Question

In Maxwell's speed distribution curve, for N₂ gas, the average of |relative velocity| between two molecules at 300 k will be :-

• A. 300 m/sec
• B. 610 m/sec
• C. 920 m/sec
• D. zero

Answer 1

Answer: (B)

$\left|V_{rel}\right|=\sqrt{V^2+V^2-2\left(V\right)\left(V\right)cos\theta }=2V\left|sin\frac{\theta }{2}\right|$
$⟨V_{rel}⟩$=$\frac{{\int }_{{}_{{}_0}}^{{}^{{}^{\pi }}}2V1|sin\frac{\theta }{2}|d\theta }{{\int }_{{}_{{}_0}}^{{}^{{}^{\pi }}}d\theta }=\frac{4V}{\pi }$
$⟨V_{rel}⟩$=$\frac{4}{\pi }{V}_{average}=\frac{4}{\pi }\sqrt{\frac{8RT}{\pi m_0}}=\frac{4}{\pi }\sqrt{\frac{8\times 8.3\times 300}{3.14\times 28\times 10^{-3}}}=606m/sec$