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At what temperature will the rms speed of oxygen molecules become just sufficient for escaping from the Earth's atmosphere? (Given: Mass of oxygen molecule (m) $= 2.76 \times 10^{-26}$ kg, Boltzmann's constant $k_B = 1.38 \times 10^{-23}\ J K^{-1}$)
Explanation
Set $\sqrt{3k_BT/m}$ equal to the escape speed 11.2 km/s and solve for T.
Detailed Solution
$V_{escape} = 11200$ m/s
Say at temperature T the rms speed attains $V_{escape}$.
So, $\sqrt{\frac{3k_BT}{m_{O_2}}} = 11200$ m/s
On solving, $T = \frac{(11200)^2 \times 2.76\times10^{-26}}{3\times 1.38\times10^{-23}} = 8.360 \times 10^{4}$ K