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In my introduction to Thermodynamics, my lecturer has described the Saturated Vapor Pressure (SVP) as an "equilibrium condition" in which the number of particles condensing and evaporating are equal. However, this doesn't click for me, since under this condition, it seems that during boiling, which occurs at SVP, the amount of vapour would not increase, and rather stay constant, which is not the case.

Furthermore, when describing cooling a vapour into a liquid, my lecturer mentioned cooling it to reduce the SVP such that it falls below the atmospheric pressure to cause it to condense. In this case, it feels slightly unintuitive as to why the vapour won't condense anyways (i.e. without cooling) due to its Partial Pressure (PP) being s.t. PP < SVP < atmospheric pressure, and also how this cooling causing condensation fits in with SVP being an equilibrium condition: does vapour condense to form water so we can have equilibrium between water and vapour, or something similar?

I understand there's likely a lot to unpack here, but I feel that my main source of misunderstanding is how the equilibrium condition stipulation and net formation of vapour/liquid don't quite click for me. My main understanding of SVP is that it is the pressure of the vapour against the surroundings, but I can't quite manage to combine the equilibrium condition with this.

I've tried many different websites, other answers and videos, but none have seemed to reconcile this misunderstanding as of yet. Thank you in advance to those who answer!

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    $\begingroup$ However, this doesn't click for me, since under this condition, it seems that during boiling, which occurs at SVP, the amount of vapour would not increase, and rather stay constant, which is not the case. This system included the whole room. Have you tried to set up an enclosed system at $100^\circ \rm C$ with the pressure inside at one atmosphere? $\endgroup$
    – Farcher
    Feb 12 at 11:14
  • $\begingroup$ Hi, thanks for the reply. I don't quite follow, sorry. An enclosed system (a beaker w/ water and air) was used to demonstrate the equilibrium condition, but I'm not quite sure how this links into boiling of water and an accumulation of vapour. $\endgroup$ Feb 12 at 11:23

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The equilibrium condition your lecturer was talking about was for an enclosed system in a "box" not for a beaker of water open to the atmosphere.

If you left a beaker of water in a room then unless the air in the room was saturated with water vapour there would be a net transfer of water from the liquid to the vapour phase.
You could of course envisage a closed room and lots of water in the beaker when eventually the air in the room was saturated with water vapour and an equilibrium state achieved, there being no net transfer between liquid and vapour.

Imagine that you have a box containing only water and water vapour at a constant temperature - the system. Eventually an equilibrium state would be achieved so that there is no net transfer of water liquid to water vapour and vica versa.
Increasing the temperature of the system would mean that there was a net change from the liquid phase to the vapour phase until an equilibrium state was reached in which there was no net transfer between the liquid and the vapour phases. The vapour pressure would have increased.
If the final temperature was $100^\circ \rm C$ then at equilibrium the vapour pressure would be one atmosphere.

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  • $\begingroup$ This is a really good answer. Thank you very much! My last question here would just be the reasoning behind the net transfer from liquid to gas in open air. Is this just the effect of evaporation in a non "box" system? $\endgroup$ Feb 12 at 11:52
  • $\begingroup$ Yes or it could be boiling if the temperature was high enough. $\endgroup$
    – Farcher
    Feb 12 at 14:26

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