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(C) According to the Kinetic Theory of Gases $(KTG)$:$1$. The $r.m.s.$ velocity of gas molecules is given by $V_{rms} = \sqrt{\frac{3RT}{M_m}}$. Since

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$A$ and $B$ only

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(C) According to the Kinetic Theory of Gases $(KTG)$:$1$. The $r.m.s.$ velocity of gas molecules is given by $V_{rms} = \sqrt{\frac{3RT}{M_m}}$. Since the temperature $T$ and the molar mass $M_m$ are the same for both vessels,the $r.m.s.$ velocity is the same. Thus,statement $A$ is correct.$2$. From the ideal gas equation $PV = NkT$,where $N$ is the number of molecules and $k$ is the Boltzmann constant,we have $P = \frac{NkT}{V}$. Since $V$,$T$,and $k$ are constant,$P \propto N$. Therefore,the ratio of pressures is $\frac{P_1}{P_2} = \frac{N_1}{N_2} = 1:4$. Thus,statement $B$ is correct.Since statements $A$ and $B$ are correct,the correct option is $C$.

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