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InterviewSolution
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A liquid mixture of benzene and toluene is composed of 1 mol of benzene and 1 mol of toluence. a.If the pressure over the mixture at 300 K is reduced, at what pressure does the first vapour form? b.What is the composition of the first trace of vapour formed? c. If the pressure is reduced further, at what pressure does the last trace of liquid disappear? d.What is the composition of the last trace of liquid? e. What will be the pressure, the composition of the liquid, and the composition of the vapour, when 1 mol of the mixture is vapourized? , Given:p_(T)@ = 32.05 mm Hg , p_(B)@ = 103 mm Hg |
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Answer» SOLUTION :a. The first vapour will be formed when the external PRESSURE becomes equal to the vapour pressure of the SYSTEM. `P=chi_(T)p_(T).^(@)+p_(B).^(@)chi_(B)` `P = 1/2 (32.05) + 1/2 (103) = 67.52 mm Hg` b. Composition of the first trace of vapour formed `chi_(T)= (P_(T).^(@)chi_(T))/(P)= (0.5 xx 32.05)/(67.52)=0.24` `chi_(B) = 1 - 0.24 = 0.76` c. The last trace of liquid will disappear when the composition of the vapour phase become `chi_(B) = 0.5` and `chi_(T) = 0.5`. The pressure at which this OCCURS can be calculated as `1/P = (chi_(T))/(P_(T)@)+(chi_(B))/(P_(B)@)=(0.5)/(32.05)+(0.5)/(103)` `P=49.01 mm Hg` d. Composition of the last trace of the liquid will be `chi_(B)=(P_(B)@chi_(B))/(P)` `0.5 =(32.05 chi_(B))/(49.01)` `chi_(B) = 0.76` and `chi_(T) = 0.24` e. `chi_(T) = 0.642`, `chi_(B) = 0.358` and `chi_(B) = 0.642`,`chi_(T) = 0.358`, `P = 57.46 mm Hg` |
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