My professor tries to demonstrate that adiabatic free expansion is an irreversible process:

  1. for and adiabatic $Q=0$;

  2. because the gas is expanding through vacuum W=0;

therefore $\Delta U=Q-W=0$

now I can put back the gas doing work on it: $W=p\Delta V$,

because $\Delta V$ is negative $W$ is negative too, so $\Delta U=-W$ we have an increment of the internal energy.

$U=U(T)$, $\Delta U=\frac32nR\Delta T$ so if the ΔU is positive $\Delta T$ is positive too,

now arrive the thing that I haven't understood, my professor said:

a) We can extract $Q=W$

b) $Q\rightarrow W$ isn't possible for the second law of thermodynamics, therefore, a free adiabatic expansion is not reversible.

I understand the second law of thermodynamics, I don't understand the passage from point a) to point b).

  • $\begingroup$ It's an irreversible process because in a free expansion the gas thermodynamic parameter pressure isn't defined. $\endgroup$
    – Gilgamesh
    Commented Feb 9, 2021 at 11:32

1 Answer 1


Your professor is saying that step b returns the gas to its original state that existed before the expansion and reversible recompression. So, overall, you have carried out a cycle (on the system), the net result of which is to take heat from a single source and produce an equal amount of work. (Incidentally, in the reversible expansion, you are supposed to be using a constant temperature reservoir, such that the internal energy doesn't change during the expansion, and the change in U is zero throughout the expansion.).

According to the second law, with a cyclic process, receiving heat from a single source at a constant temperature and converting it to work is not allowed.


In the case of the present cycle, the gas receives work from the surroundings (does negative work) and rejects heat to the surroundings. This is not prohibited by the 2nd law of thermodynamics. The 2nd law only prohibits receiving heat from a single source at constant temperature and doing an equal amount of work in a cycle. So the professor's analysis says nothing about the reversibility of the expansion.

  • $\begingroup$ "According to the second law, with a cyclic process, receiving heat from a single source at a constant temperature and converting it to work is not allowed". But doesn't the irreversible expansion followed by a reversible isothermal compress constitute a cycle? And isn't the proof that the expansion is irreversible is that entropy of the surroundings increases by $Q/T$ in the cycle? $\endgroup$
    – Bob D
    Commented Feb 9, 2021 at 13:48
  • $\begingroup$ @BobD You are correct. I was about to go back and correct my answer to reflect this, but you beat me to it. The system (the gas) experiencing the cycle has work done on it, and it rejects heat, not the other way around. I also agree that the way to show that the combined process is irreversible is that the entropy of the surroundings (and thus the universe) has increased. $\endgroup$ Commented Feb 9, 2021 at 13:58
  • $\begingroup$ Thanks Chet. I was about to post an answer but made my comment first so you might update. In any case, this example brings up something that has bothered me for some time. The Kelvin-Plank statement says no heat engine can operate in a cycle while transferring heat with a single reservoir. I've always interpreted this to mean the heat engine cannot "produce" work with a single reservoir, meaning positive work. So it seemed to me the Kelvin-Plank statement does not preclude the possibility of negative work being performed in a cycle with a singe reservoir. Is that a correct interpretation? $\endgroup$
    – Bob D
    Commented Feb 9, 2021 at 14:14
  • $\begingroup$ @BobD It would seem that way. The working fluid experiencing the cycle can not do positive work. It can always have work done on it (say by stirring) and reject heat to the surroundings. $\endgroup$ Commented Feb 9, 2021 at 15:08
  • $\begingroup$ Hmm. Stirrer work and the free expansion are both irreversible processes. I know you aren't big on extrapolating things, but I wonder if net negative work with a single reservoir in a cycle always involves an irreversible cycle, i.e., a cycle in which at least one process is irreversible. Anyway, food for thought. $\endgroup$
    – Bob D
    Commented Feb 9, 2021 at 15:15

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