This question came up when I recently answered this question Why are compressed air tanks cold? and I found this question Why does the gas get cold when I spray it? which seemed to be related, but the given answers do not really satisfy me. Also this question Why does deodorant always feel cold? is about the cooling of a deodorant spray and doesn't address the cooling of the container. A web search gave conflicting, mostly not scientifically well explained answers.

In my opinion, there are three possible cooling effects:

(1) The adiabatic expansion of the gas involves work against the external pressure thus cooling the gas. This should already work for an ideal gas.

(2) The adiabatic expansion of the gas leads to a cooling due to the Joule-Thompson effect because the gas is not an ideal gas. This is due to a change of kinetic to potential energy of the molecules.

(3) The cooling of the container is due to evaporation cooling of the liquified gas inside the container.

ad (1) The problem with answer (1) is that the adiabatic expansion occurs outside the container against the air pressure. Thus the cooling should only occur for the outside gas not for the gas in the container.

ad (2) This problem should not occur for answer (2) because the JT cooling effect already takes place inside the container upon pressure reduction. Also the cooling outside the container should occur.

ad (3) The gas is partially liquified inside the container due to the high pressure (e.g. butane) and the cooling of the container is caused by the evaporation heat of the liquid.

In the case of the aerosol spray I am now inclined to think that my answer (3) is the dominant effect in the container cooling. This does, however, not conform with the top rated answer (work due to adiabatic expansion of gas) of the mentioned identical question.

Am I right with my suspicion that a cooling by adiabatic expansion of an ideal gas cools the gas outside but not inside the container?

Is the container indeed cooled mainly by the liquid evaporation heat effect? Or is the Joule Thompson effect more important?

  • $\begingroup$ The Joule-Thompson effect is tiny at the pressures involved in aerosol cans. Note that, as I say in the linked question, in aerosol cans the dip tube dips into the liquid propellant. Try using a can of deodorant upside down to see what a difference this makes. FWIW I spent several years working for a company that made aerosol deodorant. $\endgroup$ – John Rennie Feb 14 '18 at 19:28
  • $\begingroup$ @JohnRennie _ Thank you for the linked question which I didn't know. I will look at it. $\endgroup$ – freecharly Feb 14 '18 at 19:38
  • $\begingroup$ @JohnRennie - When I understand your answer to the linked question correctly, you are attributing the cooling of the deodorant spray gas to the evaporation of the liquid propellent mostly outside the container. My question, however, concentrates on the cooling of the can itself. You point out that JT effect is comparatively small, but what about the adiabatic expansion? I was thinking of the evaporation inside the can and did not know there was a dip tube inside and in essence the fluid is expelled. $\endgroup$ – freecharly Feb 14 '18 at 23:30
  • $\begingroup$ @JonCuster - The link to this question was already given by John Renny in his comment above. As I explained in my answer to him above, the present question is about the cooling of the container not the spray itself and about which possible cooling effect dominates. Therefore, he gave a detailed answer to the present question below. $\endgroup$ – freecharly Feb 20 '18 at 16:04
  • $\begingroup$ @freecharly - the comment is an auto-generated one from flagging as duplicate. And, really, it is. A similar situation is found for propane gas tanks for camping or backyard BBQ (where ice buildup on the tank on a humid summer day is quite possible). $\endgroup$ – Jon Custer Feb 20 '18 at 16:11

Your option 3 is correct - the cooling is evaporative.

For some background see:

A typical aerosol spray can looks something like this:


(image from the Aerosols Wiki - yes, that really does exist :-)

When you use the spray a small amount of the liquid is expelled. The liquid evaporates as it passes through the nozzle and this creates the spray. The evaporation at the nozzle is why the spray is cold.

As for the can, there will be some degree of cooling due to expansion of the gas, but this will be negligible for a couple of reasons. Firstly the volume of the liquid decreases by a relatively small amount, because a small amount of liquid creates a large volume of spray. This means the gas above the can increases in volume only a small amount so the adiabatic cooling is small. Secondly the specific heat of the gas is small compared to the can and the liquid, so the gas is quickly reheated by its surroundings with very little temperature change to the walls of the can and the liquid.

The evaporative cooling happens because the pressure of the gas above the liquid is equal to the vapour pressure of the liquid propellant. When you use the spray the pressure inside the can falls, so the liquid propellant evaporates to restore the pressure to the vapour pressure of the propellant. It is the latent heat associated with this evaporation that causes the cooling of the can. The latent heat of evaporation is relatively large so the temperature drop associated with the evaporation is significant.

| cite | improve this answer | |
  • $\begingroup$ Thank you very much for your compelling argument that the cooling of the can (not only the exiting gas) is predominantly caused by the evaporative cooling of the liquid gas (also inside the container) and that adiabatic expansion effects are comparatively small! $\endgroup$ – freecharly Feb 15 '18 at 13:12
  • $\begingroup$ What is your opinion on the case of a supposed cooling of an air tank? physics.stackexchange.com/questions/386234/… There should be no liquid gas in this case. Do you think that there also a cooling of the container takes place? And if this is indeed the case, is it more by the Joule Thompson effect or just by the quasi-ideal gas adiabatic expansion work? $\endgroup$ – freecharly Feb 15 '18 at 13:23

The adiabatic expansion cooling takes place inside the container. The gas remaining within the container at any time has done work to expand against the gas that it has previously expelled from the container. This expansion has occurred at the pressure within the container, not the external pressure. So the gas remaining in the container has experienced an adiabatic reversible expansion.

Just picture an imaginary membrane at time t1 surrounding that fraction of the mass of gas that will eventually fill the container at a later time t2. The gas outside this membrane is the portion of the gas that is pushed out of the container between times t1 and t2 by the gas within the membrane. So the gas inside the membrane does adiabatic (essentially reversible) expansion work to expel the gas outside the membrane from the container.

For a reference on all this, see Fundamentals of Engineering Thermodynamics, by Moran et al, Chapter 6, Example 6.10 Considering Air Leaking from a Tank

| cite | improve this answer | |
  • $\begingroup$ Thank you for your answer. With the fictitious membrane you explain nicely how the adiabatic expansion (and thus cooling) can be understood to also occur inside the container. However, inside the container there is also the liquid in equilibrium with its saturated gas, then a cooling due to adiabatic expansion becomes less clear because there should be no change in gas pressure which should be the vapor pressure depending only on temperature. How do you see this? Also thank you for the reference. I will try to find the book. $\endgroup$ – freecharly Feb 14 '18 at 23:50
  • $\begingroup$ My inclination would be that you could still do the same trick. The contents remaining would still be cooling, and the vapor pressure would be dropping. Handling the internal energy of the material remaining would have to take into account the internal energy of both the liquid and the vapor, and the change in the mass of liquid. This doesn't seem desperately difficult to me. $\endgroup$ – Chet Miller Feb 15 '18 at 0:11
  • $\begingroup$ Also, @John Rennie has pointed out in his above comment and link to a similar question, where he explains the cooling (of only the outside gas?) by evaporation heat only, that there is a dip tube inside and that the liquid is expelled, which I was not aware of in my question and also might influence the explanation. Thanks again! $\endgroup$ – freecharly Feb 15 '18 at 0:19

Not the answer you're looking for? Browse other questions tagged or ask your own question.