Q.

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 mA and that in B is mB. 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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a

4mA=9mB

b

2mA=3mB

c

3mA=2mB

d

9mA=4mB

answer is C.

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Detailed Solution

For the gas in container AΔP=PAfinal −PAintial =nART2V−nARTVΔP=−nART2V           ........(i)For gas in container B1.5ΔP=PBfinal −PBintial =nBRT2V−nBRTV1.5ΔP=−nBRT2V    ............(ii)From Eqs. (i) and (ii), we get nB=1.5nA⇒ 2nB=3nA  or  2mB=3mA
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