# Describing evaporation without entropy -- what about Helmholtz?

I have in mind a rigid, impervious, heat-insulating vessel containing a monatomic liquid and vacuum. The vessel has an insulating forcefield inside, exactly at the surface of the liquid. It can be toggled.

Step 1: The liquid is in thermal equilibrium. The field is switched off for a very short, predetermined time. Some portion of the molecules at the surface exceed some threshold of kinetic energy, enough to escape into the vacuum.

Step 2: Now the liquid is no longer in thermal equilibrium because some of its fastest molecules have escaped. The same can be said for the vapor, except it lacks slow ones. We wait until both regain equilibrium -- they will do so separately because of the forcefield -- and then turn the field back on.

We return to step 1. This time there is some condensation as well as evaporation.

We repeat steps 1 and 2 again and again. At each iteration we can always predict the proportion of liquid that will evaporate, and the proportion of vapor that will condense, based purely on the temperatures of the liquid and vapor, the volume of the liquid, the pressure of the vapor, and the length of time the field is off. We can do this without considering changes in entropy.

But the Helmholtz energy is what determines the direction of this process, and its calculation requires knowledge of the change in entropy.

So is there a problem with my model?

• This is very similar to Maxwell's Demon.
– user7652
Commented Jul 28, 2016 at 18:23
• What is the entropy change introduced by the switched force-field? Commented Jan 25, 2017 at 21:29
• @airhuff - Hm. Very good question. Hm. Commented Jan 26, 2017 at 1:03