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Put $\beta =1/(k_{B}T)$ and define the thermal de Broglie wavelength of species $\alpha$ byIgnoring the internal degrees of freedom as stipulated, each molecule's translational canonical partition function is $V/{\lambda }_{\alpha }^3$. The two species do not interact, so their partition functions multiply; the treatment of indistinguishable particles requires a separate factorial for each species, not a factorial of the total particle number. ThusUsing $F=-k_{B}T\ln Z$ and Stirling's approximation gives the extensive Helmholtz free energyThese are ideal gas expressions, valid for $n_{\alpha }{\lambda }_{\alpha }^3\ll 1$, with $n_{\alpha }=N_{\alpha }/V$. The energy zero has been chosen so that separated CO and oxygen molecules have zero internal reference energy.

The given binding energy $ϵ$ belongs to the formation of two carbon-dioxide molecules. Relative to the reference used in (a), the internal energy offset of one carbon-dioxide molecule is therefore $-ϵ/2$, not $-ϵ$. Its mass is $m_c=m_{\mathrm{CO}}+m_{{\mathrm{O}}_2}/2$ and its thermal de Broglie wavelength is ${\lambda }_c=h/\sqrt{2\pi m_c{k}_{B}T}$, where the subscript $c$ denotes $\mathrm{C}{\mathrm{O}}_2$.

The single-molecule canonical partition function is $(V/{\lambda }_c^3)e^{\beta ϵ/2}$. For $N_c$ indistinguishable noninteracting molecules,The corresponding Helmholtz free energy, using Stirling's approximation, isThe negative sign in the last term expresses the stabilizing binding energy and ensures that low temperature favors association.

Introduce an extent of reaction $\xi$ increasing in the dissociation direction. The allowed particle-number changes conserve both carbon and oxygen atoms:At fixed temperature and volume, the resulting Helmholtz free energy change isAt an interior chemical equilibrium, both forward and backward changes are allowed, so the coefficient must vanish:Equality of the three individual chemical potentials would be wrong: a reaction exchanges species in its stoichiometric proportions rather than molecule for molecule. Stability is also explicit here, sincewhen all three species are present. Thus the equilibrium composition minimizes the free energy along the conserved-composition constraint.

Quenched disorder is randomness held fixed while the other degrees of freedom equilibrate. For a sample with couplings $J$, first compute its partition function $Z_J$ and Helmholtz free energy $-k_{B}T\log Z_J$, then average over samples. One must not silently replace ${\mathbb{E}}_J\log Z_J$ by $\log {\mathbb{E}}_J{Z}_J$.^[ref]

For finite-range interactions and bounded wall couplings, only sites within an interaction range of the wall have altered local environments. In a regular container of linear size $L$ and dimension $d$, the surface area scales as $A\sim L^{d-1}$ while volume scales as $V\sim L^d$. A boundary contribution of order $A$ therefore gives an order-$1/L$ correction to free-energy density. Away from a critical phenomenon, a regular surface contribution often has the form $F=Vf+A\gamma +\cdots$, with $\gamma$ a wall Helmholtz free energy per area. Critical finite-size rounding, edges and long-range interactions require separate care; vanishing relative boundary size does not guarantee a simple surface expansion in every regime.

An exact check uses the zero-field nearest-neighbour Ising model on a chain of $N\ge 3$ sites with fixed coupling $J>0$ and $K=J/(k_{B}T)$. The open chain has $N-1$ bonds, while the periodic ring has $N$. Their finite partition functions and dimensionless Helmholtz free energies per spin $g_N=F_N/(N{k}_{B}T)$ areFor the open chain, sum the last Ising spin successively: each removed bond contributes $2\cosh K$, leaving a factor of two. For the ring, the transfer matrix for the one-dimensional Ising model has eigenvalues $2\cosh K$ and $2\sinh K$, and its $N$th power's trace gives the displayed sum. The two ends are a constant-size boundary in one dimension, so the open-chain total correction is $k_{B}T\log \cosh K$, and its per-spin correction is exactly proportional to $1/N$.

Both sets of finite results approach the same $g_{\infty }$ at the fixed plotted $K=0.8$. Multiplying the correction by $N$ exposes the open-chain boundary term, while the ring correction vanishes exponentially at fixed finite $K$. This model's thermodynamic limit remains a real analytic function at every positive temperature: having a thermodynamic limit does not require a phase transition.

A thermodynamic system is in thermodynamic equilibrium when its macroscopic state has no spontaneous tendency to change through any exchange allowed by its constraints. For an isolated system, stable thermodynamic equilibrium maximizes entropy over the accessible states at fixed total energy and conserved material inventories. A wall can forbid an exchange, so always specify what can cross or move. Thermal, mechanical and chemical conditions answer different questions about that same constrained state. ^[ref]

For example, two compartments separated by a rigid, impermeable wall that transmits heat can reach thermal equilibrium while retaining different pressures and compositions. Allowing an unloaded piston to move adds a mechanical equilibrium condition. Allowing one chemical species to cross adds chemical potential equality for particle exchange; allowing a chemical reaction adds its chemical equilibrium condition. A fixed barrier can therefore support a constrained thermodynamic equilibrium that ceases to be an equilibrium when the barrier is removed.

Uniformity is not the definition of thermodynamic equilibrium. An isothermal atmosphere in hydrostatic equilibrium can have a common temperature while its pressure and mass density vary with height: the pressure gradient balances gravity. In an external field, the exchange condition must include the relevant potential energy; for charged matter this leads to the electrochemical potential. Relativistic gravitational corrections to thermal equilibrium are described separately by the Tolman temperature law.

A steady state only says that the macroscopic description is constant in time. A bar maintained between hot and cold thermal reservoirs has steady heat flux and produces entropy, so it is not in thermodynamic equilibrium. Even matching temperatures at one perfect-contact interface can coexist with heat transfer through the surrounding material; matching those two local values does not establish equilibrium of the whole apparatus. See thermal conduction for that spatial transport problem.

The useful simplification is that equilibrium conditions predict an endpoint without computing every collision or flow during relaxation. The first law of thermodynamics supplies the conserved quantities, and the Second law of thermodynamics supplies the direction of change. For a system held at fixed temperature and volume, the corresponding criterion is minimization of Helmholtz free energy; at fixed temperature and pressure, it is minimization of Gibbs free energy, with the material and exchange constraints still imposed.

The derivative of internal energy at fixed deformation gradient and entropy, or of specific Helmholtz free energy at fixed temperature, defines this force. Its product with the internal-variable rate contributes mechanical dissipation to the Clausius-Duhem inequality.

A free energy is a thermodynamic potential that compares possible states under specified environmental constraints. For a closed system of fixed composition, treated as a simple compressible system with only pressure-volume work, stable thermodynamic equilibrium minimizes Helmholtz free energy at fixed temperature and volume, or Gibbs free energy at fixed temperature and pressure. The choice follows the experiment: a rigid sealed vessel and a sealed cylinder with a freely moving loaded piston impose different constraints. Both may exchange heat with a thermal reservoir. See W. C. Carter, MIT thermodynamics notes, sections 2-1.14 and 2-1.19.

The useful insight is that internal energy alone cannot rank these alternatives: exchanging heat also changes the surroundings' entropy. Subtracting $TS$ packages that accounting into a property of the chosen system. Adding $pV$ additionally accounts for displacing the surroundings at fixed pressure. A free energy is a state function; supplied heat describes a transfer during a process. Under the stated constraints, supplied heat is $Q=\Delta U$ at fixed volume or $Q=\Delta H$ at fixed pressure, whereas neither $\Delta F$ nor $\Delta G$ generally equals $Q$.

A lower free energy identifies a thermodynamic preference, without specifying a timescale. A barrier can trap a system in a local minimum, producing metastability although a lower global minimum exists. The enzyme example illustrates the corresponding distinction for chemical reactions: changing the route can change the reaction rate while preserving the endpoint Gibbs free energy difference and chemical equilibrium.

The thermodynamic limit describes a sequence of larger systems, not a single enormous specimen: particle number $N$ and volume $V$ tend to infinity with number density $n=N/V$ fixed. Hold the chosen intensive and extensive thermodynamic quantities such as temperature $T$ and an applied magnetic field fixed along that sequence. In the canonical ensemble the controls are $T,N,V$; in the grand canonical ensemble one instead fixes $T$ and chemical potential, and the limiting number density is a result, possibly with a jump at phase coexistence. These are different specifications of a limiting experiment.

The useful limit is usually an intensive and extensive thermodynamic quantities normalization of the Helmholtz free energy, not a finite limit of the total extensive quantity itself:Here $Z_{N,V}$ is the canonical partition function, $k_B$ is the Boltzmann constant, $f$ is free-energy density, and $a$ is Helmholtz free energy per particle. Thus $F_{N,V}=Vf+o(V)$ when the limit exists; finite systems need not be exactly extensive quantities because of the surface correction to thermodynamic free energy.

Specify the interactions and the growing containers too. For a conventional fluid, use a fixed stable pair potential with a tempered pair potential tail, and containers whose boundary layer of any fixed thickness occupies a vanishing fraction of the volume; uniformly enlarged cubes work, a tube of fixed cross-section does not give the same bulk limit. For a finite-range Ising model, keep the lattice spacing and bond coupling fixed while enlarging all intended spatial directions. Ordinary bounded wall interactions or periodic boundary conditions can then give the same bulk free-energy density, although boundary conditions can select different spontaneous symmetry breaking states. This is a set of hypotheses, not a theorem that every interaction has a boundary-independent limit.^[ref]

The Curie–Weiss model below deliberately uses a different limiting prescription: its all-to-all coupling is divided by $N$. Without that factor its interaction energy would grow as $N^2$, rather than as an extensive quantity. The model has no spatial wall or correlation length to compare with a container; it is an exactly summable illustration of finite-size rounding, not a quantitative model of the fluid experiment below.

A state function used to describe a thermodynamic system with a chosen set of natural variables. Internal energy, enthalpy, Helmholtz free energy, and Gibbs free energy use different variables and serve different environmental constraints.

For a film described by $dU=TdS+\gamma dA$, with material content fixed, its Helmholtz free energy obeys $dF=-SdT+\gamma dA$. Equality of mixed partial derivatives gives the Maxwell relation $(\partial S/\partial A{)}_T=-(\partial \gamma /\partial T{)}_A$. If $\gamma$ depends only on temperature, write ${\gamma }^{\prime }=d\gamma /dT$; thenHere $C_A$ is the heat capacity at constant area. For $C_A>0$ and ${\gamma }^{\prime }<0$, reversible adiabatic stretching cools the film, since $dT/dA=T{\gamma }^{\prime }/C_A<0$, although positive mechanical input can increase its total internal energy. If $\gamma ={\gamma }_0-\beta T$ with constant positive $\beta$ and constant $C_A$, the isentropic process satisfies $C_A\ln (T/T_{\ast })+\beta A=\mathrm{constant}$, where $T_{\ast }$ is an arbitrary fixed reference temperature. The work convention uses the total area conjugate to $\gamma$; a two-surface film described by projected area requires the corresponding factor of two consistently throughout.

Thermoelasticity couples elasticity and temperature through a specific Helmholtz free energy $\psi (F,\theta )$. Without constraints its reversible laws are $P_{Ii}={\rho }_0{\psi }_{,F_{iI}}$ and $\eta =-{\psi }_{,\theta }$. Mixed derivatives give the same coupling tensor in the thermal stress response and the strain dependence of specific entropy, with opposite signs.

Let $fdx$ be reversible work supplied to stretch a band. Its Helmholtz free energy obeys $dF=-SdT+fdx$. Equality of mixed derivatives gives the displayed relation. An entropy decrease on stretching therefore corresponds to increasing tensile force on heating at fixed extension. The sign follows the force/work convention; a restoring force exerted by the band has the opposite sign.

For a free molecule of mass $m$ in a large three-dimensional volume $V$, replacing the translational state sum by an integral gives the displayed partition function. For identical classical molecules the many-molecule canonical partition function is $q_{\mathrm{trans}}^N/N!$, not simply $q_{\mathrm{trans}}^N$. Differentiating its Helmholtz free energy at fixed volume gives the ideal-gas translational entropy.

Use the independent spin-one-half two-level system with fixed $N$ distinguishable sites and Hamiltonian levels $E_{\pm }=\pm \mu B$, where $\mu >0$ is the magnitude of the field-axis magnetic moment and $B>0$ is a uniform externally controlled magnetic field. Assume positive temperature, no interactions between sites, and no appreciable lattice or load entropy. The two levels differ by $2\mu B$. With inverse temperature $\beta$, the canonical ensemble gives the one-site and total partition functionsHere the minus subscript denotes the lower energy $-\mu B$. Factorization of the partition function requires additive energy and independent sites; there is no particle-exchange factor $N!$ for these fixed distinguishable sites.

Differentiate the partition function, holding $B$ fixed when differentiating with respect to $\beta$. The internal energy, Helmholtz free energy, total field-axis magnetic moment $m$, and magnetic entropy areThe extensive moment $m$ is distinct from magnetization, the moment per unit volume. Direct substitution of $p_{\pm }$ into the Gibbs entropy gives the same $s(x)$ independently of the free-energy derivative.

At fixed $B$, the magnetic heat capacity isStrict decrease makes $s$ invertible on $x>0$. Constant magnetic entropy therefore fixes $x$, not $B$ or $T$ separately:The partition function may be used to label these equilibrium states even though the isolated cooling leg has no external heat bath. Actual thermalization of the lattice and load additionally requires a sufficiently fast spin-lattice relaxation; perfectly uncoupled spins by themselves cannot refrigerate an attached object.

The limiting cases are instructive. For $x\ll 1$, $s(x)=\ln 2-x^2/2+O(x^4)$, $m\simeq N{\mu }^{2}B/(k_{B}T)$, and $C_{B,\mathrm{mag}}\simeq N{k}_B{x}^2$. The moment response reproduces Curie's law. For $x\gg 1$, $s(x)\simeq (1+2x)e^{-2x}$ and the upper-state occupation is exponentially small. Taking $B\to 0$ at fixed $T>0$ gives $s\to \ln 2$; taking $T\to 0$ at fixed $B>0$ gives $s\to 0$. The origin has a path-dependent limit. Extrapolating a line of constant $x$ to $B=T=0$ does not construct a usable zero-temperature state or prove that a real material retains independent degenerate spins there.

Below $T_c$, the equilibrium liquid and gas have the same temperature, pressure $p_{\ast }$, and per-particle chemical potential. For macroscopic bulk thermodynamic phases, neglect interface contributions. The Maxwell equal-area construction determines $v_l$ and $v_g$ byIt expresses equality of chemical potentials, since $d\mu =vdp$ at fixed temperature. Equivalently, the equilibrium Helmholtz free energy per particle follows a common tangent line between the two endpoint states, lying below the homogeneous continuation between them. The construction uses that mathematical continuation to compare endpoints; it is not a physical compression path through negative isothermal compressibility states.^[ref]

Here is a reproducible calculated example at $t=0.9$. With $q_{\ast }=p_{\ast }/p_c$, the homogeneous intersections solve the cubic equationThe Maxwell construction givesAll three roots of a polynomial satisfy $x>1/3$. Only the smallest and largest are the phase coexistence endpoints; the middle one has negative isothermal compressibility. Equal pressures at three admissible values of volume therefore do not imply three stable thermodynamic phases. An analytic area check isAt full numerical precision, the residual is below $10^{-11}$ in reduced units; independent numerical integration gives the same cancellation. At $t=0.9$, the spinodal points are instead $x\simeq 0.718597189$ and $1.528504964$.

The equilibrium plateau changes the phase fraction rather than the endpoint densities. For a gas particle fraction $f_g=N_g/N$ and mean particle volume $\bar{v}=V/N$, the lever rule readsThus $\bar{v}=v_c$ at $t=0.9$ gives $f_g\simeq 0.227219$: this is a particle fraction, not the volume fraction occupied by gas. Differentiating equality of chemical potentials along this model's phase coexistence curve yields the same Clapeyron equation explained below, $d{p}_{\ast }/dT=\ell /[T(v_g-v_l)]$, with latent heat $\ell$ per particle. Both endpoint differences vanish as the Van der Waals critical point is approached; the equation's critical limit must be taken as a ratio.

The Maxwell construction presumes a pure bulk fluid at thermodynamic equilibrium. Its endpoint equality does not specify how rapidly a real compression reaches equilibrium. A homogeneous metastable sample can persist while nucleation is delayed; the spinodal marks loss of local mechanical stability in this mean-field theory, rather than the equilibrium onset of condensation. Surface tension, finite sample size and impurities can affect the observed transition. These qualifications matter when comparing a mathematical Van der Waals isotherm with a visible liquid meniscus.