Is vapour pressure a surface phenomenon?

As a fact, I know that vapour pressure doesn't depend on surface area.


Does this imply it is not a surface phenomenon also?


It shouldn't be as then it would depend on surface area of molecules and the mole fraction of liquid in a solution violating the definition of [colligative properties.](https://en.m.wikipedia.org/wiki/Colligative_properties)


If not, then why doesn't it depend on surface area exposed?


[vapour pressure doesn't depend on area](https://www.coursehero.com/file/18693336/factors-affecting-the-vapour-pressure/)




Vapour pressure is normally defined as an equilibrium phenomenon. In statistical mechanics terms it is the point where an equal number of molecules are leaving the liquid and entering the vapour and leaving the vapour and entering the liquid.


This means that, however small the exposed surface, enough molecules have gone into the vapour phase to create the equilibrium. But most systems are not observed at equilibrium. The *kinetics* of the process leading to equilibrium *clearly* depend on the surface area of the liquid as there is far more opportunity for molecules to escape into the vapour phase if the exposed area is larger. But, if you are prepared to wait for the equilibrium to be established, the surface area doesn't matter.


Hence the slight intuitive confusion: what we mostly observe depends on surface area because what we observe is mostly systems that have not yet reached equilibrium so surface area matters. But on a strict equilibrium definition surface area doesn't matter but you might have to wait a long time to see the eventual equilibrium position.


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Well, the other answer frankly seems too complicated using statistical mechanics and all, let me give you a simplified version


Imagine a beaker, when you zoom in on the surface you see all the atoms and the molecules jiggle faster with more energy than the other atoms just below it. The vapour pressure is "almost" due to these atoms, by "almost" I mean that it is these atoms which get yanked off of the liquid and create a "commotion" at the top of the beaker.But there is a limit as to how many of the molecules get yanked of the surface which is dependent on the amount of heat you provide. Now take the beaker and empty its contents to another beaker which has more surface area.


If we zoom in on the surface of the new beaker we can see that there are more atoms at the surface which now "snap" out and create an even greater "commotion" at the top of the beaker, thus creating more vapour pressure.


Now according to the definition of pressure, increasing the area should decrease the vapour pressure but it doesn't, thus we conclude and say that it is not a surface phenomenon.


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