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How is the following true?

$$\left ( \frac{∂S}{∂P} \right )_{T}=\frac{-C_{P}}{T}\left ( \frac{∂T}{∂P} \right )_{S}$$


I know that

$$C_{P}=T\left ( \frac{∂S}{∂T} \right )_{P}$$

which means the right-hand side of the first equation becomes

$$-\left ( \frac{∂S}{∂T} \right )_{P}\left ( \frac{∂T}{∂P} \right )_{S}$$

but how is that equal to $\left ( \frac{∂S}{∂P} \right )_{T}$ ?

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  • $\begingroup$ There's an excellent book by Dill and Bromberg called Molecular Driving Forces. It derives most of these relationships from first principles, and is excellent even if you're weak on the maths side! It's a fantastic introduction to thermodynamics. $\endgroup$ – user7232 Aug 21 '14 at 19:31
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It's a consequence of the triple product rule for partial derivatives of thermodynamic state variables.

In summary, if you have three interdependent (not independent) variables, then this is true:

$$ \left ( \frac{\partial{X}}{\partial{Y}} \right )_Z \left ( \frac{\partial{Z}}{\partial{X}} \right )_Y = - \left ( \frac{\partial{Z}}{\partial{Y}} \right )_X $$

Substituting in your variables, if we let $X = T$, $Y = P$, and $Z = S$, we get:

$$ - \left ( \frac{\partial{S}}{\partial{T}} \right )_P \left ( \frac{\partial{T}}{\partial{P}} \right )_S = \left ( \frac{\partial{S}}{\partial{P}} \right )_T $$

Note that the order of multiplication of the two partial derivatives doesn't matter.

Now, why this rule holds for thermodynamic variables is a little bit more complicated - but I think the wikipedia article that I linked to has a pretty good derivation if you are interested.

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