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The entropy change of the surroundings can be calculated by the equation $dS_{sur}=\frac{dq}{T_{sur}}$ regardless of the path (irreversible or reversible).

The argument is that because the surroundings may be approximated as either constant volume or constant pressure, the heat absorbed/released by it is equal to the internal energy or enthalpy change, respectively. As both are state functions, it does not matter if the path taken was reversible or not, the changes would be the same for both, so $dU_{sur,rev}=dU_{sur,irrev}$ and $dH_{sur,rev}=dH_{sur,irrev}$. However, $dH=dq$ for constant pressure, and as per definition $dS=\frac{dq_{rev}}{T}$, the initial equation follows.

This is somewhat the way my book derives the formula. It seems to be inconsistent to me because it equals $\frac{dq_{rev}}{T}=\frac{dq_{irrev}}{T}$, which some pages after the book says that is not the case and uses the fact that $dq_{rev}\geq dq_{irrev}$ to derive the Clausius inequality. I do understand the Clausius inequality and its derivation, however I do not understand why is it that it is not used to calculate the entropy change of the surrounds, only the system. The arguments for its derivation seem to apply to the first case as well.

Similar question that has no answer: Entropy change of surroundings

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In evaluating the entropy change of the surroundings, you need to first look exclusively at its initial state and its final state, before and following the irreversible process. The only difference between the initial state and the final state is a difference in enthalpy of the surroundings, equal to amount of irreversible heat that flowed to the surroundings. To get the entropy change for the surroundings, you next need to separate it entirely from the original system, and subject it to a new reversible process while in contact with its own new surroundings (i.e., a second set of surroundings). No matter what new (alternative) reversible process you devise, you will always end up with the same enthalpy change and same beginning and ending pressures, and its entropy change will always be the same, and equal to the irreversible heat it received in the original process divided by its absolute temperature.

On the other hand, for the original system, if you follow the same procedure (separating it from the original surroundings and putting it into contact with a third surroundings), the reversible process you devise will give you a different value for the integral of $dq/T_{boundary}$ than the integral of $dq/T_{boundary}$ for the original irreversible process.

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  • $\begingroup$ Could you elaborate on that separation part? I've never seen that. $\endgroup$ – user93668 May 14 at 17:47
  • $\begingroup$ That's too bad. It has annoyed me for some time that the text books out there do such a bad job of explaining how to determine the entropy change for an irreversible process. So I wrote a short primer of how to do it, which many people have found very useful: physicsforums.com/insights/grandpa-chets-entropy-recipe Please feel free to ask follow-up questions. $\endgroup$ – Chet Miller May 14 at 18:29
  • $\begingroup$ Regarding separating them: If the process was irreversible, to get the entropy change of the system and the entropy change of the surroundings, we need to devise a path that takes each of them from their initial thermodynamic equilibrium state to their final thermodynamic equilibrium state reversibly. If the original process was irreversible, there is no single path that will take them both between their initial and final states (of the irreversible process) simultaneously. So we need to separate them, and subject each of them separately to a reversible path between these states. $\endgroup$ – Chet Miller May 15 at 14:47

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