2
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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 Miller Mar 20 '17 at 11:34
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Donggyu Jang Mar 19 '17 at 14:30
  • $\begingroup$ 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? $\endgroup$ – Donggyu Jang Mar 19 '17 at 14:34
  • $\begingroup$ @DonggyuJang In classical thermodynamics, entropy is defined by $dS=\frac{\delta Q_{\text{rev}}}{T}$. en.wikipedia.org/wiki/Entropy $\endgroup$ – lucas Mar 19 '17 at 16:01
1
$\begingroup$

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 $\frac{joule}{kelvin \times sec}$.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Chet Miller Mar 20 '17 at 15:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.