SATHEE: Kinetic Theory - Result Question 12

Kinetic Theory - Result Question 12

12. 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 $(m) = 2.76 \times 10^{- 26} k g$ Boltzmann’s constant $. k_{B} = 1.38 \times 10^{- 23} J K^{- 1})$

[2018]

(a) $2.508 \times 10^{4} K$

(b) $8.360 \times 10^{4} K$

(c) $1.254 \times 10^{4} K$

(d) $5.016 \times 10^{4} K$

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Answer:

Correct Answer: 12. (b)

Solution:

(b) Let at temperature $T r m s$ speed of oxygen molecules become just sufficient for escaping from the Earth’s atmosphere

$V_{escape} = 11200 m / s$

Also, $V_{r m s} = V_{escape} = \sqrt{\frac{3 k_{B} T}{m_{O_{2}}}} = 11200 m / s$

Putting value of $K_{B}$ and $m_{O_{2}}$ we get,

$T = 8.360 \times 10^{4} K$

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Kinetic Theory - Result Question 12

12. At what temperature will the rms speed of oxygen molecules become just sufficient for escaping from the Earth’s atmosphere?

Answer:

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