Question

Consider a sample of oxygen behaving like an ideal gas. At 300 K, the ratio of root mean square (rms) velocity to the average velocity of gas molecule would be: (Molecular weight of oxygen is 32 g/mol; R = 8.3 J K⁻¹ mol⁻¹)

• A. \(\sqrt{\frac{3\pi}{8}}\)
• B. \(\sqrt{\frac{8\pi}{3}}\)
• C. \(\sqrt{\frac{8}{3}}\)
• D. \(\sqrt{\frac{3}{4}}\)

Hint:

Molecular speed ratios are independent of temperature (\(T\)) and molar mass (\(M\)). The order is always \(v_{rms}>v_{avg}>v_{mp}\).

Answer 1

The Correct Option is A

Step 1: Recall the formulas for \(v_{rms}\) and \(v_{avg}\): \[v_{rms} = \sqrt{\frac{3RT}{M}}, \quad v_{avg} = \sqrt{\frac{8RT}{\pi M}}\]
Step 2: Take the ratio: \[\frac{v_{rms}}{v_{avg}} = \frac{\sqrt{\frac{3RT}{M}}}{\sqrt{\frac{8RT}{\pi M}}} = \sqrt{\frac{3RT}{M} \times \frac{\pi M}{8RT}} = \sqrt{\frac{3\pi}{8}}\]