.12molKJmolKJRCCmVmp=1178.20molKJ绝热指数667.147.1278.201111,,molKJmolKJCCmVmp由绝热可逆方程122111pTpT解出111.667611.66721521.00010()()273.2108.71.00010pPaTTKKpPa112,21()4.40312.47(108.7273.2)vmUnCTTmolJKmolKKkJ032.9)(12,2TTnCHmp)2.2737.108(78.20403.411KKmolKJmolkJ05.150,0032.92222TQSQkJUWr（3）绝热恒外压过程n=4.403mol理想气体绝热Q=0n=4.403mol理想气体T1=273.15K,V1T2=，V2p1=1000000Pa恒外压膨胀p2=100000Pa因为过程是一个绝热不可逆过程，因此不能应用绝热可逆过程方程来确定系统的末态。但WUQ,0可适用。且理想气体绝热恒外压过程中)(12,TTnCUmV，)()(112212PnRTPnRTpVVpW外外故有：)()(111212,PTPTnRpTTnCmV外代入题给数据解得KT8.1742)2.2738.174(403.447.12113KKmolmolKJUkJ403.5kJH003.93kJUW403.53303Q12,21lnlnPmpTSnRnCpT]2.2738.174ln)314.847.12(10000.110000.1ln314.8[403.4115611KKmolKJPaPamolKJmol142.43KJ七、理想气体反应gOHHCgOHgHC52242在298K时的热力学数据如下：物质C2H4(g)H2O(g)C2H5OH(g)JmolKmS11$219.6188.72282.6kJmolfmH1$52.26－241.82－235.1⑴.,,rpmCKKKK0298373设试求时的和时的；$$⑵.若已知：物质C2H4(g)H2O(g)C2H5OH(g)11,KmolJCmp43.5633.5865.44试求373K时的$K解：⑴.1111282.6219.6188.72125.72JmolKJmolKθθrBBm298KBΔSνS11(1)52.26(1)(241.82)1(235.1)45.54fkJmolkJmolθθrBm298Km298KBΔΔHνH11145.54298(125.72)8.075TkJmolJmolkJmolθθθrrrm298Km298Km298KΔΔΔGHSlnRTθθrm298K298KΔGK3exp(/)exp{[8.07510/(8.314298)}26.03RTθθr298Km298KΔKG,,$rpmCKK0373时的,,rpmC0Q373298373298;θθθθrrrrmKmKmKmKΔΔΔΔSSHH11137337337345.54373(125.72)1.35TkJmolJmolkJmolθθθrrrmKmKmKΔΔΔGHS3373373exp(/)exp[1.3510/(8.314373)]1.54RTθθrKmKΔKG⑵1111(65.4443.5633.58)11.7JmolKJmolKθθr,mBmBΔ(B)pp,CνC因为θmrθΔ/ddΔp,mrCTH，移相积分得11121373()45.54(11.7)(373298)46.42TTkJmolKKJmolkJmolθθrrrmKm298KΔΔΔp,mHHC由TCTSp,mr/Δ/ddΔθmrθ移相积分得111123731373[125.72(11.7)]128.35298TJmolKJmolKTθθrrrmKm298KΔΔΔlnlnp,mSSC11137337337346.42373(128.35)1.45TkJmolJmolkJmol