此文档为教材配套的课后习题答案

P2 y2P

Zc20.274

0.76 0.1013 0.074MPa Zm0.2845

2-4.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 Tc 405.6 1.176 Pr PPc 0.18—普维法

∴B

0.083

0.4220.422

0.083 0.2426 1.61.6

Tr1.1760.1720.172

0.139 0.05194 Tr4.21.1764.2

B1 0.139

BPc

B0 B1 0.2426 0.25 0.05194 0.2296 RTcZ 1

BPPVBPP

1 crRTRTRTcTr

→V=1.885×10-3m3/mol

∴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 405.62

a 0.4253Pa m6 mol 2 6

64Pc64 11.28 10

b

RTc8.314 405.6 3.737 10 5m3 mol 1 6

8Pc8 11.28 10

RTa8.314 448.60.4253

2 17.65MPa 2 5 5V bV9.458 3.737 10 3.737 10

P

(3) Redlich-Kwang方程

R2Tc2.58.3142 405.62.560.5 2

a 0.42748 0.42748 8.679Pa m K mol6

Pc11.28 10

b 0.08664

P

RTc8.314 405.6 53 1

0.08664 2.59 10m mol 6Pc11.28 10

RTa8.314 448.68.679 0.5 18.34MPa 50.5 5 5

V bTVV b9.458 2.59 10448.6 9.458 109.458 2.59 10