化工热力学马沛生第二版习题答案(7)

2019-01-18 20:07

第二章 流体的p-V-T关系

12.67MPa、550K时的?H,?S,?V,?U。已知1,3-丁二烯在理想气体状态时的恒压热容

?1?52为:Cigp?22.738?2.228?10T?7.388?10T kJ?kmol?K,1,3-丁二烯的临

-1-1界常数及偏心因子为Tc=425K,pc=4.32MPa,Vc=221×10-6m?mol,?=0.193

400K,2.53MPa

3?1?H,?S

?HR1550K,12.67MPa

H2R

?S1R理想气体 400K,2.53MPa

解:

初态 Tr1?SR2

理想气体

?H,?S

igig550K,12.67MPa

4002.53?0.941,pr1??0.585 4254.3255012.67Tr2??1.294,pr2??2.929

4254.32参照图2-11,初态用第二Virial系数关系式 终态用三参数图 (1)

B(0)?0.083?0.422Tr1.6?0.083?0.422?0.941?1.6??0.382

B(1)?0.139?0.1720.172?0.139???0.083 4.24.2Tr?0.941?dB(0)0.6750.675?2.6??0.791 2.6dTr?0.941?TrdB(1)0.7220.722?5.2??0.991 5.2dTrTr?0.941?由式(3-78)得:

R?B(0)dB(0)?B(1)dB(1)??H1??pr?????????RTTdTTdTrr???r?r??0.382?0.083???0.585??0.791?0.193???0.991??

?0.941???0.941??0.8221

31

第二章 流体的p-V-T关系

H1??0.8221?8.314?400??2733.9J?mol-1

由式(3-79)得:

R?dB(0)S1dB(1)???pr???)??0.5746 ???0.585(0.791?0.193?0.991RdTr??dTrR S1??0.5746?8.314??4.7774J?mol-1?K-1 由式(2-30)和(2-31)得:

RZ1?1?Bp?1?B(0)??B(1)RT?p?????T?rr?0.585????1??0.382?0.193?0.083??0.7526?0.941?V1?Z1RT10.7526?8.314?400?63?1 ??989.21?10m?mol6p12.53?10??(2)计算理想气体的焓变和熵变

?Hig??CigpdTT1T2??550400?22.738?222.796?10?3T?73.879?10?6T2dT?222.796?10?373.879?10?6322?22.738??T2?T1???T2?T1?T2?T1323?16760J?mol-1????ig?Sig??ST??Sigp366.15Cpp1?Rln?dTp2?255.15Tig5502.53dT ???22.738?222.796?10?3T?73.879?10?6T2?40012.67T?22.002J?mol-1?K-1?8.314ln(3)由Tr2?1.294,pr2?2.929查图(2-9)和(2-10)得:Z?0??0.64,Z?1??0.20

Z2?Z?0???Z?1??0.64?0.193?0.20?0.6786 V2?Z2RT20.6786?8.314?550?63?1 ??244.91?10m?mol6p212.67?10??查图(3—4)、(3—6)、(3—8)、(3—10),分别得到:

?H?RTcR0?H???2.1,

RTcR1??0.5

?S?R

R0?S???1.2,

RR1??0.45

32

第二章 流体的p-V-T关系

由式(3-87)得:

H2HRHR?????2.1?0.193???0.5???2.197 RTcRTcRTcRH2??2.197?RTc??2.197?8.314?425??7761.22J?mol?1

R??0??1??由式(3-88)得:

S2SR?RRR??0?S???RR1??1.2?0.193???0.45???1.287

RS2??1.287?R??1.287?8.314??10.699J?mol?1?K?1

R(4)?H??H1R??Hig?H2?2733.9?16760?7761.22?11.733?103J?mol?1 R?S??S1R??Sig?S2?4.7774?22.002?10.699?16.0804J?mol?1?K?1

???????V?V2?V1??244.91?989.21??10?6??744.3?10?6m3?mol?1

?U??H???pV???H??p2V2?p1V1??11.132?103J?mol?1?11.733?103?12.67?106?244.91?10?6?2.53?106?989.21?10?6

??????3-9. 假设氯在300K、1.013×105Pa下的焓值和熵值为0,试求500K、1.013×107Pa下氯的焓值和熵值。

解:将计算分解为以下几步:

300K,0.1013MPa

?H,?S

?HR1500K,10.13MPa

H2R

?S1R理想气体 300K,0.1013MPa

SR2

理想气体

?H,?S

igig500K,10.13MPa

已知氯的临界参数为:Tc=417.15K,pc=7.711MPa,?=0.069

Tr1?Tr23000.1013?0.719,pr1??0.0131

417.157.71150010.13??1.199,pr2??1.314 417.157.711R初态压力较低,H1?0,S1R?0

33

第二章 流体的p-V-T关系

根据图2—11,末态应该使用普遍化的焓差图和熵差图进行计算, 查图(3—4)、(3—6)、(3—8)、(3—10),分别得到:

?H?RTcR0?H???1.2,

RTcR1??0.3

?S?RR0?S???0.72,

RR1??0.3

由式(3-87)得:

HRHRHR?????1.2?0.069???0.3???1.221 RTcRTcRTcHR??1.221?RTc??1.221?8.314?417.15??4233.6J?mol?1

由式(3-88)得:

??0??1??SRSR?RR??0?S???RR1??0.72?0.069???0.3???0.699

SR??0.699?R??0.699?8.314??5.811J?mol?1?K?1

查附录六,氯气的理想气体热容表达式为:

??Cig?10?3T?0.8098?10?5T2?0.5693?10?8T3?0.15256?10?11T4p?R3.056?5.3708?Hig?????CigpdTT1T2?8.314??500300?3.056?5.3708?10??3T?0.8098?10?5T2?0.5693?10?8T3?0.15256?10?11T4dT??5.3708?10?30.8098?10?53223??3.056?T?T??T?T?T?T?212121??23??8.314???8?11?0.5693?100.15256?10?4455T?T?T?T2121??45???7025.0J?mol-1??????? 34

第二章 流体的p-V-T关系

500Cpp1?Rln?dTp2?300Tigig?Sig??ST??Sigp?8.314ln??366.150.1013?8.31410.13?3255.15?3.056?5.3708?10T?0.8098?10?5T2?0.5693?10?8T3?0.15256?10?11T4?dTT5000.8098??2?3?52??3.056?ln?5.3708?10?T?T??10?T?T?2121??0.10133002?8.314ln?8.314?????10.13?0.5693?10?8??T33?0.15256?11?4??32?T1?4?10?T42?T1???38.287?17.897??20.391J?mol-1?K-1HR2?H1??H?H1?H1??Hig?HR2?0?0?7025.0?4233.6

?2791.4?J?mol?1?Sig2?S1??S?S1?SR1??S?SR2?0?0?20.391?5.811

??26.202?J?mol?1?K?1?3-10. 试用普遍化方法计算二氧化碳在473.2K、30MPa下的焓与熵。已知在相同条件下,二氧化碳处于理想状态的焓值为8377J?mol?1,熵为25.86J?mol-1?K-1。 解:需要计算该条件下二氧化碳的剩余焓和熵

已知二氧化碳的临界参数为:Tc=304.19K,pc=7.382MPa,?=0.228

Tr?473.2304.19?1.556,p30r?7.382?4.064

根据图2—11,应该使用普遍化的焓差图和熵差图进行计算, 查图(3—4)、(3—6)、(3—8)、(3—10),分别得到:

?HR?0R1??1.75,

?H?RTcRT??0.1

c?SR?0R1??0.85,

?S?RR??0.24

由式(3-87)得:

HR??0??1RT?HR??HR??1.75?0.228???0.1???1.773 cRTcRTcHR??1.773?RTc??1.773?8.314?304.19??4483.5?J?mol?1?

由式(3-88)得:

35

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