At 353 K, the vapoure pressure of pure ethylene bromide and propylene bromide are 22.93 and `16.93 K N m^(-2)`. Respectively and these compounds forms nearly ideal solution 3. moles of ethylen bromide and 2 mole of propylene bromide are equilibrated at 353 K and at a total pressure of `20.4 KN m^(-2)`.
(a) What is the composition of the liquids phase
(b) How many moles of each compound are present in the vapour phase ?

Explanation

(a) Let ethylene bromide `rarr to A`
And propylene bromide `to B`
Then from equation,
`P_(A)^(@)=22.93 N Km^(-2)`
`P_(B)^(0) 18.93 K Nm^(-2)`
`n_(A)=3` mole
`n_(B)=3` mole
`n_(B)=2` mole
Total pressure`P_(T)=20.4 K Nm^(-2)`
`:.P_(T)=P_(A)^(@)xxP_(B)^(@)X_(B)`
`=P_(A)^(@)xxP_(B)^(@)(1-X_(A))=(P_(A)^(@)-P_(B)^(@))X_(A)+P_(B)^(@)`
`rArr X_(A)=(P_(T)-P_(B)^(@))/(P_(A)^(@)-P_(B)^(@))=(20.4-16.93)/(22.93-16.93)=0.578`
`:.X_(A)=1-0.578=0.422`
(b) Let mole fraction in vapour phase `=X_(A)`
`X_(A)=(P_(A)^(0))/(P_(T))`
`X_(A)=(22.93xx0.598)/(20.4)=0.64" "...(1)`
Assuming that the no. of moles of A and B that are vaporized are a and b then
`X_(A)=(a)/(a+b)=0.64`
But composition of A in liquid phase.
`X_(A)=(3-a)/((3-a)+(2-b))=0.578`
`=(3-a)/(5-(a+b))=0.578 " "...(2)`
Solving equation (1) and (2)
`a=0.9967` mole
`b=0.537` mole