Question

Two identical containers A and B with frictionless pistons contains the same ideal gas at the same temperature and the same volume V. The mass of gas A is $m_A$ and that of B is $m_B$. The gas in each cylinder is now allowed to expand isothermally to the same final volume 2V. The change in the pressure in A and B are found to be $△P$ and $1.5△P$ respectively. Then

• A. $4m_A=9m_B$
• B. $3m_A=3m_B$
• C. $3m_A=2m_B$
• D. $9m_A=4m_B$

Answer 1

Answer: (C)

$3m_A=2m_B$

The initial pressure in the two containers will be

$P_A=\frac{n_{A}RT}{V}=m_A(\frac{RT}{MV})$

$P_B=\frac{n_{B}RT}{V}=m_B(\frac{RT}{MV})$

After isothermal expansion, the pressure will be

$P_A^{\prime }=\frac{n_{A}RT}{2V}=m_A(\frac{RT}{2MV})$

$P_B^{\prime }=\frac{n_{B}RT}{2V}=m_B(\frac{RT}{2MV})$

$\therefore -\Delta P_A=P_A-P_A^{\prime }=m_A(\frac{RT}{2MV})$

$\therefore -\Delta P_B=P_B-P_B^{\prime }=m_B(\frac{RT}{2MV})$

But, $\Delta P_A=\Delta P$ and $-\Delta P_B=1.5\Delta P$

So, $\frac{-\Delta P_A}{-\Delta P_B}=\frac{1}{1.5}⟹-\Delta P_A=\frac{-\Delta P_B}{1.5}$

$⟹m_A(\frac{RT}{2MV})=\frac{m_B}{1.5}(\frac{RT}{2MV})$

$⟹m_A=\frac{10m_B}{15}$

$\therefore 3m_A=2m_B$