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Question

Physics Question on kinetic theory

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)=2.76×1026kg(m) = 2.76 \times 10^{-26} kg Boltzmann�s constant kB=1.38×1023JK1k_B = 1.38 \times 10^{-23} J \, K^{-1})

A

1.254×104K1.254 \times 10^4 K

B

2.508×104K2.508 \times 10^4 K

C

5.016×104K5.016 \times 10^4 K

D

8.360×104K8.360 \times 10^4 K

Answer

8.360×104K8.360 \times 10^4 K

Explanation

Solution

Vescape =11200m/sV_{\text{escape }} = 11200\, m/s
Say at temperature T it attains VescapeV_{\text{escape}}
So, 3kBTmo2=11200m/s \sqrt{\frac{3k_{B}T}{m_{o_2} }} =11200 \,m/s
On solving,
T=8.360×104KT = 8.360 \times 10^{4} \,K