At NTP (Normal Temperature and Pressure), $22.4 \text{ L}$ of an ideal gas corresponds to $1 \text{ mole}$. Therefore, the molar mass of the gas is $M = 4.0 \text{ g/mol} = 4.0 \times 10^{-3} \text{ kg/mol}$. The speed of sound in a gas is given by the formula: $v = \sqrt{\frac{\gamma RT}{M}}$ Squaring both sides and solving for $\gamma$: $\gamma = \frac{v^2 M}{RT}$ Given $v = 952 \text{ ms}^{-1}$, $R = 8.31 \text{ J K}^{-1}\text{mol}^{-1}$, and $T = 273 \text{ K}$: $\gamma = \frac{(952)^2 \times 4.0 \times 10^{-3}}{8.31 \times 273} = \frac{906304 \times 0.004}{2268.63} \approx 1.6$ Alternatively, using density $\rho = \frac{\text{Mass}}{\text{Volume}} = \frac{4.0 \times 10^{-3} \text{ kg}}{22.4 \times 10^{-3} \text{ m}^3}$ and pressure $P = 1.013 \times 10^5 \text{ Pa}$: $\gamma = \frac{v^2 \rho}{P} = \frac{(952)^2 \times \left(\frac{4.0}{22.4}\right)}{1.013 \times 10^5} \approx 1.6$ The ratio of molar heat capacities is $\gamma = \frac{C_p}{C_v}$. Given the molar heat capacity at constant volume $C_v = 5.0 \text{ J K}^{-1}\text{mol}^{-1}$: $C_p = \gamma C_v = 1.6 \times 5.0 = 8.0 \text{ J K}^{-1}\text{mol}^{-1}$