An ideal gas is expanded irreversibly from 5 L to 10 L against a constant external pressure of 1 bar — Thermodynamics and Thermochemistry Chemistry Question

💡 Solution & Explanation

Isenthalpic process for ideal gas means Δ H = 0, which implies Δ T = 0 and Δ U = 0. According to the First Law: Δ U = q + w => 0 = q + w => q = -w. w = -P_ext * Δ V = -1 bar * (10 - 5) L = -5 L-bar = -5 * 100 J = -500 J. Therefore, q = +500 J.