At what temperature will the rms speed of oxygen molecules become just sufficient for escaping from the earth'

Question

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 $m = 2.76 	imes 10^{-26} 	ext{ kg}$, Boltzmann's constant $k_B = 1.38 	imes 10^{-23} 	ext{ J K}^{-1}$)

Accepted Answer

To escape Earth's atmosphere, the molecule's root mean square (rms) speed must equal the Earth's escape velocity.

1.  **Formulas:**
    *   RMS Speed ($v_{rms}$) = $\sqrt{\frac{3 k_B T}{m}}$
    *   Escape Velocity ($v_{e}$) for Earth $\approx 11.2 	ext{ km/s} = 11200 	ext{ m/s}$

2.  **Set up the equation:**
    $$ \sqrt{\frac{3 k_B T}{m}} = v_{e} $$
    Squaring both sides:
    $$ \frac{3 k_B T}{m} = v_{e}^2 $$
    $$ T = \frac{m v_{e}^2}{3 k_B} $$

3.  **Substitute values:**
    *   $m = 2.76 	imes 10^{-26} 	ext{ kg}$
    *   $v_{e} = 11200 	ext{ m/s}$
    *   $k_B = 1.38 	imes 10^{-23} 	ext{ J K}^{-1}$

$$ T = \frac{(2.76 	imes 10^{-26}) 	imes (11200)^2}{3 	imes (1.38 	imes 10^{-23})} $$
    $$ T = \frac{2.76 	imes 1.2544 	imes 10^8 	imes 10^{-26}}{4.14 	imes 10^{-23}} $$
    $$ T = \frac{3.462 	imes 10^{-18}}{4.14 	imes 10^{-23}} $$
    $$ T \approx 0.8362 	imes 10^5 	ext{ K} $$
    $$ T \approx 8.362 	imes 10^4 	ext{ K} $$

This matches Option 2.

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