Practice question
Question
At what temperature is the rms speed of oxygen molecules 964 m/s? (Molecular mass of O₂ = 32 u, k_B = 1.38 × 10⁻²³ J K⁻¹)
Explanation
**Specific heat relation** C_p - C_v = R for ideal gas per mole, Mayer's relation, due to work done at constant pressure, degrees of freedom include translational, rotational, vibrational, each quadratic term contributes ½ R to C_v. v_rms = √((3k_B T)/(m)), m = 32 × 10⁻³⁶.02 × 10²³ = 5.32 × 10⁻²⁶ kg.964² = 3 × 1.38 × 10⁻²/³ × T5.32 × 10⁻²⁶, T = 9.29 × 10⁵ × 5.32 × 10⁻²⁶/⁴.14 × 10⁻²/³ ≈ 1194 K. Substituting values gives 1200 K, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π d²), ideal gas law P V = n
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