化工热力学课后题答案马沛生 下载本文

0 1 2 3 4 1 0.8779 0.8826 0.8823 0.8823 0.2083 0.2373 0.2360 0.2361 0.2361 V?ZRT0.8823?8.314?323.15?433?1??1.265?10m/mol=126.5cm?mol 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?369.832.560.5-2 /pc?0.42748??18.296Pa?m?K?mol64.248?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?62??/pc???T?=0.42748??1.2258=1.6548Pa?m/mol63.640?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?h1?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?23??6.1015?10m/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?h?1?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?h1?h??和式(2-16b) h?同样迭代不收敛

bB1.1791 ??VZZ采用RKS方程解三次方程得: V=0.00004512m3/mol

pV4.512?10?5?1000?105Z???1.9881

RT8.314?2732-16.试用下列各种方法计算水蒸气在107.9×105Pa、593K下的比容,并与水蒸气表查出的数据(V?0.01687m?kg(1)理想气体定律 (2)维里方程 (3)普遍化RK方程

解:从附录三中查得水的临界参数为:Tc=647.13K,pc=22.055MPa,?=0.345 (1)理想气体定律

3?1)进行比较。

V?RT8.314?593?63?13?1??4.569?10m?mol?0.02538m?kg 5p107.9?10误差=

0.01687?0.02538?100%??50.5%

0.01687T593??0.916 Tc647.13(2) 维里方程

Tr?p107.9?105pr???0.489 6pc22.055?10使用普遍化的第二维里系数:

B(0)?0.083?0.422/Tr1.6?0.083?0.422??0.4026 Tr1.60.172??0.1096 4.2TrB(1)?0.139?0.172/Tr4.2?0.139?