A real gas is subjected to an adiabatic process from (2 bar, 40 L, 300 K) to (4 bar, 30 L, 300 K) ag — Thermodynamics and Thermochemistry Chemistry Question

💡 Solution & Explanation

Adiabatic implies $q=0$. $W = -P_{\text{ext}} (V_2 - V_1) = -4 \text{ bar} \times (30 - 40) \text{ L} = 40 \text{ bar L}$. $\Delta U = q + W = 40 \text{ bar L}$. Enthalpy change $\Delta H = \Delta U + \Delta(PV) = 40 + (4 \times 30 - 2 \times 40) = 40 + (120 - 80) = 80 \text{ bar L}$. Using $1 \text{ bar L} = 100 \text{ J}$, $\Delta H = 8000 \text{ J}$.