At what temperature will the rms speed of oxy | vedclass

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(B) The escape velocity from the Earth's surface is $v_{	ext{escape}} = 11200 \, m/s$.For the oxygen molecules to escape,their $rms$ speed must be eq

Vedclass · Accepted Answer

$8.360 	imes 10^4 \, K$

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(B) The escape velocity from the Earth's surface is $v_{	ext{escape}} = 11200 \, m/s$.For the oxygen molecules to escape,their $rms$ speed must be equal to the escape velocity:$v_{	ext{rms}} = \sqrt{\frac{3 k_B T}{m}} = v_{	ext{escape}}$Substituting the given values:$11200 = \sqrt{\frac{3 	imes 1.38 	imes 10^{-23} 	imes T}{2.76 	imes 10^{-26}}}$Squaring both sides:$(11200)^2 = \frac{3 	imes 1.38 	imes 10^{-23} 	imes T}{2.76 	imes 10^{-26}}$$1.2544 	imes 10^8 = \frac{4.14 	imes 10^{-23} 	imes T}{2.76 	imes 10^{-26}}$$1.2544 	imes 10^8 = 1.5 	imes 10^3 	imes T$Solving for $T$:$T = \frac{1.2544 	imes 10^8}{1.5 	imes 10^3} \approx 8.360 	imes 10^4 \, K$.

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 $k_B = 1.38 \times 10^{-23} \, JK^{-1}$)

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