A generic process is defined reversible if with its reverse is possible to come back to the orginal thermodynamic states both for the system and for its enviroment. You consider a room divided into two parts, with the gas in the initial state confined in one of the two part. At a certain instant the gas flows through a hole into the second part, and occupies all the room. We know that its internal energy doesn't change since its temperature doesn't change, as demonstrated by Joule with his experiment. In order to bring all back you need to compress the gas with a piston and after, since its temperature will increase, you will have to subtract a certain heat. You have for the free expansion: $$\Delta{U_{system}}=0=\Delta{U_{env}}$$ For the reverse: $$\Delta{U_{system}}=-|Q|+|W|=0 \rightarrow |Q|=|W| $$ It means we have to provide the same amount of heat and of work to bring all back. And I don't think it is impossible since Kelvin-Plank principle says that is impossible to create a cycle which produces net work and exchanges heat with just one reservoir. But in this case the verses of heat and work are opposite so it should be possible to create the cycle. Someone could explain where I am wrong?

  • 1
    $\begingroup$ A deleted answer links to this Quora post. $\endgroup$
    – rob
    Mar 24, 2021 at 16:58

2 Answers 2


You went wrong in interpreting the Kelvin-Plank statement of the second law. There is no violation. The statement is (the details of the statement may differ depending on the references)(italics done by me for emphasis):

"It is impossible to devise a cyclically operating heat engine, the effect of which is to absorb energy in the form of heat from a single thermal reservoir and to deliver an equivalent amount of work"

No work is done by the system in the free expansion. In this case no heat is absorbed by the system in the cycle, only rejected by the system to the surroundings as a result of the isothermal compression done by the surroundings needed to return the system to its original state. The system does not deliver work. The surroundings does work on the system. There is no violation.

You can look at this also from the perspective of the surroundings. Since the system is isolated from the surroundings during the free expansion of the system, the surroundings does not partake in that process. When the surroundings does work to compress the gas in the system, it performs work, But this work is not part of a cycle. It is simply a single process, namely, a reversible isothermal process. Net work done during a single process using heat from a single reservoir (as opposed to a sequence of processes comprising a cycle) does not violate the law.

I think there has been a misunderstanding. Summarizing my question is: "Why free expansion is said to be irreversible although (at least in my opinion) you can bring back the whole universe to its initial state?

But the whole universe has not been brought back to its original state.

The system has been brought back to its original state (including original entropy) but the state of the surroundings has now been changed. Heat has been transferred to the surroundings increasing its entropy. Thus there is a positive total entropy change of the universe (system + surroundings) making the entire process irreversible.

The following is in response to your follow up questions:

I'm trying to find another variable of the system or of the enviroment (except entropy) which has changed at the end of the cycle in order to show the irreversibility.

If by the term another "variable" you mean another property, the short answer is there is no other property that I am aware of to account for a process being irreversible than the property of entropy. This is the reason the second law and its associated property, entropy, was needed. The first law (conservation of energy) is satisfied by a process even if the process is impossible, as long as energy is conserved. The simplest example is that of natural heat flow.

We know that heat only flows naturally, or spontaneously (without external influence), from a hot body to a cold body. The reverse has never been observed to occur spontaneously, even though the reverse process does not violate the first law of thermodynamics if the heat lost by the cold body equals the heat gained by the hot body.

The example of the free expansion is more subtle. But the idea of spontaneity applies. The gas spontaneously expands from its chamber to the evacuated chamber. We would never see all the gas that expanded into the evacuated chamber spontaneously return to its original chamber. To be more precise, the probability is essentially zero. But there is no violation of the first law.

But the system comes back exactly to its initial state, I was thinking that if we consider the enviroment made by the cold reservoir, since it exchanges heat with the system, if we not consider its thermal capacity infinity then its temperature is increased. Could be this the variable which generates irreversibility in your opinion?

Good question. But if the temperature of the environment increases, and only heat transfer to the environment is involved, that would mean the internal energy of the environment would also increase. But that would be impossible in this example.

The work done by the environment on the system to return it to its original state exactly equals the heat transferred from the system to the environment (being that compression of the gas is isothermal). That means the change in internal energy of the environment has to be zero. For the entire cycle the change in internal energy of both the system and the environment is zero. This can be so regardless of whether the cycle is reversible or irreversible. The property of entropy is needed to show irreversibility.

Hope this helps.

  • $\begingroup$ Thanks for your help, but I think there has been a misunderstanding. Summarizing my question is: "Why free expansion is said to be irreversible although (at least in my opinion) you can bring back the whole universe to its initial state? " I mean, in your answer you just confirmed that the reverse process is possible since it doesn't violate the second law...Which is what I've written. So what do you think about the reversibility? $\endgroup$
    – Landau
    Aug 3, 2019 at 12:47
  • $\begingroup$ @Landau See revision to my answer to respond to this question. $\endgroup$
    – Bob D
    Aug 3, 2019 at 13:12
  • 1
    $\begingroup$ If work is provided by the environment to compress the gas into its initial state, the state of the environment changes, indicating irreversibility. Free expansion is taking place without external force. The reverse process requires an external agent, since it decreases entropy. $\endgroup$
    – p6majo
    Aug 3, 2019 at 13:17
  • $\begingroup$ @p6majo presumably your comment was directed at Landau, since it's basically what I said. $\endgroup$
    – Bob D
    Aug 4, 2019 at 23:13
  • $\begingroup$ Summarizing you say that since $\int_{cycle} \frac{\delta Q}{T}<0$ for the system then the process must be irreversible (Clausius). But I think that the problem is more complicated. I know a proof that the Clausius Inequality implies irreversibility, only for the case of quasi-static process which irreverisbility is given by friction. And this is not the case. $\endgroup$
    – Landau
    Aug 5, 2019 at 17:39

In addition to the two detailed answers, let me give you an intuitive/conceptual one:

You’re right that you can restore the condition of the gas by compressing it (using mechanical work) and cooling it (moving thermal energy from the gas to the environment). That does get the gas back to it’s starting point.

But that does not get the entire system back to it’s starting point, because the environment has converted work to heat: its entropy has gone up.

When the gas expanded, its entropy increased. The compress/cool process can move that new entropy to the environment, but it can’t make that new extra entropy go away. So the process is not reversible: there’s no way to put everything back.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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