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In a textbook for thermodynamics, it considers a situation where work is done to a system by an irreversible work source through a thermally insulating piston, and it states "any irreversible work source can be simulated by a reversible work source". It briefly explains the reason; what the work source does is simply to apply force to the piston, and therefore it does not matter how the force is applied, whether or not it is applied by an irreversible work source. I have a difficulty to fully convince myself with the statement, and can anyone kindly help me with this?

  • The textbook is a non-English one, and it is not available online as an electric file.
  • The way how it defines the work source is simply that it is any system, which is connected with the system of our interest only through a thermally insulating piston. That is, there is no heat exchange between the two system.

Let me rephrase my question: Suppose that system A is interacted with system B (irreversible work source) only with a thermally insulating piston. They exchange energy only through work, but not heat. My question is if it is possible to replace system B with a reversible work source in an indistinguishable manner, i.e., any mechanical reaction that system A receives from system B remains exactly the same.

Thank you so much for your time.

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    $\begingroup$ which textbook, page, chapter, section, etc.? $\endgroup$
    – hyportnex
    Commented Nov 19, 2023 at 15:17
  • $\begingroup$ How does your book define a work source? $\endgroup$
    – Chet Miller
    Commented Nov 19, 2023 at 15:38
  • $\begingroup$ Sorry, but I should have been clearer. Suppose that system A is interacted with system B (irreversible work source) only with a thermally insulating piston. They exchange energy only through work, but not heat. My question is if it is possible to replace system B with a reversible work source in an indistinguishable manner, i.e., any mechanical reaction that system A receives from system B remains exactly the same. $\endgroup$
    – Ketty
    Commented Nov 20, 2023 at 1:01
  • $\begingroup$ You should edit your question to include the above information. $\endgroup$
    – Michael Seifert
    Commented Nov 20, 2023 at 12:44

2 Answers 2

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It briefly explains the reason; what the work source does is simply to apply force to the piston, and therefore it does not matter how the force is applied, whether or not it is applied by an irreversible work source.

I could be wrong but I suspect the point they may be trying to make is that work does not transfer entropy. Only heat transfers entropy. So it doesn't matter if the source of the work is reversible or irreversible as long as any entropy generated by the source due to irreversible work stays with the source, since it cannot be transferred to the system by means of work.

Hope this helps

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    $\begingroup$ Doesn’t irreversible work generate entropy? $\endgroup$
    – Chet Miller
    Commented Nov 19, 2023 at 21:39
  • $\begingroup$ @ChetMiller Of course. But I was guessing they may have been talking about entropy somehow being generated in the surroundings by whatever is doing work on the system, if that makes any sense to you. I'll try an think of an example. $\endgroup$
    – Bob D
    Commented Nov 19, 2023 at 21:48
  • $\begingroup$ It isn't clear to me what the OP is talking about. $\endgroup$
    – Chet Miller
    Commented Nov 19, 2023 at 21:50
  • $\begingroup$ @ChetMiller. That may be, but do you understand my point? Namely, entropy generated in the surroundings may have no effect on what happens in the system. I've drawn an example that I will include. $\endgroup$
    – Bob D
    Commented Nov 19, 2023 at 22:11
  • $\begingroup$ I think I understand $\endgroup$
    – Chet Miller
    Commented Nov 19, 2023 at 22:18
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It briefly explains the reason; what the work source does is simply to apply force to the piston, and therefore it does not matter how the force is applied, whether or not it is applied by an irreversible work source.

I could be wrong but I suspect the point they may be trying to make is that work does not transfer entropy. Only heat transfers entropy. So it doesn't matter if the source of the work is reversible or irreversible as long as any entropy generated by the source due to irreversible work stays with the source, since it cannot be transferred to the system by means of work.

UPDATE:

This update is due to your following subsequent comment:

Sorry, but I should have been clearer. Suppose that system A is interacted with system B (irreversible work source) only with a thermally insulating piston. They exchange energy only through work, but not heat. My question is if it is possible to replace system B with a reversible work source in an indistinguishable manner, i.e., any mechanical reaction that system A receives from system B remains exactly the same.

Yes. If the irreversible work source B is neither generating entropy in A (by irreversible work on A) or transferring entropy to A (by heat), then the irreversible work source B can be replaced with a reversible work source B and A will remain the same. An example is given in the figure below.

B is an irreversible work source due to mechanical friction heating between the piston and cylinder walls. Since the heat generated by friction in B is not transferred to A, no entropy is transferred to A. Moreover, since the expansion in B is carried out very slowly (quasi-statically) the compression of the gas in A (with no mechanical friction in A) is also carried out very slowly (quasi-static) making the compression work done by B on A reversible.

If we simply remove mechanical friction from B we will have replaced an irreversible work source with a reversible work source, with no impact on A.

Hope this helps

enter image description here

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