1
$\begingroup$

The German Wikipedia reads

Das Differential $\mathrm {d} S$ ist nach Clausius bei reversiblen Vorgängen zwischen Zuständen im Gleichgewicht das Verhältnis von übertragener Wärme $\delta Q_{\mathrm {rev} }$ und absoluter Temperatur $T$: $dS=\frac{Q_{\mathrm {rev} }}{T}$

Which translates to

According to Clausius the differential $\mathrm {d} S$ for reversible processes between equilibirum states is the ratio between transmitted heat $\delta Q_{\mathrm {rev} }$ and absolute temperature $T$: $dS=\frac{Q_{\mathrm {rev} }}{T}$

This formulation seems confusing to me. Why do we need reversibility? I do not see why this shouldn't be true for quasi-static irreversible processes. We start at a state of entropy $S_1$ and by some process we reach $S_2$. As the entropy by axiom is path-independent it shouldn't matter weather the path is reversible or not.

Addendum: Many people stated in the comments that one can use a reversible process starting and resulting in the same equilibrium state, as the irreversible one. While this is true and an important concept, my question was aimed at the actual heat $\delta Q_{irev}$ that is transferred to the system during a irreversible process.

Related The actual definition of entropy

$\endgroup$
  • $\begingroup$ I think its because they are using the reversible heat Qrev not the heat to actually heat the non-reversible process $\endgroup$ – ChemEng Dec 2 '19 at 16:43
  • $\begingroup$ In addition to entropy transferred from the surroundings to the system during a process (which is described by dq/T), in an irreversible process, entropy is generated within the system, which is not accounted for by dq/T. Therefore, using dq/T for an irreversible process will give the wrong answer for the change in entropy. $\endgroup$ – Chet Miller Dec 2 '19 at 17:21
  • $\begingroup$ The title of your post should have the subscript $rev$ with $\delta Q$. $\endgroup$ – Bob D Dec 2 '19 at 18:06
  • $\begingroup$ @ChetMiller How do we account for that entropy? I always took the formula in the title as the defintion of entropy and now I'm confused on how it is really defined. It seems like theres some sort of inner degrees of freedom that are triggered during a irreversible process? $\endgroup$ – TheoreticalMinimum Dec 3 '19 at 6:43
  • $\begingroup$ @BobD I changed the title to make more clear what I was asking and added an explanation. $\endgroup$ – TheoreticalMinimum Dec 3 '19 at 6:44
1
$\begingroup$

One counterexample is a quasi static irreversible adiabatic free expansion. Here d$S>0$ and d$Q=0$, so the equality is not valid for this irreversible process.

$\endgroup$
  • $\begingroup$ How would you realize a quasi-static free expansion? By definition of q.s. the gas has to expand in a sequence of equilibria, which is not given in the case of, let's say, spontaneously removing a partition. $\endgroup$ – Nephente Dec 2 '19 at 16:04
  • $\begingroup$ You can still use assume a reversible transfer of heat process to get the entropy change for the irreversible adiabatic free expansion. In that case you can assume a reversible isothermal compression to return the system to its original state before the free expansion. The magnitude of the entropy change for the reversible isothermal compression will equal the entropy change that occurred in the free expansion. $\Delta S$ for the system will be zero when returned to its original state, but the isothermal compression will increase the entropy of the surroundings so that $\Delta S_{TOT}$ >0. $\endgroup$ – Bob D Dec 2 '19 at 16:26
  • 2
    $\begingroup$ @Nephente I imagine it as removing a series of partitions very close to each other, so the new volume is incremented in steps $\endgroup$ – Wolphram jonny Dec 2 '19 at 16:35
  • $\begingroup$ @BobD I agree with you, and perhaps I misinterpreted the question. It was not if you can find a reversible process to calculate the change in entropy, but if the equation was valid for an irreversible process, which is different to me, that is, use the change in entropy and heat transferred during that specific irreversible process.. $\endgroup$ – Wolphram jonny Dec 2 '19 at 16:38
  • $\begingroup$ @Wolphramjonny I see your point, it could be viewed that way as well. The equation is, of course, defines entropy change in terms of a reversible transfer of heat. There are many irreversible work processes between two equilibrium states not of which involves any heat transfer. But to determine what the entropy generated is we can assume a process between the states involving a reversible transfer of heat and we will obtain the entropy generated for the irreversible work process. $\endgroup$ – Bob D Dec 2 '19 at 16:50
3
$\begingroup$

Why do we need reversibility? I do not see why this shouldn't be true for quasi-static irreversible processes.

Although the definition is in terms of a reversible transfer of heat, you are correct that it is not limited to a reversible process, i.e., it applies to an irreversible process as well. Entropy is a state function or property, like internal energy. That means the difference in entropy between two equilibrium states is independent of the path or process between the states.

So if you have an irreversible process taking you between two states you can determine the entropy change of the system by assuming any convenient reversible process between the states. That will give you the entropy change for the system for the irreversible process as well since entropy is a state function.

However, if the process is irreversible, entropy is generated by the system. In order to return the system to its original state (perform a cycle) the entropy generated will need to be transferred to the surroundings making the total entropy change (system + surroundings) >0 for a complete cycle. For a reversible cycle the overall entropy change = 0.

Hope this helps.

$\endgroup$
  • $\begingroup$ "So if you have an irreversible process taking you between two states you can determine the entropy change of the system by assuming any convenient reversible process between the states." I think it is worth stressing here, for clarity, that, since heat is a path dependant property, the heat transfer on this conveinient reversible path will in general not be the same as the amount of heat transfered on the real physical irriversible process. This means we can't simply plug the measured $\delta Q$ in in place of $dQ_\mathrm{rev}$ and expect to get the right answer $\endgroup$ – By Symmetry Dec 2 '19 at 17:36
  • $\begingroup$ @BySymmetry "While heat is a path dependent property". First of all, heat is not a property. While heat is path dependent a reversible transfer of heat divided by temperature is not path dependent. If it were, it would be tantamount to saying entropy between equilibrium states is path dependent since $dS=\frac{\delta Q_{rev}}{T}$. Second of all, I never said you can simply plug in $\delta q$ for $\delta q_{rev}$. I was responding to the text of the OP's question where the subscript $rev$ is always used and not the title which incorrectly left out the $rev$ subscript. $\endgroup$ – Bob D Dec 2 '19 at 18:02
  • $\begingroup$ I don't think we disagree on anything here. When I say "Heat is a path dependant property" I mean it is a property of a path and not of the system. My comment was trying to emphasise a point rather than imply that anything you said was wrong $\endgroup$ – By Symmetry Dec 2 '19 at 18:14
  • $\begingroup$ @BySymmetry Got it, no problem. Sorry but when I hear the word property in a thermodynamics discussion it has a specific meaning to me- $U$, $P$. $V$, $T$ etc. that goes beyond the dictionary definition, which I now see is the way you were using it. $\endgroup$ – Bob D Dec 2 '19 at 18:22

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.