Benzene and toluenen form nearly ideal solutions. At `300K,P_("toluene")^(@)=30 mmHg` and `P_("benzene")^(@)=100mm Hg`. A liquid mixture is composed of `3` mol of toluene and `2` mol of benzene.
(a) If the pressure over the mixture at `300K` is reduced. At what pressure does the first vapor form ?
(b) What is the composition of the first trace of vapor 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 ?

Answer

(a) `58 mm Hg` (b) y_(b)=20//29` (c) `250//6 mm Hg` (d) `x_(b)=1//6`

Explanation

(a) `P=X_(A)P_(A)^(@)+X_(B)P_(B)^(@)`
`=(3)/(5)xx30+(2)/(5)xx100=58mm Hg`
(b) Comportition of the first trace of vapour formed
`Y_("benzene")=(X_(b)P_(b)^(0))/(P_("Total"))=(2)/(5)xx(100)/(58)=(20)/(29)`
`& Y_("toluene")=1-(20)/(29)=(9)/(29)`
(c) The last trace of liquid disapppear at `Y_(A)=3//5`
`Y_(B)=2//5`
`(1)/(P)=(Y_(A))/(P_(A)^(0))+(Y_(B))/(P_(B)^(0))`
`(1)/(P)=(3)/(5xx30)+(2)/(5xx100)` or `P=(250)/(6) mm Hg`
(d) The composition of the last trace of liquid
`X_("benzene")=(Y_("benzene")P_("Total"))/(P_(b^(@))=((2)/(5)xx(250)/(6))/(1000)=(1)/(6)`
So `X_("Total")=(5)/(6)`