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 ($m$) = $2.76 \times 10^{-26}\text{ kg}$, Boltzmann's constant $k_B = 1.38 \times 10^{-23}\text{ J K}^{-1}$) (2018)
$5.016 \times 10^4\text{ K}$
$8.360 \times 10^4\text{ K}$
$2.508 \times 10^4\text{ K}$
$1.254 \times 10^4\text{ K}$
Solution:
Escape velocity $v_e = 11.2\text{ km/s} = 11200\text{ m/s}$. RMS speed $v_{rms} = \sqrt{\frac{3k_BT}{m}}$. Equating them: $11200 = \sqrt{\frac{3 \times 1.38 \times 10^{-23} \times T}{2.76 \times 10^{-26}}}$. Solving for $T$ gives $T = 8.360 \times 10^4\text{ K}$.
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