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The equation $dS=\frac{dQ}{T}$ tells us the entropy change due to heat transfer. Likewise, the equation $dS=\frac{P}{T}dV$ tells us the entropy change due to a volume change. However, I am not sure how entropy of a system would change when work is done on it. Since work involves both a change in volume and an energy transfer, the entropy change due to work should have two terms: (1) entropy change due to energy transferred via work (2) entropy change due to volume change. (1) is where I am getting stuck. $dS=\frac{dQ}{T}$ only applies to energy transfer via heat transfer, not work. How would one calculate the entropy change due to work when there is no formula that tells you how a system's entropy change due to energy transferred by work?

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Your equation for entropy change applies only to a reversible path. It gives the wrong answer for an irreversible path. To get the entropy change for any change in thermodynamic state, you devise (dream up) a reversible path between the same two end states, and evaluate the integral of dQ/T for that reversible path. It doesn't matter whether work is done or not.

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  • $\begingroup$ Thanks for the answer. Since I can choose any path to calculate the entropy change between two states, can I treat a work process (which involves change in both energy and volume) as two separate processes with one being a pure heat transfer and the other being a pure volume change and calculate the total entropy change by just adding the entropy changes of the two separate processes? $\endgroup$ Commented Oct 3, 2021 at 23:48
  • $\begingroup$ You can't choose any path; you can choose any reversible path to calculate the entropy change. All reversible paths (and there are an infinite number of these, give exactly the value for the integral of dQ/T. The work processes for these paths can be whatever they turn out to be. But, you certainly don't need to use a direct path, and you can break the overall path into separate reversible sub-paths. $\endgroup$ Commented Oct 3, 2021 at 23:54
  • $\begingroup$ I have no idea what your are saying. Give me a specific example of two (well-defined) end states, and I will illustrate how to get the entropy change. $\endgroup$ Commented Oct 4, 2021 at 0:56

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