The molecular masses of helium and oxygen are $4$ and $32$ respectively. The ratio of r.m.s speed of helium at $327^{\circ} C$ to r.m.s speed of oxygen at $27^{\circ} C$ will be

$A$ mixture of $2$ moles of helium gas (atomic mass $= 4 \ amu$) and $1$ mole of argon gas (atomic mass $= 40 \ amu$) is kept at $300 \ K$ in a container. The ratio of the rms speeds $\left(\frac{v_{\text{rms}} \text{ (helium)}}{v_{\text{rms}} \text{ (argon)}}\right)$ is:

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

At a given temperature,if $V_{rms}$ is the root mean square velocity of the molecules of a gas and $V_s$ is the velocity of sound in it,then these are related as $\left( \gamma = \frac{C_P}{C_v} \right)$:

$A$ 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 r.m.s. speed of the molecules of a gas at $100^{\circ} C$ is $v$. The temperature at which the r.m.s. speed will be $\sqrt{3} v$ is: (in $^{\circ} C$)