Practice question
Question
What is the temperature at which the rms speed of nitrogen molecules is 1032 m/s? (Molecular mass of N₂ = 28 u, k_B = 1.38 × 10⁻²³ J K⁻¹)
Explanation
**Gas laws** Boyle, Charles, Gay-Lussac are special cases of ideal gas equation, for constant pressure volume-temperature relation V ∝ T, for constant temperature pressure-volume inverse, for constant volume pressure-temperature direct, enabling calculation of new volume from temperature ratio. v_rms = √((3k_B T)/(m)), m = 28 × 10⁻³⁶.02 × 10²³ = 4.65 × 10⁻²⁶ kg.1032² = 3 × 1.38 × 10⁻²/³ × T4.65 × 10⁻²⁶, T = 1.065 × 10⁶ × 4.65 × 10⁻²⁶/⁴.14 × 10⁻²/³ ≈ 1196 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 =
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