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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.

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

vapour pressure doesn't depend on area

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    $\begingroup$ Note that it does depend on the shape of the surface as $$p=p_{\mathrm{0}} + \frac{2\gamma}{r}$$ where $p$ is saturated vapour pressure over a spherical liquid particle of radius $r$ and surface tension $\gamma$ , p0 is the ordinary saturated vapour pressure over flat surface. $\endgroup$ – Poutnik May 22 at 8:29
  • $\begingroup$ This is why is a Pressure, isn't? $\endgroup$ – Alchimista May 22 at 9:50
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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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  • $\begingroup$ What processes do we define as "surface phenomenon"? $\endgroup$ – user226375 May 22 at 14:44

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