How can I derive this thermodynamically equivalent expression for the speed of sound?

Question

Background

In Viscous Fluid Flow by White, the speed of sound for a gas is given as $$\tag{1-71} a^2 = \frac{\partial p}{\partial \rho}\big|_s $$ where the partial derivative is computed at constant entropy.

White mentions that a thermodynamically equivalent expression is $$\tag{1-72} a^2 =\gamma \left(\frac{\partial p}{\partial \rho}\right)_T $$ where the partial derivative is evaluated at constant temperature and $\gamma$ is the ratio of specific heats (not necessarily constant) $$ \gamma = \frac{c_p}{c_v}=\frac{\partial h}{\partial T}\big|_p/\frac{\partial e}{\partial T}\big|_v $$

White writes

The student may prove as an exercise that Eqs. (1-71) and (1-72) are thermodynamically identical.

White provides this relationship before defining an ideal gas or the ideal gas law. Therefore, it's reasonable to assume that this expression can be derived strictly from thermodynamic principles without using the ideal gas assumption.

Question

Assuming an adiabatic, reversible (isentropic) process, how do I show that $\frac{\partial p}{\partial \rho}\big|_s$ is equivalent to $\gamma \left(\frac{\partial p}{\partial \rho}\right)_T$?

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Notes

As I have been trying to derive this expression, I found a few expressions which might be useful:

If $p=p(\rho,s)$, $$ dp = \left(\frac{\partial p}{\partial \rho}\right)_s d\rho + \left(\frac{\partial p}{\partial s}\right)_\rho ds $$ since $ds=0$, $$ \frac{dp}{d\rho}=\left(\frac{\partial p}{\partial \rho}\right)_s $$ If $p=p(\rho,T)$, $$ dp = \left(\frac{\partial p}{\partial \rho}\right)_T d\rho + \left(\frac{\partial p}{\partial T}\right)_\rho dT $$ so $$ \left(\frac{\partial p}{\partial \rho}\right)_s = \left(\frac{\partial p}{\partial \rho}\right)_T + \left(\frac{\partial p}{\partial T}\right)_\rho \frac{dT}{d\rho} $$ That at least gives some kind of relationship between $\left(\frac{\partial p}{\partial \rho}\right)_s$ and $\left(\frac{\partial p}{\partial \rho}\right)_T$.

Another relationship which may be useful is $$ \frac{dp}{d\rho}=\frac{p}{\rho}\frac{d h}{d e} $$ which can be derived from the 1st Law of Thermodynamics and the definition of enthalpy. The reason I think this is useful is because, if we could assume an ideal gas, then $p/\rho=\frac{\partial p}{\partial \rho}\big|_T =RT$ and $\frac{dh}{de}$ I think can be manipulated to $c_p/c_v$ somehow.

Answer 1

From the density definition $\rho\equiv m/V$, a constant-mass assumption, and the triple product rule:

\begin{align}\frac{\left.\dfrac{\partial P}{\partial \rho}\right|_S}{\left.\dfrac{\partial P}{\partial \rho}\right|_T}=\frac{\left.\dfrac{\partial P}{(-m/V^2)\partial V}\right|_S}{\left.\dfrac{\partial P}{(-m/V^2)\partial V}\right|_T}=\frac{\left.\dfrac{\partial P}{\partial V}\right|_S}{\left.\dfrac{\partial P}{\partial V}\right|_T}=\frac{-\left(\dfrac{\partial V}{\partial S}\right)^{-1}_P\left(\dfrac{\partial S}{\partial P}\right)^{-1}_V}{-\left(\dfrac{\partial V}{\partial T}\right)^{-1}_P\left(\dfrac{\partial T}{\partial P}\right)^{-1}_V}=\frac{\left.\dfrac{\partial S}{\partial T}\right|_P}{\left.\dfrac{\partial S}{\partial T}\right|_V}=\frac{c_P}{c_V}\equiv\gamma.\end{align}