Since a gas at a certain pressure exhausting into vacuum has no atmospheric pressure to push against, there shouldn't be any adiabatic cooling taking place.

But looking at the energy conservation:


$TdS = 0$ {Adiabatic process}

$PdV = 0$ {No change in volume of container}

Hence: $dU=-VdP$, leading to a lower temperature?


2 Answers 2


The gas in the container does work on the elements of gas that are moving towards the orifice and out. As the the gas remaining in the container expands, it does work and thus cools down.

One can use the dX description for any element in the container, if it does not move too violently. Increase of internal energy of the element is heat accepted minus work done (gas expands):

$$ dU = dQ - dW $$

If the process is slow, gradient of temperature is low and the heat transferred to the element will be negligible in comparison to work it does. The work done by the element is $pdV$, so the internal energy decreases:

$$ dU = -pdV < 0 $$

If the gas is ideal, decrease of energy is sufficient to conclude temperature decreases as well. The gas in the container cools down.


TdS is not zero for all adiabatic processes. It is only zero for all adiabatic reversible processes, and the process you describe is not reversible. If you have an adiabatic chamber with gas on one side of a barrier and vacuum on the other side and you suddenly remove the barrier, the gas does not do any work on the walls of the chamber and no heat comes through the walls, so the change in internal energy of the chamber contents (the gas) is zero, and its temperature remains constant. But the pressure of the gas has decreased, so its entropy has increased.

The process described by @Jan Lalinsky would occur if, rather than removing the barrier suddenly, a small pinhole were made in the barrier so that the gas escaped into the vacuum compartment very slowly. In this case, the gas in the original pressurized chamber will experience an adiabatic reversible expansion, and its temperature will drop. However, if the barrier between the chambers were adiabatic, the gas that escaped into the chamber that was originally under vacuum would actually have a temperature higher than the original gas temperature when the pressures equilibrate, since the overall change in internal energy for the combination of the two chambers would be zero. On the other hand, if the barrier were not adiabatic, and the system were allowed to re-equilibrate (both thermally and mechanically), the final state would be the same as if the barrier had been removed suddenly, and the final temperature and internal energy would match the initial temperature and internal energy of the gas before the barrier was removed or the pinhole was applied.


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