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 $A$ is $m_{A}$ and that in $B$ is $m_{B}$ . The change in the pressure in $A$ and $B$ are found to be $△ P$ and $1.5 △ P$ respectively. Then

4 m_{A} = 9 m_{B}

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2 m_{A} = 3 m_{B}

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3 m_{A} = 2 m_{B}

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9 m_{A} = 4 m_{B}

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Solution

The correct option is C $3 m_{A} = 2 m_{B}$
For gas is $A, P_{1} = (\frac{R T}{M}) \frac{m_{A}}{V_{1}}$
$P_{2} = (\frac{R T}{M}) \frac{m_{A}}{V_{2}}$
$∴ △ P = P_{1} - P_{2} = (\frac{R T}{M}) m_{A} (\frac{1}{V_{1}} - \frac{1}{V_{2}})$
Putting $V_{1} = V$ and $V_{2} = 2 V$
We get $△ P = \frac{R T}{M} \frac{m_{A}}{2 V}$
Similarly for Gas in $B, 1.5 △ P = (\frac{R T}{M}) \frac{m_{B}}{2 V}$
From eq. (I) and (II) we get
$2 m_{B} = 3 m_{A}$
Hence, (C) is correct.

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Q. 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

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 changes in the pressure in A and B are found to be

△ P

and 1.5

△ P

respectively. Then

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 $m_{A}$ 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 changes in the pressure in A and B are found to be $△ P$ and 1.5 $△ P$ respectively. Then