Practice question
Question
How much heat is required to raise the temperature of 0.4 moles of a triatomic gas by 15 K at constant volume, with no vibrational modes? (R = 8.31 J mol⁻¹ K⁻¹)
Explanation
**Ideal gas law** P V = n R T governs gas laws, at constant pressure V ∝ T, so temperature increase 300 K→600 K doubles volume 24→48 L. Charles' law quantitative prediction V₂ = V₁×(T₂/T₁), illustrating direct proportionality, absolute temperature must be in kelvin. Triatomic gas: 6 degrees of freedom, C_v = 3 R.Q = μ C_v Δ T = 0.4 × 3 × 8.31 × 15 = 149.58 J ≈ 149.6 J. Substituting values gives 149.6 J, 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.
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