At what temperature will the root mean square | vedclass

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(B) The root mean square $(rms)$ speed of a gas molecule is given by $v_{rms} = \sqrt{\frac{3k_B T}{m}}$.For the molecules to escape the Earth's atmos

Vedclass · Accepted Answer

$8.360 	imes 10^4 \, K$

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(B) The root mean square $(rms)$ speed of a gas molecule is given by $v_{rms} = \sqrt{\frac{3k_B T}{m}}$.For the molecules to escape the Earth's atmosphere,their $v_{rms}$ must be equal to the escape velocity of the Earth,$v_e = 11.2 \, km/s = 1.12 	imes 10^4 \, m/s$.Equating $v_{rms} = v_e$,we get $\sqrt{\frac{3k_B T}{m}} = v_e$.Squaring both sides: $\frac{3k_B T}{m} = v_e^2$.Solving for $T$: $T = \frac{m v_e^2}{3k_B}$.Substituting the given values: $T = \frac{(2.76 	imes 10^{-26} \, kg) 	imes (1.12 	imes 10^4 \, m/s)^2}{3 	imes 1.38 	imes 10^{-23} \, JK^{-1}}$.$T = \frac{2.76 	imes 10^{-26} 	imes 1.2544 	imes 10^8}{4.14 	imes 10^{-23}}$.$T = \frac{3.462144 	imes 10^{-18}}{4.14 	imes 10^{-23}} \approx 0.83626 	imes 10^5 \, K = 8.3626 	imes 10^4 \, K$.Rounding to the nearest option,we get $8.360 	imes 10^4 \, K$.

At what temperature will the root mean square $(rms)$ speed of oxygen molecules be sufficient to escape from the Earth's atmosphere? (Mass of an oxygen molecule $m = 2.76 \times 10^{-26} \, kg$,Boltzmann constant $k_B = 1.38 \times 10^{-23} \, JK^{-1}$,Escape velocity of Earth $v_e = 11.2 \, km/s$)

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