2022年化工热力学陈钟秀第三版1-4章答案 .pdf
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1、2-1.使用下述方法计算1kmol 甲烷贮存在体积为0.1246m3、温度为 50的容器中产生的压力: (1)理想气体方程;(2)R-K 方程;(3)普遍化关系式。解:甲烷的摩尔体积V=0.1246 m3/1kmol=124.6 cm3/mol 查附录二得甲烷的临界参数:Tc=190.6K Pc=4.600MPa Vc=99 cm3/mol =0.008(1)理想气体方程P=RT/V=8.314 323.15/124.610-6=21.56MPa (2)R-K 方程22 . 522 . 560 . 5268. 31 41 90. 60. 42 7480. 42 74 83. 2224. 610
2、ccR TaPa mKmolP53168.314190.60.086640.086642.985104.610ccRTbmmolP0.5RTaPVbTV Vb50.5558.314 323.153.22212.462.98510323.1512.46 1012.462.98510=19.04MPa (3)普遍化关系式323. 15 190. 61. 6 95rcTT T124.6 991.259rcVV V2利用普压法计算,01ZZZcrZRTPP PVcrPVZPRT654.61012.46100.21338.314323.15crrrPVZPPPRT迭代:令 Z0=1Pr0=4.687 又
3、 Tr=1.695,查附录三得: Z0=0.8938 Z1=0.4623 01ZZZ=0.8938+0.008 0.4623=0.8975 此时, P=PcPr=4.6 4.687=21.56MPa 同理,取 Z1=0.8975 依上述过程计算,直至计算出的相邻的两个Z 值相差很小,迭代结束,得Z 和 P的值。 P=19.22MPa 2-2.分别使用理想气体方程和Pitzer 普遍化关系式计算510K、2.5MPa 正丁烷的摩尔体积。已知实验值为1480.7cm3/mol。解:查附录二得正丁烷的临界参数:Tc=425.2K Pc=3.800MPa Vc=99 cm3/mol =0. 193 精
4、选学习资料 - - - - - - - - - 名师归纳总结 - - - - - - -第 1 页,共 18 页(1)理想气体方程V=RT/P=8.314 510/2.5 106=1.696 10-3m3/mol 误差:1.6961.4807100%14.54%1.4807(2)Pitzer 普遍化关系式对比参数:510 425.21.199rcTT T2. 5 3. 80. 65 79rcPP P普维法01 . 61 . 60. 4 2 20. 4 2 20. 0 8 30. 0 8 30. 2 3 2 61. 1 9 9rBT14.24.20.1720.1720.1390.1390.058
5、741.199rBT01ccBPBBRT=-0.2326+0.1930.05874=-0.2213 11crcrBPBP PZRTRT T=1-0.2213 0.6579/1.199=0.8786 PV=ZRT V= ZRT/P=0.87868.314510/2.5 106=1.49 10-3 m3/mol 误差:1.491.4807100%0.63%1.48072-3.生产半水煤气时,煤气发生炉在吹风阶段的某种情况下,76%(摩尔分数)的碳生成二氧化碳,其余的生成一氧化碳。试计算: (1)含碳量为81.38%的 100kg 的焦炭能生成1.1013MPa、303K 的吹风气若干立方米?( 2
6、)所得吹风气的组成和各气体分压。解:查附录二得混合气中各组分的临界参数:一氧化碳 (1):Tc=132.9K Pc=3.496MPa Vc=93.1 cm3/mol =0. 049 Zc=0.295 二氧化碳 (2):Tc=304.2K Pc=7.376MPa Vc=94.0 cm3/mol =0. 225 Zc=0.274 又 y1=0.24,y2=0.76 (1)由 Kay 规则计算得:0.24 132.90.76304.2263.1cmiciiTy TK0.24 3.4960.767.3766.445cmiciiPy PMPa303 263.11.15rmcmTT T0. 101 1.
7、4 450. 01 57r mc mPP P普维法利用真实气体混合物的第二维里系数法进行计算011.61.610.4220.4220.0830.0830.02989303 132.9rBT114.24.210.1720.1720.1390.1390.1336303 132.9rBT精选学习资料 - - - - - - - - - 名师归纳总结 - - - - - - -第 2 页,共 18 页016111111618.314132.90.029890.0490.13367.378 103.49610ccRTBBBP021.61.620.4220.4220.0830.0830.3417303 3
8、04.2rBT124.24.220.1720.1720.1390.1390.03588303 304.2rBT016222222628.314304.20.34170.2250.03588119.93107.37610ccRTBBBP又0.50.5132.9304.2201.068cijcicjTT TK331 31 31 31 331293.194.093.55/22cccijVVVcmmol120.2950.2740.284522cccijZZZ120.2950.2250.13722cij6/0.28458.314201.068/93.55105.0838cijcijcijcijPZRTV
9、MPa303 201.0681.507rijcijTT T0. 1 01 3 5. 0 8380. 01 99r i jc i jPP P0121.61.6120.4220.4220.0830.0830.1361.507rBT1124.24.2120.1720.1720.1390.1390.10831.507rBT01612121212126128.314201.0680.1360.1370.108339.84 105.0838 10ccRTBBBP2211112122222mBy By y By B26626630.247.3781020.240.7639.84100.76119.93108
10、4.2710/cmmol1mmB PPVZRTRTV=0.02486m3/mol V总=n V=100103 81.38%/12 0.02486=168.58m3(2) 1110.2950.240.10130.0250.2845cmZPy PMPaZ精选学习资料 - - - - - - - - - 名师归纳总结 - - - - - - -第 3 页,共 18 页2220.2740.760.10130.0740.2845cmZPy PMPaZ2-4.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
11、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 =0. 250 (1)求取气体的摩尔体积对于状态: P=2.03 MPa、T=447K 、V=2.83 m3477 405.61.176rcTT T2.03 11.280.18rcPP P普维法01.61.60.4220.4220.0830.0830.24261.176rBT14.24.20.1720.1720.1390.1390.0
12、51941.176rBT010.24260.250.051940.2296ccBPBBRT11crcrBPPVBP PZRTRTRT TV=1.885 10-3m3/mol n=2.83m3/1.885 10-3m3/mol=1501mol 对于状态:摩尔体积V=0.142 m3/1501mol=9.45810-5m3/mol T=448.6K (2)Vander Waals方程222262627278.314405.60.4253646411.28 10ccR TaPa mmolP53168.314405.63.737108811.2810ccRTbmmolP22558.314448.60.
13、425317.659.4583.737103.73710RTaPMPaVbV(3)Redlich-Kwang 方程22.522.560.5268.314405.60.427480.427488.67911.28 10ccR TaPa mKmolP53168.314405.60.086640.086642.591011.2810ccRTbmmolP0.550.5558.314448.68.67918.349.4582.5910448.69.458 109.4582.5910RTaPMPaVbTV Vb精选学习资料 - - - - - - - - - 名师归纳总结 - - - - - - -第 4
14、 页,共 18 页(4)Peng-Robinson 方程448.6 405.61.106rcTT T220.37461.542260.269920.37461.542260.250.269920.250.7433k220.50.51110.743311.1060.9247rTkT22226268.314405.60.457240.457240.92470.426211.2810cccR Ta TaTTPa mmolP53168.314405.60.077800.077802.3261011.2810ccRTbmmolPa TRTPVbV Vbb Vb510108.314448.60.42629
15、.4582.326109.4589.4582.326102.3269.4582.3261019.00MPa(5)普遍化关系式559.458 107.25 101.305rcVV V2 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、6.888 MPa条件下的体积。已知B11=14cm3/mol,B22=-265cm3/mol,B12=-9.5cm3/mol。解:2211112122222mBy By y By B2230.31420.30.79.50.7265132.58
16、/cmmol1mmB PPVZRTRTV(摩尔体积 )=4.24 10-4m3/mol 假设气体混合物总的摩尔数为n,则0.3n 28+0.7n 58=7n=0.1429mol V= nV( 摩尔体积 )=0.1429 4.24 10-4=60.57 cm32-8.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 =0. 04 (1)R-K 方程的普遍化22.522.560.5268.314126.20.427480.427481
17、.55773.394 10ccR TaPa mKmolP53168.314126.20.086640.086642.678103.39410ccRTbmmolP精选学习资料 - - - - - - - - - 名师归纳总结 - - - - - - -第 5 页,共 18 页22.5aPAR TbPBRT1.551.51.55771.5512.678108.314273AaBbRT562.67810101.3 101.19528.314273BbbPhZVZRTZZ111.5511111AhhZhBhhh、两式联立,迭代求解压缩因子Z (2)SRK 方程的普遍化273126.22.163rcTT
18、 T220.4801.5740.1760.4801.5740.040.1760.040.5427m220.50.5111110.542712.1630.25632.163rrTmTT2222.560.5268.314126.20.427480.427480.25630.39923.394 10ccR TaTPa mKmolP53168.314126.20.086640.086642.678103.39410ccRTbmmolP1.551.50.39920.39752.678108.314273AaBbRT562.67810101.3 101.19528.314273BbbPhZVZRTZZ11
19、0.39751111AhhZhBhhh、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:1PVVT,1TVkVP。试导出服从Vander Waals状态方程的和k的表达式。解: Van der waals 方程2RTaPVbV由 Z=f(x,y) 的性质1yxzzxyxyz得1TPVPVTVTP又232TPaRTVVVbVPRTVb精选学习资料 - - - - - - - - - 名师归纳总结 - - - - - - -第 6 页,共 18 页所以2321PaRTVVbVTRVb3232PRVVbVTRTVa Vb故22312PRVVbVVTRTV
20、a Vb222312TVVbVkVPRTVa Vb3-2. 某理想气体借活塞之助装于钢瓶中,压力为34.45MPa,温度为 93,反抗一恒定的外压力3.45 MPa而等温膨胀,直到两倍于其初始容积为止,试计算此过程之U、H、S、A、G、TdS、pdV、Q 和 W。解:理想气体等温过程,U=0、H=0Q=-W=21112ln 2VVVVRTpdVpdVdVRTV=2109.2 J/mol W=-2109.2 J/mol 又PPdTVdSCdPTT理想气体等温膨胀过程dT=0、PVRTPRd Sd PP222111lnlnln2SPPPSPSdSRdPRPR=5.763J/(molK) AUTS=
21、-366 5.763=-2109.26 J/(molK) GHTSA=-2109.26 J/(molK) TdSTSA=-2109.26 J/(molK) 21112ln 2VVVVRTpdVpdVdVRTV=2109.2 J/mol 3-3. 试求算 1kmol 氮气在压力为10.13MPa、温度为 773K 下的内能、焓、熵、VC、pC和自由焓之值。假设氮气服从理想气体定律。已知:(1)在 0.1013 MPa 时氮的pC与温度的关系为27.220.004187 J/ mol KpCT;(2)假定在 0及 0.1013 MPa 时氮的焓为零;(3)在 298K 及 0.1013 MPa 时
22、氮的熵为191.76J/(mol K)。3-4. 设氯在 27、0.1 MPa 下的焓、熵值为零,试求227、 10 MPa 下氯的焓、熵值。已知氯在理想气体状态下的定压摩尔热容为36231.69610.144 104.038 10J/ mol KigpCTT解:分析热力学过程精选学习资料 - - - - - - - - - 名师归纳总结 - - - - - - -第 7 页,共 18 页300K 0.1 MPa H=0S=0,真实气体,HS、500K 10 MPa ,真实气体-H1RH2R -S1RS2R300K 0.1 MPa ,理想气体11HS、500K 10 MPa ,理想气体查附录二
23、得氯的临界参数为:Tc=417K 、Pc=7.701MPa、 =0.073 (1)300K 、0.1MPa 的真实气体转换为理想气体的剩余焓和剩余熵Tr= T1/ Tc=300/417=0.719 Pr= P1/ Pc=0.1/7.701=0.013利用普维法计算01.60.4220.0830.6324rBT02.60.6751.592rrdBTdT14.20.1720.1390.5485rBT15.20.7224.014rrdBTdT又0101RrrrcrrHdBdBPBTBTRTdTdT01RrrrSdBdBPRdTdT代入数据计算得1RH=-91.41J/mol、1RS=-0.2037
24、J/( molK) (2)理想气体由300K、0.1MPa 到 500K、 10MPa 过程的焓变和熵变21500362130031.696 10.144 104.038 10TigpTHC dTTT dT=7.02kJ/mol 215003621300110ln31.69610.144 104.038 10ln0.1igTpTCPSdTRTTdTRTP=-20.39 J/( molK) (3) 500K 、10MPa 的理想气体转换为真实气体的剩余焓和剩余熵Tr= T2/ Tc=500/417=1.199 Pr= P2/ Pc=10/7.701=1.299利用普维法计算01.60.4220.
25、0830.2326rBT02.60.6750.4211rrdBTdT14.20.1720.1390.05874rBT15.20.7220.281rrdBTdT又0101RrrrcrrHdBdBPBTBTRTdTdT01RrrrSdBdBPRdTdT代入数据计算得2RH=-3.41KJ/mol、2RS=-4.768 J/( molK) 精选学习资料 - - - - - - - - - 名师归纳总结 - - - - - - -第 8 页,共 18 页H=H2-H1= H2=-1RH+1H+2RH=91.41+7020-3410=3.701KJ/mol S= S2-S1= S2=-1RS+1S+2R
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