Q. Two identical vessels $A$ and $B$ with frictionless pistons conatin the same ideal gas at the same temperature and the same volume $V$. The masses of gas in $A$ and $B$ are $m_{A}\ and\ m_{B}$ respectively. The gases are allowed to expand isothermally to same final volume $2\ V$. The change in pressures of the gas in $A$ and $B$ are found to be $\mathrm{\Delta}p\ and\ 1.5\ \mathrm{\Delta}p$ respectively. Then
A) $9m_{A} = 4m_{B}$
B) $3m_{A} = 2m_{B}$
C) $2m_{A} = 3m_{B}$
D) $4m_{A} = 9m_{B}$
Explanation
(a) Let $p_{1}\ and\ p_{2}$ are the initial and final pressures of the gas filled in $A.$ Then
$p_{1} = \frac{n_{A}RT}{V}\ \ \ and\ \ p_{2} = \frac{n_{A}RT}{2V}$
$\mathrm{\Delta}p = p_{2} - p_{1} = - \frac{n_{A}RT}{2V}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = - \left( \frac{m_{A}}{M} \right)\frac{RT}{2V}$ …(i)
where $M$ is the atomic weight of the gas.
Similarly, $1.5\mathrm{\Delta}p = - \left( \frac{m_{B}}{M} \right)\frac{RT}{2V}$ …(ii)
Dividing Eq.(ii) by Eq. (i), we get
$1.5 = \frac{m_{B}}{m_{A}}\ \ \ \ \ \ \ \ \ or\ \ \ \ \ \ \ \ \ \frac{3}{2} = \frac{m_{B}}{m_{A}}$
or $3m_{A} = 2m_{B}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.