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 $k_{B} = 1.38 \times 10^{- 23} J K^{- 1}$ )

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 \times 10^{- 26}

kg
Boltzmann's constant

k_{B} = 1.38 \times 10^{- 23} J K^{- 1}

)

Toppr Admin · Accepted Answer

Since- Vesc-11-2-xA0-km-sR-M-S- - -x221A-3KTM-xA0-x221A-3KTM-11200T-112002-xD7-2-76-xD7-10-x2212-263-xD7-1-38-xD7-10-x2212-23T-8-360-xD7-104K

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)=2.76×10−26kgBoltzmann's constant kB=1.38×10−23JK−1)2.508×104K5.016×104K8.360×104K1.254×104K

Since, Vesc=11.2 km/sR.M.S. = √3KTM √3KTM=11200T=112002×2.76×10−263×1.38×10−23T=8.360×104K

Since, $V_{e s c} = 11.2 k m / s$

R.M.S. = $\sqrt{\frac{3 K T}{M}}$

$\sqrt{\frac{3 K T}{M}} = 11200$

$T = \frac{11200^{2} \times 2.76 \times 10^{- 26}}{3 \times 1.38 \times 10^{- 23}}$

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