Two moles of an ideal gas [] is heated at constant pressure from to . The amount of heat absorbed by — Thermodynamics and Thermochemistry Chemistry Question

💡 Solution & Explanation

$q_p = \Delta H = n \int_{T_1}^{T_2} C_{p,m} dT$. $C_{p,m} = C_{v,m} + R = 20 + 0.01T + 8.314 = 28.314 + 0.01T$. $T_1 = 300 \text{ K}, T_2 = 400 \text{ K}$. Molar heat $\int_{300}^{400} (28.314 + 0.01T) dT = 28.314(100) + 0.005(400^2 - 300^2) = 2831.4 + 0.005(70000) = 2831.4 + 350 = 3181.4 \text{ J/mol}$. Total heat for 2 moles $= 2 \times 3181.4 = 6362.8 \text{ J}$.