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The vapour pressure of two pure liquids ...

The vapour pressure of two pure liquids A and B, that form an ideal solution are 100 and 900 mm Hg 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?

Text Solution

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The correct Answer is:
(a) `0.4706`, (b) `564.7` Torn (c) `0.08,0.92` (d) `675` Torr (e) `111.889`

`{:(%ofA,100,47,25,11),(8,0,,,),(%ofB,0,53,75,89),(92,100,,,):}`
Constant Temperature
(a) `Y_(A) = (P_(A)^(@)x_(A))/(P_(A)^(@)x_(A)+P_(B)^(@)(1-x_(A)))rArr0.25`
`= (300x_(A))/(300x_(A)800(1-x_(A)))`
`rArr x_(A) (800)/(1700) = 0.47, x_(B) = 0.53`
(b) `P_(tau) = P_(A)^(0) x_(A) +P_(B)^(0) x_(B) = 300 xx 0.47 +800 xx 0.53`
`= 141.2 +424`
`= 562.2` torr
(c) `760 = 300 x_(A) +800 (1-x_(A)) rArr x_(A) = (40)/(500) = 0.08`
(d) `P_(tau) = 300 xx 0.25 +800 xx 0.75 = 75 +600 = 675` tor
(e) `Y_(A) = (300 xx 0.25)/(675) = (75)/(675) = 0.11, Y_(B) = 0.89`
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