Question:
Nitrogen gas is at $300^{\circ} \mathrm{C}$ temperature. The temperature (in $\mathrm{K}$ ) at which the 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:
$\mathrm{V}_{\mathrm{rms}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}}}$
$\mathrm{V}_{\mathrm{N}_{2}}=\mathrm{V}_{\mathrm{H}_{2}}$'
$\sqrt{\frac{3 R T_{N_{2}}}{M_{N_{2}}}}=\sqrt{\frac{3 R T_{H_{2}}}{M_{H_{2}}}}$'
$\frac{573}{28}=\frac{\mathrm{T}_{\mathrm{H}_{2}}}{2} \Rightarrow \mathrm{T}_{\mathrm{H}_{2}}=40.928$
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