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I have read that vapour pressure of a mixture of two immiscible liquids is $$P=P^0_A + P^0_B$$ where A and B are immiscible liquids.

If they are immiscible, they would form separate layers. Thus since the more dense liquid would be below the less dense liquid, it would have no interface with the gas phase. Given this lack of an solid-gas interface, should the pressure due to the lower layer be zero?

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    $\begingroup$ Will the equation be more plausible if you actually achieve a mixture by stirring the two layer system strong enough to achieve a dispersion, in which the former layers are no longer observable? $\endgroup$ – Klaus-Dieter Warzecha May 17 '16 at 19:21
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    $\begingroup$ What if their densities are the same? What then is the driving force to form horizontal (rather than vertical) separation? Something else to ponder - why would a layer of B over A impact the equilibrium vapor pressure of A? (Even being 'immiscible' there is always some concentration of A in B - you can't beat the entropic contribution to the free energy). $\endgroup$ – Jon Custer May 17 '16 at 20:47
  • $\begingroup$ They don't necessarily have to form separate layers, they could just be in separate containers such as the arrangement shown here: chemistry.stackexchange.com/questions/112854/… $\endgroup$ – Karsten Theis Sep 25 at 19:35
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This is not just some vapor pressure. This is the equilibrium vapor pressure. Thermodynamics is all about equilibrium, you know. And equilibrium, roughly speaking, is what takes place in a closed container after a billion years.

Immiscible as they are, the liquids still have some solubility in each other (maybe extremely low, but anyway). Over the course of centuries, molecules of the lower liquid will slowly creep one by one through the upper layer and vaporize. Eventually they will saturate the gas phase and create the same vapor pressure as if the upper liquid were not there.

Stirring the whole mixture, as Klaus suggested, is quite likely to speed things up a bit.

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