First slide

Rms speed

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) = 2.76×10^–26 kg and Boltzmann’s constant k =1.38×10^–23 JK^–1)

Easy

A

2.508×10⁴ K
B

8.360×10⁴ K
C

1.254×10⁴ K
D

5.016×10⁴ K

Solution

V_esc= Vrms
$\Rightarrow \,11.2 \times {10^3}\, = \,\sqrt {\frac{{3kT}}{m}}$
$T = {(11.2 \times 10^{3})}^{2} \frac{m}{3 k}$
=
${(11.2 \times {10^3})^2} \times \frac{{2.76 \times {{10}^{ - 26}}}}{{3 \times 1.38 \times {{10}^{ - 23}}}}\, = \,8.36 \times {10^4}K$

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