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

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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?

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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.

answered Feb 3, 2017 by (3,625 points)

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

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