# Please help show that ((delT)/(delP))_S = (ValphaT)/C_P?

Jan 25, 2017

Okay, considering that ${\left(\frac{\partial T}{\partial P}\right)}_{S}$ contains $S$, $T$, and $P$, I would start from $S = S \left(T , P\right)$ and take the total derivative:

$\mathrm{dS} = {\left(\frac{\partial S}{\partial T}\right)}_{P} \mathrm{dT} + {\left(\frac{\partial S}{\partial P}\right)}_{T} \mathrm{dP}$ $\text{ "" } \boldsymbol{\left(1\right)}$

Now, if we divide by $\partial P$ at a constant $S$, we'd get:

${\cancel{{\left(\frac{\partial S}{\partial P}\right)}_{S}}}^{0} = {\left(\frac{\partial S}{\partial T}\right)}_{P} {\left(\frac{\partial T}{\partial P}\right)}_{S} + {\left(\frac{\partial S}{\partial P}\right)}_{T} {\cancel{{\left(\frac{\partial P}{\partial P}\right)}_{S}}}^{1}$

$\text{ "" } \boldsymbol{\left(2\right)}$

since a constant $S$ forces a derivative of $S$ with respect to any variable to go to $0$, and $\frac{\mathrm{dP}}{\mathrm{dP}} = 1$ no matter what else is held constant.

From this question, we can refer back and recall that

${\left(\frac{\partial S}{\partial T}\right)}_{P} = \frac{1}{T} {\left(\frac{\partial H}{\partial P}\right)}_{T} = {C}_{P} / T$.

$\boldsymbol{\left(3\right)}$

Now we need to figure out what ${\left(\frac{\partial S}{\partial P}\right)}_{T}$ is.

The bottom of the derivative shows $T$ and $P$, which are the natural variables of the Gibbs' free energy, so from the Maxwell relation:

$\mathrm{dG} = - S \mathrm{dT} + V \mathrm{dP}$,

we can either recall the cyclic relation ${\left(\frac{\partial S}{\partial P}\right)}_{T} = - {\left(\frac{\partial V}{\partial T}\right)}_{P}$, or rederive it. Again, from the total derivative of $G \left(T , P\right)$:

$\mathrm{dG} = {\left(\frac{\partial G}{\partial T}\right)}_{P} \mathrm{dT} + {\left(\frac{\partial G}{\partial P}\right)}_{T} \mathrm{dP}$,

showing that $- S = {\left(\frac{\partial G}{\partial T}\right)}_{P}$ and $V = {\left(\frac{\partial G}{\partial P}\right)}_{T}$. Since the order of second partial differentiation doesn't matter,

${\left(\frac{{\partial}^{2} G}{\partial P \partial T}\right)}_{P , T} = {\left(\frac{{\partial}^{2} G}{\partial T \partial P}\right)}_{T , P}$,

meaning that

$- {\left(\frac{\partial S}{\partial P}\right)}_{T} = {\left(\frac{\partial V}{\partial T}\right)}_{P}$. $\text{ "" } \boldsymbol{\left(4\right)}$

Therefore, from $\boldsymbol{\left(2\right)}$, subtract over ${\left(\frac{\partial S}{\partial P}\right)}_{T}$, plug in $\boldsymbol{\left(3\right)}$ and $\boldsymbol{\left(4\right)}$, and we have:

${\left(\frac{\partial V}{\partial T}\right)}_{P} = {C}_{P} / T {\left(\frac{\partial T}{\partial P}\right)}_{S}$

$\implies {\left(\frac{\partial T}{\partial P}\right)}_{S} = \frac{T}{C} _ P {\left(\frac{\partial V}{\partial T}\right)}_{P}$

Finally, recall that $\alpha = \frac{1}{V} {\left(\frac{\partial V}{\partial T}\right)}_{P}$, so that we have:

$\implies \textcolor{b l u e}{{\left(\frac{\partial T}{\partial P}\right)}_{S} = \frac{V \alpha T}{C} _ P}$