The rms speed of an oxygen molecule in a vessel at a particular temperature is $\left(1+\frac{5}{x}\right)^{\frac{1}{2}} v$,where $v$ is the average speed of the molecule. The value of $x$ will be: (Take $\pi=\frac{22}{7}$)

If the rms velocity of hydrogen gas at a certain temperature is $c,$ then the rms velocity of oxygen gas at the same temperature is

The molecular mass of a gas having $r.m.s.$ speed four times as that of another gas having molecular mass $32$ is

Consider an ideal gas with the following distribution of speeds:

Speed $(m/s)$	$\%$ of molecules
$200$	$10$
$400$	$20$
$600$	$40$
$800$	$20$
$1000$	$10$

$(a)$ Calculate $v_{rms}$ and hence $T$. (Given mass of one molecule $m = 3.0 \times 10^{-26} \ kg$, Boltzmann constant $k_B = 1.38 \times 10^{-23} \ J/K$)
$(b)$ If all the molecules with speed $1000 \ m/s$ escape from the system, calculate the new $v_{rms}$ and hence the new $T$.

The r.m.s. speed of gas at temperature $T$ is $2$ times the r.m.s speed at $320 \,K$. The value of the temperature $T$ is (in $\,K$)

Equal volumes of two gases are kept in different containers having densities in the ratio $1:16$. They exert equal pressures on the walls of their respective containers. Then the ratio of their r.m.s. velocities is