4.0 g  of a gas occupies 22.4 litres at NTP ( at $0^{\mathrm{o}}\mathrm{C}$).  If the speed of sound in this gas at NTP is 952 $m{s}^{-1}$, then the heat capacity at constant pressure is

(Take gas constant R = 8.3 $J{K}^{-1}$$mo{l}^{-1}$)

Since 4.0g of a gas occupies 22.4 litres at NTP, so the molecular mass of the gas is
M=4.0 g $mo{l}^{-1}$
As the speed of the sound in the gas is

$v=\sqrt{\frac{\gamma RT}{M}}$

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

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

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

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

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

By definition, 

$\gamma =\frac{C_p}{C_v} \text{ or } C_p=\gamma C_v=\frac{\gamma R}{\gamma -1}=\frac{1.6\times 8.3}{0.6}=22.13 {\mathrm{JK}}^{-1}{ \mathrm{mol}}^{-1}$