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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

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  • $\begingroup$ I think its because they are using the reversible heat Qrev not the heat to actually heat the non-reversible process $\endgroup$
    – ChemEng
    Commented Dec 2, 2019 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
    Commented Dec 2, 2019 at 17:21
  • $\begingroup$ The title of your post should have the subscript $rev$ with $\delta Q$. $\endgroup$
    – Bob D
    Commented Dec 2, 2019 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? $\endgroup$
    – user224659
    Commented Dec 3, 2019 at 6:43
  • $\begingroup$ @BobD I changed the title to make more clear what I was asking and added an explanation. $\endgroup$
    – user224659
    Commented Dec 3, 2019 at 6:44

4 Answers 4

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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.

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    $\begingroup$ "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
    Commented Dec 2, 2019 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
    Commented Dec 2, 2019 at 18:02
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    $\begingroup$ 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
    Commented Dec 2, 2019 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
    Commented Dec 2, 2019 at 18:22
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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.

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    $\begingroup$ 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
    Commented Dec 2, 2019 at 16:04
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    $\begingroup$ 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
    Commented Dec 2, 2019 at 16:26
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    $\begingroup$ @Nephente I imagine it as removing a series of partitions very close to each other, so the new volume is incremented in steps $\endgroup$
    – user65081
    Commented Dec 2, 2019 at 16:35
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    $\begingroup$ @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.. $\endgroup$
    – user65081
    Commented Dec 2, 2019 at 16:38
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    $\begingroup$ @BlueVarious A quasi-static process is just a well defined curve in the phase space. This process does not necessarily have to be reversible. $\endgroup$
    – user224659
    Commented Sep 14, 2020 at 6:36
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Just for a comment;

The discussion here seems to confuse two (three) different types of "entropy".

  • If you say that "entropy is a state quantity" this is what is called "exchanged entropy". This is uniquely determined depend on the state (U,V,N), so let's represent this as a mathematical multivariable function $S_{e}(U,V,N)$.
  • If you say that "entropy increases with decreasing irreversible processes", this is called "generated entropy". Let's express this by the symbol $S_{g}$.
  • We define the $\Delta S_{tot}$ as follows. The $\Delta S_{tot}$ represents the entropy change in the entire system via the reaction between the reaction from $(U_1,V_1,N_1)$ to $(U_2,V_2,N_2)$ ; $$\Delta S_{tot}:=S_{e}(U_2,V_2,N_2)-S_{e}(U_1,V_1,N_1)+S_{g}$$

For more details for the exchanged/generated entropies plesae this book might be helpful.

Note that since $S_{g}$ is not a state quantity, $S_{tot}$ is not a state quantity either.

This Q&A also includes other confusion. The concept of "whether this reaction can be written as a curve in U-V-N space or not" and the concept of "whether this reaction is reversible or not" are different concepts. In this sense, the reaction paths can be classified from the two different view point; "writable/unwritable" and "reversible/irreversible".

From the "writable/unwritable" viewpoint, the reaction paths can be classified into two types;

  • Reaction that cannot be written as smooth curves in the state space(U-V-N space): e.g., "adiabatic free expansion
  • A reaction that can be written as a smooth curve in the state space: the quasi-static process is pseudo-this.

The viewpoint of "reversible/irreversible", is too confused, so, I will omit the definition of this. However, there are four kinds of combinations:

  • writable and reversible,
  • writable and irreversible,
  • unwritable and reversible, and
  • unwritable and irreversible.

Probably the third one is an abstract nonsense (Such an example probably does not exist.)　

Furthermore, "adiabatic free expansion" has　also been confused with "quasi static adiabatic expansion. " In order to be "quasi static", the piston must be controlled; move a little bit, applying the brakes. This is no longer free expansion. In the process of braking the piston, an external force is applied to the system.

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    $\begingroup$ Bravo.. Excellent answer. If you don't mind could you see the thermodynamics questions on my account? I would really appreciate if you were to answer it. $\endgroup$
    – Brian
    Commented Sep 13, 2020 at 20:46
  • $\begingroup$ By the way, it would be awesome if you could give some further reading/citations to your post $\endgroup$
    – Brian
    Commented Sep 13, 2020 at 20:47
  • $\begingroup$ I don't know if I can answer that right now, but I'll try. Where do I find that question? Give me the link. $\endgroup$
    – Blue Various
    Commented Sep 13, 2020 at 20:50
  • $\begingroup$ physics.stackexchange.com/questions/579414/… I am really struggling with this one $\endgroup$
    – Brian
    Commented Sep 13, 2020 at 20:50
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Clausius' theory is about heat exchange with quasistatic volume change. When the volume is changed non quasistatically (as in free expansion or when some irreversible work is done), the question takes a different meaning. In Clausius' approach the only changes in entropy are caused by heat exchange, not irreversible volume change.

Orginally, Clausius' point about irreversibility is the following. When a system exhanges heat with the surroundings (a heat bath), we have:

  • the heat transfer is reversible if and only if $T_{surr}=T_{system}$
  • then, $\Delta S_{system} = \frac{Q}{T_{surr}}$
  • otherwise $\Delta S_{system} > \frac{Q}{T_{surr}}$

Important: "reversible/irreversible" here applies to the heat exchange, meaning to "system + surroundings". Seeing the process as reversible/irreversible for the system only does not make sense and leads to confusions.

Writing just $T$ is ambigous. Is it the temperature of the system or the temperature the surroundings? The fact is, when heat is not transfered reversibly, the temperature of the system usually does not even exist, since a gradient appears inside the system and it does not have a uniform temperature. Assuming the gradient spreads over a small volume (negligible energy), we may say the system still has constant temperature almost everywhere. Then we have $\Delta S_{system} = \frac{Q}{T_{system}}$.

As a conclusion, $T$ is meaningful for reversible processes, and then it is both the temperature of the system and the surroundings. Otherwise, $T_{system}$ does not exist during the process and $\Delta S_{system} > \frac{Q}{T_{surr}}$. When the temperature is near uniform inside the system, then $\Delta S_{system} = \frac{Q}{T_{system}}$, even for an irreversible process.

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