Let's just say we have 2 moles of monoatomic (maybe helium) ideal gas that is doing a Carnot Cycle with reservoir temperature of 300 and 750 K.

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Now, here's a simple question, what is the Gibb Free energy change from point A to point B and from point B to C in this process?

Because (do correct me if I'm wrong), I believe that we need to know the absolute value of the entropy in this case.

1 Answer 1


Generally, along a reversible path, dG=-SdT+VdP. But, for the isothermal path between A and B, $$dG=VdP=nRT\frac{dP}{P}=-nRT\frac{dV}{V}$$

  • $\begingroup$ What about from B to C? $\endgroup$
    – Tensor
    Commented Aug 16, 2022 at 11:46
  • $\begingroup$ That would involve the S at B (orC). $\endgroup$ Commented Aug 16, 2022 at 15:43
  • $\begingroup$ Yes. But how do we calculate that? I mean, its not that we need $\Delta S$, rather we need S $\endgroup$
    – Tensor
    Commented Aug 17, 2022 at 3:23
  • $\begingroup$ S is specified relative to some well-defined reference state. We don't need to know the absolute value of S in order to use it in practical calculations. $\endgroup$ Commented Aug 17, 2022 at 10:55

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