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Mixture of two liquids A and B is placed...

Mixture of two liquids A and B is placed in cylinder containing piston. Piston is pulled out isotehrmally so that volume of liquid decreases but that of vapour increases. When negligibly small amount of liquid was remaining the mole fraction of A in vapour is `0.4`. Given `P_(A)^(@) = 0.4` atm and `P_(B)^(@) = 1.2` atm at the experimental temperature. Calculate the total pressure at whcih the liquid has almost evaporated. (Assume ideal behaviour)

Text Solution

Verified by Experts

The correct Answer is:
`0.66 atm`
`P_(tau)=P_(A)^(@)X_(A)P_(B)^(@)X_(B) = P_(A)^(@) X_(A) +P_(B)^(@) (1-X_(A))`
`P_(tau) = P_(B)^(@) +X_(A) (P_(A)^(@) -P_(B)^(@))`
`Y_(A) = (P_(A)^(@) xx A)/(P_tau) = (P_(A)^(@)-A)/(P_(B)^(@)+X_(A)(P_(a)^(@) -P_(B)^(@)))`
`0.4 = (0.4X_(A))/(1.2 =0.8 X_(A))`
`1.2 = 1.8 xx A`
`X_(A) = (2)/(3)`
so `(1)/(3)`
`P_(tau) = 0.4 xx (2)/(3) +1.2 xx (1)/(3) = (2)/(3) = 0.66` atm
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