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 mA and that in B is mB. The gas in each cylinder is now allowed to expand isothermally to the same final volume 2V.The changes in the pressure in A and.B are found to be P and 1.5 P respectively. Then

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4mA = 9mB

2mA = 3mB

3mA =2mB

9mA = 3mB

answer is C.

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Detailed Solution

Process is isothermal. Therefore, T = constant, (P ∝ 1V) volume is increasing, therefore pressure will decreases.In chamber A :∆P = Pi-Pf = μARTV-μART2V = μART2V--(i)In chamber B:1.5∆P = Pi-Pf = μBRTV-μBRT2V = μBRT2V (ii)from equations (i) and (ii) μAμB = 11.5 = 23⇒ mAMmBM = 23 ⇒ 3mA = 2mB

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