Question

Consider a sample of oxygen behaving like an Ideal gas. At $300\mathrm{K}$, the ratio of root mean square ($\mathrm{rms}$) velocity to the average velocity of a gas molecule will be: (Molecular weight of oxygen is $32\mathrm{g}{\mathrm{mol}}^{-1}$; $\mathrm{R}=8.3\mathrm{J}{\mathrm{K}}^{-1}{\mathrm{mol}}^{-1}$)

• A. $\sqrt{\frac{3}{3}}$
• B. $\sqrt{\frac{8}{3}}$
• C. $\sqrt{\frac{3\mathrm{\pi }}{8}}$
• D. $\sqrt{\frac{8\mathrm{\pi }}{3}}$

Answer 1

The correct option is C

$\sqrt{\frac{3\mathrm{\pi }}{8}}$

Explanation for the correct option:

C) RMS velocity, ${\mathrm{v}}_{\mathrm{rms}}=\sqrt{\frac{3\mathrm{RT}}{\mathrm{M}}}$

where,

$\mathrm{R}$ is the Universal gas constant

$\mathrm{T}$ is the temperature

$\mathrm{M}$ is the molar mass

Average velocity, ${\mathrm{v}}_{\mathrm{avg}}=\sqrt{\frac{8\mathrm{RT}}{\mathrm{\pi M}}}$

Now,

$\frac{{\mathrm{v}}_{\mathrm{rms}}}{{\mathrm{v}}_{\mathrm{avg}}}=\frac{\sqrt{{\displaystyle \frac{3\mathrm{RT}}{\mathrm{M}}}}}{\sqrt{{\displaystyle \frac{8\mathrm{RT}}{\mathrm{\pi M}}}}}$

$\mathbf{=}\mathbf{}\sqrt{\frac{\mathbf{3}\mathbf{\pi }}{\mathbf{8}}}$

Hence option (C) is correct.