At what temperature is the rms speed of oxygen molecules 600 m/s? (Molecular mass of O₂ = 32 u, k_B = 1.38 × 10⁻²³ J K⁻¹
**RMS speed** v_rms = √(3 R T/M) = √(3 k_B T/m) where M molar mass (kg/mol), m molecular mass (kg), k_B=1.38×10⁻²/³ J/K, R=8.314 J/mol·K, T absolute temperature (K). Proportional to √T and 1/√M, lighter gases faster at same T, e.g., H₂ faster than O₂, temperature increase raises v_rms as √T. v_rms = √((3k_B T)/(m)), m = 32 × 10⁻³⁶.02 × 10²³ = 5.32 × 10⁻²⁶ kg.600² = 3 × 1.38 × 10⁻²/³ × T5.32 × 10⁻²⁶, T = 3.6 × 10⁵ × 5.32 × 10⁻²⁶/⁴.14 × 10⁻²/³ ≈ 463 K. Substituting values gives 463 K, which matches expected kinetic theory result, confirming mean free path
Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > RMS Speed and Temperature Dependence