Question:
Nitrogen gas is at $300^{\circ} \mathrm{C}$ temperature. The temperature (in $\mathrm{K}$ ) at which the $\mathrm{rms}$ speed of a $\mathrm{H}_{2}$ molecule would be equal to the rms speed of a nitrogen molecule, is_______
(Molar mass of $\mathrm{N}_{2}$ gas $28 \mathrm{~g}$ );
Solution:
(41)
Room mean square speed is given by
$v_{r m s}=\sqrt{\frac{3 R T}{M}}$
$T=$ temperature of the gas molecule
We have given $v_{\mathrm{N}_{2}}=v_{\mathrm{H}_{2}}$
$\therefore \sqrt{\frac{3 R T_{\mathrm{N}_{2}}}{\mathrm{M}_{\mathrm{N}_{2}}}}=\sqrt{\frac{3 R T_{\mathrm{H}_{2}}}{M_{\mathrm{H}_{2}}}}$
$\Rightarrow \frac{T_{\mathrm{H}_{2}}}{2}=\frac{573}{28} \Rightarrow T_{\mathrm{H}_{2}}=41 \mathrm{~K}$