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

(Given: Mass of oxygen molecule CM =2.76 \times 10^{-26} \mathrm{Kg} \text { Boltzmate }, rxn constant \mathrm{Kg}=1.38 \times 10^{-23} \mathrm{~J} \cdot \mathrm{K}^{-1})

Option: 1

5.016 \times 10^4 \mathrm{~K}


Option: 2

8.360 \times 10^4 k


Option: 3

2.508 \times 10^4 k


Option: 4

1.254 \times 10^4 k


Answers (1)

best_answer

Escape velocity of a body for esrth,

V_e=\sqrt{28 R}=\sqrt{2 \times 9.8 \times 6400 \times 10^3}=11200 \mathrm{M} / \mathrm{S}

Let us consider, at temperature TK, escape velocity of the oxygen molecule will be (V_e). Then,

\begin{aligned} & \sqrt{\frac{3 K_B T}{m}}=V_e \\ & \text { or, } \frac{3 \times 1.38 \times 10^{-23}}{2.76 \times 10^{-26}} \cdot T=(11200)^2 \end{aligned}

\text { or, } T=8.363 \times 10^4 \mathrm{~K} 

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