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Two identical containers $A$ and $B$ with frictionless pistons contain the ideal gas at the same temperature and the same volume $V$. The mass of the gas in $A$ is $m_A$ and in $B$ is $m_B$. The gas in each cylinder is now allowed to expand isothermally to the same final volume $2V$. The changes in pressure in $A$ and $B$ are found to be $\Delta p$ and $1.5\Delta P$ respectively. Then
$4m_A = 9m_B$
$2m_A = 3m_B$
$3m_A = 2m_B$
$9m_A = 4m_B$
Solution
The initial pressure in the two containers will be
$P_{A}=\frac{n_{A} R T}{V}=m_{A}\left(\frac{R T}{M V}\right)$
$P_{B}=\frac{n_{B} R T}{V}=m_{B}\left(\frac{R T}{M V}\right)$
After isothermal expansion, pressure will be
$P_{A}^{\prime}=\frac{n_{A} R T}{2 V}=m_{A}\left(\frac{R T}{2 M V}\right)$
$P_{B}^{\prime}=\frac{n_{B} R T}{2 V}=m_{B}\left(\frac{R T}{2 M V}\right)$
$\therefore-\Delta P_{A}=P_{A}-P_{A}^{\prime}=m_{A}\left(\frac{R T}{2 M V}\right)$
$\therefore-\Delta P_{B}=P_{B}-P_{B}^{\prime}=m_{B}\left(\frac{R T}{2 M V}\right)$
But $\Delta P_{A}=\Delta P$ and $-\Delta P_{B}=1.5 \Delta P$
But $\Delta P_{A}=\Delta P$ and $-\Delta P_{B}=1.5 \Delta P$
$\operatorname{So}, \frac{-\Delta P_{A}}{-\Delta P_{B}}=\frac{1}{1.5} \Rightarrow-\Delta P_{A}=\frac{-\Delta P_{B}}{1.5}$
$m_{A}\left(\frac{R T}{2 M V}\right)=\frac{m_{B}}{1.5}\left(\frac{R T}{2 M V}\right)$
or $\quad m_{A}=\frac{10 m_{B}}{15}$
$\therefore 3 m_{A}=2 m_{B}$
