1故所设温度偏低,重设T为95℃时=176.00kPa, =63.47kPa, =27.01kPa=1.11>1故所设温度偏高,重设T为91.19℃,=160.02kPa, =56.34kPa, =23.625kPa=1.0000125≈1故用平衡常数法计算该物系在100kPa时的平衡温度为91.19℃汽相组成:====0.8001 ====0.1409====0.059(2)相对挥发度法由于是理想混合物,所以, 得对于理想混合物,得=设T为80℃时,=101.29kPa, =38.82kPa, =15.63kPa故=2.61, =6.48, =/5.22, =/12.96因为=1,故=0.788又因为=100×0.788=78.8kPa,而=101.29×0.5=50.645kPa<故所设温度偏低;重设T=92℃时=163.31kPa, =57.82kPa, =24.31kPa得故=2.824, =6.718, =/5.648, =/13.436因为=1,故=0.799,=0.141, =0.0595且=100×0.799=79.9kPa,而=163.31×0.5=81.655kPa,基本相等因此,由相对挥发度计算该物系平衡温度为92℃,此时=0.799,=0.141, =0.05955.一烃类混合物含有甲烷5%、乙烷10%、丙烷30%及异丁烷55%(mol),试求混合物在25℃时的泡点压力和露点压力。解:设甲烷为1组分,乙烷为2组分,丙烷为3组分因为各组分都是烷烃,汽液相均可视为理想溶液,故符合乌拉尔定律。25℃时,=30768.14kPa, =4118.81kPa, =347.59kPa (1)泡点压力=30768.14×5%+4118.81×10%+950.31×30%+347.59×55%=2426.56kPa(2) 露点压力时由乌拉尔定律得,
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