b
since \(4.0 \mathrm{g}\) of a gas occupies \(22.4\) litres at \(NTP\), so the molecular mass of the gas is

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

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 \(C_{v}=5.0 \mathrm{JK}^{-1} \mathrm{mol}^{-1}\)

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