# ((delS)/(delV))_T = alpha/beta. Prove this relation please?

## $S = \text{entropy}$ $V = \text{volume}$ $\alpha = \text{coefficient of isobaric expansion}$ $\beta = \text{coefficient of isothermal compression}$

Jan 22, 2017

From the definitions of $\alpha$ and $\beta$:

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

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

Now, since $\alpha$ and $\beta$ both have $T$ and $P$ as their natural variables, and their definitions contain partial derivatives found within a total derivative, consider the total derivative of $V = V \left(T , P\right)$:

$\mathrm{dV} = {\left(\frac{\partial V}{\partial T}\right)}_{P} \mathrm{dT} + {\left(\frac{\partial V}{\partial P}\right)}_{T} \mathrm{dP}$

Notice how this contains both $\alpha$ and $\beta$, differing only by a constant. Now, take the whole total derivative and divide by ${\left(\partial T\right)}_{V}$:

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

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

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

This looks quite familiar. Considering what $\alpha$ and $\beta$ were:

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

$= - \frac{\alpha}{- \beta} = \frac{\alpha}{\beta}$

Now we just have to show that ${\left(\frac{\partial P}{\partial T}\right)}_{V} = {\left(\frac{\partial S}{\partial V}\right)}_{T}$.

Consider the Helmholtz free energy, the thermodynamic state function that contains the natural variables $T$ and $V$. Its total derivative is:

$\mathrm{dA} = {\left(\frac{\partial A}{\partial T}\right)}_{V} \mathrm{dT} + {\left(\frac{\partial A}{\partial V}\right)}_{T} \mathrm{dV}$

Just like the Gibbs' free energy, where we had $\mathrm{dG} = \mathrm{dH} - d \left(T S\right)$ (which closely resembles the famous equation, $\Delta G = \Delta H - T \Delta S$), we have:

$\boldsymbol{\mathrm{dA}} = \mathrm{dU} - d \left(T S\right)$

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

From the first law of thermodynamics for a reversible process in a closed system, $\mathrm{dU} = \delta {q}_{\text{rev" + deltaw_"rev}} = T \mathrm{dS} - P \mathrm{dV}$, so:

$\implies \cancel{T \mathrm{dS}} - P \mathrm{dV} - S \mathrm{dT} - \cancel{T \mathrm{dS}}$

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

which gives the Maxwell relation for the Helmholtz free energy. Depending on your professor, you may simply be able to use this relation without deriving it. Compare back to the total derivative to find:

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

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

Since state functions have exact differentials, it follows that the second-order cross-derivatives are equal:

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

So, let us take the cross-derivatives of the above expressions of $S$ and $V$:

${\left(\frac{\partial S}{\partial V}\right)}_{T} = - {\left(\frac{{\partial}^{2} A}{\partial V \partial T}\right)}_{V , T}$

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

We now see that since the second-order cross-derivatives are equal from the definition of a state function, we've established that ${\left(\frac{\partial P}{\partial T}\right)}_{V} = {\left(\frac{\partial S}{\partial V}\right)}_{T}$. This is known as a cross-derivative relationship, and you should be allowed to invoke this any time.

Therefore:

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