Claisen
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IIT-JEE Question - Solutions
Hello everyone,
I need help in understanding a question.
q) The vapour pressure of two pure liquids A and B that form an ideal solution are 300 and 800 torr respectively, at temperature T. A mixture of the
vapours of A and B for which the mole fraction of A is 0.25 is slowly compressed at temperature T. Calculate
1) the composition of the first drop of the condensate
2) the total pressure when this drop is formed
3) the composition of the solution whose normal boiling point is T
4) the pressure when only the last bubble of vapour remains
5) the composition of the last bubble
Attempt-
1) y1/(1-y1) = x1/(1-x1) * Pa/Pb
where y1 and x1 are the mole fractions in vapour phase and liquid phase respectively. Pa and Pb are the vapour pressures of pure components A and B
resp.
using this I got y1 = 0.47
2) What is actually the total pressure when first drop is formed?
Is it the sum of the pressures of the vapour phase components ?
i.e. 300*0.25 + 800*0.75 ?
The small drop cannot contribute towards any vapour pressure.
Please help
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Magpie
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Yes, I believe you have solved both 1 and 2 correctly. Raoult's Law and Dalton's Law of Partial Pressures should be sufficient to answer all these
questions.
In my Ch E curriculum, a long time ago, we first learned this in the 3rd year. This theory is the foundation for distillation column design.
Are you going to take a crack at the other questions? I will be glad to help if needed.
The single most important condition for a successful synthesis is good mixing - Nicodem
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Claisen
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Thanks for your reply Magpie. You learned this in your 3rd year ?
The 1) answer matches with the correct answer.
2) answer is wrong. The correct answer is 565 torr which is obtained by 300*0.47 + 800*0.53
300*0.25 + 800*0.75 = 675 torr is actually the answer for 4)
I don't understand this 
I will try other questions after clearing the 2)
[Edited on 21-1-2011 by Claisen]
[Edited on 21-1-2011 by Claisen]
[Edited on 21-1-2011 by Claisen]
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Magpie
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Yes, it was the beginning of the 3rd year for the application to distillation column design. Of course we had seen Dalton's Law of Partial Pressures
before; but not Raoult's Law.
Yes, I see. This is Raoult's law for the liquid droplet. Then add the partial pressures to get P.
pA = xA*PA ; pB=xB*PB; P = pA + pB
This is the same principle as the solution for 2. Only now all is in the liquid state, except that vapor bubble. Raoult's Law and Dalton's law
again.
Does this make sense?
The single most important condition for a successful synthesis is good mixing - Nicodem
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Claisen
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"Yes, I see. This is Raoult's law for the liquid droplet. Then add the partial pressures to get P.
pA = xA*PA ; pB=xB*PB; P = pA + pB "
What about the pressure due to the vapour present? Why don't we add that to the P above since the question asks Total Pressure?
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Magpie
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Quote: Originally posted by Claisen  | "Yes, I see. This is Raoult's law for the liquid droplet. Then add the partial pressures to get P.
pA = xA*PA ; pB=xB*PB; P = pA + pB "
What about the pressure due to the vapour present? Why don't we add that to the P above since the question asks Total Pressure?
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The partial pressures of the vapor components are exactly the same as those of the liquid components, as this system is in equilibrium. They do not
add; they balance each other.
The single most important condition for a successful synthesis is good mixing - Nicodem
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Claisen
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The system is slowly compressed. How will it remain in equilibrium?
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Magpie
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It is compressed only until the first tiny drop of condensate appears. The pressure is then held constant. No more vapor will condense.
The single most important condition for a successful synthesis is good mixing - Nicodem
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Claisen
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Thank you very much Magpie. I solved other parts too 
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Magpie
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Good work!
The single most important condition for a successful synthesis is good mixing - Nicodem
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