Which Potential Is Minimized? Equilibrium Criteria
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You should be comfortable with the first law of thermodynamics, the definition of entropy, and the basic Legendre transforms that define the Helmholtz and Gibbs free energies.
Why this matters
Knowing which quantity to minimise tells you exactly where a system will settle down.
In laboratory practice, we rarely hold all variables fixed. We might fix temperature and pressure, or temperature and volume. The correct thermodynamic potential depends entirely on these constraints. If you use the wrong potential, you will predict the wrong equilibrium state. This lecture maps every common experimental setup to its corresponding minimization principle, ensuring you can always identify the driving force of spontaneous change.
The Logic of Constraints
Imagine a box containing gas. If the box is perfectly insulated and rigid, the total energy and volume are locked. In this isolated case, the system evolves to maximise its entropy. However, if the box is in a water bath, the temperature is fixed by the bath, not by the system. The system can exchange energy with the bath. Here, energy is not conserved for the system alone; it is conserved for the system plus the bath. The relevant quantity to minimise is no longer energy, but a combination of energy and entropy that accounts for the heat exchange. This is the Helmholtz free energy. Similarly, if the box is flexible and in a pressure bath, the volume can change. The work done against the external pressure must be accounted for. The relevant quantity becomes the Gibbs free energy. The core idea is simple: the potential you minimise is the one whose definition matches the variables you have held constant.
Isolated Systems: Entropy Maximization
For a completely isolated system, the total energy \( U \), volume \( V \), and particle number \( N \) are fixed. The second law states that entropy never decreases. Therefore, the equilibrium state is the one with the maximum entropy \( S \). Mathematically, we require \( dS = 0 \) for a small variation and \( d^2S < 0 \) for stability. This is the most fundamental criterion, but it is rarely used directly because it is difficult to measure entropy changes directly in the lab. It serves as the theoretical anchor from which all other potentials are derived.
Fixed Temperature: Helmholtz Free Energy
Consider a system in contact with a heat reservoir at temperature \( T_0 \). The system can exchange heat but not work (fixed volume \( V \)). The total entropy change is the sum of the system's entropy change and the reservoir's entropy change. Since the reservoir is large, its temperature is constant. At fixed volume no work is done, so the heat the system takes from the reservoir is \( dU \), and the reservoir's entropy change is \( -dU/T_0 \). The total entropy change is \( dS_{\text{total}} = dS - \frac{dU}{T_0} \). For spontaneous processes, \( dS_{\text{total}} > 0 \). Rearranging this inequality gives \( dU - T_0 dS < 0 \). This quantity \( A = U - T_0 S \) is the Helmholtz free energy. Thus, at constant temperature and volume, the system minimises \( A \).
Fixed Temperature and Pressure: Gibbs Free Energy
Now allow the system to exchange work with a pressure reservoir at \( p_0 \). The system can expand or contract. The work done by the system is \( p_0 dV \). The first law for the system is \( dU = \delta Q - p_0 dV \). Substituting this into the Helmholtz condition, we get \( d(U - T_0 S) = \delta Q - p_0 dV - T_0 dS \). Using the second law \( \delta Q \leq T_0 dS \), we find \( d(U - T_0 S) \leq -p_0 dV \). Rearranging terms, \( d(U - T_0 S + p_0 V) \leq 0 \). The quantity \( G = U - T_0 S + p_0 V \) is the Gibbs free energy. Therefore, at constant temperature and pressure, the system minimises \( G \). This is the most common criterion in chemistry and biology, where reactions occur at ambient \( T \) and \( p \).
The derivation
Equilibrium as Free-Energy Minimization
For a system exchanging heat with a reservoir at fixed temperature T0 and doing work only against a fixed external pressure p0, the Clausius inequality forces the availability A = U − T0S + p0V to decrease in every spontaneous process, dA ≤ 0, so that at equilibrium the thermodynamic potential appropriate to the imposed constraints is stationary and minimal: entropy S is maximised at fixed (U,V,N); internal energy U at fixed (S,V,N); Helmholtz F = U − TS at fixed (T,V,N); and Gibbs G = U − TS + pV at fixed (T,p,N).
A common error is assuming that \( dG < 0 \) applies to all spontaneous processes. This is false. \( dG < 0 \) is strictly valid only when \( T \) and \( p \) are constant. If \( T \) or \( p \) changes during the process, you must use the general inequality involving the total entropy of the universe. The potential minimization is a constrained optimization problem; changing the constraints changes the objective function. C
Worked example
Problem. A 2.0 mol sample of an ideal gas is initially at \( T_1 = 300\ \mathrm{K} \) and \( V_1 = 10\ \mathrm{L} \). It expands isothermally and reversibly to \( V_2 = 20\ \mathrm{L} \). Calculate the change in Helmholtz free energy \( \Delta A \) and the change in Gibbs free energy \( \Delta G \). Assume \( R = 8.314\ \mathrm{J/(mol \cdot K)} \).
- Identify the process: Isothermal expansion of an ideal gas. Since \( T \) is constant, \( \Delta U = 0 \) for an ideal gas.
- Calculate the entropy change \( \Delta S \). For an ideal gas, \( \Delta S = nR \ln(V_2/V_1) \). Substituting values: \( \Delta S = 2.0 \times 8.314 \times \ln(20/10) = 16.628 \times \ln(2) \approx 11.53\ \mathrm{J/K} \).
- Calculate \( \Delta A \). By definition, \( A = U - TS \). Since \( \Delta U = 0 \) and \( T \) is constant, \( \Delta A = -T \Delta S \). \( \Delta A = -300 \times 11.53 = -3459\ \mathrm{J} \).
- Calculate \( \Delta G \). By definition, \( G = H - TS \). For an ideal gas, \( H = U + pV = U + nRT \). Since \( T \) is constant, \( \Delta H = \Delta U + nR \Delta T = 0 \). Thus, \( \Delta G = \Delta H - T \Delta S = 0 - T \Delta S = \Delta A \). So \( \Delta G = -3459\ \mathrm{J} \).
Units check. \( n \) is mol, \( R \) is \( \mathrm{J/(mol \cdot K)} \), \( T \) is K. \( nRT \) is J. \( \Delta S \) is \( \mathrm{J/K} \). \( T \Delta S \) is J. The result is in Joules, consistent with energy.
Three quick tests
- Which potential is minimized for a system at constant entropy and volume?
Answer
Internal energy \( U \). - If a reaction occurs in an open beaker at room temperature and pressure, which potential determines spontaneity?
Answer
Gibbs free energy \( G \). - Why is \( \Delta A \) equal to \( \Delta G \) for an isothermal ideal gas expansion?
Answer
Because \( \Delta U = 0 \) and \( \Delta H = 0 \) for ideal gas isothermal processes, making both potentials depend solely on \( -T \Delta S \).
Problem set for PU-203 →
Each derivation in this unit ends with five graded problems with full solutions.
Lecture notes checked line by line by the editor; every number in the worked example was recomputed.