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Q.
Gas at a pressure $P_0$ is contained in a vessel . If the masses of all the molecules are halved and their speeds are doubled, the resulting pressure P will be equal to
Kinetic Theory
Solution:
$P =\frac{1}{3} \frac{ mN }{ V _{2}} v _{ rms }^{2}$
$\therefore p \propto mv _{ rms }^{2}$
$so \frac{ P _{2}}{ P _{1}}=\frac{ m _{2}}{ m _{2}} \times\left(\frac{ v _{2}}{ v _{1}}\right)^{2}$
$=\frac{ m _{1} / 2}{ m _{1}}\left(\frac{2 v _{1}}{ v _{1}}\right)^{2}$
$=2$
$P _{2}=2 P _{1}$
$=2 P _{0}$