I'm studying for my thermodynamics exam and I came across something which really confuses me.

An infinitesimal change in entropy $ dS_{sys}$ of a system at temperature $T_{sys}$ during a reversible transformation, where $\delta Q_{rev}$ is defined as the heat going in/out the system is given by: $$ dS_{sys} = \frac{\delta Q_{rev}}{T_{sys}} $$

However, there is a statement in my book claiming that: $ dS_{sys} > \frac{\delta Q_{rev}}{T_{surr}} $

My confusion is the following: If an infinitesimal change in entropy of a system at temperature $T_{sys}$ is defined as above, how can the statement $ dS_{sys} > \frac{\delta Q_{rev}}{T_{surr}} $ be true? In order to calculate the change in entropy the path must be reversible, meaning that the temperature of the system is equal to the temperature of the surroundings, i.e. $ T_{sys} = T_{surr} $ otherwise the path isn't reversible. The statement clearly doesn't hold if my reasoning is correct.

Can someone clarify this to me because I'm really struggling with this.

  • 1
    $\begingroup$ Are you sure that in the inequality, the subscript of Q isn’t irrev? $\endgroup$ Feb 29, 2020 at 16:41
  • $\begingroup$ Like @ChetMiller pointed out, that inequality holds for irreversible processes due to the nature of not being able to extract 100% work from heat. $\endgroup$
    – Weezy
    Feb 29, 2020 at 16:54
  • $\begingroup$ Thermodynamics has lots of confusing notation - in this case, I think the $\delta Q_{\text{rev}}$ is probably meant to signify the heat if the process were reversible. I.e. the heat due to a fictitious reversible process connecting the start and end states. It is not the heat for the actual process. $\endgroup$
    – jacob1729
    Feb 29, 2020 at 18:08
  • 1
    $\begingroup$ I have the same comment as Chet. It would save us all a lot of time if you recheck the book statement. If it is as you stated, then give us the context in which the statement was made. $\endgroup$
    – Bob D
    Feb 29, 2020 at 20:32

1 Answer 1


For a general transformation between $A$ and $B$, the entropy change can be written:

$$dS_{A \to B} = \frac{dQ_{A \to B}}{T_{\mathrm{surr}}} + dS_{\mathrm{created}},$$

where $dS_{\mathrm{created}} \geq 0$ is the entropy created by irreversible processes.

For a transformation to be reversible, you need $dS_{\mathrm{created}} = 0$ and also $T_{\mathrm{surr}} = T_{\mathrm{sys}}$. In that case, $dS = \frac{dQ}{T_{\mathrm{surr}}} = \frac{dQ}{T_{\mathrm{sys}}}$ so the first inequality does not hold strictly.

However, for real macroscopic processes, $dS_{\mathrm{created}}$ is always $>0$, even by an infinitesimal amount (no transformation is fully reversible). In that case, the inequality becomes strict and $dS > \frac{dQ}{T_{\mathrm{surr}}}$. Of course it is still useful to consider adiabatic processes as they are sometimes a really good approximation to some almost reversible real transformations, and they can also be used for non-reversible process to calculate the change of entropy between two states $A$ and $B$ by considering the corresponding adiabatic path between the initial and the final states (as $dS_{\mathrm{A\to B}}$ does not dépend on the path followed).


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.