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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 kg2.76 \times 10^{-26} \text{ kg}, Boltzmann's constant kB=1.38×10−23 JK−1k_B = 1.38 \times 10^{-23} \text{ JK}^{-1})

A

5.016×104 K5.016 \times 10^4 \text{ K}

B

8.360×104 K8.360 \times 10^4 \text{ K}

C

2.508×104 K2.508 \times 10^4 \text{ K}

D

1.254×104 K1.254 \times 10^4 \text{ K}

Step-by-Step Solution

The escape velocity Vescape=11200 m/sV_{escape} = 11200 \text{ m/s}. The rms speed is given by 3kBTm=11200 m/s\sqrt{\frac{3k_B T}{m}} = 11200 \text{ m/s}. Solving for T gives T=8.360×104 KT = 8.360 \times 10^4 \text{ K}.

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