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The vapour pressures of two pure liquids...

The vapour pressures of two pure liquids `A` and `B` that form an ideal solution are `300` and 800 torr, respectively, at tempertature `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.

Text Solution

Verified by Experts

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`
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