Entropy change is defined as $\Delta S = \int \frac{\delta Q_{rev}}{T} $, where $Q_{rev}$ is heat transferred through a reversible process. I’m a little confused how this works, so I have a couple of related questions which I hope will clarify what’s going on.

  1. If a reversible process is defined as $\Delta S = 0$ then how can this integral be non-zero for a reversible process? Or is it that just $\Delta S_{Univ} = 0 = \Delta S_{sys} + \Delta S_{surr}$ and therefore $\Delta S_{sys} = -\Delta S_{surr} \iff T_{sys} = T_{surr} $, in which case

  2. How would one calculate it for an irreversible process, in particular, some heat, Q, transferred across a finite temperature difference? I would try (considering isothermal transfer between infinite temperature reservoirs) $$ \Delta S_{Univ} = Q \big( \frac{1}{T_{sys}} - \frac{1}{T_{surr}} \big) $$ But this is no longer reversible, so I need to do something else?

  3. Since entropy is a state function, is the entropy change through any process that is irreversible, but between equilibrium states, equivalent to $\int \frac{\delta Q}{T}$ through a reversible process?

  4. Is this equivalent to pretending I’m doing 2 reversible transfers to some fictitious bodies, the first taking Q reversibly from the system to the fictitious body, both at $T_{sys}$, and then magically dropping the fictitious body to $T_{surr}$ and transferring the heat Q to the surroundings? If so, why use $\delta Q$ rather than $dQ$? My understanding was that $\delta Q$ was for path-dependent processes, but there is only one fixed (adiabatic) path for reversible transfer?

Sorry for yet another entropy question, thanks in advance!


1 Answer 1


The answers to your first and third questions are "yes." Here is a tutorial on how to calculate the change in entropy of the system or the surroundings for an irreversible process: https://www.physicsforums.com/insights/grandpa-chets-entropy-recipe/. The reference also provides worked examples. Note that, in considering entropy changes for both the system and the surroundings from an irreversible process, one needs to first separate the system from the surroundings and then subject each of them separately to a reversible path. These separate reversible paths do not have to match.

It simplifies things a little to think about how entropy of a system can change. There are only two ways: entropy transfer across the boundary of the system by heat flow in and out at the boundary temperature, and entropy generation (due to irreversibility) within the system. In a reversible process, only the first mechanism is present. In both cases, the transfer across the boundary is $\int{dq/T_B}$ where $T_B$ is the boundary temperature.

Regarding questions 2 and 4 (involving ideal constant temperature reservoirs), these are somewhat limiting cases. Here, if you imagine a very thin conductive slab between the two reservoirs (with high temperature at one face of the slab and low temperature at the other face), then for the irreversible process involving the transfer of heat between the reservoirs, all the entropy generation would take place within the intervening slab, while the reservoirs would each separately experience only entropy transfer with the slab. In the limit where the slab becomes infinitesimally thin, we would obtain the system you have envisioned, but all the entropy would still have been generated at the interface, and each reservoir would only experience entropy exchange with the interface.

  • $\begingroup$ Thank you for your reply. The linked tutorial is indeed helpful. Let me assemble my questions and thoughts... I think you are saying that 4. is correct, this is equivalent to doing 2 reversible transfers to fictitious bodies? This seems similar to what you describe in the Example 1: Thermal equilibration, where the hot object has $dS = \frac{MCdT}{T}$ going from $T_h$ to $T_f$, without caring what the temperature of object 2 is, and similarly for object 2 from $T_c$ to $T_f$. $\endgroup$ Jan 5, 2021 at 3:45
  • $\begingroup$ In the case of the thin barrier, do we not need to account for thermal equilibration, as heat entering from one side must travel across to the other side, with the temperature gradient getting steeper as the barrier becomes thinner? I would imagine that this internal shuffling of energy would increase entropy, as all the heat energy being piled up on one side is not in internal equilibrium. $\endgroup$ Jan 5, 2021 at 3:46
  • $\begingroup$ In example 2, may I ask why you are using the constant-volume heat capacity during a non-constant-volume (expansion) process? Also is the denominator $C$ in eq. 6, $C_v$? I think so, but you labelled it in the numerator, so just making sure. Right before Step 3, has the $T_f$ in state 2 lost the “1-...” it had in eq. 5? Not trying to be finicky, just making sure I didn’t miss something. $\endgroup$ Jan 5, 2021 at 4:01
  • $\begingroup$ Are Ex. 2+3 saying that the $\Delta S = \int\frac{\delta q_{rev}}{T}$ is the same for any reversible path between two equilibrium states, and therefore irreversible adiabatic processes may have $\Delta S > 0$, but a reversible adiabatic process ($q_{rev} = 0$) will always have $\Delta S = 0$, no matter the reversible path constructed? $\endgroup$ Jan 5, 2021 at 4:06
  • 1
    $\begingroup$ Answer to your second question: For an ideal gas, the Cv is independent of volume, and internal energy depends only on temperature. In Eqn. 6, that should be a Cv in the denominator. I don't know what happened to the v. And, yes, the 1 got lost. I don't now what happened there either. These things were there originally. I'm going to go back and make the corrections you pointed out. $\endgroup$ Jan 5, 2021 at 13:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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