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

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$\frac{P_{0}}{2}$

$2 P_{0}$

$\frac{P_{0}}{4}$

$4 P_{0}$

answer is A.

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Detailed Solution

Using, $P = \frac{1}{3} ρ {v^{2}}_{r m s} = \frac{1}{3} \frac{m N}{V} {v^{2}}_{r m s}$

$∴ P \propto m {v^{2}}_{r m s}$

As $m$ is halved $v_{r m s}$ is doubled then $P$ becomes twice or $2 P_{0}$ .

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