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 \times 10^{-26} kg $ Boltzmann�s constant $k_B = 1.38 \times 10^{-23} J \, K^{-1}$)

• A. $1.254 \times 10^4 K $
• B. $2.508 \times 10^4 K $
• C. $5.016 \times 10^4 K $
• D. $8.360 \times 10^4 K $

Answer 1

The Correct Option is D

To determine the temperature at which the root mean square (rms) speed of oxygen molecules is sufficient for escaping Earth's atmosphere, we need to equate the rms speed to the escape speed of the Earth. Let's break down the solution step-by-step:

1. The escape speed (v_e) of Earth is given by: v_e = \sqrt{\frac{2GM}{R}}, where G = 6.674 \times 10^{-11} \, m^3 kg^{-1} s^{-2} (gravitational constant), M = 5.972 \times 10^{24} \, kg (mass of Earth), and R = 6.371 \times 10^6 \, m (radius of Earth).
2. Calculating the escape speed: v_e \approx \sqrt{\frac{2 \times 6.674 \times 10^{-11} \times 5.972 \times 10^{24}}{6.371 \times 10^6}} \approx 11186 \, m/s.
3. The rms speed (v_{rms}) of a gas molecule is given by the formula: v_{rms} = \sqrt{\frac{3k_BT}{m}}, where k_B = 1.38 \times 10^{-23} \, J K^{-1} (Boltzmann's constant), T is the temperature in Kelvin, and m = 2.76 \times 10^{-26} \, kg is the mass of an oxygen molecule.
4. Now equate v_{rms} to v_e: \sqrt{\frac{3k_BT}{m}} = v_e.
5. Square both sides to obtain: \frac{3k_BT}{m} = v_e^2.
6. Solve for T: T = \frac{mv_e^2}{3k_B}.
7. Substitute the values: T = \frac{2.76 \times 10^{-26} \times (11186)^2}{3 \times 1.38 \times 10^{-23}}.
8. Calculate the final value of T: T \approx \frac{2.76 \times (125156196)}{4.14 \times 10^{-23}} \approx 8.360 \times 10^4 \, K.

Thus, the temperature at which the rms speed of oxygen molecules becomes just sufficient for escaping from Earth's atmosphere is 8.360 \times 10^4 \, K, which matches option $8.360 \times 10^4 K $.