Question #160590
At what temperature would the average speed of helium atoms equal the escape speed from the Earth, 1.12 x 10^4 m/s, the mass of a helium atom being = 6.64310227 kg
1
Expert's answer
2021-02-07T19:17:26-0500

The average speed of helium atoms can be found from the formula:


vavg,He=8kBTπm.v_{avg,He}=\sqrt{\dfrac{8k_BT}{\pi m}}.

In order to find the temperature at which the average speed of helium atoms equals the escape speed from the Earth we must equate vavg,Hev_{avg,He} and vescv_{esc}:


vesc=8kBTπm,v_{esc}=\sqrt{\dfrac{8k_BT}{\pi m}},T=πmvesc28kB,T=\dfrac{\pi mv_{esc}^2}{8k_B},T=π6.641027 kg(1.12104 ms)281.381023 JK=2.37104 K.T=\dfrac{\pi\cdot6.64\cdot10^{-27}\ kg\cdot(1.12\cdot10^4\ \dfrac{m}{s})^2}{8\cdot1.38\cdot10^{-23}\ \dfrac{J}{K}}=2.37\cdot10^4\ K.

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