$M=4.0 \mathrm{g} \mathrm{mol}^{-1}$

As the speed of the sound in the gas is

$v=\sqrt{\frac{\gamma R T}{M}}$

where $\gamma$ is the ratio of two specific heats, $R$ is the universal gas constant and $T$ is the temperature of the gas.

$\therefore \gamma=\frac{M v^{2}}{R T}$

Here, $M=4.0 \mathrm{g} \mathrm{mol}^{-1}=4.0 \times 10^{-3} \mathrm{kg} \mathrm{mol}^{-1}$

$v=952 \mathrm{ms}^{-1}, R=8.3 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$ and $T=273 \mathrm{K}(\mathrm{at} \mathrm{NTP})$

$\because \quad \gamma=\frac{\left(4.0 \times 10^{-3} \mathrm{kg} \mathrm{mol}^{-1}\right)\left(952 \mathrm{ms}^{-1}\right)^{2}}{\left(8.3 \mathrm{JK}^{-1} \mathrm{mol}^{-1}\right)(273 \mathrm{K})}=1.6$

By definition, $\gamma=\frac{C_{p}}{C_{v}}$ or $C_{p}=\gamma C_{v}$

But $\gamma=1.6$ and $C_{v}=5.0 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$

$\therefore \quad C_{p}=(1.6)\left(5.0 \mathrm{JK}^{-1} \mathrm{mol}^{-1}\right)=8.0 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$