At what temperature will the rms speed of oxygen molecules become just sufficient for escaping from the Earth's atmosphere? [2018]

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Q.
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 $(m) = 2.76 \times 10^{- 26} kg$ , Boltzmann’s constant $k_{B} = 1.38 \times 10^{- 23} {J K}^{- 1}$ ) [2018]

1 $2.508 \times 10^{4}$ $K$

2 $8.360 \times 10^{4}$ $K$

3 $5.016 \times 10^{4}$ $K$

4 $1.254 \times 10^{4}$ $K$

Ans.
(2)

Escape velocity from the Earth's surface is

$v_{escape} = 11200 m$ $s^{- 1}$

Say at temperature $T$ , oxygen molecule attains escape velocity.

So, $v_{escape} = \sqrt{\frac{3 k_{B} T}{m_{O_{2}}}} \Rightarrow 11200 = \sqrt{\frac{3 \times 1.38 \times 10^{- 23} \times T}{2.76 \times 10^{- 26}}}$

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

Solution

Q.
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 $(m) = 2.76 \times 10^{- 26} kg$ , Boltzmann’s constant $k_{B} = 1.38 \times 10^{- 23} {J K}^{- 1}$ ) [2018]

1 $2.508 \times 10^{4}$ $K$

2 $8.360 \times 10^{4}$ $K$

3 $5.016 \times 10^{4}$ $K$

4 $1.254 \times 10^{4}$ $K$

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Solution

Q. 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)=2.76×10-26kg, Boltzmann’s constant kB=1.38×10-23J K-1) [2018]

1 2.508×104 K

2 8.360×104 K

3 5.016×104 K

4 1.254×104 K

Ans. (2) Escape velocity from the Earth's surface is vescape=11200m s-1 Say at temperature T, oxygen molecule attains escape velocity. So, vescape=3kBTmO2⇒11200=3×1.38×10-23×T2.76×10-26 On solving, T=8.360×104 K

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Q.
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 $(m) = 2.76 \times 10^{- 26} kg$ , Boltzmann’s constant $k_{B} = 1.38 \times 10^{- 23} {J K}^{- 1}$ ) [2018]

1 $2.508 \times 10^{4}$ $K$

2 $8.360 \times 10^{4}$ $K$

3 $5.016 \times 10^{4}$ $K$

4 $1.254 \times 10^{4}$ $K$