Two moles of an ideal gas ( = 5/2 R) was compressed adiabatically against constant pressure of 2 atm — Thermodynamics and Thermochemistry Chemistry Question

💡 Solution & Explanation

Adiabatic compression against constant external pressure P_ext = 2 atm. Initial state: T1 = 350 K, P1 = 1 atm, n = 2. V1 = nRT1/P1 = 2 * R * 350 / 1 = 700 R. Δ U = w => n * $C_{v,m}$ * (T2 - T1) = -P_ext * (V2 - V1). Since V2 = nRT2/P2 = 2 * R * T2 / 2 = R * T2, we have: 2 * (5/2 R) * (T2 - 350) = -2 * (R * T2 - 700 R). Dividing by R: 5 * (T2 - 350) = -2 * T2 + 1400 => 5 T2 - 1750 = -2 T2 + 1400 => 7 T2 = 3150 => T2 = 450 K. Work done w = Δ U = n * $C_{v,m}$ * (T2 - T1) = 2 * (5/2 R) * (450 - 350) = 5 R * 100 = 500 R.