It is said that reversible processes do more work than irreversible ones. A graph as the one below showing an irreversible and reversible isothermal expansion is used to justify this.

enter image description here

However, I don't see how this is a proof or an intuitive explanation. Why wouldn't it be possible for the gas to, say, expand to $\text{V}_2$ at constant $\text{P}_1$, then drop to $\text{P}_2$. Since the gas in this case begins and ends at the same temperature, shouldn't it also be a valid irreversible isothermal expansion? In this case, though, the work done would be clearly greater than the reversible path, since the area below it would be greater.

What am I missing?

  • $\begingroup$ For processes between the same initial and final thermodynamic equilibrium states, there can be reversible paths that give less work. than an irreversible path. $\endgroup$ Nov 7, 2023 at 23:07
  • $\begingroup$ How then should I interpret the statement that reversible processes do more work than irreversible ones? I hear this a lot. $\endgroup$ Nov 8, 2023 at 0:02
  • $\begingroup$ Imagine. a gas expanding against p_ext < p $\endgroup$
    – Poutnik
    Nov 8, 2023 at 4:53
  • $\begingroup$ How you interpret and judge statements like those are up to you. I would say that there would have to be more constraints placed on the processes used in such a comparison. Here is a reference to a thread in another forum giving specific examples of reversible paths between the same two end states in which, for some of the reversible paths considered, the reversible work is less than the work for the irreversible path: physicsforums.com/threads/… $\endgroup$ Nov 8, 2023 at 11:45

1 Answer 1


You're right to be confused. The diagram you show is neither a proof nor an intuitive explanation. It can't be a proof because the statement "reversible processes do more work than irreversible ones" is not true without significant additional, unstated assumptions.

It's this sort of thing that makes thermodynamics incredibly confusing to students.

It's hard to say what assumptions your teacher/textbook/whatever wanted to add here. But I'll give you a pair of assumptions that do the job, that might be close to what they had in mind:

  1. We aren't considering processes that require putting work into the system, even temporarily.
  2. The only heat source available is the environment, which has the same temperature $T$ as the system's initial and final temperature.

Together, these assumptions prevent you from increasing the temperature above $T$. That means your process is forced to follow a path that lies at or below the reversible isothermal expansion, and hopefully that makes it intuitive that the total work done by any process following these rules must be less than or equal to the reversible process.

There are certainly irreversible paths that do more work, but they are disallowed by the assumptions. You gave one example; it includes an isobaric expansion that requires adding heat, violating assumption 2.


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