One mole of an ideal gas at $STP$ is heated in an insulated closed container until the average speed of its molecules is doubled. Its pressure would therefore increase by what factor?

Derive the equation of $v_{rms}$ in terms of the mass of a molecule.

$A$ cubical box with porous walls containing an equal number of ${O_2}$ and ${H_2}$ molecules is placed in a large evacuated chamber. The entire system is maintained at constant temperature $T$. The ratio of the ${v_{rms}}$ of ${O_2}$ molecules to that of the ${v_{rms}}$ of ${H_2}$ molecules,found in the chamber outside the box after a short interval,is:

Consider a mixture of gas molecules of types $A, B$,and $C$ having masses $m_{A} < m_{B} < m_{C}$. The ratio of their root mean square speeds at normal temperature and pressure is:

Equal volumes of two gases,having their densities in the ratio of $1: 16$,exert equal pressures on the walls of two containers. The ratio of their rms speeds $(c_1: c_2)$ is

The $rms$ speed of a molecule of a diatomic gas is $v$. When the temperature is doubled,the molecule dissociates into two atoms. The new $rms$ speed of the atom is: