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 $T = 293 \, K$]

Let $A$ and $B$ be two gases and given: $\frac{T_A}{M_A} = 4 \cdot \frac{T_B}{M_B}$,where $T$ is the temperature and $M$ is the molecular mass. If $C_A$ and $C_B$ are the $r.m.s.$ speeds,then the ratio $\frac{C_A}{C_B}$ will be equal to:

The root mean-square $(rms)$ speed of oxygen molecules $(O_2)$ at a certain absolute temperature is $\upsilon$. If the temperature is doubled and the oxygen gas dissociates into atomic oxygen,the $rms$ speed would be

The molecules of a given mass of a gas have root mean square speeds of $100 \ m/s$ at $27^{\circ} \ C$ and $1.00 \ \text{atm}$ pressure. What will be the root mean square speeds of the molecules of the gas at $127^{\circ} \ C$ and $2.0 \ \text{atm}$ pressure?

$A$ 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:

What is the ${v_{rms}}$ of gas molecules in equilibrium?