derivative w.r.t. conjugate variables (Thermodynamics)

Question

Is there a general relation to derivatives of natural variables in thermodynamics with respect to their conjugate variable?

e.g. \(\left(\frac{\partial S}{\partial T}\right)_p =\) ?

something like the Maxwell relations?

Answer 1

Well, yes. But the list is very big. For your example we have that

$$ \left( \frac{\partial P }{ \partial T } \right)\Big|_{S} = -  \left( \frac{\partial P }{ \partial S } \right)\Big|_{T}  \left( \frac{\partial S }{ \partial T } \right)\Big|_{P}$$

but the LHS due to Maxwell's relations is the same as $ \left( \frac{\partial S}{ \partial V } \right)\Big|_{P}$. But for this guy we have that

$$ \left( \frac{\partial P }{ \partial V } \right)\Big|_{S} = -  \left( \frac{\partial P }{ \partial S } \right)\Big|_{V}  \left( \frac{\partial S }{ \partial V } \right)\Big|_{P}$$

and from Maxwell we have that $-  \left( \frac{\partial P }{ \partial S } \right)\Big|_{V} =  \left( \frac{\partial T }{ \partial V } \right)\Big|_{S} $. Now for the RHS of this we have

 $$ \left( \frac{\partial T }{ \partial V } \right)\Big|_{S} = -  \left( \frac{\partial T }{ \partial S } \right)\Big|_{V}  \left( \frac{\partial S }{ \partial V } \right)\Big|_{T}$$

Finally, again from Maxwell we know that $\left( \frac{\partial S }{ \partial V } \right)\Big|_{T} = \left( \frac{\partial P }{ \partial T } \right)\Big|_{V} = -\frac{\alpha}{\kappa}$ where

$$ \alpha = \frac{1}{V}\left( \frac{\partial V}{\partial T} \right)\Big|_{P}  $$ 

and

$$ \kappa = -\frac{1}{V}\left( \frac{\partial V}{\partial P} \right)\Big|_{T}  $$ 

I hope this helps.

Comments

Thanks, yes it's helpful :) the trick u use to split the partial derivatives didn't found it's way to my mind...