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E.g., the change in entropy of a reversible process 0, then irreversible process should have 0 change in entropy. This question arose because I couldn't understand why generally work done by reversible process is greater than work done by irreversible process, but w(reversible) < w(irreversible) in compression of gas (magnitude of work done is more in irreversible compression). I'm in Highschool, would be helpful if you could clarify my questions in in simple terms.

NOTE: Question has been edited. ( Entropy of reversible process is 0 - has been changed to - the change in entropy of a reversible process 0. Subsequent counterpart(irreversible process) should have 0 entropy - has been changed to - then irreversible process should have 0 change in entropy )

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  • $\begingroup$ Please articulate your understanding of how to determine the change in entropy for an irreversible process. $\endgroup$ Dec 30, 2022 at 16:22
  • $\begingroup$ Hello sir! To be honest I only know that change in entropy can be found through ΔG=ΔH−TΔS and ΔS = Q/T. I do not know any other way to determine change in entropy for an irreversible process. I'm a student and I'm still in the basics, would be really helpful if you could explain why irreversible process generates additional entropy. Thank you for taking time in answering my questions. $\endgroup$
    – Avaneesh B
    Dec 31, 2022 at 6:19
  • $\begingroup$ Your equation for $\Delta S$ is incorrect. It should read $\Delta S=\int{\frac{dq_{rev}}{T}}$ $\endgroup$ Dec 31, 2022 at 12:12

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For eg, the entropy of a reversible process (is) 0,

In order to respond to @Nasu answer you should edit it to say "change in entropy". The way you word it now suggests entropy is itself a process, which it is not (as I believe @Nasu is trying to tell you).

That said, the change in entropy of the system for a reversible process can be zero, greater than zero, or less than zero, depending on whether there is (1) no heat transfer (e.g., reversible adiabatic process), (2) heat transfer into the system (e.g., reversible isothermal expansion process), or (3) heat transfer out of the system (e.g., reversible isothermal compression), respectively. It is the total entropy change of the system plus its surroundings that is zero for any reversible process, and greater than zero for an irreversible process.

then its subsequent counterpart (irreversible process) should have 0 entropy.

It's not clear to me what you mean by "its subsequent counterpart". Subsequent to what? Counterpart of what?

why generally work done by reversible process is greater than work done by irreversible process, but w(reversible) < w(irreversible) in compression of gas (magnitude of work done is more in irreversible compression)

When one says, in general, that reversible work is greater than irreversible work, one is speaking of the work work done by the system, i.e., more "positive" work. For the gas, expansion work is positive, compression work negative, when using the first law version $\Delta U=Q-W$.

If the magnitude of the irreversible compression is greater than the magnitude of the reversible compression, the irreversible work is more negative than the reversible work. That's the same thing as saying the reversible work is more positive than the irreversible work.

To illustrate, consider the two compression paths from point 2 to point 1 in the figure below, one being reversible and the other irreversible. The reversible path is a reversible isothermal compression of an ideal gas. The irreversible path consists of a sudden increase in external pressure (so sudden there is no time for heat transfer), followed by a rapid compression at constant external pressure until equilibrium is re-established at state 1.

Both paths involve negative work. The magnitude of the work done for each path is the area under the path. The magnitude of the irreversible work is greater than the reversible work, meaning the work done by the system is more positive for the reversible path than the irreversible path.

Hope this helps.

enter image description here

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  • $\begingroup$ Great explanation Sir! I meant that if the change in entropy of a reversible process is 0, then the change in entropy of a irreversible process is also 0. I do understand that reversible includes both forward and backward hence its change in entropy is 0 but I came across an answer where they had mentioned that the some additional entropy is generated in irreversible process and is lost. I can't understand why there's additional generation of entropy in irreversible process, thanks a lot for your explanation! $\endgroup$
    – Avaneesh B
    Dec 31, 2022 at 5:58
  • $\begingroup$ @AvaneeshB the change in entropy of the SYSTEM for any process, reversible or irreversible, is zero going from state 1 to 2 and back to 1 because entropy is a state property. But the irreversible process generates entropy that must be transferred to the surroundings in the form of heat. So there’s an increase in entropy of the surroundings for the irreversible process. $\endgroup$
    – Bob D
    Dec 31, 2022 at 12:09
  • $\begingroup$ So for a reversible reaction going from state 1 to 2, the change in entropy isn't zero because it isn't returning back to state 1? you're trying to say that for all the reactions the change in entropy is zero, irrespective of reversible or irreversible? The generated entropy is lost so that the zero mark is maintained in irreversible process? Once again thank you for taking your time in answering my questions. $\endgroup$
    – Avaneesh B
    Dec 31, 2022 at 14:53
  • $\begingroup$ Sorry for disturbing you, why's there no work done on the gas in adiabatic condition? I came across some answers where they mentioned that it is because there's no external pressure acting on the gas making the change in internal energy to be '0', but if the walls of the system is adiabatic then the change in the internal energy of the system is equal (=) to work done. Shouldn't work done be 0 in this case too? $\endgroup$
    – Avaneesh B
    Dec 31, 2022 at 15:21
  • $\begingroup$ @AvaneeshB One comment at a time, please. In response to the first comment: “So for a reversible reaction going from state 1 to 2, the change in entropy isn't zero because it isn't returning back to state 1? “ No. I’m saying the change entropy is zero in going from 1 to 2 and back again to 1 no matter what the process(es). As I said in my answer, the change in entropy (of the system) in going from state 1 to state 2 can be zero, greater than zero, or less than zero. It is only zero for a reversible adiabatic process. $\endgroup$
    – Bob D
    Dec 31, 2022 at 15:29
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The entropy is a state parameter. The work is a process parameter. This means that "the entropy of a process" is meaningless. In a process you have a change in entropy, between the initial and final states. If the initial and final states are the same, the change is the same, no matter how you get between the states. Maybe it helps if you think about potential energy, another state parameter.

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  • $\begingroup$ I think by your claim " that 'the entropy of a process' is meaningless." you have just thrown out all of 20th century thermodynamical (not thermostatics) theory of irreversible processes whose main aim is to quantify entropy specifically in irreversible processes, see, Onsager, Truesdell, Eckart, Bridgman, Prigogine, Coleman,... $\endgroup$
    – hyportnex
    Dec 30, 2022 at 15:02
  • $\begingroup$ You are confusing the change (variation) of entropy in a process with the "entropy of the process". $\endgroup$
    – nasu
    Dec 30, 2022 at 17:02
  • $\begingroup$ Thank you for pointing out my mistake sir! Your answer gave me some clarification on entropy, came to understanding I have to be more clear. I'm a bit confused about how some additional entropy is generated in irreversible process and that is lost, hence reversible process is more efficient. Would be helpful if you could throw light on why there's additional generation of entropy in irreversible process. $\endgroup$
    – Avaneesh B
    Dec 31, 2022 at 6:02
  • $\begingroup$ If the two processes are between the same two states the change in entropy is the same. There is no additional entropy. If there is a difference then the states are different. You may start from the same initial state but end in two distinct final states by the two processes $\endgroup$
    – nasu
    Dec 31, 2022 at 16:22
  • $\begingroup$ Got it, thank you! $\endgroup$
    – Avaneesh B
    Dec 31, 2022 at 18:38
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The issue is not that when comparing two processes, say $\mathfrak P_r$ and $\mathfrak P_i$, both starting at the equilibrium state $\mathfrak S_0$ and both ending at the another equilibrium sate $\mathfrak S_1$ they have the same entropy, but when calculating the end result of $\mathfrak P_r$ that progresses reversibly we know how to calculate its state parameters, specifically its entropy even for the other irreversible process starting and ending at the same equilibrium states as the reversible one. And that is because the equilibrium sates are by definition the same.

The reason for being able to calculate the parameters of the reversible process easier is because the external, i.e., environmental, parameters are exactly the same during the process as the system's internal equilibrium parameters are. This direct link is lost in an irreversible process but when they both land in the same equilibrium state that link is reestablished.

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