化工热力学详细答案2————————————————————————————————作者：————————————————————————————————日期：3/16化工热力学第二章作业解答2.1试用下述三种方法计算673K，4.053MPa下甲烷气体的摩尔体积，（1）用理想气体方程；（2）用R-K方程；（3）用普遍化关系式解（1）用理想气体方程（2-4）V＝RTP＝68.3146734.05310＝1.381×10-3m3·mol-1（2）用R-K方程（2-6）从附录二查的甲烷的临界参数和偏心因子为Tc＝190.6K，Pc＝4.600Mpa，ω＝0.008将Tc，Pc值代入式（2-7a）式（2-7b）22.50.42748ccRTap22.560.42748(8.314)(190.6)4.610＝3.224Pa·m6·K0.5·mol-20.0867ccRTbp60.08678.314190.64.610＝2.987×10-5m3·mol-1将有关的已知值代入式（2-6）4.053×106＝58.3146732.98710V－0.553.224(673)(2.98710)VV迭代解得V＝1.390×10-3m3·mol-1(注：用式2-22和式2-25迭代得Z然后用PV=ZRT求V也可)（3）用普遍化关系式6733.53190.6rTTTc664.053100.8814.610rPPPc因为该状态点落在图2-9曲线上方，故采用普遍化第二维里系数法。由式（2-44a）、式（2-44b）求出B0和B1B0＝0.083－0.422/Tr1.6＝0.083-0.422/(3.53)1.6＝0.0269B1＝0.139－0.172/Tr4.2＝0.139－0.172/(3.53)4.2＝0.138代入式（2-43）010.02690.0080.1380.0281BPcBBRTc由式（2-42）得Pr0.881110.02811.0073.53BPcZRTcTrV＝1.390×10-3m3·mol-12.2试分别用（1）VanderWaals,（2）R-K，（3）S-R-K方程计算273.15K时将CO2压缩到比体积为550.1cm3·mol－1所需要的压力。实验值为3.090MPa。解：从附录二查得CO2得临界参数和偏心因子为4/16Tc＝304.2KPc＝7.376MPaω＝0.225（1）VanderWaals方程2RTaPVbV式中222764ccRTap2227(8.314)(304.2)647.376＝3.658×105MPa·cm6·mol-28ccRTbp＝8.314304.287.376＝42.86cm3·mol-1则得8.314273.15550.142.86P－523.65810(550.1)＝3.268Mpa误差％＝3.0903.2683.090×100％＝－5.76％（2）R-K方程0.5()RTapVbTVVb22.50.42748ccRTap＝22.50.42748(8.314)(304.2)7.376＝6.466×106MPa·cm6·K0.5·mol-20.0867ccRTbp＝0.08678.314304.27.376＝29.71cm3·mol-1则得8.314273.15550.129.71P－60.56.46610(273.15)(550.1)(550.129.71)＝3.137Mpa误差％＝3.0903.1373.090×100％＝－1.52％（3）S-R-K方程()()RTaTPVbVVb式中220.42748cccRTaTaTTp20.5()1'(1-)TmTr22'0.4801.5740.176=0.480+1.5740.225-0.1760.2250.8252m5/16得2273.15()10.82521-1.088304.2T2250.42748(8.314)(304.2)1.0884.033107.376aTMPa·cm6·mol-2又0.0867ccRTbp＝0.08678.314304.27.376＝29.71cm3·mol-1将有关的值代入S-R-K程，得8.314273.15550.129.71P－54.03310550.1(550.129.71)＝3.099Mpa误差％＝3.0903.0993.090×100％＝－0.291％比较（1）、（2）与（3）结果，说明Vanderwaals方程计算误差较大，S-R-K方程的计算精度较R-K方程高。2.3试用下列各种方法计算水蒸气在10.3MPa和643K下的摩尔体积，并与水蒸气表查出的数据（V=0.0232m3·kg-1）进行比较。已知水的临界常数及偏心因子为：Tc=647.3K，Pc=22.05MPa，ω＝0.344。（a）理想气体方程；（b）R-K方程；（c）普遍化关系式。解:（a）理想气体方程V=RT/P=8.314×10-3×643/10.3=0.519m3·kmol-1=0.0288m3·kg-1误差％＝0.02320.0288100%0.0232＝－24.1％（b）R-K方程为便于迭代，采用下列形式的R-K方程：1.5111ahZhbRTh--------(A)式中bbphVZRT---------(B)22.50.42748RTcaPc322.50.42748(8.31410)(647.3)22.05=14.29MPa·m6·K0.5kmol-20.08664RTcbPc30.086648.31410647.322.05=0.02115m3·kmol-11.5abRT=31.514.290.02115(8.31410)(643)=4.984bRT=30.021158.31410643=3.956×10-3MPa-1将上述有关值分别代入式（A）和（B）得：6/1614.98411hZhh--------(C)33.9561010.3hZ=0.04075Z--------(D)利用式（C）和式（D）迭代求解得：Z=0.8154因此ZRTVP＝30.81548.3141064310.3=0.4232m3·kmol-1=0.02351m3·kg-1误差％＝0.02320.02351100%0.0232＝－1.34％(c)普遍化关系式6430.993647.3TTrTc10.30.46722.05PPrPc...
