A gas at a pressure $P_{0}$ is contained in a vessel. If the masses of all the molecules are halved and their velocities doubled, the resulting pressure $P$ would be equal to :

4 P_{0}

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2 P_{0}

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P_{0}

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\frac{P_{0}}{2}

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Solution

The correct option is B $2 P_{0}$
$P = \frac{1}{3} \frac{m N}{V} {v^{2}}_{r m s}$

Thus, $P \propto m {v^{2}}_{r m s}$

So, $\frac{P_{2}}{P_{1}} = \frac{m_{2}}{m_{1}} \times {(\frac{v_{2}}{v_{1}})}^{2}$

$\frac{P_{2}}{P_{1}} = \frac{\frac{m_{1}}{2}}{m_{1}} {(\frac{2 v_{1}}{v_{1}})}^{2} = 2$

$P_{2} = 2 P_{1} = 2 P_{o}$

Hence, the correct option is $B$ .

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A gas at a pressure P0 is contained in a vessel. If the masses of all the molecules are halved and their velocities doubled, the resulting pressure P would be equal to :

The correct option is B 2P0P=13mNVv2rmsThus, P ∝ mv2rmsSo, P2P1=m2m1×(v2v1)2 P2P1=m12m1(2v1v1)2=2P2=2P1=2PoHence, the correct option is B.

A gas at a pressure $P_{0}$ is contained in a vessel. If the masses of all the molecules are halved and their velocities doubled, the resulting pressure $P$ would be equal to :