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 cylinder is now allowed to expand isothermally to the same final volume $2\,V$. The changes in the pressure in $A$ and $B$ are found to be $\Delta p$ and $1.5 \, \Delta p$, respectively. Then

• 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)

$3\, m_A = 2\, m_B$

Answer 2

Process is isothermal. Therefore, $T =$ constant. Volume is increasing; therefore, pressure will decrease $\left(p \propto \frac{1}{V}\right)$. In chamber $A$, $\Delta p = \left(p_{A}\right)_{i} -\left(p_{A}\right) _{f} = \frac{n_{A}RT}{V} - \frac{n_{A}RT}{2V}$. $= \frac{n_{A}RT}{2V}\quad...\left(i\right)$ In chamber $B$, $1.5 \Delta p = \left(p_{B}\right)_{i} -\left(p_{B}\right) _{f} = \frac{n_{B}RT}{V} - \frac{n_{B}RT}{2V}$. $= \frac{n_{B}RT}{2V}\quad ...\left(ii\right)$ From Eqs. $\left(i\right)$ and $\left(ii\right)$, $\frac{n_{A}}{n_{B}} = \frac{1}{1.5} = \frac{2}{3}$; $\frac{m_{A }/M}{m_{B}/M} = \frac{2}{3}$ or $\frac{m_{A}}{m_{B}} = \frac{2}{3} \therefore 3m_{A}^{ }= 2m_{B}$