# Nitrogen gas at $300^{\circ}$temp. The temp(in K) at which the rms speed of $H_{2}$  molecule would be equal to the rms speed of a nitrogen molecule is (Molar mass of $N_{2}$ gas is 28gm). Upto 2 digit after decimal Option: 1 40.94 Option: 2 81 Option: 3 32 Option: 4 273

$V_{rms}=\sqrt{\frac{3RT}{M}}$

For N2 $T_1=300+273.15=573.15 K \ and \ \ M_1=28$

For H2 $\frac{V_{N_2}}{V_{H_2}}=1=\sqrt{\frac{T_1}{T_2}*\frac{M_2}{M_1}}=\sqrt{\frac{573.15}{T}*\frac{2}{28}}\\T=40.94 K$

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