The vapoure pressure of two pure liquids A and B, that from an ideal solution are 100 and 900 torr respectively at temperature T. This liquid solution of A and B is composed of 1 mole of A and 1 mole of B. What will be the pressure, when 1 mole of mixture has been vaporized?

Answer

`300 mm Hg`

Explanation

Let `n_(B)` mole of B present in 1 mole of mixture that has been vaporized.
Thus, `Y_(B) = n_(B)`
`X_(B)=1-n_(B)`
`P=P_(A)^(@)X_(A)+P_(B)^(@)X_(B)=P_(A)^(@)+X_(B)(P_(B)^(@)-P_(A)^(@))`
`X_(B)=(P-P_(A)^(@))/(P_(B)^(@)-P_(A)^(@))=1-n_(B)` ....(i)
`Y_(B) = (P_(B)^(@)X_(B))/(P) = gt n_(B) = (P_(B)^(@)(1-n_(B)))/(P)`
`n_(B)P=P_(B)^(@)-n_(B)P_(B)^(@)`
`n_(B) = (P_(B)^(@))/(P_(B)^(@)+P)` .....(ii)
from equation (i) and (ii)
`- (P_(B)^(@))/(P_(B)^(@)+P) = (P-P_(A)^(@))/(P_(B)^(@)-P_(A)^(@)) = (P-P_(A)^(@))/(P_(B)^(@)-P_(A)^(@))`
on solving
`P = sqrt(P_(A)^(@)P_(B)^(@)) = sqrt(100xx900) implies 300` mm Hg