InterviewSolution
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InterviewSolution
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The vapour pressures of two pure liquids `A` and `B` that form an ideal solution are `300` and 800 torr, respectively, at tempertature `T`. Calculate a. The composition of the first drop of the condensate. b.The total pressure when this drop is formed. c. The composition of the solution whose normal boiling point is `T`. d. The pressure when only the last bubble of vapour remains. e. Composition of the last bubble. |
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Answer» Given `p_(A)@ = 300` torr, `chi_(A)^(l) = 0.25, chi_(B)^(l) = 1 - 0.25 = 0.75` a. By the condensation of only one drop, we can assume that the composition of the vapour remanns the same. `chi_(A)^(V) = (p_(A)@chi_(A)^(l))/p_(T)` and `chi_(B)^(V) = (p_(B)@chi_(B)^(l))/p_(T)` or `chi_(A)^(V)/ chi_(B)^(V) = p_(A)@chi_(A)/p_(B)@(1-chi_(A))` Putting various known values, we get `chi_(A)^(V) = 0.111` and `chi_(B)^(V) = 0.888` b. `p=p_(A)@chi_(A)^(V) + p=p_(B)@chi_(B)^(V)` = `300 xx 0.11+ 800 xx 0.888` `= 743.7` c. `760 = 300chi_(A) + 800chi_(B)` `chi_(A) = 0.08` and `chi_(B) = 0.92` d. When only the last bubble of vapour remains, we can assume the composition of vapour is now the composition of the condensate.n Hence, `P= (0.25 xx 300) + 0.75 xx 800 = 675` torr e. Composition of last bubble `(chi_(A) = p_(A)@chi(A))/P = (0.25 xx 300 )/675 = 0.11` `chi_(B) = 0.89` |
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