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 gas contained 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. 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

For gas in A, $P_{1} = \left( \frac{m_{A}}{M} \right)\frac{RT}{V_{1}}$

$P_{2} = \left( \frac{m_{A}}{M} \right)\frac{RT}{V_{2}}$

$\therefore\Delta P = P_{2} - P_{1} = \left( \frac{RT}{M} \right)m_{A}\left( \frac{1}{V_{1}} - \frac{1}{V_{2}} \right)$

Putting$V_{1} = VandV_{2} = 2V$

We get $\Delta P = \left( \frac{RT}{M} \right)\frac{m_{A}}{2V}$

Similarly for Gas in B, $1.5\Delta P = \left( \frac{RT}{M} \right)\frac{m_{B}}{2V}$

From eq. (I) and (II) we get 2$m_{B}$ = 3$m_{A}$

Hence (3) is the correct.