Timeline for Is $dS=\frac{\delta Q_{irev}}{T}$ true for non-reversible processes?
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| Sep 14, 2020 at 9:17 | comment | added | Blue Various | @TheoreticalMinimum Because both $δQ_{rev}(U,V,N)$ and $δQ_{rev}(U,V,N)/T$ are always state quantities, it is possible to calculate them if we record the U,V,N,T,P,μ along the reaction path, even if the reaction is irreversible; of cause we suppose that the reaction path could be written as a smooth curve in the state space, So, the notion of the "reversible reaction" may be less important, IMO. | |
| Sep 14, 2020 at 6:38 | comment | added | user224659 | The issue which I had at the time I asked the question is most likely the same as you do: Many books are imprecise in the definition of irreversible and non-quasi static processes. There is often a lack of a proper definition of "reversibility". Anyways it should not be defined by "quasi-static" as some authors do. | |
| Sep 14, 2020 at 6:36 | comment | added | user224659 | @BlueVarious A quasi-static process is just a well defined curve in the phase space. This process does not necessarily have to be reversible. | |
| Sep 13, 2020 at 23:27 | comment | added | Blue Various | Additionally, "Quasi static irreversible adiabatic free expansion" you wrote is a nonsense concept. | |
| Sep 13, 2020 at 23:25 | comment | added | Blue Various | A "quasi-static adiabatic change" is quasi-static and reversible because it is an adiabatic change with constant entropy in the course of the reaction. | |
| Sep 13, 2020 at 23:23 | comment | added | user65081 | @BlueVarious you should read again my description in the previous comments | |
| Sep 13, 2020 at 23:19 | comment | added | Blue Various | Irreversible adiabatic free expansion is indeed irreversible. But it is not quasi-static. Because it is not possible to draw the reaction path of irreversible adiabatic free expansion on U-V-N space. It is via non-equilibrium state where U, V and N cannot be defined at any time before equilibrium is reached. | |
| Sep 13, 2020 at 20:50 | comment | added | user65081 | @BlueVarious the process I described is both quasi-static and irreversible, what part makes you think it is not so? | |
| Sep 13, 2020 at 20:43 | comment | added | Blue Various | The questioner's (@TheoreticalMinimum) question contains a lot of confusion. But he/she is asking about whether "dS=δQ_rev/T" holds for "quasi-static but not reversible processes". But your answer is about "examples that are neither quasi-static nor reversible". | |
| Sep 13, 2020 at 20:33 | comment | added | user65081 | @BlueVarious sorry, I read the answer too fast. Yes, I mean that, quasi-static. I did not clarify it, but by that I meant a process in which you open a new compartment that is only differentially larger than the original, and after an infinite number of such processes (that is, mini free expansions) you get the final finite volume you want. | |
| Sep 13, 2020 at 20:27 | comment | added | Blue Various | So, what is the quasi static of " quasi static irreversible adiabatic free expansion" you means? Is this a typo? | |
| Sep 13, 2020 at 19:56 | comment | added | user65081 | @BlueVarious it is an irreversible process, I never meant it to be quasi-static, and it is not required either | |
| Sep 13, 2020 at 19:49 | comment | added | Blue Various | Is adiabatic free expansion a quasi-static process? What the "quasi static irreversible adiabatic free expansion” means? Free but quasi static ? | |
| Dec 3, 2019 at 6:45 | comment | added | user224659 | @Wolphramjonny Indeed you interpreted the question just right. | |
| Dec 3, 2019 at 6:35 | vote | accept | CommunityBot | ||
| Dec 2, 2019 at 16:50 | comment | added | Bob D | @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. | |
| Dec 2, 2019 at 16:38 | comment | added | user65081 | @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.. | |
| Dec 2, 2019 at 16:35 | comment | added | user65081 | @Nephente I imagine it as removing a series of partitions very close to each other, so the new volume is incremented in steps | |
| Dec 2, 2019 at 16:26 | comment | added | Bob D | 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. | |
| Dec 2, 2019 at 16:04 | comment | added | Nephente | 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. | |
| Dec 2, 2019 at 16:00 | history | answered | user65081 | CC BY-SA 4.0 |