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Author Topic: How can we calculate the vapour pressure of a mixture of liquids?  (Read 5153 times)

Atomic-S

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If at a certain temperature we know the vapour pressure of a chemical upon its liquid in a sealed container, and likewise at the same temperature we know the vapour pressure of another chemical upon its liquid in a separate sealed container, then if the 2 chemicals are miscable without losing their molecular identities (that is to say, for all relative proportions of the chemicals for which this is true), what can we say about the vapour pressure of the 2 chemicals mixed together in a sealed container; and does the answer have anything to do with the heat of solution?
« Last Edit: 19/10/2010 08:15:30 by chris »

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http://en.wikipedia.org/wiki/Raoult's_law
but remember that not all mixtures obey it.

lightarrow

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Ok, but that law is valid in case of no heat of solution, I'd say. The existence of a non zero enthalpy of solution means that the interaction between the fluids' molecules are not negligible and the vapor pressure of the solution should be lower. So to the OP's question "the answer have anything to do with the heat of solution?", I'd say yes.
« Last Edit: 19/10/2010 17:25:57 by lightarrow »

Atomic-S

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Do we have a formula for giving the deviations of the pressures from ideal, straightforwardly in terms of the heat of solution?  Such as, the partial pressure
Pi of a constituent present in the molar fractional amount Xi
being given by

       Pi = XiP*iK

P*i being the partial pressure over the pure consituent, and K being a correction that is a function of the heat of solution only and is the same for all constituents, or is it more complicated than that? (For the moment I have ignored problems in the gas phase that would be occasioned by deviations of the vapor from ideal gas behavior, this approximation being acceptable if we are for the moment assuming low vapor pressures)
« Last Edit: 20/10/2010 04:59:37 by Atomic-S »

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I'm sure k will depend on x, t and probably other things too.

 

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