解：由附录三查得甲烷的临界参数为：Tc＝190.56K，pc＝4.599MPa，?＝0.011 （1）利用理想气体状态方程pV?RT得：

V?RT8.314??273.15?50??43?13?1??1.433?10m?mol?143.3cm?mol 6p18.745?10m?M?V总125?16??13.95g V143.3（2）RK方程

p?RTa?0.5 V?bTV(V?b)式中：

0.42748??8.314???190.56?a?0.42748R2Tc2.5/pc＝＝3.2207Pa?m6?K0.5?mol-2 64.599?1022.5b?0.08664RTc/pc＝0.08664?8.314?190.56＝2.985?10?5m3?mol?1 64.599?10ap3.2207?18.745?106A?22.5＝＝0.4653 22.5RT?8.314???323.15?bp2.985?10?5?18.745?106B?＝＝0.2083

RT8.314?323.15按照式（2-16a）Z?1A?h?1?h???＝?2.2342??? 1?hB?1?h?1?h?1?h?和式（2－16b） h?bB0.2083 ??VZZZ 1 0.8779 0.8826 0.8823 0.8823 h 0.2083 0.2373 0.2360 0.2361 0.2361 迭代计算，取初值Z＝1，迭代过程和结果见下表。 迭代次数 0 1 2 3 4 V?ZRT0.8823?8.314?323.15??1.265?10?4m3/mol＝126.5cm3?mol?1 6p18.745?10m?M?V总125?16??15.81g V126.5可见，用RK方程计算更接近实验值。

2－13．欲在一个7810cm3的钢瓶中装入1kg的丙烷，且在253.2℃下工作，若钢瓶的安全工作压力为10MPa，问是否安全？

解：查得丙烷的临界性质为：Tc＝369.83K，pc＝4.248MPa，?＝0.152

n?m1000??22.727mol M44V总7810?10?6V???343.63?10?6m3?mol?1

n22.727使用RK方程： p?首先用下式计算a，b：

RTa?0.5 V?bTV(V?b)a?0.42748RT22.5c8.3142?369.832.5/pc?0.42748??18.296Pa?m6?K0.5?mol-2 64.248?10??b?0.08664RTc/pc?0.08664?代入RK方程得：p?9.870MPa

8.314?369.83?53?1 ?6.2771?10m?mol64.248?10非常接近于10MPa，故有一定危险。

2－14．试用RKS方程计算异丁烷在300K，3.704×105Pa时的饱和蒸气的摩尔体积。已知实验值为V?6.081?10m?mol。

解：由附录三查得异丁烷的临界参数为：Tc＝407.8K，pc＝3.640MPa，?＝0.177

?33?1Tr?T/Tc?300/407.8?0.7357

m?0.480?1.574??0.176?2?0.480?1.574?0.177?0.176?0.1772?0.7531?(T)??1?m(1?Tr0.5)???1?0.7531?1?0.73570.5???1.2258

22a?T??a???T??0.4278RT22c22?8.314???407.8?/pc???T?＝0.42748??1.2258＝1.6548?Pa?m6?/mol263.640?10b?0.08664RTc/pc＝0.08664?8.314?407.8/3.640?106?8.0700?10?5m3/mol

??ap1.6548?3.704?105A?22＝＝0.09853 22RT?8.314???300?bp8.0700?10?5?3.704?105B?＝＝0.01198

RT8.314?300按照式（2-16a）Z?1A?h?1?h????8.2245??＝? 1?hB?1?h?1?h?1?h?和式（2－16b） h?bB0.01198 ??VZZZ 1 0.9148 0.9070 0.9062 0.9061 0.9061 h 0.01198 0.01310 0.01321 0.01322 0.01322 0.01322 迭代计算，取初值Z＝1，迭代过程和结果见下表。 迭代次数 0 1 2 3 4 5 V?ZRT0.9061?8.314?300??6.1015?10?2m3/mol 6p3.704?10?2误差 ?6.031?6.1015??10/6.031?10?2??1.2%

2－15．试分别用RK方程及RKS方程计算在273K、1000×105Pa下，氮的压缩因子值，已知实验值为Z＝2.0685。

解：由附录三查得氮的临界参数为：Tc＝126.10K，pc＝3.394MPa，?＝0.040 （1）RK方程

a?0.42748RT22.5c0.42748??8.314???126.10?60.5-2/pc＝＝1.5546Pa?m?K?mol 63.394?1022.5b?0.08664RTc/pc＝0.08664?8.314?126.10?53?1＝2.6763?10m?mol 63.394?10ap1.5546?100?106A?22.5＝＝1.8264 22.5RT?8.314???273?bp2.6763?10?5?1000?105B?＝＝1.1791

RT8.314?273按照式（2-16a）Z?1A?h?1?h????1.5489??＝? 1?hB?1?h?1?h1?h??和式（2－16b） h?bB1.1791 ??VZZZ 2 1.862 2.1260 1.6926 0.8823 h 0.58955 0.6332 0.5546 0.6966 0.2361 迭代计算，取初值Z＝2，迭代过程和结果见下表。 迭代次数 0 1 2 3 4 …….. 迭代不收敛，采用RK方程解三次方程得： V=0.00004422m3/mol

pV4.422?10?5?1000?105Z???1.9485

RT8.314?273RKS方程

Tr?T/Tc?273/126.1?2.1649

m?0.480?1.574??0.176?2?0.480?1.574?0.040?0.176?0.0402?0.5427?(T)??1?m(1?Tr0.5)???1?0.5427?1?2.16490.5???0.5538

22a?T??a???T??0.4278RT22c22?8.314???126.1?/pc???T?＝0.42748??0.5538＝0.076667?Pa?m6?/mol63.394?10b?0.08664RTc/pc＝0.08664?8.314?126.1/3.394?106?2.6763?10?5m3/mol

??ap0.076667?1000?105A?22＝＝1.4882 22RT?8.314???273?bp2.6763?10?5?1000?105B?＝＝1.1791

RT8.314?273按照式（2-16a）Z?1A?h?1?h????1.2621??＝? 1?hB?1?h?1?h?1?h?和式（2－16b） h?bB1.1791?? VZZ