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So, I know that $\oint \frac{\delta Q}{T}\le0$, that's the inequality and it's equal to 0 for reversible and less than 0 for irreversible. But at the same time I know $\Delta S=\oint \frac{\delta Q_{rev}}{T_{surr}}$ and $\frac{\delta Q_{rev}}{T}\ge \frac{\delta Q}{T}$ coming from the Planck-Kelvin statement of the second law of thermodynamics. But doesn't this just mean that $\Delta S \ge \oint \frac{\delta Q}{T} \le0$, which looks kinda useless. It means that entropy might increase, might still be 0 or might even still decrease as long as the cyclic integral is even lower. How does this show that entropy always increases in an irreversible process? Or does it show that it can decrease as long as the net increase is greater in the surroundings? And if so, how do we prove (again, mathematically) that that is the case and entropy isn't just decreasing in the whole Universe?

Thanks for the help in advance!

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  • $\begingroup$ This is a common issue with understanding of the Clausius inequality. The temperature in the Clausius inequality is the temperature of the environment (or really, boundary) of the system, not the system itself. See here. $\endgroup$
    – march
    Jun 5, 2022 at 16:24

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You mention an expression of the second law that holds only for systems going through a cyclic transformation (due to the $\oint$ notation). Since you also mention the universe, which has no reason to be going through a cyclic process, I'll answer without this assumption.

Entropy doesn't always increase in an irreversible process, that's a shortcut that leads to wrong conclusions.

You can write the variation of entropy for a system as:

$$\Delta S=S_e+S_c$$

with $S_e$ the entropy exchanged with the environment and $S_c$ the entropy created in the process. The second law states that:

$$S_c\geqslant 0$$

A process can be irreversible ($S_c>0$) but lead to decreasing entropy if $S_e<-S_c$ (loosely speaking, it's losing thermic energy, restoring more order than is lost by the irreversible process) so that $\Delta S<0$.

Assuming the universe is an isolated system, $S_e=0$ so $\Delta S=S_c$. The second law then yields $\Delta S\geqslant 0$: the total entropy of the universe doesn't decrease.

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  • $\begingroup$ Thank you so much for the quick answer! But I wanted to ask if we imagine the case I gave as isolated form the entire Universe, the equations still uphold, so $S_e=0$, but still nothing in the math actually says $\Delta S>0$ for an irreversible process, just that it's bigger than the cyclic integral, which is already negative and not 0 for an irreversible process. So shouldn't the second law of thermodynamics say that $\Delta S>\oint \frac{\delta Q}{T}$ for an irreversible process in an isolated system and that's it? Where does the strictly positive part come in? $\endgroup$
    – Momchi
    Jun 5, 2022 at 13:10
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For a cyclic process on a closed system, $\Delta S=\oint{\frac{\delta Q_{rev}}{T_{syst,rev}}}=0$. So, for a cyclic process on a closed system, the Clausius inequality tells us that $$\Delta S=0\ge \oint{\frac{\delta Q}{T_I}}$$where $T_I$ is the temperature at the interface between the system and its surroundings.

For any process on a closed system between thermodynamic equilibrium states A and B, the Clausius inequality reads: $$\Delta S=\int_A^B{\frac{\delta Q_{rev}}{T_{syst,rev}}}\ge \int_A^B{\frac{\delta Q}{T_I}}$$ Based on all this, I don't see what your doubt is.

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  • $\begingroup$ Thank you for the answer! I agree with what you wrote, I have no doubts there. My question was more trying to ask why entropy always has to increase in an isolated system. All the math says is that entropy is higher than the cyclic integral, but the cyclic integral is already negative, so what is stopping the integral from being -2 and entropy being -1 in an irreversible process in that case? Purely mathematically I don't see it unless I get into statistical mechanics or philosophy of science. $\endgroup$
    – Momchi
    Jun 5, 2022 at 17:50
  • $\begingroup$ So your real question is "how do we know for sure that the Clausius inequality is always correct?" Back in the day, Clausius concluded this empirically, based on extensive observations. As you pointed out, now we have more clear confirmation of this from statistical thermodynamics. $\endgroup$ Jun 5, 2022 at 18:26

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