TBR: Vapor Pressure Example 7.14

[justadream]

TBR Book II page 85 (top of page)



They provide a solution with 1/3 moles of methanol and 2/3 moles of ethanol.

VP of pure methanol = 150 torr. VP of pure ethanol = 120 torr.



Total VP = VP (methanol) + VP (ethanol)

= (1/3)(150) + (2/3)(120)

= 130



Now my question is: technically, should the VP of methanol be slighter greater than ethanol (since methanol has a lower BP and thus should occupy slightly more of the total VP than what its relative fraction in the solution is)? If so, then the total VP of the solution might not equal 130 anymore, right*?



*In fact, since methanol is more volatile, the real VP should be slightly higher than 130?

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[BerkReviewTeach]

The vapor pressure of pure methanol is higher than the vapor pressure of pure ethanol, which is accounted for in the math. It's basically a weighted average solution.

Pure ethanol would be 120 torr and pure methanol would be 150 torr. A 1:1 mixture would be 135 torr. Because the solution is richer in ethanol than methanol, it is less than 135 torr but not as low as 120 torr. So your thought is valid, but that has already been taken into account using 150 and 120 in the answer solution.