Two identical containers A and B have frictionless pistons. They contain the same volume of an ideal gas at the same temperature. The mass of the gas in A is m_Aand that in B is m_B. The gas in each cylinder is now allowed to expand isothermally to double the initial volume. The change in the pressure in A and B, respectively, is $∆$ p and 1.5 $∆$ p. Then

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$4 m_{A} = 9 m_{B}$

$2 m_{A} = 3 m_{B}$

$3 m_{A} = 2 m_{B}$

$9 m_{A} = 4 m_{B}$

answer is C.

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Detailed Solution

For the gas in container A

$\begin{array}{l} ΔP = {(P_{A})}_{final} - {(P_{A})}_{intial} = \frac{n_{A} RT}{2 V} - \frac{n_{A} RT}{V} \\ ΔP = - \frac{n_{A} RT}{2 V} . . . . . . . . (i) \end{array}$

For gas in container B

$\begin{array}{l} 1 .5 ΔP = {(P_{B})}_{final} - {(P_{B})}_{intial} = \frac{n_{B} RT}{2 V} - \frac{n_{B} RT}{V} \\ 1 .5 ΔP = - \frac{n_{B} RT}{2 V} . . . . . . . . . . . . (ii) \end{array}$