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 ?

Answer

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

Explanation

(a) `P=X_AP_A^0+X_BP_B^0=3/5xx30+2/5xx100=58 mmHg`
Composition of the first trace of vapour formed
`Y_("benzene")=(X_bP_b^0)/P_("total")=2/5xx100/58=20/29&Y_("toluence")`
`=1-20/29=9/29`
The last trace of liquid disappear 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^@=(2/5xx250/6)/100=1/6`
so `X_("toluence")=5/6`