Question

Two identical containers $A$ and $B$ with frictionless pistons contain the same ideal gas at the same temperature and the same volume $V$. The mass of the gas in $A$ is $m_{A}$ and that in $B$ is $m_{B}$. The gas in each container is now allowed to expand isothermally to the same final volume $2 V$. The changes in the pressure in $A$ and $B$ are to be found $\Delta p$ and $1.5 \Delta p$ respectively, then relation for masses will be

• A. $4 m_{A}=9 m_{B}$
• B. $2 m_{A}=3 m_{B}$
• C. $3 m_{A}=2 m_{B}$
• D. $9 m_{a}=4 m_{B}$

Answer 1

Answer: (C)

For isothermal expansion of an ideal gas
$p V=$ constant
$\Rightarrow p \Delta V+V \Delta p=0$
For container $A$,
$p_{i A}(V)+V(\Delta p)=0$
$\Rightarrow p_{i A}=-\frac{V \Delta P}{V}=-\Delta p$ ...(i)
where, $p$ is initial pressure.
For container $B$,
$p_{i B}(V)+V(15 \Delta p)=V$
$\Rightarrow p_{i B}=-15 \Delta p$ ...(ii)
From Eqs. (i) and (ii), we have
$p_{i A}=\frac{p_{i B}}{15}=\frac{2}{3} p_{i B}$
$\Rightarrow m_{A}=\frac{2}{3} m B$
$\Rightarrow 3 m_{A}=2 m B[\because p \propto m]$