A gas is filled in a container at pressure $P_{0}$ . If the mass of molecules is halved and their rms speed is doubled, then the resultant pressure would be.

2 P_{0}

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4 P_{0}

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\frac{P_{0}}{4}

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\frac{P_{0}}{2}

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Solution

The correct option is C $2 P_{0}$
Using, $P = \frac{1}{3} ρ v_{r m s}^{2} = \frac{1}{3} \frac{m N}{V} v_{r m s}^{2}$
$∴ P \propto m v_{r m s}^{2}$
As m is halved and $v_{r m s}$ is doubled then P becomes twice or $2 P_{0}$ .

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A gas is filled in a container at pressure P0. If the mass of molecules is halved and their rms speed is doubled, then the resultant pressure would be.

The correct option is C 2P0Using, P=13ρv2rms=13mNVv2rms∴P∝mv2rmsAs m is halved and vrms is doubled then P becomes twice or 2P0.

A gas is filled in a container at pressure $P_{0}$ . If the mass of molecules is halved and their rms speed is doubled, then the resultant pressure would be.

The correct option is C $2 P_{0}$
Using, $P = \frac{1}{3} ρ v_{r m s}^{2} = \frac{1}{3} \frac{m N}{V} v_{r m s}^{2}$
$∴ P \propto m v_{r m s}^{2}$
As m is halved and $v_{r m s}$ is doubled then P becomes twice or $2 P_{0}$ .