化工热力学第三版陈钟秀课后习题答案.docx

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1、第二章第二章3、温度为 50的容器中产生的压力：1理想气体方程；2方程；3普遍化关系式。解：甲烷的摩尔体积 0.1246 m3/1124.63查附录二得甲烷的临界参数：190.6K4.600993(1)理想气体方程8.314323.15/124.610-6(2)方程22.522.560.5268.314190.60.427480.427483.2224.6 10ccR TaPa mKmolP53168.314 190.60.086640.086642.985 104.6 10ccRTbmmolP0.5RTaPVbT V Vb50.5558.314323.153.22212.462.985103

2、23.1512.46 1012.462.98510(3)普遍化关系式323.15 190.61.695rcTT T124.6 991.259rcVV V2利用普压法计算，01ZZZcrZRTPPPVcrPVZPRT654.6 1012.46 100.21338.314323.15crrrPVZPPPRT迭代：令Z0=10=4.687又 1.695，查附录三得：Z0=0.8938Z101ZZZ=0.8938+0.0080.4623=0.8975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，迭代完毕，得 Z 和 P 的值。3。解：查附录二得正丁烷的临界参数：

3、425.2K3.800993=0.1931理想气体方程8.314510/106=10-3m3误差：1.6961.4807100%14.54%1.48072普遍化关系式比照参数：510 425.21.199rcTT T2.5 3.80.6579rcPP P普维法01.61.60.4220.4220.0830.0830.23261.199rBT 14.24.20.1720.1720.1390.1390.058741.199rBT 01ccBPBBRT11crcrBPBP PZRTRT T 0.87868.314510/2.5106=1.4910-3m3误差：1.491.4807100%0.63%1

4、.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 100 的焦炭能生成 1.1013、303K的吹风气假设干立方米？2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：3=0.3=0.又 y1=0.24，y2(1)由规那么计算得：0.24 132.90.76304.2263.1cmiciiTy TK0.243.4960.767.3766.445cmiciiPy PMPa303 263.11.15rmcmTT T0.101 1.4450.0157rmcmPP P普维法利用

5、真实气体混合物的第二维里系数法进展计算011.61.610.4220.4220.0830.0830.02989303 132.9rBT 114.24.210.1720.1720.1390.1390.1336303 132.9rBT016111111618.314 132.90.029890.0490.13367.378 103.496 10ccRTBBBP 021.61.620.4220.4220.0830.0830.3417303 304.2rBT 124.24.220.1720.1720.1390.1390.03588303 304.2rBT 016222222628.314304.20.

6、34170.225 0.03588119.93 107.376 10ccRTBBBP 又0.50.5132.9304.2201.068cijcicjTT TK331 31 31 31 331293.194.093.55/22cccijVVVcmmol120.2950.2740.284522cccijZZZ120.2950.2250.13722cij6/0.2845 8.314201.068/93.55 105.0838cijcijcijcijPZ RTVMPa303 201.0681.507rijcijTT T0.1013 5.08380.0199rijcijPP P0121.61.6120.

7、4220.4220.0830.0830.1361.507rBT 1124.24.2120.1720.1720.1390.1390.10831.507rBT01612121212126128.314201.0680.1360.1370.108339.84 105.0838 10ccRTBBBP 2211112122222mBy By y By B26626630.247.378 1020.240.7639.84 100.76119.93 108 4.27 10/cmmol 1mmB PPVZRTRT 3V总1001033(2)1110.2950.240.10130.0250.2845cmZPy

8、PMPaZ2220.2740.760.10130.0740.2845cmZPy PMPaZ33压缩到 0.142 m3，假设压缩后温度 448.6K，那么其压力为假设干？分别用下述方法计算：1 方程；2方程；3方程；4普遍化关系式。解：查附录二得33=0.250(1)求取气体的摩尔体积对于状态 m3477 405.61.176rcTT T2.03 11.280.18rcPP P普维法01.61.60.4220.4220.0830.0830.24261.176rBT 14.24.20.1720.1720.1390.1390.051941.176rBT010.24260.25 0.051940.2

9、296ccBPBBRT 11crcrBPPVBP PZRTRTRT T 1.88510-3m33/1.88510-3m31501对于状态：摩尔体积 0.142 m3/150110-5m3(2)方程222262627278.314405.60.42536464 11.28 10ccR TaPa mmolP53168.314405.63.737 1088 11.28 10ccRTbmmolP22558.314448.60.425317.659.4583.737103.737 10RTaPMPaVbV(3)方程22.522.560.5268.314405.60.427480.427488.67911

10、.28 10ccR TaPa mKmolP53168.314405.60.086640.086642.59 1011.28 10ccRTbmmolP0.550.5558.314448.68.67918.349.4582.5910448.69.458 109.4582.5910RTaPMPaVbT V Vb(4)方程448.6 405.61.106rcTT T220.37461.542260.269920.37461.542260.250.269920.250.7433k 220.50.51110.74331 1.1060.9247rTkT 22226268.314405.60.457240.4

11、57240.92470.426211.28 10cccR Ta TaTTPa mmolP53168.314405.60.077800.077802.326 1011.28 10ccRTbmmolP a TRTPVbV Vbb Vb510108.314448.60.42629.4582.326109.4589.4582.326102.3269.4582.3261019.00MPa(5)普遍化关系式559.458 107.25 101.305rcVV V2 适用普压法，迭代进展计算，方法同 1-132-6.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.

12、888 条件下的体积。B11=143，B222653，B129.53。解：2211112122222mBy By y By B2230.31420.3 0.79.50.7265132.58/cmmol 1mmB PPVZRTRT V(摩尔体积)=4.2410-4m3假设气体混合物总的摩尔数为 n，那么 nV(摩尔体积)=0.14294.2410-4=60.573解：适用的普遍化形式查附录二得3的临界参数：126.2K3.394=0.041方程的普遍化22.522.560.5268.314126.20.427480.427481.55773.394 10ccR TaPa mKmolP53168.

13、314 126.20.086640.086642.678 103.394 10ccRTbmmolP22.5aPAR TbPBRT1.551.51.55771.5512.678 108.314273AaBbRT562.678 10101.3 101.19528.314273BbbPhZVZRTZZ111.5511111AhhZhBhhh、两式联立，迭代求解压缩因子 Z2方程的普遍化273 126.22.163rcTT TEMBEDEquation.DSMT4220.4801.5740.1760.4801.5740.040.1760.040.5427m 220.50.5111110.542712.

14、1630.25632.163rrTmTT 2222.560.5268.314126.20.427480.427480.25630.39923.394 10ccR TaTPa mKmolP53168.314 126.20.086640.086642.678 103.394 10ccRTbmmolP1.551.50.39920.39752.678 108.314273AaBbRT562.678 10101.3 101.19528.314273BbbPhZVZRTZZ110.39751111AhhZhBhhh、两式联立，迭代求解压缩因子 Z第三章第三章3-1.物质的体积膨胀系数和等温压缩系数k的定义

15、分别为：1PVVT，1TVkVP。试导出服从 状态方程的和k的表达式。解：方程2RTaPVbV由()的性质1yxzzxyxyz 得1TPVPVTVTP 又232TPaRTVVVbVPRTVb所以2321PaRTVVbVTRVb 3232PRVVbVTRTVa Vb故22312PRVVbVVTRTVa Vb222312TVVbVkVPRTVa Vb 3-2.某理想气体借活塞之助装于钢瓶中，压力为 34.45，温度为 93，对抗一恒定的外压力 3.45 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=021112

16、ln2VVVVRTpdVpdVdVRTV又PPdTVdSCdPTT理想气体等温膨胀过程 0、PVRTPRdSdPP 222111lnlnln2SPPPSPSdSRdPRPR =5.763(K)AUT S 3665.7632109.26(K)GHT SA (K)TdST SA(K)21112ln2VVVVRTpdVpdVdVRTV、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。：时氮的pC及温度的关系为27.220.004187 J/mol KpCT；2假定在 0及 时氮的焓为零；3在 298K 及 时氮的熵为 191.76(K)。3-4.设氯在 27、下的焓、熵值为零，试求 227、10 下氯的焓、熵值。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300K 0.1 MPa H=0S=0，真实气体，