Skip to content

Question

A gas mixture has equal numbers of oxygen and argon molecules at 300 K. What is the ratio of their rms speeds? (Molecular mass: O₂ = 32 u, Ar = 39.9 u)

Options

Choose one · Correct answer highlighted

Explanation

**Molecular mass and density** relation ρ = P M/(R T) allows density calculation, ideal gas law also P = n k_B T where n number density, molecular mass determines mass per molecule, density increases with pressure and decreases with temperature, inverse T dependence. v_rms ∝ (1)/(√(m)), v_O₂v_Ar = √(m_Ar)m_O₂.v_O₂v_Ar = √((39.9)/(32)) ≈ √(1.247) ≈ 1.117. Substituting values gives 1.12:1, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π d²), ideal gas law P V = n R T and v_rms = √(3 R T/M) relations.

Discussion

Comments

0 comments

No comments yet. Be the first to start the discussion.