Question

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

• A. $7.0 \, JK^{-1} \, mol^{-1}$
• B. $8.5 \, JK^{-1} \, mol^{-1}$
• C. $8.0 \, JK^{-1} \, mol^{-1}$
• D. $7.5 \, JK^{-1} \, mol^{-1}$

Answer 1

Answer: (C)

Since $4.0 \,g$ of a gas occupies $22.4$ litres at NTP, so the molecular mass of the gas is
$M=4.0 \,g \, mol ^{1}$
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, $M=4.0 \,g \, mol ^{-1}=4.0 \times 10^{-3} \,kg \,mol ^{-1}$,
$v=952 \,ms ^{-1}, R=8.3 \, JK ^{-1} \, mol ^{-1}$
and $T=273 \,K$ (at NTP)
$\therefore \gamma=\frac{\left(4.0 \times 10^{-3} \,kg \,mol ^{-1}\right)\left(952 \, ms ^{-1}\right)^{2}}{\left(8.3 \,JK ^{-1} \, mol ^{-1}\right)(273 \, K )}=1.6$
By definition,
$\gamma=\frac{C_{p}}{C_{v}} $ or $C_{p}=\gamma C_{v}$
But $\gamma=1.6$ and $C_{\gamma}=5.0 \, JK ^{-1} \,mol ^{-1}$
$\therefore C_{p} =(1.6)\left(5.0 JK ^{1} mol ^{-1}\right) $
$=8.0 \,JK ^{-1} \, mol ^{-1}$