2
$\begingroup$

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 $\gamma \left(\frac{\partial p}{\partial \rho}\right)_T$?


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.

$\endgroup$
2
  • $\begingroup$ For an ideal gas, $R=(c_p-c_v)$ $\endgroup$
    – RC_23
    May 16, 2023 at 14:09
  • 1
    $\begingroup$ It looks like you have the right idea by expanding a differential term with different parameters held constant. You can find more guidance here. $\endgroup$ May 16, 2023 at 14:47

1 Answer 1

2
$\begingroup$

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}

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.