熱力學(xué)作業(yè)題答案

1、第二章第二章21.使用下述方法計(jì)算1kmol甲烷貯存在體積為0.1246m3、溫度為50℃的容器中產(chǎn)生的壓力：（1）理想氣體方程；（2）RK方程；（3）普遍化關(guān)系式。解：甲烷的摩爾體積V=0.1246m31kmol=124.6cm3mol查附錄二得甲烷的臨界參數(shù)：Tc=190.6KPc=4.600MPaVc=99cm3molω=0.008(1)理想氣體方程P=RTV=8.314323.15124.6106=21.56MPa(2)RK方程

2、22.522.560.5268.314190.60.427480.427483.2224.610ccRTaPamKmolP?????????53168.314190.60.086640.086642.985104.610ccRTbmmolP?????????∴??0.5RTaPVbTVVb????????50.5558.314323.153.22212.462.98510323.1512.461012.462.98510????????

3、????=19.04MPa(3)普遍化關(guān)系式＜2323.15190.61.695rcTTT???124.6991.259rcVVV???∴利用普壓法計(jì)算，01ZZZ???∵crZRTPPPV??∴crPVZPRT?654.61012.46100.21338.314323.15crrrPVZPPPRT????????迭代：令Z0=1→Pr0=4.687又Tr=1.695，查附錄三得：Z0=0.8938Z1=0.4623=0.89380.0

4、080.4623=0.897501ZZZ???此時(shí)，P=PcPr=4.64.687=21.56MPa同理，取Z1=0.8975依上述過程計(jì)算，直至計(jì)算出的相鄰的兩個(gè)Z值相差很小，迭代結(jié)束，得Z和P的值?！郟=19.22MPa24.將壓力為2.03MPa、溫度為477K條件下的2.83m3NH3壓縮到0.142m3，若壓縮后溫度448.6K，則其壓力為若干？分別用下述方法計(jì)算：（1）VerWaals方程；（2）RedlichKwang方程

5、；（3）Peng??????220.50.51110.743311.1060.9247rTkT???????????????????????22226268.314405.60.457240.457240.92470.426211.2810cccRTaTaTTPammolP?????????????53168.314405.60.077800.077802.3261011.2810ccRTbmmolP??????????∴??????a

6、TRTPVbVVbbVb????????????510108.314448.60.42629.4582.326109.4589.4582.326102.3269.4582.32610???????????????19.00MPa?(5)普遍化關(guān)系式∵＜2適用普壓法，迭代進(jìn)行計(jì)算，方法同1559.458107.25101.305rcVVV???????1（3）27：答案：3cm第三章第三章33.試求算1kmol氮?dú)庠趬毫?0.13MPa

7、、溫度為773K下的內(nèi)能、焓、熵、、和自由焓之值。VCpC假設(shè)氮?dú)夥睦硐霘怏w定律。已知：（1）在0.1013MPa時(shí)氮的與溫度的關(guān)系為；pC??27.220.004187JmolKpCT???（2）假定在0℃及0.1013MPa時(shí)氮的焓為零；（3）在298K及0.1013MPa時(shí)氮的熵為191.76J(molK)。答案：8272KJKmol14703KJKmol181.4JKmolK22.13KJKmolK30.45JKmolK125

8、507KJKmol38.試估算純苯由0.1013MPa、80℃的飽和液體變?yōu)?.013MPa、180℃的飽和蒸汽時(shí)該過程的、V?和。已知純苯在正常沸點(diǎn)時(shí)的汽化潛熱為3.733Jmol；飽和液體在正常沸點(diǎn)下的體積為95.7H?S?cm3mol；定壓摩爾熱容；第二維里系數(shù)??16.0360.2357JmolKigpCT???。2.4310mol???????31B=78cmT解：1.查苯的物性參數(shù)：Tc=562.1K、Pc=4.894MPa