Two moles of an ideal gas (γ = 1.4) was allowed to expand reversibly and adiabatically from 1 L, 527 — Thermodynamics and Thermochemistry Chemistry Question

💡 Solution & Explanation

Initial temp T1 = 527 + 273 = 800 K. Initial vol V1 = 1 L, final vol V2 = 32 L. Γ = 1.4, so γ - 1 = 0.4. For reversible adiabatic expansion: T1 * V1^(γ-1) = T2 * V2^(γ-1) => 800 * 1^0.4 = T2 * 32^0.4 => 800 = T2 * (2^5)^0.4 = T2 * 2^2 = T2 * 4 => T2 = 200 K. Molar enthalpy change Δ H_m = $C_{p,m}$ * Δ T. Since γ = 1.4 (diatomic), $C_{p,m}$ = 3.5 * R. So Δ H_m = 3.5 R * (200 - 800) = 3.5 R * (-600) = -2100 R.