2
$\begingroup$

From thermodynamic identity you get: $$\left(\frac{\partial U}{\partial V}\right)_{S,N} = -P$$ But with Helmholtz free energy $F = U-TS$, we can also get pressure from this equation: $$\left(\frac{\partial F}{\partial V}\right)_{T,N} = -P.$$ In which cases should you use free energy version?

$\endgroup$
2
  • 2
    $\begingroup$ Both of them are equivalent. You use the one which best suits your needs. $\endgroup$
    – RedGiant
    Nov 28, 2021 at 15:58
  • 2
    $\begingroup$ As indicated in the index of your function, the first identity apply for S=constant and N=constant while the second one apply for T=constant and N=constant $\endgroup$ Nov 28, 2021 at 16:25

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.