发布时间 : 星期日 文章第三章 化工热力学习题解答更新完毕开始阅读5ca3fb25bcd126fff7050b36
真实气体 127℃ 2.53MPa ΔH，ΔS，ΔU，ΔV 真实气体 277℃ 12.67MPa ① ④ 理想气体 127℃ 2.53MPa 理想气体 127℃ 理想气体 277℃ ②12.67MPa ③ 12.67MPa （1）127℃，2.53MPa下真实气体转变成理想气体
查表知，Tc=425K, Pc=4.327MPa，ω=0.195 Tr?400.15?0.94 p?2.53?0.585
r4254.327查图知用普遍化维利系数法计算。
B0=0.083?0.422=?0.383 1.6Tr B1=0.139?0.172=?0.084
4.2Tr Bpc?B0??B1??0.383?0.195???0.084???03994
RTc
Z1?ppVBp?1??1?r?B0??B1?RTRTTr0.585??0.383?0.195?0.084??0.75140.942.53?10
?1? V1?ZRT?0.7514?8.314?400.15?9.8813?10?4 m3?mol-1
6p dB0dTr? 0.675?0.793Tr2.6 0.722?0.996Tr5.2
dB1dTr???dB0B0??dB1B1??H1R? ??pr????????0.826????RTdTTdTr??rTr????r H1R??0.826RT1??0.826?8.314?400.15??2748.22 kJ?kmol-1?dB0S1RdB1???pr??????0.5742RdTdT?rr?S1R??0.5742?8.314??4.774 kJ?kmol-1?k-1
（2） 理想气体恒温加压
?HT?0
5
?ST??Rln12.67
??13.39 kJ?kmol-1?K-12.53（3） 理想气体恒压升温
T21id?H??222.796?10?3??550.152?400.152??p??T1CpdT?22.738??550.15?400.15??21?73.879?10?6??550.153?400.153?3?16788 kJ?kmol-1 ?S??300CpdT?22.738?ln550.15?222.796?10?3?550.15?400.15
??p?273.15T400.151?73.879?10?6???550.152?400.152??35.393 kJ?kmol-1?K-12id（4） 理想气体转变为真是气体
Tr?550.1512.67?1.3 pr??2.912 4254.327用普遍化压缩因子法计算，
查图可知 Z0?0.64 Z1?0.2
(HR) (HR)??0.5??2.1RTcRTcR' (S)??1.2 (SR)
??0.45RR00'Z2?Z0??Z1?0.64?0.195?0.2?0.672
V2?Z2RT20.672?8.314?550.15??2.882?10?4 m3?mol-1 6p212.67?100' (H2R)(HR)(HR)?????2.198RTcRTcRTc S2RR??1.2?0.195?(?0.45)??1.288
?H2R??2.198?8.314?425??7766.5 kJ?kmol-1 S2R??1.288?8.314??10.708 kJ?kmol-1?K-1
故 ?V?V2?V1??2.882?9.8813??10?4??6.999 m3?mol?1
idR-1 ?H= H1R??HT??Hidp?(?H2)?11769.7 kJ?kmolidR-1-1 ?S= S1R??ST??Sidp?(?S2)?14.0378 kJ?kmol?K ?U??H??(pV)?11769.7?(12.67?106?2.882?10?4?2.53?106?9.8813?10?4) ?10618.3 kJ?kmol-1
3-6 计算氨的热力学性质时，通常把0℃饱和液氨的焓规定为418.6kJ·kg-1，此时的饱和
6
蒸气压为0.43 MPa，汽化热为21432kJ·kmol-1，饱和液氨的熵为4.186 kJ·kg-1·K-1，试由此基准态数据求：
(1)1.013MPa，300K气氨的焓和熵； (2)30.4MPa，500 K气氨的焓和熵。 解：（1）设计过程如下：
273.15K0.43MPa饱和液氨①273.15K0.43MPa气氨②273.15K0.43MPa理想气氨③300K1.013MPa理想气氨 ④300K1.013MPa气氨
① 273K，0.43MPa下液氨汽化
-1-1?HV?21432 kJ?kmol-1 ?SV?78.462 kJ?kmol?K
② 273K，0.43MPa下真实气体转变成理想气体 查表知，Tc=405.6K, Pc=11.28MPa，ω=0.250
T?273.15?0.673 pr?0.43?0.0381 r405.611.28查图知用普遍化维利系数法计算。
B0=0.083?0.422=?0.712 1.6Tr B1=0.139?0.172=?0.769
4.2Tr dB0dTr? 0.675?1.89Tr2.6 dB1 0.772??5.66dTrTr5.2??dB0B0? ?dB1B1??H1R? ??pr?Tr????????0.119????RTc?dTrTr????dTrTr? ?dB0S1RdB1???pr?????0.126?RdTr??dTr故 H1R??0.119?RTC??0.119?8.314?405.6??401.287 kJ?kmol-1
S1R??0.126?R??0.126?8.314??1.048 kJ?kmol-1?k-1
③ 273K，0.43MPa下理想气体变化为300K，1.013MPa的理想气体 查表已知 Cid?27.31?0.02383T?1.707?10?5T2?1.185?10?8T3
pT21id?Hp??CiddT?27.31??300?273.15???0.02383??3002?273.152??pT12
11 ?1.707?10?5??3003?273.153???1.185?10?8??3004?273.154?34-1 ?961.585 kJ?kmol 7
?S??1.013300?27.31?ln?0.02383??300?273.15? 273.15T0.43273.1511 ?1.707?10?5???3002?273.152??1.185?10?8???3003?273.153?231.013 ?8.314?ln0.43 ??3.766 kJ?kmol-1?K-1idp300idCpdT?Rln④ 300K，1.013MPa的理想气体变化为300K，1.013MPa的真实气体
Tr? 3001.013?0.740Pr??0.0898405.611.280.422
??0.61.6Tr查图知用普遍化维利系数法计算。
B0?0.083?
B'?0.139?0.172
??0.474.2Tr dB00.6750.675?2.6??1.477dTrTr0.742.6 dB'dTr? 0.7220.722??3.456Tr5.20.745.2
?dB0B0H2RdB1B1???prTr?(?)??(?)?RTcdTrTr??dTrTr??0.6?0.47?????0.898?0.74???1.477???0.25??13.456??0.74?0.74????????0.219? H2R??0.219?8.314?405.6??738.5 kJ?kmol-1
?dB0S2RdB1???Pr?????0.898?(1.477?0.25?3.456)??0.210?RdTr??dTrS2R??0.210?8.314??1.746 kJ?kmol-1?K-1
又因 H0?418.6?17?7116.2 kJ?kmol-1
S0=4.186?17=71.162 kJ?kmol-1?K-1
故 H= H0?(?H1R)?(H2R)??HV??Hid?28370 kJ?kmol-1
S= S0?(?S1R)?(S2R)??SV??S?143.1 kJ?kmol-1?K-1
id(2) 同理可求出30.4MPa，500 K气氨的焓和熵。 过程①和②的结果与上述相同 过程③的焓变和熵变为：
?Hidp??1idCpdT?27.31??500?273.15??0.02383???5002?273.152?273.15211 ?1.707?10?5???5003?273.153??1.185?10?8???5004?273.154?34-1 ?3080.56 kJ?kmol500 8