Two identical containers A and B with frictionless positions 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 of B is mB. The gas in each cylinder is now allowed to expand isothermally to the same final volume 3V. The change in the pressure in A and B are found to be ∆P and 2.5 ∆P respectively. Then

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3mA=2mB

2mA=3mB

5mA=2mB

5mA=3mB

answer is C.

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Detailed Solution

T = Constant, Pα1VFor Chamber A, ΔP=Pi−Pt=nARTV−nART3V=2nART3V ------(I)For Chamber B, 2.5ΔP=Pi−Pt=nBRTV−nBRT3V=23nsRTV ------(II)From equations (I) and (II)nAns=12.5=25⇒mA/MmB/M=25⇒5mA=2mB

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