Two moles of an ideal gas () are compressed adiabatically against a constant external pressure of 2 — Thermodynamics and Thermochemistry Chemistry Question

💡 Solution & Explanation

Since the process is adiabatic, $q = 0$ and $W = \Delta U = nC_v(T_2 - T_1) = 2(2.5R)(T_2 - 350) = 5R(T_2 - 350)$. Irreversible work is also $-P_{ext}(V_2 - V_1) = -2(\frac{nRT_2}{P_2} - \frac{nRT_1}{P_1}) = -2R(T_2 - 700)$. Equating the two expressions: $5R(T_2 - 350) = -2R(T_2 - 700)$. Solving gives $7T_2 = 3150 \Rightarrow T_2 = 450$ K. Finally, $W = 5R(450 - 350) = 500R$.