An ideal gas at pressure $P_{0}$ in a vessel. If the masses of all the molecules are halved and their RMS speeds doubled, the resulting pressure P will be:

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 $2 P_{0}$
As we know,

$U_{r m s} = \sqrt{\frac{3 R T}{M}} = \sqrt{\frac{3 P V}{M}}$

$U_{1} = \sqrt{\frac{3 P_{0} V}{M}}$

$2 U_{1} = \sqrt{\frac{3 P V}{M / 2}}$

by dividing these equations,

$∴ \frac{1}{2} = \sqrt{\frac{P_{0}}{2 P}}$

$2 P = 4 P_{0}$

$P = 2 P_{0}$

Hence, the correct option is $B$

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