The vapour pressures of two pure liquids `A` and `B` that form an ideal solution are `300` and 800 torr, respectively, at tempertature `T`. A mixture of the vapours of A and B for which the mole fraction of A is 0.25 is slowly compressed at temperature T. Calculate
a. The composition of the first drop of the condensate.
b.The total pressure when this drop is formed.
c. The composition of the solution whose normal boiling point is `T`.
d. The pressure when only the last bubble of vapour remains.
e. Composition of the last bubble.

Explanation

Given
`p_(A)@ = 300` torr,
`chi_(A)^(l) = 0.25, chi_(B)^(l) = 1 - 0.25 = 0.75`
a. By the condensation of only one drop, we can assume that the composition of the vapour remanns the same.
`chi_(A)^(V) = (p_(A)@chi_(A)^(l))/p_(T)` and `chi_(B)^(V) = (p_(B)@chi_(B)^(l))/p_(T)`
or `chi_(A)^(V)/ chi_(B)^(V) = p_(A)@chi_(A)/p_(B)@(1-chi_(A))`
Putting various known values, we get
`chi_(A)^(V) = 0.111` and `chi_(B)^(V) = 0.888`
b. `p=p_(A)@chi_(A)^(V) + p=p_(B)@chi_(B)^(V)`
= `300 xx 0.11+ 800 xx 0.888`
`= 743.7`
c. `760 = 300chi_(A) + 800chi_(B)`
`chi_(A) = 0.08` and `chi_(B) = 0.92`
d. When only the last bubble of vapour remains, we can assume the composition of vapour is now the composition of the condensate.n
Hence, `P= (0.25 xx 300) + 0.75 xx 800 = 675` torr
e. Composition of last bubble
`(chi_(A) = p_(A)@chi(A))/P = (0.25 xx 300 )/675 = 0.11`
`chi_(B) = 0.89`