An ideal mixture of liquids A and B with 2 moles of A and 2 moles of B has a total vapour pressure of 1 atm at a certain temperature. Another mixture with 1 mole of A and 3 moles of B has a vapour pressure greater than 1 atm. But if 4 moles of C are added to second mixture, the vapour pressure comes down to 1 atm. Vapour pressure of `C, P_(C)^(@)=0.8` atm. Calculate the vapour pressure of pure A and B :

Options

A. `P_(A)^(@)=1.4atm,P_(B)^(@)=0.7atm`
B. `P_(A)^(@)=1.2atm,P_(B)^(@)=0.6atm`
C. `P_(A)^(@)=1.4atm,P_(B)^(@)=0.6atm`
D. `P_(A)^(@)=0.6atm,P_(B)^(@)=1.4atm`

Answer

4

Explanation

`(P_(A)^(@))/(2)+(P_(B)^(@))/(2)=1 "atm "implies" "P_(A)^(@)+P_(B)^(@)=2"atm"`
`(P_(A)^(@))/(4)+(3P_(B)^(@))/(8)gt"1 atm "implies" "P_(A)^(@)+3P_(B)^(@)gt1"atm"`
`& " "(P_(A)^(@))/(8)+(3P_(B)^(@))/(8)+(4P_(C)^(@))/(8)=1"atm "implies" "P_(A)^(@)+3P_(B)^(@)+4P_(C)^(@)=8"atm"`
So, `P_(A)^(@)+3P_(B)^(@)=(8-4xx0.8)"atm"=4.8"at"`
Hence `P_(B)^(@)=1.4 "atm"`
`P_(A)^(@)=0.6"atm"`