The temperature of a gas is $-68^{\circ} C$. To what temperature should it be heated,so that the r.m.s. velocity of the molecules be doubled (in $^{\circ} C$)?

At what temperature is the root mean square velocity of gaseous hydrogen molecules equal to that of oxygen molecules at $47^{\circ}C$ (in $; K$)?

An ideal gas in a closed container is heated so that the final rms speed of the gas particles increases by $2$ times the initial rms speed. If the initial gas temperature is $27^{\circ} C$, then the final temperature of the ideal gas is : (in $^{\circ} C$)

Out of hydrogen and air,which will have a higher root mean square speed $(v_{\text{rms}})$?

Let the r.m.s. velocity of a molecule of a given mass of gas be $C_{1}$ at temperature $27^{\circ} C$. When the temperature is increased to $327^{\circ} C$,the r.m.s. velocity is $C_{2}$. Then the ratio $\frac{C_{2}}{C_{1}}$ is

Six molecules of a gas in a container have speeds $2 \,m/s, 5 \,m/s, 3 \,m/s, 6 \,m/s, 3 \,m/s$ and $5 \,m/s$. The root mean square (r.m.s.) speed is: (in $\,m/s$)