Equilibrium of a System in Contact with a Reservoir: Gibbs Free Energy

Consider a system A in contact with a large reservoir A', maintained at constant temperature \(T_0\) and pressure \(P_0\).

The system and reservoir together form an isolated composite system \(A^*\), which exchanges neither energy nor volume with its surroundings.

Fig. : A thermodynamic system A in thermal and mechanical contact with a large reservoir A'. The combined system A* is isolated.

Let \(E\), \(S\), and \(V\) represent the internal energy, entropy, and volume of system A, respectively. The corresponding quantities for reservoir A' are denoted by \(E'\), \(S'\), and \(V'\).

Since the composite system is isolated, its total internal energy remains constant:

\[ E^*=E+E'=\text{constant}. \]

Taking the differential,

\[ dE+dE'=0. \]

Therefore,

\[ \boxed{dE'=-dE} \tag{1} \]

Similarly, the total volume of the composite system remains constant:

\[ V^*=V+V'=\text{constant}. \]

Differentiating,

\[ dV+dV'=0. \]

Hence,

\[ \boxed{dV'=-dV} \tag{2} \]

Thus, any increase in the energy or volume of system A is accompanied by an equal decrease in the corresponding quantity of reservoir A'.

Entropy of the Composite System

Suppose \(\Omega\) and \(\Omega'\) represent the numbers of accessible microstates of system A and reservoir A', respectively, compatible with the specified macrostate and the conservation constraints.

For two weakly interacting systems, the total number of accessible microstates of the composite system is

\[ \boxed{ \Omega^*=\Omega\Omega' } \tag{3} \]

According to Boltzmann's entropy relation,

\[ S=k_B\ln\Omega \]

where \(k_B\) is Boltzmann's constant.

For the composite system,

\[ S^*=k_B\ln\Omega^*. \]

Substituting equation (3),

\[ S^* = k_B\ln[\Omega\Omega']. \]

\[ S^* = k_B\ln\Omega + k_B\ln\Omega'. \]

Therefore,

\[ \boxed{ S^*=S+S' } \tag{4} \]

Thus, the total entropy of the composite system is equal to the sum of the entropies of the system and reservoir.

According to the second law of thermodynamics, the entropy of an isolated system cannot decrease during a spontaneous process.

Therefore, for the composite system,

\[ \boxed{ \Delta S^*\geq0 } \tag{5} \]

From equation (4),

\[ \Delta S^*=\Delta S+\Delta S'. \]

Hence,

\[ \boxed{ \Delta S+\Delta S'\geq0 } \tag{6} \]

At thermodynamic equilibrium, the entropy of the isolated composite system is maximum with respect to permitted variations.

Entropy Change of the Reservoir

Since reservoir A' is very large, its temperature \(T_0\) and pressure \(P_0\) remain constant during the interaction.

Assume that only pressure–volume work is performed and that no particles are exchanged between the system and reservoir.

For a simple compressible reservoir, the fundamental thermodynamic relation is

\[ \Delta U'=T_0 \Delta S'-P_0 \Delta V'. \]

Rearranging,

\[ T_0 \Delta S'=\Delta U'+P_0 \Delta V'. \]

Therefore,

\[ \boxed{ \Delta S'= \frac{\Delta U'+P_0 \Delta V'}{T_0} } \tag{7} \]

\[ \Delta S' = \frac{-\Delta U-P_0 \Delta V}{T_0}.   \because eq. (1) and (2) \]

Thus,

\[ \boxed{ \Delta S' = -\frac{\Delta E+P_0\Delta V}{T_0} } \tag{8} \]

Derivation of Gibbs Free Energy

The entropy change of the isolated composite system is

\[ \Delta S^*=\Delta S+\Delta S'. \]

Substituting equation (8),

\[ \Delta S^* = \Delta S - \frac{\Delta E+P_0\Delta V}{T_0}. \]

Taking \(T_0\) as the common denominator,

\[ \Delta S^* = \frac{ T_0\Delta S-\Delta E-P_0\Delta V }{T_0}. \]

Rearranging the numerator,

\[ \Delta S^* = -\frac{ \Delta E+P_0\Delta V-T_0\Delta S }{T_0}. \]

Since \(T_0\) and \(P_0\) are constant,

Therefore,

\[ \boxed{ \Delta S^* = -\frac{ \Delta(E+P_0V-T_0S) }{T_0} } \tag{9} \]

We now define a thermodynamic potential

\[ \boxed{ \mathcal{G} = E+P_0V-T_0S } \tag{10} \]

where \(E\), \(V\), and \(S\) are the thermodynamic properties of system A, while \(T_0\) and \(P_0\) are the fixed temperature and pressure of the reservoir.

The function \(\mathcal{G}\) is the thermodynamic potential appropriate to a system interacting with a constant-temperature, constant-pressure reservoir.

When the system is in thermal and mechanical equilibrium with the reservoir,

\[ T=T_0,\qquad P=P_0. \]

Consequently, the potential \(\mathcal{G}\) becomes the ordinary Gibbs free energy:

\[ \boxed{ G=E+PV-TS } \tag{11} \]

Since enthalpy is defined as

\[ H=E+PV, \]

the Gibbs free energy may also be written as

\[ \boxed{ G=H-TS } \tag{12} \]

The SI unit of Gibbs free energy is the joule (J).

3.7.7 Gibbs Free Energy and the Criterion for Spontaneous Processes

From equation (13) and the definition of \(\mathcal{G}\),

\[ \boxed{ \Delta S^* = -\frac{\Delta\mathcal{G}}{T_0} } \tag{13} \]

According to the second law,

\[ \Delta S^*\geq0. \]

Therefore,

\[ -\frac{\Delta\mathcal{G}}{T_0} \geq0. \]

Since the absolute temperature of the reservoir is positive,

\[ -\Delta\mathcal{G}\geq0. \]

Hence,

\[ \boxed{ \Delta\mathcal{G}\leq0 } \tag{14} \]

Thus, for a spontaneous process in a system interacting with a constant-temperature, constant-pressure reservoir, the appropriate thermodynamic potential cannot increase.

For processes occurring at constant system temperature and pressure, this criterion takes the familiar form

\[ \boxed{ (\Delta G)_{T,P}\leq0 } \tag{15} \]