0
$\begingroup$

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.

enter image description here

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.
$\endgroup$

1 Answer 1

1
$\begingroup$

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

$\endgroup$
4
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.