化工热力学(第三版)陈钟秀课后习题答案-医药卫生

化工热力学(第三版)陈钟秀课后习题答案-医药卫生此文档为教材配套的课后习题答案第二章2-1.使用下述方法计算1kmol甲烷储藏在体积为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）宽泛化关系式。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=8.314×323.15/124.6×106=21.56MPa-(2)R-K方程R2Tc2.5a0.427Pc0.086642.58.314190.260.464.6106Pa3.2m22K0.5mol2RTc8.314190.65310.086642.98510mmol6Pc4.610∴PRTa0.512.462.985105323.150.512.4610512.462.985105=19.04MPa(3)宽泛化关系式TrTTc323.5190.6Z0Z16V1.9r5Vc991.259＜2∴利用普压法计算，Z∵∴ZRTPcPrVPVZcPrRTP65PV4.61012.4610ZcPrPr0.2133PrRT8.314323.15迭代：令Z0=1→Pr0=4.687又Tr=1.695，查附录三得：Z0=0.8938Z1=0.4623ZZ0Z1=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宽泛化关系式计算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此文档为教材配套的课后习题答案（1）理想气体方程V=RT/P=8.314×510/2.5×106=1.696×-103m3/mol偏差：1.6961.4807100%14.54%1.48072）Pitzer宽泛化关系式比较参数：TrTc*.21.199PrPc2.3.80.00.*.60.6―普维法579∴B00.083B10.139BPcB0B1=-0.2326+0.193×0.05874=-0.2213RTcZ1BPBPP1crRTRTcTr=1-0.2213×0.6579/1.199=0.8786∴

PV=ZRT→V=ZRT/P=0.8786×8.314

×510/2.5

×106=1.49

×-310m3/mol

偏差：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状况下，76%（摩尔分数）的碳生成二氧化碳，其他的生成一氧化碳。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：查附录二得混杂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：TcmyiTci0.24132.90.76304.2263.1KiPcmyiPci0.243.4960.767.3766.445MPaiTrmTcm263.11.15PrmPcm0.11.445利用真实气体混杂物的第二维里系数法进行计算―普维法0157B100.0830.1390.13364.2Tr4.211B10.139此文档为教材配套的课后习题答案B11RTc108.314132.916BB0.029890.0490.13367.378101116Pc13.49610B20.0830.1390.035884.2Tr4.2304.221B20.139B22又TcijRTc208.314304.21B22B20.34170.2250.03588119.931066Pc27.37610TciTcj0.5132.9304.230.5201.068K3Vc1Vc293.1394.01Vcij93.55cm3/mol22Zc1Zc20.2950.2740.28452220.2950.225cij10.13722ZcijPcijZcijRTcij/Vcij0.28458.314201.068/93.551065.0838MPa∴0.1390.10834.2Tr4.21.*01990.B120.0831B120.139B12RTc1208.314201.06816B1212B120.1360.1370.108339.84106Pc125.08381022Bmy1B112y1y2B12y2B220.2427.37810620.240.7639.841060.762119.9310684.27106cm3/mol∴Zm1BmPPVRTRTV=0.02486m3/mol∴V总=nV=100×103×81.38%/12×0.02486=168.58m3(2)P1y1PZc10.2950.240.10130.025MPaZm0.2845此文档为教材配套的课后习题答案P2y2PZc20.2740.760.10130.074MPaZm0.28452-4.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）宽泛化关系式。解：查