4.0 , g of a gas occupies 22.4 , L at NTP. Th | vedclass

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(B) Since $4.0 \, g$ of a gas occupies $22.4 \, L$ at $NTP$,the molar mass of the gas is $M = 4.0 \, g \, mol^{-1} = 4.0 	imes 10^{-3} \, kg \, mol^{

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$8.0$

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(B) Since $4.0 \, g$ of a gas occupies $22.4 \, L$ at $NTP$,the molar mass of the gas is $M = 4.0 \, g \, mol^{-1} = 4.0 	imes 10^{-3} \, kg \, mol^{-1}$.The speed of sound in an ideal gas is given by $v = \sqrt{\frac{\gamma R T}{M}}$.Rearranging for $\gamma$,we get $\gamma = \frac{M v^2}{R T}$.Given $v = 952 \, m s^{-1}$,$R = 8.3 \, J K^{-1} mol^{-1}$,and $T = 273 \, K$ at $NTP$:$\gamma = \frac{(4.0 	imes 10^{-3} \, kg \, mol^{-1}) 	imes (952 \, m s^{-1})^2}{(8.3 \, J K^{-1} mol^{-1}) 	imes (273 \, K)} \approx \frac{3625.216}{2265.9} \approx 1.6$.We know that $\gamma = \frac{C_p}{C_v}$,so $C_p = \gamma C_v$.Given $C_v = 5.0 \, J K^{-1} mol^{-1}$,we have $C_p = 1.6 	imes 5.0 \, J K^{-1} mol^{-1} = 8.0 \, J K^{-1} mol^{-1}$.

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