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The entropy change of the Universe for the free expansion of an ideal gas is calculated by replacing the irreversible path between the initial and final equilibrium states with a reversible, isothermal path. My question is that in computing the entropy change of the Surroundings, do we consider the entropy change of the Surrounding of the irreversible system, or the replacement reversible isothermal system ?

As discussed here: Ideally, how to achieve isothermal expansion of an ideal gas?

One means of achieving isothermal expansion is to place the system in contact with a heat reservoir which can supply the requisite heat. However, if we include the entropy change of this reservoir, it is a non-zero contribution to the entropy change of the Universe. While the entropy change of the surroundings for the original irreversible process is zero.

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3 Answers 3

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The change in entropy of the reservoir is the opposite of the change in the gas, because for a reversible process the total change in entropy is zero. When you replace the irreversible process by a reversible one, you are calculating the change in entropy in your system only, because you are using the fact that entropy is a function of state. In the free expansion, as you said, there is no change in the external entropy, but in other irreversible processes there might be. This change is not included in the calculation, and has to be calculated and added to the change in the system if you want the total change in entropy of the universe.

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  • $\begingroup$ I said that in the original (irreversible) process there is no change in the entropy of the surroundings (consider the gas being surrounded by an adiabatic wall). There is however, a change in the entropy of the surroundings in the reversible case because heat is transferred from a reservoir to keep the gas in an isothermal state during the expansion. $\endgroup$
    – Frost
    Feb 5, 2020 at 18:27
  • $\begingroup$ Yes, and that change has the same magnitude (and opposite sign) than the change in the isothermic gas. $\endgroup$
    – user65081
    Feb 5, 2020 at 18:35
  • $\begingroup$ To summarize: Are you suggesting that we use the entropy change in the surroundings from the irreversible or the reversible system, when computing the entropy change of the Universe ? $\endgroup$
    – Frost
    Feb 5, 2020 at 18:41
  • $\begingroup$ you can use either, but it is usually easier to calculate the one in the system $\endgroup$
    – user65081
    Feb 5, 2020 at 18:47
  • $\begingroup$ This is inconsistent: Both answers lead to different values of the change in entropy of the Universe. $\endgroup$
    – Frost
    Feb 5, 2020 at 18:48
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You need to understand the difference between entropy change and entropy generation.

As @Wolfram jonny pointed out for a reversible isothermal process there is an overall change in entropy (system + surroundings) of zero. The entropy lost/gained by the system exactly equals the entropy gained/lost by the surroundings.

For an irreversible process entropy is generated in the system. To return the system to its original state (complete a cycle) that entropy has to be transferred to the surroundings. The only way to transfer entropy is by heat transfer.

This is what happens in the case of the free expansion of an ideal gas. Take a perfectly insulated rigid vessel with a partition in between. In one half is an ideal gas. In the other half a vacuum. An opening is made in the partition so that the ideal gas expands into the evacuated half of the vessel. Since the vessel is rigid and perfectly insulated, the system is isolated from the surroundings and there is no heat or work crossing the boundary. There is no change in the internal energy of the system and, for an ideal gas, no change in temperature, $T$.

However, the free expansion is clearly an irreversible process. The gas will not spontaneously return to its original half of the vessel. Therefore, entropy has been generated in the system. Since the system is isolated from the surroundings, there is no entropy change in the surroundings. To return the system to its original state, the entropy generated within the system during the irreversible expansion has to be transferred to the surroundings. In order to calculate the entropy generated in the system during the expansion, a convenient process is a reversible isothermal compression. This process transfers the entropy $\frac{Q}{T}$, which was generated in the system, out of the system and to the surroundings, returning the system to its original state.

Overall for the irreversible cycle we have:

$$\Delta S_{sys}=0$$

$$\Delta S_{sur}=+\frac{Q}{T}$$

$$\Delta S_{univ}=\Delta S_{sys}+\Delta S_{sur}= +\frac{Q}{T}$$

Hope this helps.

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The way it works is this: After the irreversible process is complete, you first separate the original system from the original surroundings (that were involved in the irreversible process), and then subject each of them separately to new (alternative) reversible processes using a new surroundings for each of them (i.e., a second surroundings for the system and a third surroundings for the original surroundings). So in the reversible process for the original surroundings, you are treating it as a "system." The entropy change for the original system is equal to the entropy change for its separate reversible process between the same two end states, and the entropy change for the original surroundings is equal to the entropy change for its separate reversible process between the same two end states. The entropy change for the universe in the irreversible process is equal to the sum of the entropy changes for the original system and original surroundings.

For more details on how to determine the entropy change for an irreversible process, see the following link: https://www.physicsforums.com/insights/grandpa-chets-entropy-recipe/

In the free expansion example you gave, there is no entropy change for the surroundings in its irreversible process, and, to get the entropy change for the universe, the entropy change for the system (only) is added to this; the entropy change for the second surroundings is not included.

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  • $\begingroup$ The puzzle is that in the original system, the gas and the surroundings are thermally isolated. However, in the reversible case, the surroundings do thermally interact with the gas. It is not clear, then, why "the entropy change for the second surroundings is not included" when computing the entropy change of the Universe for the irreversible case. $\endgroup$
    – Frost
    Feb 5, 2020 at 19:00
  • $\begingroup$ Like I said, you are separating the original system from the original surroundings, and then devising separate reversible processes for each separately, using a second surroundings and a third surroundings. In the reversible paths for each, the entropy change of the 2nd and 3rd surroundings are discarded. Only the entropy change of the original system and the original surroundings are included in calculating the entropy change of the universe for the irreversible process. $\endgroup$ Feb 5, 2020 at 19:21
  • $\begingroup$ Please understand that the reversible paths used to evaluate the entropy changes of the system and the surroundings for the irreversible process do not need to bear any resemblance whatsoever to the actual irreversible path. And separating the original system from the original surroundings in specifying separate alternative reversible paths for each is perfectly allowable. This is a consequence of entropy being an equilibrium physical property of the material comprising the system (and the surroundings), rather than a characteristic of any specific process. $\endgroup$ Feb 5, 2020 at 23:56

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