# Please help me to derive that ((delS)/(delT))_P = C_P/T and ((delS)/(delT))_V = C_V/T?

Jan 25, 2017

Okay, so I will assume you know or can figure out the Maxwell relations. Those are a common starting point. Furthermore, you have by definition that ${C}_{P} = {\left(\frac{\partial H}{\partial T}\right)}_{P}$ and ${C}_{V} = {\left(\frac{\partial U}{\partial T}\right)}_{V}$, so you do not have to derive those.

Recall that $G = H - T S$. Then, the derivative is:

$\mathrm{dG} = \mathrm{dH} - d \left(T S\right)$

$= \mathrm{dH} - S \mathrm{dT} - T \mathrm{dS}$
(use product rule)

Now, for the first derivation, you can divide through by $\partial T$ at a constant $P$ to generate two derivatives that can be figured out, and the third one is what you're looking for:

${\left(\frac{\partial G}{\partial T}\right)}_{P} = {\left(\frac{\partial H}{\partial T}\right)}_{P} - {\cancel{S {\left(\frac{\partial T}{\partial T}\right)}_{P}}}^{1} - T {\left(\frac{\partial S}{\partial T}\right)}_{P}$

$\text{ "" } \boldsymbol{\left(1\right)}$

where the third term goes to $1$ because $T$ does not change with respect to $T$; it is one-to-one with itself.

By definition, ${\left(\frac{\partial H}{\partial T}\right)}_{P} = {C}_{P}$. $\text{ "" } \boldsymbol{\left(2\right)}$

Next, recall that for the natural variables $T$ and $P$ correspond to the Gibbs' free energy, so that the Maxwell relation is:

$\mathrm{dG} = - S \mathrm{dT} + V \mathrm{dP}$ $\text{ "" "" } \boldsymbol{\left(3\right)}$

We do not yet know what ${\left(\frac{\partial G}{\partial T}\right)}_{P}$ is, but using the Maxwell relation, we get:

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

Therefore. plugging $\boldsymbol{\left(4\right)}$ and $\boldsymbol{\left(2\right)}$ into $\boldsymbol{\left(1\right)}$:

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

It follows that if ${C}_{P} = T {\left(\frac{\partial S}{\partial T}\right)}_{P}$, it must be that $\textcolor{b l u e}{{\left(\frac{\partial S}{\partial T}\right)}_{P} = {C}_{P} / T}$.

A similar process follows for deriving ${C}_{V} / T = {\left(\frac{\partial S}{\partial T}\right)}_{V}$. Using:

$A = U - T S$,

take the derivative to get:

$\mathrm{dA} = \mathrm{dU} - S \mathrm{dT} - T \mathrm{dS}$

Almost like before, divide by $\partial T$ at a constant $V$ instead of $P$ to generate two derivatives that can be figured out, and the third one is what you're looking for:

${\left(\frac{\partial A}{\partial T}\right)}_{V} = {\left(\frac{\partial U}{\partial T}\right)}_{V} - S {\cancel{{\left(\frac{\partial T}{\partial T}\right)}_{V}}}^{1} - T {\left(\frac{\partial S}{\partial T}\right)}_{V}$

We know by definition that ${\left(\frac{\partial U}{\partial T}\right)}_{V} = {C}_{V}$, ${\left(\frac{\partial T}{\partial T}\right)}_{V} = 1$, and that $A$ is a function of the natural variables $T$ and $V$.

So from the Maxwell relation:

$\mathrm{dA} = - S \mathrm{dT} - P \mathrm{dV}$

we have that:

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

so plugging back into the main equation, we get:

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

Since ${C}_{V} = T {\left(\frac{\partial S}{\partial T}\right)}_{V}$, it follows that $\textcolor{b l u e}{{\left(\frac{\partial S}{\partial T}\right)}_{V} = {C}_{V} / T}$.