In Maxwell's speed distribution curve,for N_2 | vedclass

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(B) The relative velocity between two molecules with velocity $V$ is given by $\left|V_{	ext{rel}}ight| = \sqrt{V^2 + V^2 - 2V^2 \cos 	heta} = 2V

Vedclass · Accepted Answer

$606$

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(B) The relative velocity between two molecules with velocity $V$ is given by $\left|V_{	ext{rel}}ight| = \sqrt{V^2 + V^2 - 2V^2 \cos 	heta} = 2V \sin(	heta/2)$.The average relative velocity is $\langle V_{	ext{rel}} angle = \frac{\int_{0}^{\pi} 2V \sin(	heta/2) d	heta}{\int_{0}^{\pi} d	heta} = \frac{4V}{\pi}$.Since the average velocity of a molecule is $\langle V angle = \sqrt{\frac{8RT}{\pi M}}$,the average relative velocity is $\langle V_{	ext{rel}} angle = \frac{4}{\pi} \sqrt{\frac{8RT}{\pi M}}$.Given $R = 8.314 \, J/(mol \cdot K)$,$T = 300 \, K$,and $M = 28 	imes 10^{-3} \, kg/mol$ for $N_2$:$\langle V_{	ext{rel}} angle = \frac{4}{\pi} \sqrt{\frac{8 	imes 8.314 	imes 300}{\pi 	imes 28 	imes 10^{-3}}} \approx \frac{4}{3.14} 	imes 476.5 \approx 606 \, m/s$.

In Maxwell's speed distribution curve,for $N_2$ gas,the average relative velocity (in $m/s$) between two molecules at $300 \, K$ is:

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