Practice question

**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 = 2 × 10⁻³⁶.02 × 10²³ = 3.32 × 10⁻²⁷ kg.2000² = 3 × 1.38 × 10⁻²/³ × T3.32 × 10⁻²⁷, T = 4 × 10⁶ × 3.32 × 10⁻²⁷/⁴.14 × 10⁻²/³ ≈ 321 K. Substituting values gives 321 K, which matches expected kinetic theory result, confirming mean free path