Helmholtz Free Energy: The Energy You Can Actually Use

A cylinder of compressed gas, a charged battery, a stretched rubber band and a crystal on the edge of melting all hold energy. A natural question follows: how much of it can be turned into useful work? The obvious answer is "all of it", and the obvious answer is wrong. Some energy is always tied up in the disordered motion of molecules, and the second law of thermodynamics says you cannot get it back as work for free. The Helmholtz free energy is the quantity that does this bookkeeping. It tells you how much energy is "free" to do work when a system sits at a fixed temperature, and it tells you which way a process will go on its own when both temperature and volume are held fixed.

The definition is short:

F = U − TS

Here U is the internal energy of the system, T is its absolute temperature in kelvin, and S is its entropy. Chemists usually write the same quantity as A (from the German Arbeit, "work"), and the International Union of Pure and Applied Chemistry (IUPAC) recommends the symbol A and the name "Helmholtz energy". Physicists more often write F. They are the same thing. This article uses F, but you will see A in most chemistry books.

This article builds the idea from the ground up. It starts with intuition, then the history, then the math, then a set of worked numerical examples, and finally the places where the Helmholtz energy earns its keep, from equations of state for water to the theory of magnets and the statistics of molecules. Along the way it clears up some of the common confusions, especially the difference between Helmholtz and Gibbs free energy.

The intuition: energy minus the part you can't have

Total energy is not usable energy

Imagine a box of gas at room temperature. Its internal energy U is the sum of the kinetic energies of all the molecules plus any energy stored in the forces between them. For one mole of a monatomic gas such as argon at 25 °C, U (measured from the state where the atoms are at rest) is about 3.7 kilojoules. That is a real amount of energy, enough to lift a one-kilogram mass about 380 metres.

But you cannot simply take that energy out and use it. The molecules are moving in random directions. To extract work you need some ordered process, such as letting the gas push a piston. And the moment you try that while keeping the gas at a constant temperature, something interesting happens: heat flows in from the surroundings to keep the temperature up, the gas expands, and the work you get out depends not on U but on how much the entropy changes.

Entropy as a tax

Entropy is a measure of how many microscopic arrangements are consistent with what you see at the large scale. A gas spread over a larger volume has more possible arrangements than the same gas squeezed into a corner, so it has higher entropy. The second law says that the total entropy of an isolated system never goes down.

The term TS in F = U − TS can be read as a kind of tax. At temperature T, the energy TS is the part of the internal energy that is, loosely speaking, committed to the system's disorder. Helmholtz himself drew this distinction between "free" energy, which is available for conversion into work, and "bound" energy, which is not. The free energy is what is left after the tax.

This reading is a helpful picture, but it is only a picture. TS is not a separate store of energy sitting somewhere inside the system. The precise statement is about changes: when a system goes from one state to another at constant temperature, the decrease in F sets an upper limit on the work you can get out.

A ball rolling downhill, with a thermometer attached

A familiar idea from mechanics is that a ball rolls downhill to the lowest point of potential energy. Thermodynamic potentials generalise this. For a system that can exchange heat with surroundings at a fixed temperature, and whose volume is held fixed, the quantity that plays the role of "height of the hill" is F. Left alone, such a system moves toward lower F, and it comes to rest (equilibrium) at the minimum of F.

The interesting part is that F has two pieces pulling in different directions. Lowering U favours tight, ordered, low-energy arrangements: atoms bonded into a crystal, a protein folded up, a magnet with all its tiny magnetic moments aligned. Raising S favours loose, disordered, spread-out arrangements: atoms flying around as a gas, a protein flopping as a random chain, magnetic moments pointing every which way. Temperature sets the exchange rate between the two. At low T the entropy term matters little and energy wins, so you get crystals and magnets. At high T the entropy term dominates and disorder wins, so you get gases and demagnetised iron. Melting, boiling and the loss of magnetism at the Curie point are all places where the balance tips.

That one sentence, "minimise U − TS", is behind an enormous amount of physics and chemistry.

Where the idea came from

The nineteenth-century problem of chemical affinity

In the middle of the nineteenth century, chemists wanted a rule for predicting whether a reaction would go. The French chemist Marcellin Berthelot and the Danish chemist Julius Thomsen championed a simple rule, often called the Thomsen–Berthelot principle: reactions that give off heat (exothermic reactions) are the ones that happen spontaneously, and the heat released measures the "affinity" driving the reaction.

The rule works surprisingly often at room temperature, which is why it was taken seriously. But it fails for reactions that absorb heat and still proceed by themselves. Dissolving some salts in water, for example, makes the water colder, and yet the salt dissolves. Something other than heat release had to be involved.

Gibbs and Helmholtz

Two people independently found the answer. The American physicist Josiah Willard Gibbs, working at Yale, published his long memoir "On the Equilibrium of Heterogeneous Substances" in the Transactions of the Connecticut Academy in parts between 1875 and 1878. It introduced the thermodynamic potentials and the conditions for chemical and phase equilibrium in great generality, but it appeared in a little-read journal and was written in a dense mathematical style.

Hermann von Helmholtz, the German physician and physicist already famous for his work on the conservation of energy, physiology and acoustics, approached the problem from the direction of electrochemistry. In 1882 he presented "Die Thermodynamik chemischer Vorgänge" ("The thermodynamics of chemical processes") to the Prussian Academy of Sciences in Berlin. The first part was submitted on 2 February 1882; a second part, concerned with galvanic polarisation, followed in July 1882, and a third, describing experiments on zinc chloride and calomel cells, in 1883.

Helmholtz's central point was that the electrical work a battery can deliver is not equal to the heat the same chemical reaction would release if you simply let it run in a beaker. The difference is related to the entropy change of the reaction and the temperature. He argued that it is the free energy, not the total energy released as heat, that determines the direction in which chemical affinity acts. That quantity, U − TS, now carries his name.

The relation between the free energy and its temperature dependence, now called the Gibbs–Helmholtz equation, also appears in Helmholtz's 1882 paper. The fact that it carries Gibbs's name too reflects the way the two lines of work were later merged, partly through Wilhelm Ostwald, who translated Gibbs's papers into German.

Names and symbols

The names have shifted over time. Older literature calls F the "Helmholtz function", the "work function", or simply the "free energy". American chemists in the early twentieth century, following G. N. Lewis, sometimes used F for what is now called the Gibbs free energy, which caused real confusion. IUPAC now recommends:

When reading an old book, always check the definition rather than the letter.

The math, one step at a time

The fundamental relation for internal energy

For a simple system (a fixed amount of a single substance, with only pressure-volume work), the first and second laws combine into one equation for a small reversible change:

dU = T dS − P dV

In words: the internal energy rises when heat T dS flows in, and falls when the system does work P dV by expanding. This equation says that the "natural variables" of U are S and V. If you know U as a function of S and V, you know everything about the system's equilibrium thermodynamics.

The trouble is that S is not something you can set with a knob in the laboratory. Temperature is. You can put a sample in a water bath and hold its temperature fixed; you cannot easily hold its entropy fixed. It would be much more convenient to have a function whose natural variables are T and V.

The Legendre transform

The trick that does this is called a Legendre transform. Define

F = U − TS

and take its differential:

dF = dU − T dS − S dT

Substitute dU = T dS − P dV:

dF = −S dT − P dV

The T dS terms cancel. F is now naturally a function of T and V. That is the whole point of subtracting TS: it swaps the roles of S and T as the independent variable.

If the number of particles N can also change, a term for the chemical potential μ is added:

dF = −S dT − P dV + μ dN

Since F is a state function (its value depends only on the current state, not on the history), these relations hold for any change between equilibrium states, not just for reversible ones.

Reading off properties

Because dF = −S dT − P dV + μ dN, the partial derivatives of F give the other quantities directly:

• Entropy: S = −(∂F/∂T) at constant V and N
• Pressure: P = −(∂F/∂V) at constant T and N
• Chemical potential: μ = (∂F/∂N) at constant T and V

The second of these is especially useful. If you can compute F as a function of volume, you get the pressure by taking a slope. That is how equations of state are built from microscopic models.

A further consequence comes from the fact that mixed second derivatives do not depend on order. Differentiating S with respect to V, and P with respect to T, gives one of the Maxwell relations:

(∂S/∂V) at constant T = (∂P/∂T) at constant V

The left side, how entropy changes when you expand a system at constant temperature, is hard to measure directly. The right side, how pressure rises when you heat a sealed container, is easy. The Maxwell relation lets you trade one for the other.

Recovering energy from F

You can also go back the other way. Since S = −∂F/∂T, the energy is

U = F + TS = F − T (∂F/∂T)

which can be rearranged into the Gibbs–Helmholtz form:

U = −T² ∂(F/T)/∂T (at constant V)

This says that if you know how F/T varies with temperature, you know the energy. It is the route by which measured free energies at several temperatures are turned into energies, and vice versa.

Why F tells you the direction of change

Now for the principle that makes F so important. Consider a system in thermal contact with a large heat bath at temperature T, so that its temperature is fixed, and suppose its volume is also fixed. Let the system undergo some change, reversible or not, and let it deliver some work W to the outside world (other than by changing volume, since volume is fixed, this could be electrical work, for instance).

Energy conservation says that the heat Q taken from the bath satisfies ΔU = Q − W. The second law, applied to system plus bath, says the total entropy cannot decrease. The bath gives up heat Q at temperature T, so its entropy changes by −Q/T. The system's entropy changes by ΔS. Therefore

ΔS − Q/T ≥ 0, which means Q ≤ TΔS.

Combine this with Q = ΔU + W:

ΔU + W ≤ TΔS, which rearranges to W ≤ −(ΔU − TΔS) = −ΔF.

So the work you can extract is at most the decrease in Helmholtz free energy. The equality holds only for a perfectly reversible process. Any real process wastes some of the available free energy as extra entropy.

Two special cases follow:

1. Maximum work. At constant temperature, −ΔF is the maximum work obtainable. Strictly, if the volume is allowed to change and the system's temperature starts and ends equal to the bath temperature, −ΔF bounds the total work, including the work of expansion. If the volume is held constant, −ΔF bounds the remaining "other" work, such as electrical work.
2. Spontaneity. If no work is extracted at all (W = 0), the inequality becomes ΔF ≤ 0. A system at constant temperature and volume can only move toward lower F. Equilibrium is reached when F can go no lower: the minimum of F.

This is the precise version of the ball-rolling-downhill picture. The heat bath is doing something subtle here. When the system lowers its energy by releasing heat, the bath's entropy goes up. When the system increases its own entropy, it may need to absorb heat and lower the bath's entropy. F = U − TS is the single number that combines both effects so you can look only at the system and still apply the second law correctly.

Worked examples

The following examples use R = 8.314 J/(mol·K) for the gas constant and k = 1.381 × 10⁻²³ J/K for the Boltzmann constant. Calculations were done to more digits and then rounded.

Example 1: Letting a gas double its volume

Take 1 mole of an ideal gas at 298.15 K (25 °C) and let it expand isothermally from volume V to 2V.

For an ideal gas, the internal energy depends only on temperature, so at constant T, ΔU = 0.

The entropy change for isothermal expansion of an ideal gas is ΔS = nR ln(V₂/V₁) = (1)(8.314)(ln 2) = 8.314 × 0.6931 ≈ 5.763 J/K.

Then

ΔF = ΔU − TΔS = 0 − (298.15)(5.763) ≈ −1718 J.

So the Helmholtz energy falls by about 1.72 kJ. That is the maximum work the gas can do on a piston during the expansion, and it is achieved only if the expansion is carried out slowly and reversibly. You can check this against the textbook result for reversible isothermal work, W = nRT ln(V₂/V₁) = 8.314 × 298.15 × 0.6931 ≈ 1718 J. They match, as they should.

Notice the remarkable thing here: the internal energy of the gas did not change at all, yet the gas did 1.7 kJ of work. Where did the energy come from? From the heat bath. Heat equal to TΔS ≈ 1718 J flowed into the gas from the surroundings and came straight back out as work. The decrease in F was driven entirely by the increase in entropy. This is the cleanest possible illustration of why total energy is the wrong quantity to look at.

If instead you just open a valve and let the gas rush into an evacuated second chamber of equal size (a "free expansion"), the gas does no work at all. ΔF is still −1718 J, because F is a state function and the start and end states are the same. The difference is that the free energy has been wasted: the process generated entropy without extracting work. This contrast between the reversible and the free expansion is the heart of the inequality W ≤ −ΔF.

Example 2: A tenfold expansion at 300 K

For 1 mole of ideal gas expanding from 1 litre to 10 litres at 300 K:

ΔF = −nRT ln(V₂/V₁) = −(8.314)(300)(ln 10) = −(8.314)(300)(2.3026) ≈ −5743 J.

About 5.7 kJ is available. Going the other way, compressing the gas from 10 litres back to 1 litre at 300 K, raises F by 5.7 kJ, which is the minimum work you must do on the gas. A compressor that does more than this (as all real ones do) is wasting some of its input as extra heat.

Example 3: Mixing two gases in a fixed box

Put 1 mole of helium in one half of a rigid box held in a heat bath and 1 mole of argon in the other, both at 298.15 K and the same pressure. Remove the partition. Each gas spreads to fill twice its original volume. Treating both as ideal gases, each one's entropy rises by R ln 2, so

ΔS_mix = 2R ln 2 ≈ 11.53 J/K, ΔU = 0 (ideal gases, constant T), ΔF = −TΔS_mix = −(298.15)(11.53) ≈ −3437 J.

The negative ΔF tells you that mixing is spontaneous at constant temperature and volume, which is what everyone observes: gases never spontaneously un-mix. It also tells you something less obvious. With a pair of ideal semipermeable membranes (one passing only helium, the other only argon), you could in principle extract up to about 3.4 kJ of work from the mixing. Reversing the process, separating air into oxygen and nitrogen, for example, requires at least this kind of minimum work. The free-energy calculation gives the thermodynamic floor for the energy cost of gas separation, which is why it appears in discussions of carbon capture and industrial air separation.

Example 4: A two-level system

Statistical mechanics gives a way to calculate F from first principles, which is explained in more detail below. Here is a preview with the simplest possible system: a particle that can be in a ground state with energy 0 or an excited state with energy ε.

The partition function (a sum over states of exp(−energy/kT)) is

Z = 1 + exp(−ε/kT)

and the free energy is F = −kT ln Z. Take the temperature where kT equals ε exactly. Then

Z = 1 + e⁻¹ = 1.3679, F = −kT ln(1.3679) = −0.3133 kT.

The average energy is U = ε × (probability of being excited) = ε × e⁻¹/(1 + e⁻¹) = 0.2689 ε = 0.2689 kT.

The entropy follows from S = (U − F)/T = (0.2689 + 0.3133) k = 0.5822 k.

As a check: the maximum possible entropy for a two-state system, reached when both states are equally likely, is k ln 2 = 0.693 k. At kT = ε the particle spends about 27% of its time excited, not 50%, so the entropy should be below the maximum. It is. And F is negative even though U is positive, because the entropy term TS = 0.5822 kT outweighs U = 0.2689 kT. This little model is the basis for understanding paramagnets, defects in crystals and many other systems where things can be "on" or "off".

Example 5: The free energy of argon gas from scratch

For a monatomic ideal gas of N atoms each of mass m, statistical mechanics gives (using Stirling's approximation for large N)

F = −NkT [ ln( V / (N Λ³) ) + 1 ]

where Λ = h / √(2π m k T) is the thermal de Broglie wavelength and h is Planck's constant. Λ is roughly the quantum "size" of an atom moving with thermal speed.

For argon (molar mass 39.948 g/mol) at 298.15 K and a pressure of 1 bar (10⁵ Pa):

• Mass of one atom: m = 0.039948 kg / 6.022 × 10²³ ≈ 6.634 × 10⁻²⁶ kg.
• Thermal wavelength: Λ ≈ 1.600 × 10⁻¹¹ m, about 0.16 ångström.
• Volume per atom: V/N = kT/P ≈ 4.116 × 10⁻²⁶ m³.
• Ratio: (V/N)/Λ³ ≈ 1.0 × 10⁷, and its natural logarithm is about 16.12.

So F per mole ≈ −RT(16.12 + 1) = −(8.314)(298.15)(17.12) ≈ −42.4 kJ/mol, measured relative to atoms at rest.

From this one formula you can get everything else. The pressure, P = −∂F/∂V = NkT/V, is just the ideal gas law. The entropy, S = −∂F/∂T, gives the Sackur–Tetrode equation; for argon under these conditions it predicts S ≈ 154.8 J/(mol·K). The experimentally measured standard molar entropy of argon is 154.8 J/(mol·K) as well. That agreement, between a calculation using only Planck's constant, the atom's mass and the temperature and a value obtained from careful heat-capacity measurements down toward absolute zero, was one of the early triumphs of quantum statistical mechanics.

As a final check, U = F + TS = −42.4 kJ + (298.15)(0.1548) kJ ≈ 3.72 kJ, which equals (3/2)RT, the familiar kinetic energy of a mole of monatomic gas.

Example 6: Helmholtz versus Gibbs for burning methane

Consider the combustion of methane at 298.15 K with liquid water as a product:

CH₄(g) + 2 O₂(g) → CO₂(g) + 2 H₂O(l)

Using standard Gibbs energies of formation from common tables (CO₂: −394.4 kJ/mol; H₂O liquid: −237.1 kJ/mol; CH₄: −50.5 kJ/mol; O₂: 0), the standard Gibbs energy change is

ΔG° ≈ −394.4 + 2(−237.1) − (−50.5) ≈ −818 kJ per mole of methane.

The Helmholtz and Gibbs energies differ by PV: G = F + PV. For a reaction, the change in PV comes mainly from the change in the number of moles of gas, Δn_gas, and treating the gases as ideal, Δ(PV) ≈ Δn_gas RT. Here three moles of gas (one CH₄ and two O₂) become one mole (CO₂), so Δn_gas = −2, and

ΔF ≈ ΔG − Δn_gas RT = −818 − (−2)(2.479) ≈ −813 kJ.

The difference is less than 1%. That is typical: for most reactions the Helmholtz and Gibbs energies are close, and they differ only by a few kJ per mole of gas created or destroyed. The choice between them is mostly about which conditions you hold fixed, as explained next.

Helmholtz versus Gibbs: which one, and when?

The two free energies answer the same type of question under different constraints.

The Gibbs energy wins in ordinary chemistry because most reactions happen in vessels open to the atmosphere, at essentially constant pressure. If a reaction makes gas, it pushes back the atmosphere and does a little PV work that is unavoidable and useless. G automatically accounts for that, so −ΔG directly gives the useful work, such as the electrical work of a battery.

The Helmholtz energy wins in other settings:

• Theoretical physics and statistical mechanics. When you model a system microscopically, the natural thing to specify is the number of particles, the volume they occupy and the temperature. That leads directly to F, through the canonical partition function described below. Most theoretical calculations produce F first, and G is obtained afterward by adding PV.
• Solids. For solids, the PV term at ordinary pressures is tiny (a mole of iron occupies about 7 cubic centimetres, so PV at 1 bar is less than 1 J), and F and G are nearly the same. Theorists often work with F because it is easier to compute.
• Closed, rigid vessels. Explosions and reactions in sealed bombs happen at nearly constant volume, with pressure changing enormously. Here F is the right potential, which is one reason the Helmholtz energy shows up in explosives research.
• Equations of state. Modern reference equations for fluids are written as formulas for F in terms of temperature and density, for reasons explained below.

The bridge to statistical mechanics

The canonical partition function

The deepest reason the Helmholtz energy matters is that it connects directly to the microscopic world. Consider a system with fixed N, V and T, in contact with a heat bath. Statistical mechanics says that the probability of finding it in a particular microscopic state r with energy E_r is

P_r = exp(−E_r / kT) / Z

where Z, the canonical partition function, is the sum of exp(−E_r / kT) over every possible microstate. The exponential factor is the Boltzmann factor: high-energy states are exponentially unlikely, but at higher temperature they become less suppressed.

From this, the average energy is U = −∂(ln Z)/∂β, where β = 1/(kT). And the free energy is

F = −kT ln Z

This is one of the most important formulas in physics. It says the Helmholtz energy is (minus kT times) the logarithm of a single number that sums over all the system's states. Everything else follows by differentiation: entropy from the temperature derivative, pressure from the volume derivative, heat capacity from a second derivative.

Example 4 above was a direct application of this formula. Example 5, the argon gas, used the same formula with Z worked out for free particles in a box.

Why the logarithm makes sense

There is a nice way to see why F = −kT ln Z. Suppose the system could only occupy states all with the same energy E, and there were Ω of them. Then Z = Ω exp(−E/kT), and

F = −kT ln Ω + E = E − T(k ln Ω).

But k ln Ω is exactly Boltzmann's formula for entropy. So F = E − TS, just as it should be. The partition function automatically does the competition between energy and entropy: states of low energy get a big Boltzmann factor, while groups of many states get a big count. The logarithm turns the product into the sum E − TS.

Free energy as a minimum over probability distributions

There is also a variational version of the minimum principle. For any trial probability distribution p_r over the microstates, define a trial free energy

F_trial = Σ p_r E_r + kT Σ p_r ln p_r

(the first term is the average energy, and the second is −T times the Gibbs entropy of the trial distribution). One can prove that F_trial is never smaller than the true F, and that it equals F exactly when p_r is the Boltzmann distribution. In other words, nature picks the probability distribution that minimises free energy.

This turns into a practical tool called the Gibbs–Bogoliubov inequality. For a complicated system you cannot solve, choose a simpler model you can solve, compute its free energy plus a correction for the difference in average energy, and the result is guaranteed to be an upper bound on the true F. Then adjust the simple model to make the bound as low as possible. This is the logic behind mean-field theory, one of the most widely used approximations in condensed-matter physics, chemistry and materials science.

Applications

Phase transitions

Phase diagrams are maps of where one phase has a lower free energy than another. At fixed temperature and volume, the stable phase is the one with the lowest F. When two phases have equal free energy per particle (more precisely, equal chemical potential and pressure), they can coexist.

A classic tool is the "common tangent construction". Plot F per mole against volume per mole at a fixed temperature for a fluid that can be liquid or gas. If the curve has a region where it bends the wrong way (concave down), the uniform fluid is unstable there and would lower its free energy by splitting into two phases. Drawing a straight line that touches the curve at two points gives the volumes of the coexisting liquid and gas, and the slope of the line gives (minus) the vapour pressure. This is the free-energy version of Maxwell's equal-area rule for the van der Waals equation.

Magnets and the Curie point

In a ferromagnet such as iron, the atomic magnetic moments prefer to line up, which lowers the energy. Lining up reduces entropy, though. At low temperatures the energy wins and the material is magnetised. Above the Curie temperature (about 770 °C for iron), the entropy term wins and the spontaneous magnetisation disappears. Landau's theory of phase transitions writes F as a power series in the magnetisation and asks where the minimum sits; as temperature rises through the Curie point, the minimum moves from a nonzero magnetisation to zero. This free-energy picture is the starting point for the modern theory of critical phenomena.

Rubber elasticity

When you stretch a rubber band, most of the restoring force is not from stretched chemical bonds. It comes from entropy. A rubber molecule is a long chain that can wiggle into an enormous number of shapes. Pulling it straighter reduces the number of available shapes and lowers its entropy. At constant temperature, the tension f in the band is related to the change in Helmholtz energy with length L:

f = (∂F/∂L) at constant T = (∂U/∂L) − T(∂S/∂L)

For an ideal rubber, the energy term is close to zero and the force is almost entirely the entropic term, −T(∂S/∂L). Two observable consequences follow, and you can feel both. A stretched rubber band held at constant length pulls harder when heated (because the force is proportional to T), and a rubber band stretched quickly against your lip feels warm (because squeezing out entropy releases heat). Both are signatures of an entropy-dominated free energy.

Equations of state for real fluids

Engineers need extremely precise properties of water, steam, refrigerants, natural gas and carbon dioxide. The modern approach writes the Helmholtz energy as a function of temperature and density, F(T, ρ), fitted to thousands of measurements. Every other property, including pressure, entropy, enthalpy, heat capacity and the speed of sound, comes out as a derivative of this one function, so they are all automatically consistent with each other.

The international standard for water used in science is the IAPWS-95 formulation, issued by the International Association for the Properties of Water and Steam in 1995. It is exactly such a Helmholtz-energy equation. Reference equations for many other fluids follow the same design.

Why Helmholtz rather than Gibbs? Because temperature and density are good variables for both liquid and vapour, and F(T, ρ) is a smooth function across the whole range, including near the critical point. G(T, P) would have a kink at the boiling line, which makes it awkward to fit with one formula.

Computer simulation and drug design

Molecular simulations of proteins, drugs and materials often aim to compute free-energy differences. Is this drug molecule more stable bound to the protein or floating in water? Which crystal structure will this pharmaceutical adopt? Which of two mutations makes an enzyme more stable? These questions are about which state has lower free energy, not lower energy, because entropy (for example, the freedom of a molecule to tumble) matters as much as bonding.

Simulations performed at fixed N, V and T compute Helmholtz free-energy differences directly. Methods with names like thermodynamic integration and free-energy perturbation are built on the statistical-mechanics formula F = −kT ln Z, cleverly rewritten so that the ratio of two partition functions, which is what matters for a difference in F, can be estimated from samples of a simulation.

Batteries and fuel cells

The connection to electrochemistry that Helmholtz himself pursued is still central. The maximum electrical work a cell can deliver per mole of reaction is the decrease in free energy (Gibbs for cells open to the atmosphere, Helmholtz for sealed rigid ones), and the cell voltage is that work divided by the charge transferred. Since the free-energy change differs from the heat of reaction by TΔS, a cell can deliver more electrical energy than the heat its reaction would release in a beaker (if ΔS is positive, the cell absorbs heat from its surroundings while running) or less (if ΔS is negative, it gives off heat even under ideal operation). This was exactly the kind of observation that broke the Thomsen–Berthelot principle.

Common misconceptions

"Free energy is free"

The word "free" does not mean "costless" or "available from nowhere". It means "available to be converted into work", as opposed to "bound". Devices advertised as producing "free energy" in the popular sense have nothing to do with the thermodynamic quantity, and the Helmholtz energy, if anything, is the tool that shows such devices cannot work: no cyclic process at constant temperature can deliver net work, because F returns to its starting value after a cycle and so ΔF = 0.

"F is the energy that is left over"

It is tempting to picture F as a portion of the internal energy that sits in some reservoir, waiting to be spent, while TS sits in another. That is not quite right. F can be negative, as in Examples 4 and 5, and its absolute value depends on the choice of energy zero. What has physical meaning is the change in F between two states at the same temperature. Example 1 makes the point vividly: the gas did work while its internal energy stayed constant, and the energy came from the heat bath, not from some "free" part of U.

"A spontaneous process always lowers the energy"

This is the Thomsen–Berthelot idea, and it is wrong in general. Spontaneous processes at fixed T and V lower F, not U. The energy can go up, provided the entropy goes up enough. Ice melting above 0 °C absorbs heat and raises its energy, yet happens spontaneously, because the liquid's higher entropy wins at that temperature.

"ΔF < 0 means it will happen quickly"

Free energy says nothing about speed. A mixture of hydrogen and oxygen at room temperature has a much lower free energy as water, but it can sit unchanged for years until a spark sets it off. Diamond at room temperature and pressure has slightly higher free energy than graphite, but diamonds do not turn into pencil lead on any human timescale. Thermodynamics tells you where the bottom of the valley is. Kinetics, the study of energy barriers and reaction rates, tells you how quickly you can get there.

"Helmholtz and Gibbs are interchangeable"

They are closely related and often numerically close, as Example 6 shows, but they apply to different constraints. Using ΔG to predict the behaviour of a sealed rigid container, or ΔF for a reaction in an open flask at atmospheric pressure, can give the right sign in most cases but the wrong number, and near equilibrium even the sign can be wrong. Pick the potential whose natural variables are the ones you are actually holding fixed.

"Constant temperature means no heat flows"

Isothermal processes typically involve plenty of heat flow; that is how the temperature is kept constant. In Example 1, about 1.7 kJ of heat flowed into the gas. The free-energy framework is useful precisely because it accounts for this heat exchange with the surroundings without making you track the surroundings explicitly.

Limits of the concept

The Helmholtz energy is an equilibrium idea. It is defined for systems in thermal equilibrium, where temperature and entropy have their usual meanings. For a system far from equilibrium (a flame, a living cell taken as a whole, a turbulent fluid), one can often apply free-energy reasoning to small pieces that are locally near equilibrium, but the whole system does not have a single well-defined F.

The simple formulas also assume the system is large. For very small systems, a few molecules or a single protein, fluctuations become significant. Individual runs of a process can briefly deliver more work than −ΔF, although on average they cannot. Modern results in non-equilibrium statistical mechanics, such as the Jarzynski equality (1997), make this precise by relating the average of exp(−W/kT) over many repetitions of a process to exp(−ΔF/kT). Experiments that stretch single RNA molecules with optical tweezers have used this relation to measure free-energy differences from irreversible pulls.

Finally, F depends on which variables you hold fixed and which kinds of work you allow. If a system can do magnetic work, electrical work or elastic work, extra terms appear in dF, and you have to be clear about which form you are using. The core idea, energy minus temperature times entropy, stays the same.

Frequently asked questions

What are the units of Helmholtz free energy?

It is an energy, so its SI unit is the joule. In chemistry it is usually quoted per mole, in kJ/mol. In statistical mechanics it is often expressed in units of kT; at room temperature, kT is about 4.1 × 10⁻²¹ J, or about 2.5 kJ/mol when multiplied by Avogadro's number.

Is F an extensive quantity?

Yes. If you double the size of a system (twice the particles in twice the volume at the same temperature), F doubles, because U and S both double. This is why it is often quoted per mole or per particle.

Why is the T in F = U − TS the temperature of the surroundings?

In the derivation of W ≤ −ΔF, the T comes from the heat bath. For a system in equilibrium with its bath, the two temperatures are the same, so the distinction rarely matters. It matters for processes where the system's temperature temporarily differs from the bath's, as in a sudden compression. The inequality still holds, as long as the initial and final states of the system are at the bath temperature.

Can ΔF be positive?

Yes, for a process that you drive by doing work on the system, such as compressing a gas or charging a battery. The minimum work required is ΔF. A process with ΔF > 0 cannot happen spontaneously at constant T and V, but it can happen if something else supplies the work.

What happens to F at absolute zero?

As T approaches zero, the TS term vanishes and F approaches U, the ground-state energy. This is consistent with the third law of thermodynamics, which says that the entropy of a perfect crystal approaches a constant (conventionally zero) as T approaches zero.

How does Helmholtz energy relate to the "grand potential" and other potentials?

All the thermodynamic potentials are related by Legendre transforms. Starting from U(S, V, N): subtract TS to get F(T, V, N); add PV to F to get G(T, P, N); subtract μN from F to get the grand potential Ω(T, V, μ), which is used for open systems that exchange particles. Each is minimised at equilibrium under its own set of fixed variables.

Why do physicists write F and chemists write A?

Historical habit. The letter A comes from the German Arbeit and is what IUPAC recommends. Many physics textbooks, including the classic statistical mechanics texts, use F. When reading any source, check the definition: in some older American chemistry books, F means the Gibbs energy instead.

Is the Helmholtz energy related to information?

There is a close connection. Since F = U − TS and entropy can be measured in bits (k ln 2 per bit), erasing one bit of information at temperature T requires at least kT ln 2 of work, a result known as Landauer's principle. At room temperature that is about 2.9 × 10⁻²¹ J per bit, far below what present-day computers use, but a genuine lower limit set by free-energy accounting.

Summary

The Helmholtz free energy F = U − TS combines energy and entropy into one number that does two jobs. First, its decrease at constant temperature sets the maximum work a system can deliver. Second, at constant temperature and volume, a system evolves toward lower F and settles at its minimum. Helmholtz introduced it in 1882 to explain why the heat of a chemical reaction is not the same as its ability to do work, correcting the old belief that only heat-releasing reactions can proceed on their own.

Mathematically, F is the Legendre transform of internal energy that swaps entropy for temperature, giving dF = −S dT − P dV + μ dN and letting pressure, entropy and chemical potential be read off as slopes. Microscopically, F = −kT ln Z links thermodynamics to the sum over all the states of a system, and the argon example shows how well that link works: a calculation from Planck's constant and an atom's mass reproduces a measured entropy to within the uncertainty of the measurement.

The same quantity sits behind phase diagrams, the vanishing of magnetism at the Curie point, the pull of a rubber band, the reference equation for the properties of water, and the free-energy calculations used to design drugs. Understanding it means understanding that the universe does not simply roll toward lower energy. It rolls toward the best trade-off between energy and disorder, at an exchange rate set by temperature.

Sources

• Helmholtz free energy (Wikipedia)
• Thermodynamik chemischer Vorgänge (Wikipedia)
• 22.1: Helmholtz Energy, Physical Chemistry (LibreTexts)/22%3A_Helmholtz_and_Gibbs_Energies/22.01%3A_Helmholtz_Energy)
• 12.3: Helmholtz Free Energy, Heat and Thermodynamics by J. B. Tatum (LibreTexts)/12%3A_Free_Energy/12.03%3A_Helmholtz_Free_Energy)
• Gibbs–Helmholtz equation (Wikipedia)
• Sackur–Tetrode equation (Wikipedia)
• IAPWS Release on the IAPWS Formulation 1995 for water and steam