JEE Main 2022 · 25 June · morning · Single correct · +4 / −1
The relation between root mean square speed (vᵣₘₛ) and most probable sped (vₚ) for the molar mass M of oxygen gas molecule at the temperature of 300 K will be :
- (a) vᵣₘₛ = √(2/3) vₚ
- (b) vᵣₘₛ = √(3/2) vₚ
- (c) vᵣₘₛ = vₚ
- (d) vᵣₘₛ = √(1/3) vₚ
Answer and solution
Answer (b) vᵣₘₛ = √(3/2) vₚ
Solution
Physics sir · 3 steps
Both speeds are √(kRT/M) with a different k, so divide them and T and M cancel.
- Write both speeds. For an ideal gas, vᵣₘₛ = √(3RT/M) and vₚ = √(2RT/M). They differ only in the number in front of RT/M.
vᵣₘₛ/vₚ = √((3RT/M)/(2RT/M)) = √(3/2)
- Data drops out. R, T and M cancel in the ratio, so the 300 K and the molar mass of oxygen do not change the answer. The same relation holds for every gas at every temperature.
- So. vᵣₘₛ = √(3/2) vₚ ≈ 1.22 vₚ, which is option (b).
The idea to keep. Remember the order vₚ : v_(avg) : vᵣₘₛ = √2 : √(8/π) : √3. The rms speed is always the largest, and any ratio of these speeds is independent of gas and temperature.
Answer from the NTA final answer key. Solution by PrepWiser.