Practice question

**Pressure-temperature relation** at constant volume Gay-Lussac law P ∝ T, for V constant, P₁/T₁ = P₂/T₂, if T doubles from 300 K to 600 K P doubles, e.g., P₁=1 atm at 300 K P₂=2 atm at 600 K, no work done, ΔU = n C_v ΔT = Q. For solids, C ≈ 3R , but here C_v = 24.4 , and Δ V ≈ 0 , so C_p ≈ C_v .However, typically C_p - C_v = R , but for solids in PDF context, C is given directly.Since C = 24.4 is molar specific heat, C_p ≈ C_v = 24.4 J mol⁻¹ K⁻¹ (negligible Δ