So, with the internal energy in terms of entropy and volume


since it is a total differential, from my understanding, the following should be true

$dU=\left(\frac{\partial U}{\partial S}\right)_VdS+\left(\frac{\partial U}{\partial V}\right)_SdV$

resulting in

$\left(\frac{\partial U}{\partial V}\right)_S=-p$

Now, is there a difference between $\left(\frac{\partial U}{\partial V}\right)_S$ and $\left(\frac{\partial U}{\partial V}\right)_T$? Because my text book says the following relation is true (without proving it):

$\left(\frac{\partial U}{\partial V}\right)_T=-p+T\left(\frac{\partial S}{\partial V}\right)_T$

What exactly causes the extra term?

  • 1
    $\begingroup$ The varible you are keeping constant always matters. In both cases, you are "dividing" the total differential $dU$ by the infinitesimal $dV$. However, if you keep entropy constant as in the first case, $dS=0$ and the term vanishes as expected. But keeping temperature constant does not stop entropy from changing (we can absorb heat at constant temperature), and therefore the term should be accounted for. $\endgroup$ Jul 10, 2018 at 11:50
  • $\begingroup$ I am not quite clear what you are asking. Are you asking "does the difference between $\left(\frac{\partial U}{\partial V}\right)_S$ and $\left(\frac{\partial U}{\partial V}\right)_T$ matter?" or "what does the term $T\left(\frac{\partial S}{\partial V}\right)_T$ mean physically?" or "how do I derive the equation for $\left(\frac{\partial U}{\partial V}\right)_T$?", or something else entirely? $\endgroup$ Jul 10, 2018 at 11:57
  • $\begingroup$ @GabrielGolfetti Aaah thanks I think I understand it now. I just found the whole partial derivative thing very confusing, especially since "dividing" isn't considered the mathematically correct way of doing it. $\endgroup$
    – Keno
    Jul 10, 2018 at 12:04
  • $\begingroup$ @BySymmetry I guess it was mostly about deriving the equation, and what exactly the difference between $\left(\frac{\partial U}{\partial V}\right)_S$ and $\left(\frac{\partial U}{\partial V}\right)_T$ is. $\endgroup$
    – Keno
    Jul 10, 2018 at 12:05
  • $\begingroup$ In your book, I suppose they did the same thing with dS in terms of dV and dT that you did for dU in terms of dS and dV. $\endgroup$ Jul 10, 2018 at 12:12

1 Answer 1


The derivation goes as follows:


$dS=\left(\frac{\partial S}{\partial T}\right)_VdT+\left(\frac{\partial S}{\partial V}\right)_TdV\Rightarrow dU=T\left(\frac{\partial S}{\partial T}\right)_VdT+T\left(\frac{\partial S}{\partial V}\right)_TdV-pdV$

Now, $dU$ isn't expressed in terms of $dS$ anymore, but in terms of $dT$ instead, which makes it possible to determine $\left(\frac{\partial U}{\partial V}\right)_T$:

$dU=\left(\frac{\partial U}{\partial V}\right)_TdV+\left(\frac{\partial U}{\partial T}\right)_VdT=\left[T\left(\frac{\partial S}{\partial V}\right)_T-p\right]dV+T\left(\frac{\partial S}{\partial T}\right)_VdT$

resulting in the equation

$\left(\frac{\partial U}{\partial V}\right)_T=T\left(\frac{\partial S}{\partial V}\right)_T-p$


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