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 32g/mol; R = 8.3 J K⁻¹ mol⁻¹)

Solution

Physics sir · 4 steps

Both speeds have the same √(RT/M), so divide them and only the numbers in front survive.

1. Write both speeds. For an ideal gas, vᵣₘₛ = √(3RT/M) and v_(avg) = √(8RT/(π M)).

   vᵣₘₛ/v_(avg) = √((3RT/M)/(8RT/(π M))) = √(3π/8)

2. Data not needed. R, T and M cancel completely. The ratio is the same for every ideal gas at every temperature, so the 300 K, 32 g/mol and 8.3 are only there to distract you.
3. Quick check. √(3π/8) ≈ √1.18 ≈ 1.09. That is just above 1, as expected, because the rms speed is always slightly larger than the average speed.
4. So. The ratio is √(3π/8), option (a).

The idea to keep. Remember vᵣₘₛ : v_(avg) : vₘₚ = √3 : √(8/π) : √2. Any ratio of these speeds is a pure number, independent of gas and temperature.

Answer from the NTA final answer key. Solution by PrepWiser.