As we know that, the rms speed of molecules of a gas
$$
\begin{aligned}
&c=\sqrt{\frac{3 p}{\rho}}=\sqrt{\frac{3 R T}{M}} \\
&\therefore \frac{p}{\rho}=\left(\frac{R T}{M}\right) \quad \quad \because \frac{p}{\rho}=\frac{R T / V}{M / V}
\end{aligned}
$$
where $M=$ molar mass of the gas.
Speed of sound wave in gas,
$$
v=\sqrt{\frac{\gamma p}{\rho}}=\sqrt{\frac{\gamma R T}{M}}
$$
On dividing Eq. (i) by Eq. (ii), we get
$$
\frac{\rho}{v}=\left[\frac{\sqrt{\frac{3 R T}{M}}}{\sqrt{\frac{\gamma R T}{M}}}\right]
$$
$$
\frac{c}{v}=\sqrt{\frac{3 R T}{M} \times \frac{M}{\gamma R T}} \Rightarrow \frac{c}{v}=\sqrt{\frac{3}{\gamma}}
$$
where $\gamma$ = adiabatic constant for diatomic gas
$$
\gamma=\frac{7}{5} \quad\left[\text { since } \gamma=\frac{C_p}{C_v}\right]
$$
So, $\frac{c}{v}=\sqrt{\frac{3}{7 / 5}}=\sqrt{\frac{15}{7}}=$ constant
Hence, $\frac{c}{v}=$ constant