Question

$4.0g$ of a gas occupies $22.4$ litres at NTP. The specific heat capacity of the gas at constant volume is $5.0J{K}^{-1}mo{l}^{-1}$. If the speed of sound in this gas at NTP is $952m{s}^{-1}$, then the heat capacity at constant pressure is (Take gas constant $R=8.3J{K}^{1}mo{l}^1$ )

• A. $7.0J{K}^{-1}mo{l}^{-1}$
• B. $8.5J{K}^{-1}mo{l}^{-1}$
• C. $8.0J{K}^{-1}mo{l}^{-1}$
• D. $7.5J{K}^{-1}mo{l}^{-1}$

Answer 1

Answer: (C)

Since $4.0g$ of a gas occupies $22.4$ litres at NTP, so the molecular mass of the gas is
$M=4.0gmo{l}^1$
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.
$\therefore \gamma =\frac{M{v}^2}{RT}$
Here, $M=4.0gmo{l}^{-1}=4.0\times 10^{-3}kgmo{l}^{-1}$,
$v=952m{s}^{-1},R=8.3J{K}^{-1}mo{l}^{-1}$
and $T=273K$ (at NTP)
$\therefore \gamma =\frac{\left(4.0\times 10^{-3}kgmo{l}^{-1}\right){\left(952m{s}^{-1}\right)}^2}{\left(8.3J{K}^{-1}mo{l}^{-1}\right)(273K)}=1.6$
By definition,
$\gamma =\frac{C_p}{C_v}$ or $C_p=\gamma C_v$
But $\gamma =1.6$ and $C_{\gamma }=5.0J{K}^{-1}mo{l}^{-1}$
$\therefore C_p=(1.6)\left(5.0J{K}^{1}mo{l}^{-1}\right)$
$=8.0J{K}^{-1}mo{l}^{-1}$