化工热力学(第三版)陈钟秀课后习题答案解析

范文范例指导学习word版本整理分享第二章2-1.使用下述方法计算1kmol甲烷贮存在体积为0.1246m3、温度为50℃的容器中产生的压力：（1）理想气体方程；（2）R-K方程；（3）普遍化关系式。解：甲烷的摩尔体积V=0.1246m3/1kmol=124.6cm3/mol查附录二得甲烷的临界参数：Tc=190.6KPc=4.600MPaVc=99cm3/molω=0.008(1)理想气体方程P=RT/V=8.314×323.15/124.6×10-6=21.56MPa(2)R-K方程22.522.560.5268.314190.60.427480.427483.2224.610ccRTaPamKmolP53168.314190.60.086640.086642.985104.610ccRTbmmolP∴0.5RTaPVbTVVb50.5558.314323.153.22212.462.98510323.1512.461012.462.98510=19.04MPa(3)普遍化关系式323.15190.61.695rcTTT124.6991.259rcVVV＜2∴利用普压法计算，01ZZZ∵crZRTPPPV∴crPVZPRT654.61012.46100.21338.314323.15crrrPVZPPPRT迭代：令Z0=1→Pr0=4.687又Tr=1.695，查附录三得：Z0=0.8938Z1=0.462301ZZZ=0.8938+0.008×0.4623=0.8975此时，P=PcPr=4.6×4.687=21.56MPa同理，取Z1=0.8975依上述过程计算，直至计算出的相邻的两个Z值相差很小，迭代结束，得Z和P的值。∴P=19.22MPa2-2.分别使用理想气体方程和Pitzer普遍化关系式计算510K、2.5MPa正丁烷的摩尔体积。已知实验值为1480.7cm3/mol。解：查附录二得正丁烷的临界参数：Tc=425.2KPc=3.800MPaVc=99cm3/molω=0.193范文范例指导学习word版本整理分享（1）理想气体方程V=RT/P=8.314×510/2.5×106=1.696×10-3m3/mol误差：1.6961.4807100%14.54%1.4807（2）Pitzer普遍化关系式对比参数：510425.21.199rcTTT2.53.80.6579rcPPP—普维法∴01.61.60.4220.4220.0830.0830.23261.199rBT14.24.20.1720.1720.1390.1390.058741.199rBT01ccBPBBRT=-0.2326+0.193×0.05874=-0.221311crcrBPBPPZRTRTT=1-0.2213×0.6579/1.199=0.8786∴PV=ZRT→V=ZRT/P=0.8786×8.314×510/2.5×106=1.49×10-3m3/mol误差：1.491.4807100%0.63%1.48072-3.生产半水煤气时，煤气发生炉在吹风阶段的某种情况下，76%（摩尔分数）的碳生成二氧化碳，其余的生成一氧化碳。试计算：（1）含碳量为81.38%的100kg的焦炭能生成1.1013MPa、303K的吹风气若干立方米？（2）所得吹风气的组成和各气体分压。解：查附录二得混合气中各组分的临界参数：一氧化碳(1)：Tc=132.9KPc=3.496MPaVc=93.1cm3/molω=0.049Zc=0.295二氧化碳(2)：Tc=304.2KPc=7.376MPaVc=94.0cm3/molω=0.225Zc=0.274又y1=0.24，y2=0.76∴(1)由Kay规则计算得：0.24132.90.76304.2263.1cmiciiTyTK0.243.4960.767.3766.445cmiciiPyPMPa303263.11.15rmcmTTT0.1011.4450.0157rmcmPPP—普维法利用真实气体混合物的第二维里系数法进行计算011.61.610.4220.4220.0830.0830.02989303132.9rBT114.24.210.1720.1720.1390.1390.1336303132.9rBT范文范例指导学习word版本整理分享016111111618.314132.90.029890.0490.13367.378103.49610ccRTBBBP021.61.620.4220.4220.0830.0830.3417303304.2rBT124.24.220.1720.1720.1390.1390.03588303304.2rBT016222222628.314304.20.34170.2250.03588119.93107.37610ccRTBBBP又0.50.5132.9304.2201.068cijcicjTTTK331313131331293.194.093.55/22cccijVVVcmmol120.2950.2740.284522cccijZZZ120.2950.2250.13722cij6/0.28458.314201.068/93.55105.0838cijcijcijcijPZRTVMPa∴303201.0681.507rijcijTTT0.10135.08380.0199rijcijPPP0121.61.6120.4220.4220.0830.0830.1361.507rBT1124.24.2120.1720.1720.1390.1390.10831.507rBT∴01612121212126128.314201.0680.1360.1370.108339.84105.083810ccRTBBBP2211112122222mByByyByB26626630.247.3781020.240.7639.84100.76119.931084.2710/cmmol∴1mmBPPVZRTRT→V=0.02486m3/mol∴V总=nV=100×103×81.38%/12×0.02486=168.58m3(2)1110.2950.240.10130.0250.2845cmZPyPMPaZ范文范例指导学习word版本整理分享2220.2740.760.10130.0740.2845cmZPyPMPaZ2-4.将压力为2.03MPa、温度为477K条件下的2.83m3NH3压缩到0.142m3，若压缩后温度448.6K，则其压力为若干？分别用下述方法计算：（1）VanderWaals方程；（2）Redlich-Kwang方程；（3）Peng-Robinson方程；（4）普遍化关系式。解：查附录二得NH3的临界参数：Tc=405.6KPc=11.28MPaVc=72.5cm3/molω=0.250(1)求取气体的摩尔体积对于状态Ⅰ：P=2.03MPa、T=447K、V=2.83m3477405.61.176rcTTT2.0311.280.18rcPPP—普维法∴01.61.60.4220.4220.0830.0830.24261.176rBT14.24.20.1720.1720.1390.1390.051941.176rBT010.24260.250.051940.2296ccBPBBRT11crcrBPPVBPPZRTRTRTT→V=1.885×10-3m3/mol∴n=2.83m3/1.885×10-3m3/mol=1501mol对于状态Ⅱ：摩尔体积V=0.142m3/1501mol=9.458×10-5m3/molT=448.6K(2)VanderWaals方程222262627278.314405.60.4253646411.2810ccRTaPammolP53168.314405.63.737108811.2810ccRTbmmolP22558.314448.60.425317.659.4583.737103.73710RTaPMPaVbV(3)Redlich-Kwang方程22.522.560.5268.314405.60.427480.427488.67911.2810ccRTaPamKmolP53168.314405.60.086640.086642.591011.2810ccRTbmmolP0.550.5558.314448.68.67918.349.4582.5910448.69.458109.4582.5910RTaPMPaVbTVVb范文范例指导学习word版本整理分享(4)Peng-Robinson方程∵448.6405.61.106rcTTT∴220.37461.542260.269920.37461.542260.250.269920.250.7433k220.50.51110.743311.1060.9247rTkT22226268.314405.60.457240.457240.92470.426211.2810cccRTaTaTTPammolP53168.314405.60.077800.077802.3261011.2810ccRTbmmolP∴aTRTPVbVVbbVb510108.314448.60.42629.4582.326109.4589.4582.326102.3269.4582.3261019.00MPa(5)普遍化关系式∵559.458107.25101.305rcVVV＜2适用普压法，迭代进行计算，方法同1-1（3）2-6.试计算含有30%（摩尔分数）氮气（1）和70%（摩尔分数）正丁烷（2）气体混合物7g,在188℃、6.888MPa条件下的体积。已知B11=14cm3/mol，B22=-265cm3/mol，B12=-9.5cm3/mol。解：2211112122222mByByyByB2230.31420.30.79.50.7265132.58/cmmol1mmBPPVZRTRT→V(摩尔体积)=4.24×10-4m3/mol假设气体混合物总的摩尔数为n，则0.3n×28+0.7n×58=7→n=0.1429mol∴V=n×V(摩尔体积)=0.1429×4.24×10-4=60.57cm32-8.试用R-K方程和SRK方程计算273K、101.3MPa下氮的压缩因子。已知实验值为2.0685解：适用EOS的普遍化形式查附录二得NH3的临界参数：Tc=126.2KPc=3.394MPaω=0.04（1）R-K方程的普遍化22.522.560.5268.314126.20.427480.427481.55773.39410ccRTaPamKmolP5