In classical thermodynamics, the heat energy ($q$) absorbed or released by a closed gaseous system u

Question

In classical thermodynamics, the heat energy ($q$) absorbed or released by a closed gaseous system undergoing a process is generally a path-dependent quantity. However, under specific operational constraints, the heat exchanged becomes equal to the change in a state function, specifically the enthalpy change ($\Delta H$). Under which of the following exact sets of thermodynamic conditions is the statement '$q = \Delta H$' unconditionally true for a closed system?

Professor PC Thomas Classes · Answer

By the first law of thermodynamics, the infinitesimal change in internal energy of a closed system is $dU = dq + dw$. If the only work involved is pressure-volume work against an external pressure ($dw = -P_{	ext{ext}} dV$), then at constant pressure where $P_{	ext{ext}} = P_{	ext{system}} = P = 	ext{constant}$, we have: $dU = dq_P - P dV$. Integrating between initial state 1 and final state 2 gives: $U_2 - U_1 = q_P - P(V_2 - V_1) \implies q_P = (U_2 + PV_2) - (U_1 + PV_1)$. Since enthalpy is defined as $H = U + PV$, this reduces to $q_P = H_2 - H_1 = \Delta H$. Therefore, heat exchanged equals enthalpy change strictly at constant pressure with only $PV$ work involved.