I will place my two questions infront of you guys. Let's consider the liquid is water.

1) " How and why exactly does water vapor in air condense at the surface of water? Is vapor in air condensing at some rates all the time at all temperatures without a water surface? What are the factors influencing the rate of condensation? "
( I heard condensation near the surface occurs because vapor molecules at the surface lose energy by getting attracted to the surface water molecules and losing kinetic energy from striking them and form bonds with them and become water. )

2) Say, air is at relative humidity of 100%, which as I understand, means rate of evaporation of water is equal to rate of condensation of vapor and that there is zero net evaporation of water. It implies that after reaching a certain level of concentration of water vapor in air (vapor pressure), even if I increase the rate of evaporation to whatever amount by supplying heat to the water, rate of condensation will get equal to that. I understand this point is also called as dew point.

MY QUESTION explicitly is, why does the amount of increased fraction of molecules escaping the given water surface area will be equal to the increased fraction of water vapor molecules that are gonna get condensed? What's the reason for this saturation point? Also, does this mean if I place water at low temperature i.e for which rate of evaporation is less than the rate of condensation of fully saturated air (RH=100%) above it, will water level rise?

I'm not a physics college student. Just a curious guy trying to make sense of nature in simple mathematical models. So, please explain as simple as you can guys.

  • $\begingroup$ You appear to be trying to use kinetics (rates) to understand thermodynamics (energetics). This is a mistake. $\endgroup$
    – Jon Custer
    Commented Sep 24, 2018 at 16:39
  • $\begingroup$ Sorry but I'm not sure what you mean. Is that a sarcastic remark on the tags I put for my question ? $\endgroup$ Commented Sep 24, 2018 at 16:46
  • $\begingroup$ No, not at all. Thermodynamics deals with things like the Gibbs free energy of phases, from which one can calculate phase stability and relative phase fractions under specific conditions. Kinetics deals with how fast (if at all) one might get to thermodynamic stability. Visualizing thermodynamic stability in terms of molecule kinetics is going to get you into trouble. $\endgroup$
    – Jon Custer
    Commented Sep 24, 2018 at 16:51
  • $\begingroup$ I can see your point, Jon. I'm waiting for other answers to clear my thinking. $\endgroup$ Commented Sep 24, 2018 at 17:24
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    $\begingroup$ Related (a very nice answer!): physics.stackexchange.com/a/773781/226902 $\endgroup$
    – Quillo
    Commented Jul 27, 2023 at 19:31

1 Answer 1


I think your intuition is good. You have described accurately what the processes of evaporation and condensation look like on a molecular level. The only problem I see is the assumption that you are free to arbitrarily increase the evaporation rate without simultaneously causing a change in the relative humidity.

At a given temperature, the rate at which water molecules escape the surface of liquid water per unit area is a constant, unaffected by the relative humidity of the air above it or any other factors. The rate at which water molecules are captured by the surface does depend on the humidity: if there are more water molecules floating around above the liquid water, it stands to reason that more of them will get caught. This rate (the rate at which water molecules are captured) is less than the rate at which they escape if the relative humidity is less than 100%. As evaporation adds water vapor, eventually the rates converge, and you've reached equilibrium.

Some of your difficulties seem to stem from situations involving warm water in contact with cold air, or vice versa. This is not really a possible situation on a microscopic level. The sharp temperature gradient at the surface will more or less instantly smooth out.

  • $\begingroup$ Thanks for the answer, Ben. I will wait for other answers and reflect back on yours. $\endgroup$ Commented Sep 24, 2018 at 17:25

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