化工热力学答案详解第三版VIP免费

实用标准文档精彩文案化工热力学课后答案（第三版）陈钟秀编著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22.522.560.5268.314190.60.427480.427483.2224.610ccRTaPamKmolP53168.314190.60.086640.086642.985104.610ccRTbmmolP∴0.5RTaPVbTVVb50.5558.314323.153.22212.462.98510323.1512.461012.462.98510=19.04MPa(3)普遍化关系式323.15190.61.695rcTTT124.6991.259rcVVV＜2∴利用普压法计算，01ZZZ crZRTPPPV∴crPVZPRT实用标准文档精彩文案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（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：510425.21.199rcTTT2.53.80.6579rcPPP—普维法∴01.61.60.4220.4220.0830.0830.23261.199rBT14.24.20.1720.1720.1390.1390.058741.199rBT实用标准文档精彩文案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%1.48072-3.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：0.24132.90.76304.2263.1cmiciiTyTK0.243.4960.767.3766.445cmiciiPyPMPa303263.11.15rmcmTTT0.1011.4450.0157rmcmPPP—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实用标准文档精彩文案114.24.210.1720.1720.1390.1390.1336303132.9rBT016111111618.314132.90.029890.0490.13367.378103.49610ccRTBBBP021.61.620.4220.4220.0830.0830.3417303304.2rBT124.24.220.1720.1720.1390.1390.03588303304.2rBT016222222628.314304.20.34170.2250.03588119.93107.37610ccRTBBBP又0.50.5132.9304.2201.068cijcicjTTTK331313131331293.194.093.55/22cccijVVVcmmol120.2950.2740.284522cccijZZZ120.2950.2250.13722cij6/0.28458.314201.068/93.55105.0838cijcijcijcijPZRTVMPa∴303201.0681.507rijcijTTT0.10135.08380.0199rijcijPPP0121.61.6120.4220.4220.0830.0830.1361.507rBT1124.24.2120.1720.1720.1390.1390.10831.507rBT∴01612121212126128.314201.0680.1360.1370.108339.84105.083810ccRTBBBP2211112122222mByByyByB26626630.247.3781020.240.7639.84100.76119.931084.2710/cmmol∴1mmBPPVZRTRT→V=0.02486m3/mol∴V总=nV=100×103×81.38%/12×0.02486=168.58m3实用标准文档精彩文案(2)1110.2950.240.10130.0250.2845cmZPyPMPaZ2220.2740.760.10130.0740.2845cmZPyPMPaZ2-4.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...