发布时间 : 星期日 文章化工热力学答案(第三版).更新完毕开始阅读c4e9b63126d3240c844769eae009581b6bd9bda0
(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?1.176 Pr?PPc?2.0311.28?0.18—普维法
∴B0?0.083?B1?0.139?0.4220.422?0.083???0.2426 1.61.6Tr1.1760.1720.172?0.139??0.05194 Tr4.21.1764.2BPc?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?10b?RTc8.314?405.6??3.737?10?5m3?mol?1 68Pc8?11.28?10P?RTa8.314?448.60.4253?2???17.65MPa 2?5?5V?bV?9.458?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
∴k?0.3746?1.54226??0.26992?2?0.3746?1.54226?0.25?0.26992?0.252?0.7433
??T????1?k?1?T0.5r???2???1?0.7433??1?1.1060.5???2?0.9247
R2Tc28.3142?405.62a?T??ac??T??0.45724??T??0.45724??0.9247?0.4262Pa?m6?mol?2 6Pc11.28?10b?0.07780RTc8.314?405.6?53?1?0.07780??2.326?10m?mol Pc11.28?106∴P?a?T?RT ?V?bV?V?b??b?V?b?8.314?448.60.4262??9.458?2.326??10?59.458??9.458?2.326??10?10?2.326??9.458?2.326??10?10? ?19.0 M0Pa(5) 普遍化关系式
∵ Vr?VVc?9.458?10?57.25?10?5?1.305＜2 适用普压法，迭代进行计
算，方法同1-1（3）
2-6.试计算含有30%（摩尔分数）氮气（1）和70%（摩尔分数）正丁烷（2）气体混合物7g,在188℃、6.888MPa条件下的体积。已知B11=14cm3/mol，B22=-265cm3/mol，B12=-9.5cm3/mol。 解：Bm?y12B11?2y1y2B12?y22B22
?0.32?14?2?0.3?0.7???9.5??0.72???265???132.58cm3/mol
Zm?1?BmPPV→V(摩尔体积)=4.24×10-4m3/mol ?RTRT假设气体混合物总的摩尔数为n，则 0.3n×28+0.7n×58=7→n=0.1429mol
∴V= n×V(摩尔体积)=0.1429×4.24×10-4=60.57 cm3
2-8.试用R-K方程和SRK方程计算273K、101.3MPa下氮的压缩因子。已知实验值为2.0685 解：适用EOS的普遍化形式
查附录二得NH3的临界参数：Tc=126.2K Pc=3.394MPa ω=0.04
（1）R-K方程的普遍化
R2Tc2.58.3142?126.22.5a?0.42748?0.42748?1.5577Pa?m6?K0.5?mol?2 6Pc3.394?10b?0.08664RTc8.314?126.2?0.08664?2.678?10?5m3?mol?1 6Pc3.394?10A?aPR2T2.5
B?bP RTAa1.5577???1.551 BbRT1.52.678?10?5?8.314?2731.5BbbP2.678?10?5?101.3?1061.1952??∴h??? ①
ZVZRTZ?8.314?273ZZ?1A?h?1?h?????1.551??? ② 1?hB?1?h?1?h?1?h?①、②两式联立，迭代求解压缩因子Z （2）SRK方程的普遍化
Tr?TTc?273126.2?2.163m?0.480?1.574??0.176?2?0.480?1.574?0.04?0.176?0.042?0.5427
??T??2211?1?m?1?Tr0.5????1?0.5427??1?2.1630.5???0.2563 ??Tr?2.163?R2Tc28.3142?126.22.560.5?2 a?0.42748???T??0.42748?0.2563?0.3992Pa?m?K?mol6Pc3.394?10b?0.08664RTc8.314?126.2?53?1?0.08664?2.678?10m?mol 6Pc3.394?10Aa0.3992???0.3975 BbRT1.52.678?10?5?8.314?2731.5BbbP2.678?10?5?101.3?1061.1952??∴h??? ①
ZVZRTZ?8.314?273ZZ?1A?h?1?h?????0.3975??? ② 1?hB?1?h?1?h1?h??①、②两式联立，迭代求解压缩因子Z 第三章
3-1. 物质的体积膨胀系数?和等温压缩系数k的定义分别为：
??1??V???V??T?P?V?。试导出服从Vander Waals状态方程的?和k的表，k??1???V??P?T达式。
解：Van der waals 方程P??z?由Z=f(x,y)的性质???RTa?2V?bV
?T??P???V??????????P??V???TP??T? ??1?V???x???y?????????1得 ??x?y??y?z??z?x又
2aRT??P?????23??V?TV?V?b?
R ??P??????T?VV?b