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The vapor pressure of two immiscible liquids (A and B) is the sum of their vapor pressures in their pure liquid form.

Intuitively, I don't see how this is possible. If the liquids are being stirred continuously, there will be a few particles of liquid A and a few particles of liquid B on the surface of the liquid.

Isn't this the same case as two miscible liquids? How is this any different? I understand that the liquids have no interactions with each other whatsoever so that can change the situation but I still can't wrap my head around it.

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    $\begingroup$ The EQUILIBRIUM vapor pressure depends on the chemical potential of the liquid. A pure liquid with less exposure will establish the same VP but take longer to attain it. There is an effect for very small droplet size[nanodrops] where the VP is actually higher than for larger drops or bulk material. $\endgroup$
    – jimchmst
    Commented Apr 23 at 20:33
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    $\begingroup$ If the liquids are being stirred continuously... then you could just as well have them in two smaller containers and there would be no difference. $\endgroup$
    – Mithoron
    Commented Apr 24 at 0:44
  • $\begingroup$ Before asking, you may find useful reviewing the targetted search results like site:stackexchange.com OR site:libretexts.org OR site:wikipedia.org vapor pressure immiscible liquids $\endgroup$
    – Poutnik
    Commented Apr 24 at 9:12
  • $\begingroup$ @Mithoron I did check out that thread but unfortunately, it didn't answer my question. But your explanation did. Thanks! :D $\endgroup$
    – hi hi
    Commented Apr 24 at 14:18
  • $\begingroup$ @Poutnik Thanks for the suggestion. I will keep it in mind when I ask my next question. :D (sorry it's my first time asking a question here) $\endgroup$
    – hi hi
    Commented Apr 24 at 14:19

1 Answer 1

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Assume a scenario where two immiscible liquids are in a closed flask, with one floating on top of the other. In this situation, if the system is kept still, only the vapor pressure of the liquid on top will be measured because it's the only one in contact with the vapor phase.

However, if the mixture is agitated or stirred, both liquids will be broken up into drops, and at any given time, there will be drops of both liquids on the surface. This means that both liquids contribute to the overall vapor pressure of the mixture.

When the mixture is agitated, both liquids are in equilibrium with their vapors, and the total vapor pressure is simply the sum of the individual vapor pressures. This is described by the equation:

Total vapor pressure = p°A + p°B

Where p°A and p°B refer to the saturated vapor pressures of the pure liquids A and B, respectively. This equation is independent of the amount of each liquid present, as long as there's enough of each for both to exist in equilibrium with their vapor.

While the concept of vapor pressure applies similarly to both immiscible and miscible liquids, there are important distinctions between the behavior of these two types of mixtures when it comes to vapor pressure and boiling points.

In both cases, the vapor pressure of a mixture is determined by the partial pressures of each component. However, in immiscible liquids, the components do not mix at a molecular level, so each component maintains its own distinct vapor pressure regardless of the presence of the other. In contrast, miscible liquids form a homogeneous mixture, meaning the individual components dissolve into each other at a molecular level, resulting in a combined vapor pressure that is not simply the sum of the individual vapor pressures. Instead, it's determined by the vapor pressure of the mixture as a whole, which can be higher or lower than the sum of the individual vapor pressures depending on the interactions between the components.

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