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时间:2017-05-06
《《高等化工热力学》习题及参考答案.doc》由会员上传分享,免费在线阅读,更多相关内容在教育资源-天天文库。
1、1.1如果流体服从vanderWaals方程,请导出A和G的偏离函数以及逸度系数的表达式。解:1.2请根据下列丙酮(1)-氯仿(2)溶液的实验数据(35.17℃)计算(1)以纯液态为标准态,丙酮的活度及活度系数,(2)以无限稀释为参考态,氯仿的活度及活度系数。x20.00000.05880.12320.29100.42320.51430.66350.79970.91751.000p1/kPa45.9343.0939.9030.7023.2818.0010.535.001.730.0p2/kPa0.0001.232.727
2、.3911.8515.7122.6929.9235.6139.08解:(1)以纯液态为标准态,则(1)以无限稀释为参考态,则x20.00000.05880.12320.29100.42320.51430.66350.79970.91751.000p1/kPa45.9343.0939.9030.7023.2818.0010.535.001.730.0p2/kPa0.0001.232.727.3911.8515.7122.6929.9235.6139.0810.93820.86870.66840.50690.39190.22
3、930.10890.0377010.99680.99080.94270.87870.80690.68130.54350.456600.03150.06960.18910.30320.40200.58060.76560.91121=0.500.53530.56490.64980.71650.78160.87510.95740.993111.00001.07051.12991.29971.43301.56331.75011.91471.98632.00001.3已知对45℃时四氯化碳(1)-乙腈(2)混合物的平衡气相组成,总
4、蒸气压及混合热如下,请分别计算:(1)活度,与的关系曲线,(2)45℃时及与的关系曲线,(3),及与的关系曲线。x100.1280.3170.4140.6310.8211y100.4000.5210.5600.6080.6701P/kPa27.5040.5348.0148.6549.0947.1034.50△H/J/mol04147458609307360解:(1)以纯液态为标准态,由表知,则x10.00000.1280.3170.4140.6310.8211.000x210.8720.6830.5860.3690.17
5、90y10.0000.4000.5210.5600.6080.6701y210.6000.4790.4400.392.03300P27.540.5348.0148.6549.0947.1034.5p1/kPa016.2125.0127.2429.8531.5634.5p2/kPa27.524.3223.0020.7718.817.53000.450.720.790.870.91110.880.840.760.680.640(2)恒温下,x10.000.1280.3170.4140.6310.8211.000J/mol0-
6、541-594-646-591-4640J/mol095513391506152112000(3)x10.000.1280.3170.4140.6310.8211.000J/mol0624106011501153779004147458609307360J/mol0-210-315-290-223-4301.4已知NaCl水溶液中NaCl的浓度mB及平均离子活度系数如下,请求出相应的xB及,并请讨论,这些数据可说明什么问题?mB0.0010.010.10.51.02.04.06.00.96490.89600.77840.
7、68110.65690.66760.79320.9862解:mB0.0010.010.10.51.02.04.06.00.96490.89600.77840.68110.65690.66760.79320.98620.0000180.00020.00180.00880.01740.03360.06290.08880.96490.89630.78120.69340.68050.71570.90741.19922.1由和,导出巨正则系综热力学函数关系式.:又由于:或者:由麦克斯韦变换关系式:2.3在1atm,300K,V=1
8、cm3条件下,求Ne的分子配分函数q,此值有无量纲?并请求出正则配分函数(以表示)。(已知mH=1.660×10-27kg,MNe=20.179)解:Ne为单原子分子:只有平动加电子配分函数无量纲2.4(1)请证明(2)请证明(式中),并由此说明,尽管可以忽略,但是一个不小的量。解:(1)∵,∴(2)若,大于04.2
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