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I am a high school student and I am very confused in a topic about "vapor pressure of liquid". Why doesn't a change in atmospheric pressure affect vapor pressure? I think intuitively that vapor pressure should decrease on increasing the atmospheric pressure. Say I have an open bowl of water at temperature T and 1atm pressure and the vapor pressure at temperature T is P1. Now I increase the atmospheric pressure over the bowl keeping the temperature same (we can do it by pushing more air by some external means towards liquid surface). Because a higher atmospheric pressure is pushing on the liquid, doesn't that mean less evaporation and hence less vapor pressure for same temperature?

But on the other hand, I also think that vapor pressure is defined when there is an equilibrium between the liquid and vapor phase of water and equilibrium constants are only affected by temperature. Therefore, if vapor pressure changes with a change in external pressure, than it would mean the equilibrium constant would have to change?

Please prove me wrong (if I am) with some intuitive explanation at a high school level.

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  • $\begingroup$ The vapor pressure does not depend on the pressure. It only depends on the temperature. $\endgroup$
    – Maurice
    Jul 5 at 19:19
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    $\begingroup$ @Maurice The vapor pressure is proportional to the gas pressure. What is pressure independent is the saturated vapor pressure, and event that is not exactly true. $\endgroup$
    – Poutnik
    Jul 5 at 19:36
  • $\begingroup$ Beside reading the answer, you have to see it as an attitude of the liquid... $\endgroup$
    – Alchimista
    Jul 6 at 7:37
  • $\begingroup$ Please check also many of additional posts linked from the duplicate. This question has been asked many times before in different guises. $\endgroup$
    – Buck Thorn
    Jul 6 at 10:20
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The equilibrium is independent on the pressure above the liquid because the mechanism of creating it has nothing to do with the other gases above the liquid

It may seem intuitive that a higher gas pressure above a liquid would "push" liquid vapour back into the liquid. But not if you understand how gases work or how a liquid/vapour equilibrium is established.

First, the molecules of gases (unless very far from ideality) don't interact much. So there is no "push". Most of the space in a gas if free space and, while molecules bounce off each other, that's about it for their interactions. Also worth noting is that those interactions redistribute the kinetic energy among all the gas molecules in a specific statistical distribution.

Second, the equilibrium between a liquid and its vapour is reached when the number of liquid molecules having enough energy to escape the liquid, match the number of gas molecules having insufficient energy to stay in the gas (or, think of the number of molecules striking the liquid surface being equal to the number escaping from the liquid surface because they have enough energy to escape). The only role any gas molecules have here is to distribute the kinetic energy evenly in the gas according to the statistical distribution for a given temperature. That distribution is independent of the content of the gas (all gases are–approximately–the same).

Since the equilibrium between vapour and gas only depends on the distribution of kinetic energy in the liquid and in the vapour, there is no role for the other gases present or for pressure. The same equilibrium between the two states will be achieved in a vacuum or in a high-pressure vessel.

To be fair this is an approximation. This point of view ignores the relationship between boiling and pressure (but that doesn't depend on the other gases).

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  • $\begingroup$ Could you enlighten me why you want to keep a question that was asked like a dozen times and it doesn't make it any better? We don't need so many dupes, by any means. $\endgroup$
    – Mithoron
    Jul 5 at 23:36
  • $\begingroup$ @Mithoron I didn’t check for dupes. If it is, close it and I’ll agree. $\endgroup$
    – matt_black
    Jul 6 at 0:30
  • $\begingroup$ Say I have a bowl of water let in a box having air at 1 atm. So sooner an equilibrium will get establish between its vapors and liquid then we call it saturation vapor pressure, now I have a 2nd identical bowl kept I a box having very less air (or vaccum) we know at room temperature water will start boiling so more vapors will form so how can we say that vapor pressure will remain the same as in previous case? $\endgroup$ Jul 6 at 3:54
  • $\begingroup$ @VirenderBhardwaj boiling occurs when the pressure is lower than the equilibrium vapour pressure. What that equilibrium pressure is doesn't change. $\endgroup$
    – matt_black
    Jul 6 at 8:42
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I assume you consider the saturated vapor pressure at 100% relative humidity. (As at 50% rel.humidity, vapor pressure = 0.5 saturated vapor pressure).

Be aware that ability to contain certain amount of vapor, having respective vapor pressure, is property of space, not of a gas.

In first approximation, presence of air or other gas has no effect of evaporation equilibrium.

In the second approximation, the saturated vapor pressure very slightly increases with the total pressure.

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