# Does evaporation decrease entropy?

I have a container of water in thermal equilibrium (there are no temperature fluctuations inside the container).

Some molecules will evaporate out of the container of water thus decreasing its temperature. This is clearly a spontaneous process. But doesn't this mean that the entropy of the system is decreasing? The molecules that evaporated were clearly in thermal equilibrium with the other water molecules before. But now they gained some energy and the other water molecules lost energy.

How can this be?

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What about entropy of the environment? – evil999man Mar 6 '14 at 8:01
It seems you are ignoring that the law "entrpy always increases or at most stays constant" applies only to closed systems. When a system is open then entropy can decrease. Crystals coming out of a solution have decreased entropy, since they are so highly ordered. The entropy solution+crystal has increased though from the large number of microstates. Every live thing from the monocell cultures to humans live by decreasing entropy within their volume, but they are open systems. – anna v Mar 6 '14 at 11:30
@annav: That would serve as a nice answer ;-) – Waffle's Crazy Peanut Mar 6 '14 at 18:43

No, in fact you could even view the spontaneous evaporation as being driven by the fact that it increases entropy.

Basically what's happening is the liquid particles have random speeds (with distribution characterized by temperature), and they bump into each other. Every once in a while, two particles near the interface will collide in just such a way that one of the particles will escape the liquid and transition to the gas phase. The particle that remains in the liquid will loose energy in the process, thereby slightly lowering the temperature of the liquid. However, generally a gas has considerably more specific entropy (entropy per particle) than liquid at fixed temperature/pressure [1]. So, the overall entropy of the system increases.

[1] Compare for example the molar entropy for solid, liquid, and gas phases of water in these tables from Wikipedia.

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Temperature is a macroscopic concept, so you're bound to run into some problems when you apply it on a molecular scale (what does temperature and equilibrium (or for that matter, friction) even mean on such a small level?).

A thermal equilibrium does not mean all the molecules have the same energy. The distribution of their energies looks like this (normal distribution):

Here $\mu$ is the average energy and $\sigma$ describes the range of energies. For example 68% of molecules have an energy between $E = \mu - \sigma$ and $E = \mu + \sigma$.

Say that for example the energy necessary to evaporate is $E = \mu + 2 \sigma$. That means in your fluid everything to the right of $x =\mu +2 \sigma$ is gone. Due to random collisions, energy is redistributed and it will once again form a normal distribution. But because the molecules with highest energies were lost, the peak of this distribution will be at smaller $x$ than the original one. In a nutshell that means the liquid or gas has cooled.

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What question are you trying to answer ...? The term "entropy" doesn't even appear in your post. – chase Mar 6 '14 at 8:36

It depends on how you define your system or your control volume. If only the container is considered then indeed the entropy has decreased due to cooling. On the other hand if you account for the container plus the escaped vapour the entropy has increased, as the randomness of the molecules in the vapour state is larger than compared to in liquid state at the same temperature. Instead of randomness you can picture entropy as the number (actually the logarithm) of possible states in the system that lead to the same macroscopic observables (as T, p, etc.).

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