化工热力学第三版习题答案

《化工热力学》（第三版）习题参考答案58页第2章2-1求温度673.15K、压力4.053MPa的甲烷气体摩尔体积。解：（a）理想气体方程133610381.110053.415.673314.8molmpRTVRTpV（b）用R-K方程①查表求cT、cp；②计算a、b；③利用迭代法计算V。133113301103896.110381.1molmVmolmVbVVTbVabpRTVbVVTabVRTpiiiii（c）用PR方程步骤同（b），计算结果：1331103893.1molmVi。（d）利用维里截断式2.416.1010172.0139.0422.0083.0111rrrrrrrrccTBTBTpBTpBTpRTBpRTBpRTpVZ查表可计算rp、rT、0B、1B和Z由13310391.1molmpZRTVRTpVZ2-4V=1.213m3，乙醇45.40kg，T=500.15K，求压力。解：（a）理想气体状态方程MPaVRTMmVnRTp383.3213.115.500314.84640.45（b）用R-K方程a0.42748R2TC2.5PC28.039b0.08664RTCPC0.058MPapkmolmnVVbVVTabVRTpmmmm759.2229.146/40.45213.113（c）用SRK方程计算（d）用PR方程计算（e）用三参数普遍化关联MPaBVRTpBBBpTRBBBBBRTBpBVRTpRTBpRTpVZkmolmnVVkmolMmnmccccmmm779.2267.0635.0,057.0,361.01229.1987.0213.1987.0464.451011010132-7计算T=523.15K，p=2MPa的水蒸气的Z和V解：（a）用维里截断式221pVCRTpVBRTpRTVVCVBRTpVZ采用迭代法计算V=2.006之后求得Z=0.923（d）利用维里截断式2.416.1010172.0139.0422.0083.0111rrrrrrrrccTBTBTpBTpBTpRTBpRTBpRTpVZ查表可计算rp、rT、0B、1B可得到Z=0.932；由13310025.2molmpZRTVRTpVZ（c）水蒸气表9223.015.523314.800592.2200000592.21811144.011144.01313RTVpZkmolmkgmV92页第三章3-4丁二烯13R8.314T1127273.15T2227273.15P12.53106PaP212.67106PaTc425Pc4.326106Pa0.181CpT()22.738222.796103T73.879106T2利用三参数压缩因子计算方法，查图表，得到压缩因子：Tr1T1TcTr2T2TcTr10.942Tr21.177Pr1P1PcPr2P2PcPr10.585Pr22.929Z10.677Z20.535VZ2RT2P2Z1RT1P1V7.146104m3mol1H2RRTcP2Pc0.0830.1391.097T2Tc1.60.894T2Tc4.2H2R8.475103H1RRTcP1Pc0.0830.1391.097T1Tc1.60.894T1Tc4.2H1R2.704103HT1T2TCpT()dH2RH1RH5.028103Jmol1S2RRP2Pc0.675T2Tc2.60.722T2Tc5.2S2R12.128S1RRP1Pc0.675T1Tc2.60.722T1Tc5.2S1R4.708ST1T2TCpT()TdRlnP2P1S2RS1RS3.212Jmol1K13-7：解：113333261.510261.5381200010551.110095.2SV,,TVV121KkgJKkgkPamVdppSTVpSppTpTp注意：JkPam3310163091.110619.110551.1261.5270kgkJpVSTH或者156336.10901081.310210551.127010095.21121kgJdpVTHLpp3-9解：乙腈的Antonie方程为kPactps85.241/24.32717258.14ln（1）60℃时，乙腈的蒸气压kPappss813.48888.385.2416024.32717258.14ln（2）乙腈的标准沸点ctct375.81605.485.241/24.32717258.14100ln（3）20℃、40℃和标准沸点时的汽化焓molkJcHmolkJcHmolkJcHtTHRTHtRTHdTpds/72.32375.81;/57.3340;/09.342085.241314.824.327185.24124.3271ln22222117页第四章4-131h12u2gzqwz3mh23003230()103Jkg104kg3600sq0m2.778kgs104kg3600s2.778kgsu1202502109Jkgu109Jkgh2.583106Js12mu21.65104Jsgzm81.729Jsh12mu2gzm2.567106Jsw2.567106Js2.567106Wwc2.5832.5672.567100%wc0.623%4-2方法一：h12u2gzqwu13R8.314h12u2wu20.07520.252u1u20.27T2353.15T1593.15HCpmhT2T1HR2HR1HR1R647.3Tr00Pr0PrdB00B00Tr00.344dB10B10Tr0dHR1576.771HR2R647.3Tr10Pr1PrdB01B01Tr10.344dB11B11Tr1dHR256.91经计算得1103.35KmolJCpmh体积流速为：132210132.02075.0314.32/smduV摩尔流速为：1015.41500000/15.593314.80132.0/smolpRTVVVnm根据热力学第一定律，绝热时Ws=-△H，所以HnCpmhT2T1nHR2HR1Ws4.0158.40810356.91576.771()3.167104W方法二：根据过热蒸汽表，内插法应用可查得35kPa、80℃的乏汽处在过热蒸汽区，其焓值h2=2645.6kJ·kg-1；1500kPa、320℃的水蒸汽在过热蒸汽区，其焓值h1=3081.5kJ·kg-1；wh12u22u122645.63081.54.464103435.904kJkg1按理想气体体积计算的体积VRTP8.314593.1515000003.288103m3mol1N4.015mols0.0132m3s13.288103m3mol14.015molsw435.90418N3.15104W4-6解：二氧化碳T1303.15R8.314P11.5106PaP20.10133106PaTc304.2Pc7.357106Pa0.225CpT()45.3698.688103T9.619105T2H1RRTcP1Pc0.0830.1391.097T1Tc1.60.894T1Tc4.2H2RT2RTcP2Pc0.0830.1391.097T2Tc1.60.894T2Tc4.2通过112TCHTpmhR迭代计算温度，T2=287.75KCpmsT1T2TCpT()TdlnT2T1HT1T2TCpT()dH2RT2H1R1.822108Jmol1S1RRP1Pc0.675T1Tc2.60.722T1Tc5.2S2RRP2Pc0.675T2Tc2.60.722T2Tc5.2ST1T2TCpT()TdRlnP2P1S2RS1R21.801Jmol1K14-7解：T1473.15R8.314P12.5106PaP20.20106PaTc305.4Pc4.88106Pa0.098CpT()9.403159.837103T46.234106T2S1RRP1Pc0.675T1Tc2.60.722T1Tc5.2S2RT2RP2Pc0.675T2Tc2.60.722T2Tc5.2经迭代计算（参考101页例题4-3）得到T2=340.71K。H1RRTcP1Pc0.0830.1391.097T1Tc1.60.894T1Tc4.2H2RT2RTcP2Pc0.0830.1391.097T2Tc1.60.894T2Tc4.2HT1T2TCpT()dH2RT2H1R8.32725103Jmol1。146页第五章5-1：b5-2:c5-4:a5-5:a5-1：解：可逆过程熵产为零，即0050syssysfsysgSTSSSS。5-2：解：不可逆过程熵产大于零，即00505TSTSSSSsyssysfsysg。即系统熵变可小于零也可大于零。5-4：解