I have a mono-atomic ideal gas, expanding from a smaller volume V1 to a larger volume V2, inside a piston. If the expansion is done slowly, so the process is reversible, I understand how to calculate the work done, etc., for a system that is either thermally isolated or connected to a reservoir.

However, I'm having some difficulty understanding how an irreversible process of expansion takes place, if the gas just expands from V1 to V2 due to the removal of a partition connecting to the gas to a vacuum of volume V2-V1. If there is no heat reservoir, I understand there is no work done, so the final and initial temperatures are the same, with increasing entropy. We do not worry about the intermediate states, as our normal rules don't apply anyway.

What happens if during the expansion of the gas into the vacuum, the system is connected to a reservoir, maintained at the same temperature T as the initial temperature of the gas? Does it make sense to say the presence of the reservoir makes no difference to the previous case? The way I see it, there is still no heat transfer, so the system's behaviour is essentially identical to normal free expansion. Is this accurate?

  • $\begingroup$ I agree with you. The reservoir is there to be in thermal equilibrium with the system. If there is no equilibrium, becuase of the process, there will be no thermal equilibrium with the font neither. It still just a free expasion. $\endgroup$
    – Gilgamesh
    Feb 14, 2021 at 5:49

1 Answer 1


The way I see it, there is still no heat transfer, so the system's behaviour is essentially identical to normal free expansion. Is this accurate?

There is no net heat transfer if the uninsulated vessel is immersed in an ideal thermal reservoir whose temperature is the same as the gas before expanding.

Although during the expansion there will be temperature gradients in the gas, reflecting the effects of internal expansion, recompression and viscous friction heating, resulting in heat transfers across the boundary in both directions, after the expansion the gas will once again be in thermal equilibrium with the reservoir. So overall, $\Delta T=0$. Since the internal energy of an ideal gas depends only on temperature, $\Delta U=0$. Finally since there is no boundary work, from the first law the overall net heat transfer between the gas and reservoir, $Q=0$.

So the end result is free expansion in a vessel in contact with a thermal reservoir of the same temperature is equivalent to free expansion in an insulated vessel.

Hope this helps.

  • $\begingroup$ During the irreversible expansion process, there may be heat flow across the boundary with a constant temperature reservoir at T, but, in the end, the net heat flow will be zero. $\endgroup$ Feb 14, 2021 at 13:07
  • $\begingroup$ @ChetMiller Let's say the gas is initially in the left side of the vessel. So does that mean during the expansion the temperature of the gas on the left side drops below that of the reservoir as it does work pushing gas in front to the evacuated side resulting in heat transfer into the vessel from the reservoir , whereas the gas on the right side gets compressed increasing its temperature over that of the reservoir transferring an equal amount of heat to the reservoir, so overall $Q=0$, $\Delta U=0$ and $\Delta T=0$. Is that what you are saying? $\endgroup$
    – Bob D
    Feb 14, 2021 at 14:28
  • $\begingroup$ This makes sense. Thanks for such a clear answer. $\endgroup$
    – user6314
    Feb 14, 2021 at 14:29
  • $\begingroup$ I guess the only thing that bothers me is that the compression of the gas on the right is sufficient to raise its temperature above that of the reservoir. $\endgroup$
    – Bob D
    Feb 14, 2021 at 14:30
  • $\begingroup$ @BobD I'm not sure exactly how all this plays out during the process, except that initially, there has to be some cooling. Also, it is not only the recompression that heats the gas back up; there is also viscous heating. $\endgroup$ Feb 14, 2021 at 14:35

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