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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=0.1246 m3/1kmol=124.6 cm3/mol

查附录二得甲烷的临界参数：Tc=190.6K Pc=4.600MPa Vc=99 cm3/mol ω=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

R2Tc2.58.3142?190.62.560.5?2 a?0.42748?0.42748?3.222Pa?m?K?mol6Pc4.6?10b?0.08664RTc8.314?190.6?0.08664?2.985?10?5m3?mol?1 6Pc4.6?10∴P??RTa?0.5 V?bTV?V?b?8.314?323.153.222?

?12.46?2.985??10?5323.150.5?12.46?10?5?12.46?2.985??10?5 =19.04MPa (3) 普遍化关系式

Tr?TTc?323.15190.6?1.695Vr?VVc?124.699?1.259＜2

∴利用普压法计算，Z?Z0??Z1

ZRT?PcPr VPV∴Z?cPr

RT∵P?1 / 29

PV4.6?106?12.46?10?5cZ?Pr?Pr?0.2133Pr

RT8.314?323.15迭代：令Z0=1→Pr0=4.687 又Tr=1.695，查附录三得：Z0=0.8938 Z1=0.4623

0.4623=0.8975 Z?Z0??Z1=0.8938+0.008×

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=19.22MPa

2-2.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=425.2K Pc=3.800MPa Vc=99 cm3/mol ω=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：

1.696?1.4807?100%?14.54%

1.4807（2）Pitzer普遍化关系式

对比参数：Tr?TTc?510425.2?1.199Pr?PPc?2.53.8?0.6579—普维法 ∴B0?0.083?B1?0.139?0.4220.422?0.083???0.2326 Tr1.61.1991.60.1720.172?0.139???0.05874 4.24.2Tr1.199BPc?B0??B1=-0.2326+0.193×0.05874=-0.2213 RTc2 / 29

Z?1?BPBPP?1?cr=1-0.2213×0.6579/1.199=0.8786 RTRTcTr∴PV=ZRT→V= ZRT/P=0.8786×8.314×510/2.5×106=1.49×10-3 m3/mol 误差：

1.49?1.4807?100%?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=132.9K Pc=3.496MPa Vc=93.1 cm3/mol ω=0.049 Zc=0.295

二氧化碳(2)：Tc=304.2K Pc=7.376MPa Vc=94.0 cm3/mol ω=0.225 Zc=0.274 又y1=0.24，y2=0.76 ∴(1)由Kay规则计算得：

Tcm??yiTci?0.24?132.9?0.76?304.2?263.1K

iPcm??yiPci?0.24?3.496?0.76?7.376?6.445MPa

iTrm?TTcm?303263.1?1.15Prm?PPcm?0.1011.445?0.0157—普维法

利用真实气体混合物的第二维里系数法进行计算

B10?0.083?0.4220.422?0.083???0.02989 1.61.6Tr1?303132.9?0.1720.172?0.139??0.1336 4.2Tr4.2?303132.9?11B1?0.139?3 / 29

B11?RTc108.314?132.91B??B??0.02989?0.049?0.1336???7.378?10?6 ?111?6?Pc13.496?100B2?0.083?0.4220.422?0.083???0.3417 1.61.6Tr2?303304.2?0.1720.172?0.139???0.03588 4.2Tr4.2?303304.2?21B2?0.139?B22?RTc208.314?304.21B2??2B2??0.3417?0.225?0.03588???119.93?10?6 ??6?Pc27.376?100.50.5又Tcij??TciTcj???132.9?304.2??201.068K

?Vc113?Vc123??93.113?94.013?Vcij?????93.55cm3/mol ??22????33Zc1?Zc20.295?0.274??0.2845 22???20.295?0.225?cij?1??0.137

22Zcij?Pcij?ZcijRTcij/Vcij?0.2845?8.314?201.068/?93.55?10?6??5.0838MPa

∴Trij?TTcij?303201.068?1.507Prij?PPcij?0.10135.0838?0.0199

0B12?0.083?0.4220.422?0.083???0.136 1.61.6Tr121.5070.1720.172?0.139??0.1083 4.2Tr4.21.507121B12?0.139?01?6∴B12?RTc12?B12 ??12B12?0.136?0.137?0.1083??39.84?10????8.314?201.0686Pc125.0838?102Bm?y12B11?2y1y2B12?y2B22?0.242???7.378?10?6??2?0.24?0.76???39.84?10?6??0.762???119.93?10?6???84.27?10?6cm3/mol∴Zm?1?BmPPV?→V=0.02486m3/mol RTRT∴V总=n V=100×103×81.38%/12×0.02486=168.58m3 (2)P1?y1PZc10.295?0.24?0.1013?0.025MPa Zm0.28454 / 29

P2?y2PZc20.274?0.76?0.1013?0.074MPa Zm0.28452-4.将压力为2.03MPa、温度为477K条件下的2.83m3NH3压缩到0.142 m3，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）Vander Waals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=405.6K Pc=11.28MPa Vc=72.5 cm3/mol ω=0.250 (1) 求取气体的摩尔体积

对于状态Ⅰ：P=2.03 MPa、T=447K、V=2.83 m3

Tr?TTc?477405.6?1.176Pr?PPc?2.0311.28?0.18—普维法

∴B0?0.083?B1?0.139?0.4220.422?0.083???0.2426 Tr1.61.1761.60.1720.172?0.139??0.05194 4.24.2Tr1.176BPc?B0??B1??0.2426?0.25?0.05194??0.2296 RTcZ?1?BPPVBPP??1?cr→V=1.885×10-3m3/mol RTRTRTcTr∴n=2.83m3/1.885×10-3m3/mol=1501mol

对于状态Ⅱ：摩尔体积V=0.142 m3/1501mol=9.458×10-5m3/mol T=448.6K

(2) Vander Waals方程

27R2Tc227?8.3142?405.626?2 a???0.4253Pa?m?mol664Pc64?11.28?105 / 29