2-1.使用下述方法计算1kmol甲烷贮存在体积为、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= m3/1kmol= cm3/mol
查附录二得甲烷的临界参数：Tc= Pc= Vc=99 cm3/mol ω=
(1) 理想气体方程
P=RT/V=××10-6=
(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?53?1?0.08664?2.985?10m?mol 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 =
(3) 普遍化关系式
Tr?TTc?323.15190.6?1.695 Vr?VVc?124.699?1.259＜2 ∴利用普压法计算，Z?Z0??Z1
ZRT?PcPr VPV∴ Z?cPr
RT∵ P?
PV4.6?106?12.46?10?5cZ?Pr?Pr?0.2133Pr
RT8.314?323.15迭代：令Z0=1→Pr0= 又Tr=，查附录三得：Z0= Z1=
Z?Z0??Z1=+×=
此时，P=PcPr=×=
同理，取Z1= 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=
2-2.分别使用理想气体方程和Pitzer普遍化关系式计算510K、正丁烷的摩尔体积。已知实验值为mol。
解：查附录二得正丁烷的临界参数：Tc= Pc= Vc=99 cm3/mol ω= （1）理想气体方程
V=RT/P=×510/×106=×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?1.199 Pr?PPc?2.53.8?0.6579—普维法 ∴ B0?0.083?0.4220.422?0.083???0.2326 1.61.6Tr1.199B1?0.139?0.1720.172?0.139???0.05874 Tr4.21.1994.2BPc?B0??B1=+×= RTcZ?1?BPBPP?1?cr=×= RTRTcTr∴ PV=ZRT→V= ZRT/P=××510/×106=×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%的100kg的焦炭能生成、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= Pc= Vc= cm3/mol ω= Zc= 二氧化碳(2)：Tc= Pc= Vc= cm3/mol ω= Zc= 又y1=，y2=
∴(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?1.15 Prm?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?B11?RTc108.314?132.91B1??1B1??0.02989?0.049?0.1336???7.378?10?6 ??6?Pc13.496?100B2?0.083?0.4220.422?0.083???0.3417 1.6Tr1.6?303304.2?20.1720.172?0.139???0.03588 4.24.2Tr2?303304.2?1B2?0.139?B22?RTc208.314?304.21?6B??B??0.3417?0.225?0.03588??119.93?10 ???222?6Pc27.376?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?1.507 Prij?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?∴B12?RTc1208.314?201.0681B12??12B12???0.136?0.137?0.1083???39.84?10?6 ??6Pc125.0838?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=mol ?RTRT∴V总=n V=100×103×%/12×= (2) P1?y1PP2?y2PZc10.295?0.24?0.1013?0.025MPa Zm0.2845Zc20.274?0.76?0.1013?0.074MPa Zm0.28452-4.将压力为、温度为477K条件下的压缩到 m3，若压缩后温度，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= Pc= Vc= cm3/mol ω=
(1) 求取气体的摩尔体积
对于状态Ⅰ：P= MPa、T=447K、V= m
Tr?TTc?477405.6?1.176 Pr?PPc?2.0311.28?0.18—普维法
3
∴B0?0.083?0.4220.422?0.083???0.2426 1.61.6Tr1.176B1?0.139?0.1720.172?0.139??0.05194 Tr4.21.1764.2BPc?B0??B1??0.2426?0.25?0.05194??0.2296 RTcZ?1?BPPVBPP-33
??1?cr→V=×10m/mol RTRTRTcTr∴n=×10-3m3/mol=1501mol
对于状态Ⅱ：摩尔体积V= m3/1501mol=×10-5m3/mol T=
(2) Vander Waals
方程
27R2Tc227?8.3142?405.626?2 a???0.4253Pa?m?mol664Pc64?11.28?10b?RTc8.314?405.6?53?1??3.737?10m?mol 68Pc8?11.28?10P?RTa8.314?448.60.4253?2???17.65MPa 2?5?5V?bV9.458?3.737?10???3.737?10?(3) Redlich-Kwang方程
R2Tc2.58.3142?405.62.5a?0.42748?0.42748?8.679Pa?m6?K0.5?mol?2 6Pc11.28?10b?0.08664P?RTc8.314?405.6?0.08664?2.59?10?5m3?mol?1 6Pc11.28?10RTa8.314?448.68.679?0.5???18.34MPa ?50.5?5?5V?bTV?V?b??9.458?2.59??10448.6?9.458?10?9.458?2.59??10(4) Peng-Robinson方程
∵Tr?TTc?448.6405.6?1.106
化工热力学第三版答案陈钟秀
