The effect of temperature on Maxwell's speed distribution is correctly shown by

If the oxygen $(O_2)$ has a root mean square velocity of $C \ m/s$,then the root mean square velocity of hydrogen $(H_2)$ at the same temperature will be:

Let $\bar{v}$,${v_{rms}}$ and ${v_p}$ respectively denote the mean speed,root mean square speed,and most probable speed of the molecules in an ideal monoatomic gas at absolute temperature $T$. The mass of a molecule is $m$. Then

The $N_2$ molecule is $14$ times heavier than the $H_2$ molecule. At what temperature will the $rms$ speed of $H_2$ molecules be equal to the $rms$ speed of $N_2$ molecules at $27^{\circ}C$?

The absolute temperature of a gas is determined by

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 $T$.