# Does evaporation do work on the surroundings?

Suppose it is a dry (effectively no water vapor in the ambient air ) and hot day ($$T_{ambient}\approx 40\ C$$). My body temperature on this day is $$T_{body}\approx 37C$$. As a cooling mechanism, my body begins to sweat releasing pure water (to a good approximation at least) onto the surface of my skin. This water (I will refer to sweat as water from now on) is at the same temperature as my body initially. The molecules in this water roughly obey a Maxwell-Boltzmann distribution and hence a sizable fraction of the water molecules have enough energy to escape and enter the vapor phase. When these high energy molecules escape into the vapor phase, the water on my skin cools to some temperature $$T_{reduced}$$ which is lower than my body temperature ($$T_{reduced}). Now because of this temperature gradient between my body and the water, heat flows from my body which is at the higher temperature to the water which is at the reduced temperature, heating the water whilst cooling my body at the same time.

Now my question can be stated as follows: If we take the water molecules (both those in the liquid and vapour phase) as a system, does this system do work on the surroundings when evaporation occurs? That is, when the most energetic of liquid phase molecules break free into the gaseous phase, do they push against the atmosphere and as a result do work on it? Secondly, if the answer to the first question is yes, can this be assumed to be the reason the water cools during evaporation? That is, the system (initially only liquid water) partly evaporates and hence expands against the atmosphere doing work on it. This work reduces the energy of the system and hence its temperature. After this reduction in temperature, heat flows from the human body to the cooled system heating it up and causing the cycle to repeat.

I ask these two questions because if there is no expansion work or if the expansion work is not the cause of the temperature reduction, then the evaporation of sweat on a body at a temperature lower than that of the surroundings (as in this case where $$T_{body} = 37C < T_{ambient} =40C$$) appears to be an example of a system colder than its surroundings spontaneously cooling without doing work on the surroundings. This would be a violation of the second law. So am I correct in thinking that the evaporation of sweat leads to the system (all sweat molecules, both vapour and liquid) doing work on the surroundings as it expands against the atmosphere and it is this work which reduces the internal energy of the system which reduces its temperature?

• Consider that water spontaneously evaporates (which cools it) in a vacuum, removing work from the scenario entirely. The Second Law isn't violated because the newly created gas phase has much more entropy than the liquid phase. Jun 5, 2021 at 14:20
• @Chemomechanics I had actually considered this already and was going to include it but my question seemed too long already. In your case of evaporation in/against a vacuum , the isolated systems internal energy will not change but the temperature will still drop because kinetic energy is converted to potential during the evaporation. I understand this. But evaporation of sweat still irks me because it seems as if its an example of a system colder than its surroundings spontaneously cooling without doing work on the surroundings. Surely such a process cannot occur? Jun 5, 2021 at 14:26
• It can't occur for a single-phase system. Jun 5, 2021 at 14:28
• Thanks, okay so if I have a small system at temperature $T_{low}$ immersed in a heat bath at $T_{high}$ where $T_{high}>T_{low}$, It is entirely plausible that the small system could spontaneously start decreasing in temperature if a phase change occurs within the small system? My intuitions (which are obviously incorrect) tell me that such a system should never be able to decrease. Jun 5, 2021 at 14:43
• The systems will evolve to maximize the total entropy. Jun 5, 2021 at 14:51

Does evaporation do work on the surroundings?

Actually, apparently it does, at least according to the Hyperphysics website: http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/phase.html#hvap. To quote from the site:

"In the process of vaporization of water, a large amount of energy must be added to overcome the remaining cohesive forces between the molecules and an additional amount of energy goes into PdV work to expand the gas from its very small liquid volume to the volume occupied by the resulting vapor".

It goes on to deduce that the $$PdV$$ work required to expand water (perspiration) into its gaseous form at 37°C is 34.2 calories/gm, 6.8 calories/gm less than at 100°C.

The analysis is rather lengthy, so I leave it to you to dig further.

Hope this helps.

If we take the water molecules (both those in the liquid and vapour phase) as a system, does this system do work on the surroundings when evaporation occurs? That is, when the most energetic of liquid phase molecules break free into the gaseous phase, do they push against the atmosphere and as a result do work on it?

Indeed there is work done on the atmosphere (the part without the newly added water), because this body as a whole expands a little in the gravity field when adding new particles. This is somewhat non-intuitive because the newly added water molecule is much smaller than the empty volume it has around it in the air, so one tends to imagine the water molecules just diffuse into empty space between the air molecules and the air molecules aren't affected.

But this is not so; adding new particles actually causes the air to expand. There are two ways to understand this.

First, if the water is boiling in the atmosphere, this work is done when bubble of vapor is still expanding in the liquid. Since it is expanding against atmospheric pressure + hydrostatic pressure of water above it, it does some work on the atmosphere (since liquid water is almost incompressible). Then when the bubble gets out in in the atmosphere, it starts cooling down from 100 Celsius to lower temperature and the vapor also diffuses into the atmosphere. This means some work goes back from atmosphere to the vapor, but not as much as was initially put during boiling, so net result is that positive work is done on the atmosphere.

Second, if evaporation happens at temperature of the atmosphere (so the change of temperature of the water is zero), there is no boiling, only diffusion from the water surface. There is still work done on the atmosphere. This is because atmosphere is quite a rare gas that obeys law of ideal gas. For an imaginary pocket of air that contains $$N$$ particles, volume of this pocket is:

$$V = \frac{NkT}{p}.$$

This means volume of a gas pocket is determined by pressure $$p$$, temperature $$T$$ and total number of particles $$N$$ in it. If the air in the pocket changes from state $$T,p,N_1$$ to state $$T,p,N_2$$ where $$N_2 > N_1$$, its volume has to increase. This happens when the water molecules push the air molecules away during collisions/scattering events and eventually mix with them.

if the answer to the first question is yes, can this be assumed to be the reason the water cools during evaporation? That is, the system (initially only liquid water) partly evaporates and hence expands against the atmosphere doing work on it. This work reduces the energy of the system and hence its temperature. After this reduction in temperature, heat flows from the human body to the cooled system heating it up and causing the cycle to repeat.

Yes, evaporation means gas expansion into open atmosphere. Any gas expansion into open atmosphere is associated with that expanding gas getting colder (possibly even colder than the atmosphere) and the rest of the atmosphere getting hotter(however because of its size, this is not easily observed with open atmosphere, but can be observed when expanding into closed vessel).

However, this cooling effect acts directly only on the vapour, not on the remaining liquid. This liquid does not expand and does not cool down by doing work. However, the remaining liquid still cools down, because

1. by losing particles it is losing energy at a higher rate than the rate that would keep the temperature constant; this is because evaporated molecule carries away more than average energy per molecule in the remaining liquid;

2. the colder vapour-rich air may cool down the liquid by heat conduction and accelerating evaporation.