Any reversible process can be described as a sum of many infinitesimally small Carnot cycles, so $\oint {dS} = \oint {\frac{{dQ}}{T}} = 0 % MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbWexLMBbXgBd9gzLbvyNv2CaeHb5MDXbpmVaibaieYlf9irVe % eu0dXdh9vqqj-hEeeu0xXdbba9frFj0-OqFfea0dXdd9vqaq-JfrVk % FHe9pgea0dXdar-Jb9hs0dXdbPYxe9vr0-vr0-vqpWqaaeaabiGaci % aacaqabeaadaqaaqaafaGcqaaaaaaaaaWdbeaadaWdfaqaaiaadsga % caWGtbaaleqabeqdcqWIr4E0cqGHRiI8aOGaeyypa0Zaa8qbaeaada % WcaaqaaiaadsgacaWGrbaabaGaamivaaaaaSqabeqaniablgH7rlab % gUIiYdGccqGH9aqpcaaIWaaaaa!5091! $ holds. It means the integral is independent of path it takes so the entropy S is a state variable. Such a path-independence is only true so reversible process, in strickly speaking. Then...is the entropy S not state variable for irreversible process?
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$\begingroup$ Your problem is that you really don't know how to determine the change in entropy of a system that has been subjected to an irreversible process. If I'm wrong about this, please tell us how you would do it. $\endgroup$– Chet MillerCommented Mar 20, 2017 at 11:34
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2 Answers
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Entropy is a property of the system and it does not depend on the process that system experiences. Also, it does not depend on how you measure it. For irreversible process $\oint {\frac{{dQ}}{T}}$ is not equal to zero, but the system has a property called entropy at any instance that its change is depend only on initial and final states of the system.
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$\begingroup$ Thanks for giving me a comment. I was trying to prove that the entropy S is really a state variable by showing its differential dS has a property that its integral is path-independent. I thought there is a simple rule to determine which variable is a state variable; If a differential dG is exact differential so its integral is path-independent, then its integral is equal to G(re) - G(rs) where re and rs are an end and starting point of a path, respectively. So...if the integral of dS is not fully path-independent, I thought S is no longer state variable. $\endgroup$ Commented Mar 19, 2017 at 14:30
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$\begingroup$ As you mentioned, the integral of dS is not 100 % fully path-independent due to the presence of irreversible process path, how can we prove that S is truly a state variable? Is irreversible path supposed not to be taken into account when we're talking about path-independence of the integral of a differential? $\endgroup$ Commented Mar 19, 2017 at 14:34
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$\begingroup$ @DonggyuJang In classical thermodynamics, entropy is defined by $dS=\frac{\delta Q_{\text{rev}}}{T}$. en.wikipedia.org/wiki/Entropy $\endgroup$– lucasCommented Mar 19, 2017 at 16:01
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In an irreversible process, the system itself generates entropy that has to be added to the terms $\frac{\delta Q}{T}$ representing heat transfer from/to the outside. For example, if a current $I$ passes through a resistor $R$ at temperature $T$ the rate of entropy generation is $\frac{I^2R}{T}$ as measured in units of $\frac{joule}{kelvin \times sec}$.
answered Mar 19, 2017 at 15:04
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$\begingroup$ I think it's worth mentioning that, in the case of an irreversible process, the terms $\delta Q/T$ are evaluated at the temperature $T=T_B$ of the boundary between the system and the outside, where the heat transfer occurs. $\endgroup$ Commented Mar 20, 2017 at 15:21