Q 11-12-120JEE MainJEE Main 2020 (5 Sep, Shift 2)Easy
Nitrogen gas is at $300\ ^\circ\text{C}$ temperature. The temperature (in K) at which the rms speed of an $\text{H}_2$ molecule would be equal to the rms speed of a nitrogen molecule is ______. (Molar mass of $\text{N}_2$ gas is $28$ g)
Numerical value type. Enter your answer.
Answer: 41
$v_{\text{rms}} = \sqrt{3RT/M}$, so equal speeds need equal $T/M$.
$T_{N_2} = 300 + 273 = 573$ K.
$$\frac{T_{H_2}}{2} = \frac{573}{28} \Rightarrow T_{H_2} = \frac{573}{14} \approx 41\ \text{K}$$
Solution by Sreeraj P, M.Sc Physics