Equilibrium Conditions for a System at Constant Pressure in Contact with a Heat Reservoir: Enthalpy

Consider a closed thermodynamic system \(A\) consisting of a gas enclosed in a cylinder fitted with a frictionless movable piston. The system is in thermal contact with a large heat reservoir maintained at temperature \(T_0\).

The external pressure \(P_0\) remains constant. When the system absorbs heat from the reservoir, the gas may expand and perform pressure–volume work against the surroundings.

Suppose the system absorbs heat \(Q\) and undergoes a change in volume \(\Delta V\).

The work done by the gas against the constant external pressure is

\[ \Delta W=P_0\Delta V. \]

From the first law of thermodynamics,

\[ \Delta U=Q-\Delta W. \]

Therefore,

\[ Q=\Delta U+P_0\Delta V. \]

Since \(P_0\) is constant,

\[ Q=\Delta(U+P_0V). \]

A thermodynamic state function called enthalpy is defined as

\[ \boxed{H=U+PV} \tag{1} \]

For initial and final equilibrium states at the same constant pressure \(P=P_0\), and when only pressure–volume work occurs,

\[ \boxed{Q_P=\Delta H} \tag{2} \]

Thus, the change in enthalpy equals the heat absorbed by a closed system at constant pressure, provided no other forms of work are performed.

Entropy Change of the Combined System

The heat reservoir supplies heat \(Q\) to the system at temperature \(T_0\).

Therefore, the entropy change of the reservoir is

\[ \Delta S'=-\frac{Q}{T_0}. \]

Using equation (3.13.2),

\[ \boxed{ \Delta S'=-\frac{\Delta H}{T_0} } \tag{3} \]

The total entropy change of the system and its surroundings is

\[ \Delta S_{\mathrm{total}} = \Delta S+\Delta S'. \]

Substituting equation (3),

\[ \Delta S_{\mathrm{total}} = \Delta S-\frac{\Delta H}{T_0}. \]

Rearranging,

\[ \Delta S_{\mathrm{total}} = -\frac{\Delta H-T_0\Delta S}{T_0}. \]

Since \(T_0\) is constant,

\[ \boxed{ \Delta S_{\mathrm{total}} = -\frac{\Delta(H-T_0S)}{T_0} } \tag{4} \]