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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.

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    $\begingroup$ Are you sure that in the inequality, the subscript of Q isn’t irrev? $\endgroup$ – Chet Miller Feb 29 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 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 at 18:08
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    $\begingroup$ 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 at 20:32
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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).

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