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?
Given:p_(T)@ = 32.05 mm Hg , p_(B)@ = 103 mm Hg

Explanation

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`