If the temperature of gas molecules is raised from $127^{\circ} C$ to $527^{\circ} C$,the ratio of the r.m.s. speed of the molecules is respectively:

$A$ vessel is partitioned into two equal halves by a fixed diathermic separator. Two different ideal gases are filled in the left $(L)$ and right $(R)$ halves. The rms speed of the molecules in the $L$ part is equal to the mean speed of the molecules in the $R$ part. Then the ratio of the mass of a molecule in the $L$ part to that of a molecule in the $R$ part is

The $rms$ speed of an ideal diatomic gas at temperature $T$ is $v$. When the gas dissociates into atoms,its new $rms$ speed becomes double. At what temperature does the dissociation of the gas occur?

The molecules of a given mass of a gas have a $r.m.s.$ velocity of $200 \, m/s$ at $27^{\circ}C$ and $1.0 \times 10^5 \, N/m^2$ pressure. When the temperature is $127^{\circ}C$ and pressure is $0.5 \times 10^5 \, N/m^2$,the $r.m.s.$ velocity in $m/s$ will be

When the temperature of a gas is raised from $27^{\circ} C$ to $90^{\circ} C$,the increase in the rms velocity of the gas molecules is: (in $\%$)

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