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 $2V$. The change in the pressure in A and B are found to be $\Delta$p and $1.5$ $\Delta$p, respectively. Then.

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

Answer 1

Answer: (C)

$3m_A=2m_B$

Process is isothermal. Therefore, T$=$constant.

Volume is increasing therefore, pressure will decrease $(P\propto \frac{1}{V})$

In chamber A, $\Delta p=(p_A{)}_i-(p_A{)}_f=\frac{n_{A}RT}{V}-\frac{n_{A}RT}{2V}$

$=\frac{n_{A}RT}{2V}$ ..(i)

In chamber B, $1.5$ $\Delta p=(p_B{)}_i-(p_B{)}_f=\frac{n_{B}RT}{V}-\frac{n_{B}RT}{2V}$

$=\frac{n_{B}RT}{2V}$ ..(ii)

From Eqs. (i) and (ii),

$\frac{n_A}{n_B}=\frac{1}{1.5}=\frac{2}{3}$; $\frac{m_A/M}{m_B/M}=\frac{2}{3}$ or $\frac{m_A}{m_B}=\frac{2}{3}$

$\therefore 3m_A=2m_B$.