[Physics] Relation between speed of sound and compressibility


We know that
$c^2=\frac{\partial p}{\partial ρ}$

The adiabatic compressibility is defined as: $\beta_S=-\frac{1}{V}\frac{\partial V}{\partial p}$ such that the subscript "S" stands for "adiabatic"

How can I show that $c^2=\frac{1}{\rho \beta_S}$ ?

I tried replacing $V$ by $\frac{m}{\rho}$ but I get for $\beta_S=-\rho \frac{\partial \frac{1}{\rho}}{\partial p}$

Best Answer

Hint: use the chain rule:

$$\beta_S=-\rho \frac{\partial \frac{1}{\rho}}{\partial p}=-\rho\frac{\partial \frac{1}{\rho}}{\partial \rho} \frac{\partial {\rho}}{\partial p}$$

No need to use $PV=nRT$, which does not hold for non-ideal gases.

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