The velocity of the molecules of a gas at temperature $120\,K$ is $v$. At what temperature in $K$ will the velocity be $2v$?

The root mean square velocity of molecules of a gas is $200 \,m/s$. What will be the root mean square velocity of the molecules, if the molecular weight is doubled and the absolute temperature is halved?

The respective speeds of the molecules are $1, 2, 3, 4$ and $5 \ km/sec$. The ratio of their $r.m.s.$ velocity and the average velocity will be

The $r.m.s.$ speed of the molecules of a gas at a pressure $10^5 \ Pa$ and temperature $0^\circ C$ is $0.5 \ km \ sec^{-1}.$ If the pressure is kept constant but temperature is raised to $819^\circ C,$ the velocity will become ........ $km \ sec^{-1}.$

The root mean square speed of smoke particles of mass $5 \times 10^{-17} \, kg$ in their Brownian motion in air at $NTP$ is approximately $....... \, mm \, s^{-1}$. [Given $k = 1.38 \times 10^{-23} \, J \, K^{-1}$ and $T = 293 \, K$]

$A$ flask contains argon and chlorine in the ratio of $2:1$ by mass. The temperature of the mixture is $27^{\circ}\text{C}$. The ratio of root mean square speed of the molecules of the two gases $\left(\frac{v_{\text{rms}}^{\text{Ar}}}{v_{\text{rms}}^{\text{Cl}}}\right)$ is : (Atomic mass of argon $= 40.0 \text{u}$ and molecular mass of chlorine $= 70.0 \text{u}$)