Helmholtz Free Energy, Enthalpy, and Gibbs Free Energy

A beaker of reactants sitting on a bench does not minimize its energy — if it did, nothing would ever melt, dissolve, or mix, since those processes usually raise the energy. Yet ice melts above $0^\circ\mathrm{C}$ , salt dissolves, and gases mix spontaneously. The resolution is that a system in contact with its surroundings does not minimize energy; it minimizes a free energy, a quantity that balances the system's energy against the entropy it can borrow from the environment. Choosing the right free energy for the situation is the single most useful skill in applied thermodynamics. Which free energy? That depends on what the surroundings hold fixed. A system in a rigid, sealed box in a thermal bath is held at constant temperature and volume: it minimizes the Helmholtz free energy $F = U - TS$ . A system open to the atmosphere sits at constant temperature and pressure: it minimizes the Gibbs free energy $G = U - TS + PV$ . A process that is run at constant pressure exchanges heat equal to the change in enthalpy $H = U + PV$ . Each of these is a Legendre transform of the internal energy, re-expressed in the variables the environment actually controls, and each is minimized precisely when its natural variables are the ones held fixed. The deep reason these are the right quantities is that minimizing a system's free energy is identical to maximizing the total entropy of the system plus its reservoir. When you eliminate the reservoir from the second law — using the fact that heat $\delta Q$ leaving a bath at temperature $T$ costs it $\delta Q/T$ of entropy — the total-entropy-maximum principle collapses into a free-energy-minimum principle for the system alone. The $-TS$ term is not a bookkeeping trick; it is the entropy the environment gains, expressed in the system's own variables. This is why free energies govern everything practical. The direction of a chemical reaction is set by the sign of $\Delta G$ ; the boiling point is where the Gibbs energies of liquid and vapor cross; the stability of a phase is the convexity of its free energy; the maximum useful work extractable is the free-energy drop. Energy tells you what is conserved; free energy tells you what will happen.

Key takeaways

A system in contact with its surroundings minimizes a free energy, not its energy, because it can trade energy for entropy with the environment.
Helmholtz $F = U - TS$ (fixed $T, V$ ), Gibbs $G = U - TS + PV$ (fixed $T, P$ ), enthalpy $H = U + PV$ (fixed $S, P$ ) — each a Legendre transform of $U$ in the controlled variables.
Minimizing the system's free energy is exactly maximizing the total entropy of system + reservoir: $\Delta S_{\text{tot}} = -\Delta F/T$ .
Enthalpy is the heat exchanged at constant pressure: $\Delta H = Q_P$ ; Gibbs energy sets the direction of reactions and phase equilibria.
The maximum useful (non- $PV$ ) work a process can deliver is the drop in the appropriate free energy.
Key idea: each thermodynamic potential is minimized under exactly the constraints matching its natural variables, because free-energy minimization is total-entropy maximization with the reservoir eliminated.

The core insight

The free energies $F(T,V,N)$ , $G(T,P,N)$ , and enthalpy $H(S,P,N)$ are Legendre transforms of the internal energy that are each minimized at equilibrium under the constraints their natural variables describe — because minimizing a system's free energy is precisely maximizing the total entropy of system plus reservoir, with the reservoir eliminated via $\Delta S_{\text{tot}} = -\Delta F/T$ .

Where this lives today

The free-energy concept is the operational core of an enormous range of modern science and engineering. In chemistry and chemical engineering, tabulated standard Gibbs energies of formation $\Delta G_f^\circ$ let one predict, for essentially any reaction, whether it proceeds and to what equilibrium constant ( $\Delta G^\circ = -RT\ln K$ ); this single relation underlies industrial process design, electrochemistry, and corrosion science. In materials science, the CALPHAD method stores the Gibbs energy of every phase and computes multicomponent phase diagrams by minimizing the total $G$ — the direct descendant of Gibbs's common-tangent construction. In biology, free energy is the currency of life: the hydrolysis of ATP releases about $-30\ \mathrm{kJ/mol}$ of Gibbs energy under cellular conditions, and this quantum powers muscle contraction, active transport, and biosynthesis; enzymes work by lowering the free-energy barrier without changing the overall $\Delta G$ . In physics, the Helmholtz free energy $F = -k_BT\ln Z$ is the direct bridge from the statistical-mechanical partition function to all thermodynamics, and free-energy landscapes are the language of protein folding, glassy dynamics, and nucleation. Whenever a system is coupled to a reservoir — which is nearly always — its behavior is read off a free-energy surface, not an energy one.

The physical picture

Why does a system minimize $F = U - TS$ rather than $U$ ? Because it is not alone. Sitting in a thermal bath, the system can dump energy into the bath or draw energy from it, and every joule $\delta Q$ that flows into the bath raises the bath's entropy by $\delta Q/T$ . The second law cares about the total entropy — system plus bath — and when you rewrite that total-entropy-maximum condition in terms of the system's own variables, the bath's entropy contribution becomes exactly the $-TS$ term. Minimizing $F$ is maximizing total entropy in disguise. The $-TS$ term is a competition. Lowering $U$ is favorable; raising $S$ is favorable; but they often pull in opposite directions. At low temperature the energy term dominates and the system freezes into its lowest-energy configuration; at high temperature the $TS$ term wins and disorder prevails. The melting temperature is simply where the two balance — where $\Delta F = \Delta U - T\Delta S$ changes sign.

Figure

Free energy is a competition: F = U − TS balances low energy against high entropy

Which free energy you use is dictated entirely by what the surroundings clamp. Hold $T$ and $V$ fixed (a sealed rigid vessel in a bath): the equilibrium is the minimum of $F(T,V,N)$ . Hold $T$ and $P$ fixed (open to the atmosphere): the equilibrium is the minimum of $G(T,P,N)$ . This is not a matter of taste — it is forced by the second law, because the reservoir you couple to determines which combination of $U$ , $TS$ , and $PV$ measures the total entropy.

Figure

Which potential is minimized: match the potential to the controlled variables

The core equations

The three potentials are Legendre transforms of the internal energy, each subtracting off the conjugate product for the variable being traded away. Start from the Helmholtz free energy, the transform on $S$ :

F = U - TS \qquad (1)

where: $F$ — Helmholtz free energy, natural variables $(T,V,N)$ ; $U$ — internal energy; $TS$ — the reservoir's entropy contribution expressed in system variables. $F$ is minimized at equilibrium for fixed $T, V, N$ . Its differential shows why $T, V, N$ are its natural variables — those are the quantities whose differentials multiply the coefficients:

\mathrm{d}F = -S\,\mathrm{d}T - P\,\mathrm{d}V + \mu\,\mathrm{d}N \qquad (2)

where: $-S = (\partial F/\partial T)_{V,N}$ , $-P = (\partial F/\partial V)_{T,N}$ , $\mu = (\partial F/\partial N)_{T,V}$ . Notice $S$ is now recovered as a derivative, so $F$ loses no information despite depending on $T$ instead of $S$ . The enthalpy is the transform on $V$ , and it has a direct operational meaning — the heat exchanged at constant pressure:

H = U + PV, \qquad \mathrm{d}H = T\,\mathrm{d}S + V\,\mathrm{d}P + \mu\,\mathrm{d}N \qquad (3)

where: $H$ — enthalpy, natural variables $(S,P,N)$ . At constant $P$ and $N$ , $\mathrm{d}H = T\,\mathrm{d}S = \delta Q_{\text{rev}}$ , so $\Delta H = Q_P$ : enthalpy is the heat content at constant pressure. This is why reaction enthalpies are what calorimeters measure. The Gibbs free energy transforms on both $S$ and $V$ , and is the master potential of chemistry:

G = U - TS + PV = H - TS = \mu N, \qquad \mathrm{d}G = -S\,\mathrm{d}T + V\,\mathrm{d}P + \mu\,\mathrm{d}N \qquad (4)

where: $G$ — Gibbs free energy, natural variables $(T,P,N)$ . For a single component the Euler relation collapses $G = \mu N$ , so the chemical potential is the molar Gibbs energy. $G$ is minimized at fixed $T, P, N$ , and its sign governs reaction spontaneity. The unifying principle is that free-energy minimization equals total-entropy maximization with the reservoir eliminated:

\Delta S_{\text{tot}} = \Delta S_{\text{sys}} + \Delta S_{\text{res}} = -\frac{\Delta F}{T} \qquad (5)

where: at fixed $T, V, N$ the reservoir change is $\Delta S_{\text{res}} = -\Delta U_{\text{sys}}/T$ , so $\Delta S_{\text{tot}} = \Delta S_{\text{sys}} - \Delta U_{\text{sys}}/T = -\Delta F/T$ . Maximizing $S_{\text{tot}}$ is therefore identical to minimizing $F$ .

Derivation AFree-energy minimization from total-entropy maximization

The goal is to show that the free-energy-minimum principle is not an independent postulate but a direct consequence of the second law once you account for the reservoir. We prove that for a system exchanging heat with a bath at temperature $T$ at fixed volume, maximizing the total entropy (system + bath) is mathematically identical to minimizing the system's Helmholtz free energy $F = U - TS$ . The same argument with a pressure reservoir gives the Gibbs energy. This is the reason the $-TS$ term appears at all: it is the bath's entropy change, translated into the system's variables.

Step 1. Write the total entropy as $S_{\text{tot}} = S_{\text{sys}} + S_{\text{bath}}$ . The bath is enormous, at fixed temperature $T$ ; any heat $Q$ that flows into the system comes out of the bath, changing the bath entropy by $\Delta S_{\text{bath}} = -Q/T$ . Why: a reservoir at fixed $T$ has $\Delta S = \Delta U_{\text{bath}}/T$ by definition, and $\Delta U_{\text{bath}} = -Q$ .

Step 2. At fixed system volume (no work), the first law gives $Q = \Delta U_{\text{sys}}$ , so $\Delta S_{\text{bath}} = -\Delta U_{\text{sys}}/T$ . Therefore $\Delta S_{\text{tot}} = \Delta S_{\text{sys}} - \Delta U_{\text{sys}}/T$ . Why: the heat drawn from the bath is exactly the system's energy increase when no work is done.

Step 3. Factor out $-1/T$ : $\Delta S_{\text{tot}} = -\dfrac{1}{T}\big(\Delta U_{\text{sys}} - T\,\Delta S_{\text{sys}}\big) = -\dfrac{\Delta F_{\text{sys}}}{T}$ , where $F = U - TS$ evaluated at the bath temperature. Why: the combination $\Delta U - T\Delta S$ is precisely the change in the Helmholtz free energy, so the total entropy is $-\Delta F/T$ .

Step 4. Since $T > 0$ , maximizing $S_{\text{tot}}$ is identical to minimizing $F$ . The equilibrium of a system at fixed $T, V, N$ is therefore the minimum of its Helmholtz free energy — a system-only criterion equivalent to the second law. Repeating with a pressure reservoir (allowing $PV$ work) adds $+PV$ and gives the Gibbs energy $G$ minimized at fixed $T, P, N$ . Why: the reservoir has been eliminated, turning a global entropy statement into a local free-energy statement.

Go deeper · Derivation B · 1 minDifferentials of the potentials and the ideal-gas free energy

Derivation BDifferentials of the potentials and the ideal-gas free energy

The goal is to derive the differentials $\mathrm{d}F$ , $\mathrm{d}H$ , $\mathrm{d}G$ from the definitions and the fundamental relation, confirming that each potential's natural variables are exactly those appearing as differentials — and then to compute a concrete, dimensionful free energy for the ideal gas so the abstractions become numbers. This is where the sign structure ( $-S\,\mathrm{d}T$ , $+V\,\mathrm{d}P$ ) that feeds all four Maxwell relations is pinned down.

Step 1. Differentiate $F = U - TS$ : $\mathrm{d}F = \mathrm{d}U - T\,\mathrm{d}S - S\,\mathrm{d}T$ . Insert $\mathrm{d}U = T\,\mathrm{d}S - P\,\mathrm{d}V + \mu\,\mathrm{d}N$ ; the $T\,\mathrm{d}S$ terms cancel, leaving $\mathrm{d}F = -S\,\mathrm{d}T - P\,\mathrm{d}V + \mu\,\mathrm{d}N$ . Why: the cancellation of $T\,\mathrm{d}S$ is exactly what trades the natural variable $S$ for $T$ .

Step 2. Similarly $\mathrm{d}H = \mathrm{d}U + P\,\mathrm{d}V + V\,\mathrm{d}P = T\,\mathrm{d}S + V\,\mathrm{d}P + \mu\,\mathrm{d}N$ (the $P\,\mathrm{d}V$ terms cancel), and $\mathrm{d}G = -S\,\mathrm{d}T + V\,\mathrm{d}P + \mu\,\mathrm{d}N$ . Why: each transform cancels one term of $\mathrm{d}U$ and introduces the conjugate differential.

Step 3. Read off the constant-pressure meaning of enthalpy: at fixed $P, N$ , $\mathrm{d}H = T\,\mathrm{d}S = \delta Q_{\text{rev}}$ , so $\Delta H = Q_P$ . Why: with $V\,\mathrm{d}P = 0$ , the only change in $H$ is the reversible heat, which is why calorimeters at atmospheric pressure measure $\Delta H$ .

Step 4. Compute the ideal-gas Helmholtz free energy from $F = -Nk_BT[\ln(n_Q/n) + 1]$ , where $n = N/V$ and $n_Q = (2\pi m k_B T/h^2)^{3/2}$ is the quantum concentration. For $1\ \mathrm{mol}$ of argon at $T = 300\ \mathrm{K}$ , $V = 22.4\ \mathrm{L}$ : $Nk_BT = RT = 2494\ \mathrm{J}$ , $\ln(n_Q/n) \approx 16.0$ , giving $F \approx -42.5\ \mathrm{kJ}$ . Cross-check: $S = (U - F)/T$ with $U = \tfrac32 RT$ gives $S_m \approx 154\ \mathrm{J\,mol^{-1}K^{-1}}$ , the measured argon entropy. Why: the abstract $F = U - TS$ produces a real, measurable entropy for a real gas — the theory closes.

Limiting cases

The potentials specialize according to what is held fixed:

Case	Result	Meaning
Fixed $T, V, N$	equilibrium minimizes $F = U - TS$	Helmholtz — sealed rigid vessel in a bath.
Fixed $T, P, N$	equilibrium minimizes $G = U - TS + PV$	Gibbs — open to atmosphere; chemistry, phases.
Fixed $S, P, N$	equilibrium minimizes $H = U + PV$	Enthalpy — adiabatic constant-pressure process.
Constant- $P$ heating	$\Delta H = Q_P = \int C_P\,\mathrm{d}T$	Enthalpy is the heat content; calorimetry measures it.
Single component	$G = \mu N$ (Euler)	Chemical potential = molar Gibbs energy.
$T \to 0$	$F \to U$ , $G \to H$	The $TS$ term vanishes; free energy → energy (third-law regime).

Common trap“Spontaneous processes always lower the energy”

No — spontaneous processes lower the appropriate free energy, not the energy. Many everyday spontaneous processes actually raise the internal energy: ice melting, ammonium nitrate dissolving (the pack gets cold), a gas expanding into vacuum. They happen because they raise entropy enough that $\Delta F = \Delta U - T\Delta S$ (or $\Delta G$ ) is negative even though $\Delta U > 0$ . Conversely, at low temperature the $-T\Delta S$ term is small and energy-lowering dominates — which is why water freezes when it is cold enough. The sign of $\Delta G$ , not $\Delta U$ , tells you whether a process at fixed $T, P$ will proceed. Confusing the two is the single most common thermodynamics error.

Worked examples

1 · The ideal-gas free energy and the entropy it hides (confidence builder)

Compute the Helmholtz free energy of $1\ \mathrm{mol}$ of argon at $T = 300\ \mathrm{K}$ , $V = 22.4\ \mathrm{L}$ , and extract the entropy from $F = U - TS$ . Show the entropy matches the measured value.

Step 1. Thermal energy scale: $Nk_BT = RT = (8.314)(300) = \mathbf{2494\ J}$ . The internal energy of a monatomic ideal gas is $U = \tfrac32 RT = \mathbf{3741\ J}$ .

Step 2. Quantum concentration: $n_Q = (2\pi m k_B T/h^2)^{3/2}$ . For argon ( $m = 6.63\times10^{-26}\ \mathrm{kg}$ ) at $300\ \mathrm{K}$ , $n_Q \approx 2.4\times10^{32}\ \mathrm{m^{-3}}$ , while the actual density $n = N_A/V = 6.02\times10^{23}/0.0224 = 2.69\times10^{25}\ \mathrm{m^{-3}}$ . So $n_Q/n \approx 8.9\times10^{6}$ , deeply classical.

Step 3. Free energy: $F = -RT[\ln(n_Q/n) + 1] = -2494\,[16.0 + 1] = \mathbf{-4.25\times10^{4}\ J} = -42.5\ \mathrm{kJ}$ . Strongly negative because the large accessible phase space (entropy) dominates the modest energy.

Step 4. Extract entropy: $S = (U - F)/T = (3741 - (-42\,483))/300 = 46\,224/300 = \mathbf{154\ J\,mol^{-1}K^{-1}}$ . This is the Sackur–Tetrode / measured molar entropy of argon (154.8 J mol $^{-1}$ K $^{-1}$ ) — the abstract free energy delivered the right real number.

2 · Reaction direction from ΔG (exam level)

A reaction has $\Delta H = +30\ \mathrm{kJ/mol}$ (endothermic) and $\Delta S = +120\ \mathrm{J\,mol^{-1}K^{-1}}$ . At what temperature does it become spontaneous at constant pressure, and is it spontaneous at $T = 300\ \mathrm{K}$ ?

Step 1. Spontaneity at constant $T, P$ requires $\Delta G = \Delta H - T\Delta S < 0$ . The crossover is where $\Delta G = 0$ : $T^\star = \Delta H/\Delta S$ .

Step 2. Compute: $T^\star = 30\,000/120 = \mathbf{250\ K}$ . Above $250\ \mathrm{K}$ the $-T\Delta S$ term outweighs the positive $\Delta H$ and the reaction proceeds; below it, it does not.

Step 3. At $T = 300\ \mathrm{K}$ : $\Delta G = 30\,000 - 300(120) = 30\,000 - 36\,000 = \mathbf{-6000\ J/mol} = -6\ \mathrm{kJ/mol}$ . Negative — spontaneous, despite being endothermic ( $\Delta H > 0$ ).

Step 4. Interpretation: this is entropy-driven spontaneity — the reaction absorbs heat yet proceeds because it creates enough disorder. At $200\ \mathrm{K}$ , $\Delta G = 30\,000 - 200(120) = +6000\ \mathrm{J/mol} > 0$ , and the reaction would run backward. The sign of $\Delta G$ , set by the $U$ -vs- $TS$ competition, is the whole story — exactly why $G$ , not $U$ or $H$ alone, governs chemistry.

Origins · 1 minfree energy and how this was discovered

free energy

concept · maximum available work

The concept of "free" energy — the portion of a system's energy actually available to do useful work at fixed temperature — emerged in the 1870s–1880s from the recognition that not all of the internal energy is convertible. Hermann von Helmholtz introduced the free energy $F = U - TS$ in 1882, calling the bound part $TS$ the energy unavailable for work; Josiah Willard Gibbs had already introduced the potential now bearing his name in his 1875–78 memoirs on heterogeneous equilibrium. The word "free" captures the split: $F$ is the maximum work extractable at constant temperature, while $TS$ is locked away as heat that must be exchanged with the reservoir. This split is the operational meaning of the potentials — a reaction can deliver at most $-\Delta G$ of non-expansion work — and it is why fuel cells, batteries, and biological ATP hydrolysis are all rated by free-energy changes rather than by enthalpies.

History

1865

— Clausius defines entropy and writes dU = T dS − P dV, the foundation of the potentials.

1873

— Gibbs introduces the "available energy" and thermodynamic surfaces in his graphical papers.

1875

— Gibbs, On the Equilibrium of Heterogeneous Substances, defines G, μ, and the potentials.

1882

— Helmholtz introduces the free energy F = U − TS ("freie Energie") in Zur Thermodynamik chemischer Vorgänge.

1884

— Le Chatelier states the equilibrium-shift principle governed by free-energy minimization.

1906

— Nernst's heat theorem fixes the entropy constant, giving absolute free energies.

Going further

References. Callen, Thermodynamics and an Introduction to Thermostatistics, 2nd ed., chs. 5–6 · Pathria & Beale, Statistical Mechanics, 3rd ed., ch. 3 · Reif, Fundamentals of Statistical and Thermal Physics, chs. 5, 8 · Kardar, Statistical Physics of Particles, ch. 1 · Atkins & de Paula, Physical Chemistry (free energies and reaction spontaneity) · Sethna, Statistical Mechanics, ch. 6. The reservoir-elimination result $\Delta S_{\text{tot}} = -\Delta F/T$ follows from straightforward algebraic factoring. The ideal-gas Helmholtz free energy $F = -Nk_BT[\ln(n_Q/n)+1]$ with $n_Q = (2\pi m k_BT/h^2)^{3/2}$ gives, for $1\ \mathrm{mol}$ argon at $300\ \mathrm{K}$ , $22.4\ \mathrm{L}$ : $F = -42.5\ \mathrm{kJ}$ and recovered molar entropy $S_m = 154\ \mathrm{J\,mol^{-1}K^{-1}}$ (measured $154.8$ ). The reaction example ( $\Delta H = +30\ \mathrm{kJ/mol}$ , $\Delta S = +120\ \mathrm{J\,mol^{-1}K^{-1}}$ , $T^\star = 250\ \mathrm{K}$ , $\Delta G(300) = -6\ \mathrm{kJ/mol}$ ) is direct arithmetic. $R = 8.314\ \mathrm{J\,mol^{-1}K^{-1}}$ .

Builds on: Thermodynamic Potentials and the Fundamental Relation · Legendre Transforms · Second Law of Thermodynamics and Entropy · First Law of Thermodynamics and Internal Energy. Leads to: Maxwell Relations · Euler Relation and Gibbs–Duhem Equation · Temperature, Pressure, and Chemical Potential as Entropy Derivatives · Helmholtz Free Energy from the Partition Function.

What exactly is "free" about free energy?

It is the part of the internal energy that is available to do useful work at constant temperature. When a process happens in contact with a bath at temperature $T$ , an amount $T\Delta S$ of energy must be exchanged with the bath as heat and is not available for work; the remainder, the change in $F = U - TS$ (or $G$ at constant pressure), is the maximum useful work you can extract. So "free" means "free to be converted into work," as opposed to the "bound" energy $TS$ that is locked into heat exchange with the surroundings. This is why a battery or fuel cell is rated by $\Delta G$ : that is the electrical work it can deliver, not the full enthalpy $\Delta H$ of the reaction.

Why is there a different free energy for different constraints — can't one potential do everything?

Each potential is minimized only when its own natural variables are the ones held fixed, and different experiments fix different variables. In a sealed rigid vessel you control $T$ and $V$ , so equilibrium is the minimum of $F(T,V,N)$ ; open to the atmosphere you control $T$ and $P$ , so it is the minimum of $G(T,P,N)$ . You can use any potential in principle — they all contain the same information — but if you use the wrong one you must track the reservoir explicitly and the equilibrium is no longer a simple minimum. Choosing the potential whose natural variables match your constraints turns a messy total-entropy calculation into a clean "minimize this function," which is the whole practical point.

How can an endothermic reaction (ΔH > 0) be spontaneous?

Because spontaneity at constant $T, P$ is governed by $\Delta G = \Delta H - T\Delta S$ , not by $\Delta H$ alone. If the reaction increases entropy enough (large positive $\Delta S$ ), the $-T\Delta S$ term can overcome a positive $\Delta H$ and make $\Delta G < 0$ . Dissolving ammonium nitrate is a classic example: it absorbs heat (the solution gets cold, $\Delta H > 0$ ) yet dissolves spontaneously because the ions gain so much entropy on dispersing. This is entropy-driven spontaneity. The crossover temperature is $T^\star = \Delta H/\Delta S$ : above it the reaction goes, below it it reverses. Nature minimizes free energy, and entropy is half of that ledger.

Is enthalpy just "energy plus PV" — what does the PV term physically mean?

The $PV$ term is the energy cost of making room for the system against the surrounding pressure. When a system at constant pressure grows by a volume $\Delta V$ , it must push the atmosphere back, doing work $P\Delta V$ on the surroundings. The enthalpy $H = U + PV$ bundles the internal energy with this "room-making" energy, which is exactly why the heat exchanged at constant pressure equals $\Delta H$ rather than $\Delta U$ : part of the heat you supply goes into expansion work against the atmosphere, and $H$ accounts for it automatically. For processes at constant volume the $PV$ term is irrelevant and $\Delta U = Q_V$ ; at constant pressure (most chemistry, most of biology) the relevant heat is $\Delta H$ .

How does the Helmholtz free energy connect to the partition function in statistical mechanics?

Through the single most important bridge relation in equilibrium statistical mechanics: $F = -k_BT\ln Z$ , where $Z = \sum_i e^{-E_i/k_BT}$ is the canonical partition function. This one equation converts a microscopic sum over states into a macroscopic free energy, and from $F$ every thermodynamic quantity follows by the differentials derived here: $S = -\partial F/\partial T$ , $P = -\partial F/\partial V$ , $\mu = \partial F/\partial N$ , and $U = F + TS = -\partial \ln Z/\partial\beta$ . So the free energies of this article are exactly where thermodynamics and statistical mechanics meet: statistical mechanics computes $F$ from a Hamiltonian via $Z$ , and thermodynamics reads off the rest. This is developed fully in the canonical-ensemble articles.

Why does the free energy reduce to the energy as temperature goes to zero?

Because the entropy term $TS$ vanishes. As $T \to 0$ , both $F = U - TS \to U$ and $G = H - TS \to H$ , since $S$ approaches a finite constant (the third law) while $T \to 0$ . Physically, near absolute zero there is no thermal energy to "borrow" entropy from the surroundings, so the entropy competition disappears and the system simply minimizes its energy — it settles into the ground state. This is why crystalline order, superconductivity, and other low-temperature ordered phases emerge: at low $T$ the $-TS$ term can no longer stabilize disorder, and the energy-minimizing configuration wins. The whole "energy vs. entropy" balance of free energy is a finite-temperature phenomenon.