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so i'm following a derivation of the statement in the title and it goes the following way. Consider we have a system and we look at two processes between two states $(1,2)$ and $(2,1)$ where path $(1,2)$ is reversible and $(2,1)$ is irreversible. So we have: $$\oint dS=\int_{1}^{2} dS + \int_{2}^{1} dS \le 0$$ Since $(1,2)$ is a reversible process we can write in the following way: $$\int_{2}^{1}dS +(S_2-S_1) \le 0 \\ \Rightarrow \int_{2}^{1}dS \le (S_1-S_2) $$ $$\Rightarrow S_{irr} + \int_{2}^{1}dS = (S_1-S_2)$$ Where $S_{irr}$ is the bonus increase in entropy due to the irreversability of the process. So now if we say that the system is isolated $(\delta Q=0)$: $$\int_{2}^{1}dS=\int_{2}^{1}\frac{\delta Q}{T}=0$$ $$\Rightarrow S_{irr} = (S_1-S_2)$$ My question here is why when we assume that the system is isolated don't we say that the other integral is also going to result in zero $(S_2-S_1=0)$?

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Most of your analysis seems incorrect to me. For the combination of reversible and irreversible cycle you describe, $$\Delta S_{cycle}=\Delta S_{1,2}+\Delta S_{2,1}=0$$That is, for a system which undergoes a cycle, the initial and final states of the overall cycle are identical, and the entropy change is zero. If the irreversible part of the cycle is adiabatic or isolated, $$\Delta S_{2,1}=\sigma>0$$where $\sigma$ is the entropy generation during the irreversible part of the path. So we have $$\Delta S_{1,2}+\sigma=0$$or $$\Delta S_{1,2}=\int_1^2{\frac{dQ}{T}}=-\sigma<0$$This tells us that the reversible part of the path cannot be adiabatic or isolated, and some heat transfer must occur over this part of the path. There is no possible cyclic path comprised of an irreversible- and a reversible adiabatic segments.

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  • $\begingroup$ Thanks a lot! Just one question. What about ideal Carnot cycle? isn't the 2nd and 4th part reversible adiabatic? $\endgroup$ May 22, 2023 at 22:42
  • $\begingroup$ The typical Carnot cycle is not irreversible for any of the 4 parts. The entropy changes of the 1st and 3rd isothermal parts sum to zero, and the entropy changes of the 1st and 3rd parts are each zero. $\endgroup$ May 22, 2023 at 23:04
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Because in an isolated system you can have a spontaneous process, this can happen by removing a constraint within the otherwise isolated system itself. For example, you can have a rigid adiabatic wall separating two fluids at different temperatures and/or pressures and the equilibrium is maintained while the wall is in place. Now remove the wall and observe a spontaneous relaxation process during which the temperatures and pressures equalize. With the spontaneous equalization comes an irreversible entropy production that is characteristic to the process itself. As long as the spontaneous process is not so violent that macroscopic quantities cannot even be defined, at every instant/step/stage of the spontaneous process you have a system entropy dependent on the state of the system and a positive entropy production dependent on the process.

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