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This question is inspired by Reif Problem 5.5. Note that it is not a homework problem and, even if it were, my question only loosely relates to it.

A vertical cylinder contains $N$ molecules of a monatomic ideal gas and is closed off at the top by a piston of mass $M$ and area $A$. The acceleration due to gravity is $g$. The heat capacities of the piston and cylinder are negligibly small, and any frictional forces between the piston and the cylinder walls can be neglected. The whole system is thermally insulated. Initially, the piston is clamped in position so that the gas has a volume $V$ and a temperature $T$. The piston is now released and after some oscillations, comes to rest in a final equilibrium situation corresponding to a larger volume of the gas. Does the entropy increase?

Now the answer, intuitively, is yes, since the system finds a new equilibrium precisely because that equilibrium is entropy-maximizing (has the most microstates).

But consider the following reasoning (and please tell me why it's wrong). While the particular process described is not quasi-static, we can imagine a corresponding quasistatic process which takes us to the prescribed final state. Since entropy is a function of state, the change in entropy will be precisely the same. Now for such a quasistatic process, we are now able to write that $dS=dQ/T=0$ because the system is thermally insulated. Thus we seem to conclude that there is no change in entropy.

I have two questions then. The first is about the aforementioned system, and the second is about "calculating with quasistatic processes" more generally.

(1) Was my reasoning flawed in that $dq\neq0$ in the imagined quasistatic process case? That is to say, is it the case that the imagined quasistatic process which takes us to our final equilibrium cannot have $dQ=0$ (i.e. there is no quasistatic, thermally insulated process with which we can reach our final state)? If so, how can I see this?

(2) In general, when am I allowed to calculate with quasistatic processes? Are there general guidelines for constructing valid quasistatic processes which take us to a desired final state?

Edit: after thinking a little more, I think it is fair to conclude that the gas itself gains no entropy (as evidenced by my argument), but that the universe itself does gain entropy (the entropy of the environment increases). This still begs the question though: if $Q=0$ here, and the environment is a heat reservoir so that $\Delta S = Q/T$ then the environment should also not gain entropy?

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  • $\begingroup$ If I can prove to you that it is impossible to connect the same two equilibrium states with a reversible and irreversible adiabatic process for an ideal gas, would you accept my answer? $\endgroup$
    – Bob D
    Jun 3, 2022 at 21:16
  • $\begingroup$ @BobD Yes, but by reversible I would appreciate us being clear about what is meant. In my question I speak about a process which is quasistatic for the system, but not reversible (isentropic) for the universe. $\endgroup$
    – EE18
    Jun 4, 2022 at 2:17
  • $\begingroup$ I don’t know what you mean. By definition a reversible process is one in which both the system and the surroundings (I.e. the universe) return to their original state. $\endgroup$
    – Bob D
    Jun 4, 2022 at 6:26
  • $\begingroup$ That's fair enough, but the question was about a quasistatic process. That no such reversible process exists follows from the fact that the system changed its equilibrium at all when the constraint was removed. $\endgroup$
    – EE18
    Jun 4, 2022 at 10:02
  • $\begingroup$ Let's find out what we agree on. Do you agree that the non quasi static process you described generates entropy in the system? $\endgroup$
    – Bob D
    Jun 4, 2022 at 14:54

4 Answers 4

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The piston is now released and after some oscillations, comes to rest in a final equilibrium situation corresponding to a larger volume of the gas. Does the entropy increase?

Yes. Because the process is not quasi static, entropy increases. But the increase is not due to entropy transfer which can only occur when there is heat transfer. It is instead due to entropy generated by the sudden drop in pressure and the ensuing rapid expansion and oscillations.

While the particular process described is not quasi-static, we can imagine a corresponding quasi static process which takes us to the prescribed final state.

The quasi static process you imagine cannot be adiabatic because a process that is both quasi static (and frictionless) and adiabatic is by definition isentropic (constant entropy), whereas the process you are replacing has generated entropy. So in order for your quasi static process to connect the two same states there needs to be entropy transfer to the system in the form of heat from the surroundings equal to the entropy generated in the irreversible adiabatic process it is replacing.

(1) Was my reasoning flawed in that $dq\neq0$ in the imagined quasistatic process case?

Yes it was flawed. The imagined quasi static process needs to involve heat transfer from the surroundings. Such heat transfer cannot occur for any adiabatic process.

(2) In general, when am I allowed to calculate with quasistatic processes? Are there general guidelines for constructing valid quasistatic processes which take us to a desired final state?

You are allowed to use any reversible (quasi static and frictionless) between two states to calculate the change in entropy between the two states, since entropy is a state function. For helpful guidelines on how to construct a valid process, see the following: https://www.physicsforums.com/insights/grandpa-chets-entropy-recipe/

Edit: after thinking a little more, I think it is fair to conclude that the gas itself gains no entropy (as evidenced by my argument), but that the universe itself does gain entropy (the entropy of the environment increases).

There can be no increase in entropy of the environment (surroundings) unless there is heat transfer from the system to the surroundings. Only heat can transfer entropy. Work does not transfer entropy. This means there is never a change in entropy of the surroundings as a result of any adiabatic system process (quasi static or otherwise). Likewise an increase in entropy of an adiabatic system can never be the result of entropy transfer from the surroundings. The increase in entropy of an adiabatic system has to be all generated entropy.

Bottom Line: Entropy is generated between the initial and final equilibrium state in the non quasi static process you presented. In order to return to the initial state this entropy needs to be transferred to the environment so that the total change in system entropy is zero. Entropy can only be transferred to the surroundings in the form of heat. Therefore no adiabatic path exists that can return the system to its initial state.

As a final comment it can be shown, without an entropy argument, that a reversible (quasi static) and irreversible adiabatic process starting at the same initial equilibrium state cannot wind up in the same final equilibrium state. If the final equilibrium pressures are the same (which is the case in your example) you will find that the final temperatures and volumes will not be the same. See my answer to the following:

Reversible vs. Irreversible Expansion

Hope this helps.

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  • $\begingroup$ Thank you. This was very helpful. Especially the article/recipe you linked. $\endgroup$
    – EE18
    Jun 4, 2022 at 19:34
  • $\begingroup$ You’re welcome. Credit for the recipe goes to Chet Miller who provided the other answer (which I upvoted but for some inexplicable reason someone else downvoted) $\endgroup$
    – Bob D
    Jun 4, 2022 at 19:54
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The answer to your first question is, as you suspected, that there is no adiabatic reversible path between the same initial and final states for an adiabatic irreversible process. We know this because the entropy change of the irreversible adiabatic process is positive (entropy generated) and the entropy change for a reversible adiabatic path is zero. You are encouraged to consider various adiabatic irreversible process (such as the example you gave) and to try to devise an adiabatic reversible path that gives the same entropy change. You will, of course, not be able to do so.

The answer to your second question it that there are an infinite number of reversible (non-adiabatic) paths that give the same entropy change as for the adiabatic irreversible process. They all give exactly the same entropy change. Therefore, it is best to devise a reversible path which is simple to work with in terms of determining the integral of dQ/T.

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  • $\begingroup$ Sorry, I'm very confused with the terminology you use in this answer. For example, you talk about reversible but non-adiabatic paths. By adiabatic do you mean thermally insulated? And, if so, then isn't a reversible process necessarily adiabatic? $\endgroup$
    – EE18
    May 27, 2022 at 2:12
  • $\begingroup$ Also, you assert that there is no such process, but can you explain why? Why can't I imagine just letting the piston move slightly up towards the final state, the entire time ensuring equilibrium is maintained while no heat is exchanged? $\endgroup$
    – EE18
    May 27, 2022 at 2:30
  • $\begingroup$ See my edits in the question if possible. $\endgroup$
    – EE18
    May 27, 2022 at 2:58
  • $\begingroup$ By adiabatic, I mean thermally insulated. A reversible process is not necessarily adiabatic. You can have an isothermal reversible path also, for example. A reversible path consists of a continuous sequence of thermodynamic equilibrium states. $\endgroup$ May 27, 2022 at 10:04
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    $\begingroup$ @ChetMiller Chet, good luck with that! $\endgroup$
    – Bob D
    Jun 3, 2022 at 21:02
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Generally speaking, for a infinitesimal step of a quasistatic transformation, you have:

$$dS=\frac{\delta Q}{T_\text{ext}}+\delta S_c$$

where $T_\text{ext}$ is the temperature of the environment around the system, and $\delta S_c$ is the created entropy during the process (always positive, due thermodynamics' second law).

If the system is thermally isolated and there's no heat source inside the system, then yes $\delta Q=0$.

That leaves $\delta S_c$, however. You are indeed free to consider the imaginary transformation that shares the same initial and final states as the real one while being also quasistatic, but you can't add a reversible hypothesis.

For an ideal gas, you should be able to compute $dS$ separately, and then check its sign. Since it's equals to $\delta S_c$, it has to be positive.

Designing an "equivalent" transformation (imaginary process with the same initial and final states) is a very convenient tool, but you have to make sure that the additional properties to assign to it are self-consistent, and consistent with the real transformation (another example: with a real process where temperature rises up, you can decide that the equivalent process be quasistatic, but not isothermal).

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Now for such a quasistatic process, we are now able to write that $dS=dQ/T=0$ because the system is thermally insulated. Thus we seem to conclude that there is no change in entropy.

You are making the (understandably) common mistake of thinking that if there is no heat transfer, there can be no change in entropy, because of the way entropy change is defined ($dS=dQ_{rev}/T$).

But entropy can be generated without any heat transfer. @Chet Miller describes mechanisms for generating entropy that do not involve heat transfer in his answer to the following post: What makes procesess internally reversible?

Since entropy is a state function which is independent of the path between states, you can calculate the entropy change by assuming any convenient reversible path that connects the two states, per the first law, and applying the entropy definition. The assumed path need not bear any resemblance to the actual path. That means in this example the convenient reversible path you choose can (and will) involve heat transfer.

Hope this helps.

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  • $\begingroup$ I'm not quite sure it does, although I really appreciate the answer. Maybe we can flesh this out. I describe a reversible (it seems you and Chet use reversible to refer to what I call quasistatic -- i.e. moving through a series of equilibrium states) path which is thermally insulated and so $dQ=0$. Why then can I not conclude that $dS=0$ for my gas? $\endgroup$
    – EE18
    May 27, 2022 at 13:31
  • $\begingroup$ I believe that the extra entropy you refer to is ultimately manifested in an entropy increase of the surroundings (in the real irreversible process originally described in the process). But the entropy change of the gas in particular can be calculated by my "reversible" path. $\endgroup$
    – EE18
    May 27, 2022 at 13:33
  • $\begingroup$ @EE18 Regarding your first comment, Chet Miller has already pointed out that you can connect the same two equilibrium states with a reversible and irreversible path. Regarding your second comment, In order to reverse the path to go back to the original state then, yes, the entropy generated will need to be transferred to the surroundings in the form of heat. $\endgroup$
    – Bob D
    May 27, 2022 at 13:42
  • $\begingroup$ @EE18 Keep in mind that although the change in entropy between two equilibrium states is the same for a reversible and irreversible process, the components of the entropy change will be different for a reversible and irreversible path. For the reversible process the change will be only entropy transfer due to reversible heat transfer. For an irreversible process the change in entropy is some combination of entropy transfer and entropy generated. For an irreversible adiabatic process all of the change in entropy is entropy generated because there is no heat transfer. $\endgroup$
    – Bob D
    May 27, 2022 at 13:48
  • $\begingroup$ @EE18 On my first comment, I obviously meant "can't" not "can" connect the same two equilibrium states. $\endgroup$
    – Bob D
    May 27, 2022 at 17:49

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