At constant temperature,the vapour density of two gases are in the ratio of $6 : 9$. Compare their $v_{rms}$.

At what temperature will the $rms$ speed of oxygen molecules become just sufficient for escaping from the Earth's atmosphere $?$ (Given: Mass of oxygen molecule $(m) = 2.76 \times 10^{-26} \, kg$,Boltzmann's constant $k_B = 1.38 \times 10^{-23} \, JK^{-1}$)

The speeds of gas molecules are $2, 3, 4, 5, 6 \ km/s$. The $rms$ speed is ............. $km/s$.

Temperature of an ideal gas, initially at $27^{\circ} C$, is raised by $6^{\circ} C$. The rms velocity of the gas molecules will

Column-$I$ represents the formula for ${v_{rms}}$ and Column-$II$ represents the corresponding condition (phenomena). Match them correctly:

Column-$I$	Column-$II$
$(a)$ ${v_{rms}} = \sqrt {\frac{3P}{\rho}}$	$(i)$ For $1 \text{ mole ideal gas}$
$(b)$ ${v_{rms}} = \sqrt {\frac{3RT}{M_0}}$	$(ii)$ For one molecule of gas
$(c)$ ${v_{rms}} = \sqrt {\frac{3{k_B}T}{m}}$	$(iii)$ On the basis of kinetic theory

The $RMS$ velocity of oxygen molecules at $NTP$ is $0.5 \,km/s$. The $RMS$ velocity for the hydrogen molecule at $NTP$ is (in $\,km/s$)