JEE Main · 2019 · 08 Apr 2019 · Shift 2 · Heat & Thermodynamics

Set the root-mean-square speed of a hydrogen molecule equal to the escape speed from earth. The rms speed is $v_{rms}=\sqrt{\frac{3RT}{m}}$ and the escape speed is $v_{esc}=\sqrt{2g{R}_e}$. Equating the squares, $\frac{3RT}{m}=2g{R}_e$, so $T=\frac{2g{R}_{e}m}{3R}$, where $m$ is the molar mass of hydrogen, $2\times 10^{-3}\text{kg}$. Putting in $g=10$, $R_e=6.4\times 10^6\text{m}$ and $R=k_B{N}_A$, the temperature works out to roughly $10^4\text{K}$. Such a huge value reflects how fast a very light molecule would have to move to escape earth's gravity.