The $N_2$ molecule is $14$ times heavier than the $H_2$ molecule. At what temperature will the $rms$ speed of $H_2$ molecules be equal to the $rms$ speed of $N_2$ molecules at $27^{\circ}C$?

The $rms$ speed of a gas at room temperature of $27^{\circ}C$ is found to be $412 \ m/s$. The gas is:

At which temperature will the velocity of $O_2$ molecules be equal to the velocity of $N_2$ molecules at $0 ^\circ C$?

$N$ molecules each of mass $m$ of gas $A$ and $2N$ molecules each of mass $2m$ of gas $B$ are contained in a vessel which is maintained at temperature $T$. The mean square velocity of the molecules of gas $B$ is denoted by $V_2^2$ and the mean square of the $X$-component velocity of the molecules of gas $A$ is denoted by $V_1^2$,then $\frac{V_1}{V_2}$ is

$Assertion :$ The root mean square and most probable speeds of the molecules in a gas are the same.
$Reason :$ The Maxwell distribution for the speed of molecules in a gas is symmetrical.

At a pressure of $24 \times 10^5 \, \text{dyne/cm}^2$,the volume of $O_2$ is $10 \, \text{litre}$ and its mass is $20 \, \text{gm}$. The $r.m.s.$ velocity will be ....... $m/s$.