I read that adiabatic changes in a closed system can never decrease the entropy.

To try to understand why, I thought of two scenerios: reversible processes, and irreversible processes.

For a reversible process where $$ds = dq_{rev}/T$$

any process done adiabatically means that the heat transfer is 0, and therefore this equality becomes 0. So far, what I read holds up.

However, for an irreversible process, I'm not sure why it can't decrease entropy. Can anyone prove to me why an irreversible adiabatic process on a closed system will always increase the entropy of that system?

Please note, this isn't a homework problem that I'm trying to get an answer for; I'm trying to understand the ideas inside-out and came across this statement online, which has made me curious.


I just thought of a possibility of why. According to the Clausius inequality:

$$ds ≥ dq_{irrev}/T$$

So if there is an adiabatic change, dq (and q) equals 0. So, if ds ≥ 0, the entropy change is positive or zero.

However, assuming this is correct, this is just the mathematical explanation/proof. Can anyone provide a conceptual and intuitive explanation of why?


1 Answer 1


In an irreversible expansion or compression, viscous forces (related to how quickly the gas is deforming) are significant, and result in a conversion of part of the available mechanical energy into internal energy (and thus temperature). This is sometimes referred to as "viscous heat generation" (even though no actual heat is transferred). But the increase in temperature from this viscous heat generation translates into entropy generation. The viscous forces are negligible in a slow reversible process, so negligible viscous heat generation and entropy generation occurs.

  • $\begingroup$ Hmm, so this is the real physics behind the idea, but the math I wrote in my post is correct? $\endgroup$
    – F16Falcon
    Oct 9, 2019 at 21:03
  • 1
    $\begingroup$ The math you wrote is of course correct, since it is a mathematical statement of the 2nd law of thermodynamics, except that, in Eqn.2, I would change the greater than or equal to a >. $\endgroup$ Oct 9, 2019 at 22:39

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