First-Order Phase Transitions

Boil a kettle and the water's temperature stalls at 100 °C no matter how hard you heat it: every joule you pour in goes not into raising the temperature but into ripping liquid apart into vapour. That stall — a plateau on the heating curve, a slug of energy swallowed at fixed temperature — is the fingerprint of a first-order phase transition. In Ehrenfest's classification, a transition is first order when a first derivative of the relevant free energy is discontinuous. For the Gibbs free energy $G(T,p)$ the first derivatives are the entropy $S=-(\partial G/\partial T)_p$ and the volume $V=(\partial G/\partial p)_T$ ; a jump in either one is the transition. The chemical potential $\mu=G/N$ stays continuous — the two phases meet with equal $\mu$ — but its slope kinks, and that kink is the entropy discontinuity. The entropy jump $\Delta S$ is not an abstraction: multiply it by the temperature and you get the latent heat $L=T\,\Delta S$ , the energy released or absorbed while the phases interconvert. Because two distinct phases genuinely coexist along a line in the $(T,p)$ plane — liquid beside gas, solid beside liquid, up-domain beside down-domain — the order parameter, density, and entropy all jump discontinuously across it. The correlation length stays finite; there are no diverging fluctuations, no critical exponents (except at an isolated critical end point where the coexistence line terminates). This article owns the structural claim that defines the class: a first-order transition is the crossing of two free-energy minima that stay separated in order-parameter space, forcing a discontinuous jump and a latent heat, with metastable superheating and supercooling as generic companions.

Key takeaways

A first-order transition is a discontinuity in a first derivative of the free energy — entropy $S$ , volume $V$ , or magnetization $M$ jumps across it, while $G$ (and $\mu$ ) stays continuous.
The entropy jump gives the latent heat $L=T\,\Delta S$ , energy exchanged at fixed $T$ as one phase converts to the other — the plateau on a heating curve.
Two phases coexist along a line, meeting at equal chemical potential $\mu_1(T,p)=\mu_2(T,p)$ ; the slope of that line is set by Clausius–Clapeyron.
The free energy has two competing minima; the transition fires when they become degenerate, and metastability (superheating, supercooling) is unavoidable because the minima are separated by a barrier.
The correlation length stays finite — no critical divergences — which is exactly what distinguishes first-order from continuous transitions.
Key idea: a first-order transition is two free-energy minima exchanging global stability while staying apart in order-parameter space, so the order parameter and entropy jump discontinuously and a latent heat $L=T\,\Delta S$ is released.

The core insight

A first-order transition happens when two free-energy minima exchange which one is lowest while remaining separated in order-parameter space. The order parameter, entropy, and density therefore jump discontinuously at the transition, the two phases can coexist at equal chemical potential, and the entropy jump appears as latent heat $L=T\,\Delta S$ .

Where this lives today

First-order transitions are the workhorse of materials engineering and Earth science: the latent heat of fusion is why phase-change materials in building walls and spacecraft buffer temperature swings, storing energy at a fixed melting point. The metastability that accompanies them — supercooling, superheating, nucleation past a barrier — governs cloud formation (supercooled droplets seed rain), steel quenching (trapping a metastable phase for hardness), and the explosive boiling of superheated liquids. In cosmology, the early universe is thought to have passed through first-order transitions (electroweak, QCD) whose bubble nucleation could seed gravitational waves now being hunted by pulsar-timing arrays and LISA. The same two-competing-minima mathematics reappears wherever a system must choose discontinuously between states.

The physical picture

Picture the Gibbs free energy per particle, $g(T,p)$ , as a landscape over the order parameter — say the density $\rho$ of a fluid. Well above the boiling line only one valley exists (gas); well below, only the other (liquid). In between, both valleys are present, and the phase that is realized is simply the deeper one. A first-order transition is the instant the two valleys sit at exactly the same depth. On one side the liquid valley wins, on the other the gas valley wins, and at the crossing the system can sit in either — the two phases coexist. Crucially the valleys are separated: there is a hump between them, so the order parameter must leap across a gap rather than slide continuously. That leap is the discontinuity. The cleanest way to see the transition is to plot $g$ against temperature at fixed pressure. Each phase contributes its own nearly straight line, $g_i(T)\approx e_i - T s_i$ , with slope $-s_i$ set by that phase's entropy. The two lines cross at the transition temperature $T_{\text{tr}}$ . Below $T_{\text{tr}}$ the low-entropy (ordered) line lies lower; above it the high-entropy line does. The physical free energy is the lower envelope — and at the crossing it has a kink: its slope, which is $-S$ , jumps by $-(s_2-s_1)$ . That kink is the whole phenomenon. A continuous transition, by contrast, has the two curves meet tangentially, with no slope discontinuity.

Figure

Gibbs free energy g(T) at fixed pressure: two phase branches (blue ordered, low entropy; red disordered, high entropy) cross at T_tr. The physical g is the lower envelope (thick), which has a KINK at T_tr where the slope −S jumps. A smooth tangential meeting (grey dashed) would be continuous instead.

The second picture is the free-energy landscape itself as temperature is tuned through $T_{\text{tr}}$ . Below the transition the ordered minimum is the global one; a shallower disordered minimum survives as a metastable state. As $T$ rises the two minima trade depths, passing through exact degeneracy at $T_{\text{tr}}$ , and above it the disordered minimum wins. Because a barrier separates them, a system prepared in the now-metastable well does not instantly relax — it superheats or supercools until either a fluctuation kicks it over the barrier or the barrier vanishes. This is why you can chill pure water below 0 °C or superheat it in a clean microwave mug: the stable phase exists, but the system is trapped behind the hump.

Figure

Landau-type free energy with a cubic term, f(m)=a m² − c m³ + b m⁴, as temperature falls (a decreasing). A second minimum at m≠0 appears, deepens, becomes degenerate with m=0 at the transition, then wins — the order parameter JUMPS from 0 to m* discontinuously (first order), with a barrier giving metastability.

The core equations

The Ehrenfest classification rests on the first derivatives of the Gibbs free energy. Holding the natural variables $(T,p)$ , entropy and volume are recovered by differentiation, and a first-order transition is defined by a discontinuity in one of them.

S=-\left(\frac{\partial G}{\partial T}\right)_p,\qquad V=\left(\frac{\partial G}{\partial p}\right)_T

where: $G(T,p)$ is the Gibbs free energy, $S$ the entropy, $V$ the volume. Physically, a jump $\Delta S$ or $\Delta V$ across the transition means the two phases store energy and occupy space differently. Because $S$ and $V$ are first derivatives of $G$ , their discontinuity is the "first order" of the name. At the transition the two phases genuinely coexist, which pins the thermodynamic condition. Coexistence means neither phase can lower the total free energy by converting into the other, so their chemical potentials are equal.

\mu_1(T,p)=\mu_2(T,p)\quad\Longleftrightarrow\quad g_1(T,p)=g_2(T,p)

where: $\mu_i=g_i=G_i/N_i$ is the Gibbs free energy per particle (chemical potential) of phase $i$ . Physically, this is why $\mu$ itself is continuous through the transition even though its slope is not: the two phases meet at a shared value of $\mu$ , and it is the derivative $\partial\mu/\partial T=-s$ that jumps. The entropy discontinuity is the latent heat. Because the free energies are equal but their temperature-slopes differ, the entropy jump follows from differentiating the coexistence condition, and multiplying by temperature gives the heat exchanged.

L=T\,\Delta S=T\,(s_2-s_1)

where: $L$ is the latent heat (per particle or per mole), $\Delta S=s_2-s_1$ the entropy difference between the two coexisting phases, and $T=T_{\text{tr}}$ the transition temperature. Physically, $L$ is the energy that must be supplied (or is released) to convert one phase entirely into the other at fixed $T$ — the flat plateau on a heating curve. For melting and vaporization the higher-temperature phase has the larger entropy, so $\Delta S>0$ and $L>0$ : heat is absorbed on the way up.

Derivation ALatent heat from the coexistence condition

We want to prove that the latent heat is fixed entirely by the entropy discontinuity, $L=T\,\Delta S$ , and to see precisely why $G$ can be continuous while its first derivative jumps. The strategy is to sit on the coexistence line where $g_1=g_2$ , differentiate the difference of the two Gibbs free energies, and read off both the continuity of $g$ and the jump in its temperature-derivative $-s$ . This turns the qualitative "two curves crossing" picture into an exact thermodynamic identity, and it is the identity that Clausius–Clapeyron will later differentiate a second time.

Step 1. Model each phase near the transition by its Gibbs free energy per particle $g_i(T)=e_i-T s_i$ at fixed pressure, where $e_i$ and $s_i$ are (locally constant) energy and entropy parameters. The physical free energy is the lower of the two, $g=\min(g_1,g_2)$ . Why this form: near $T_{\text{tr}}$ each branch is smooth, so a first-order Taylor expansion with slope $-s_i=(\partial g_i/\partial T)_p$ captures the entropy of that phase.

Step 2. Locate the transition by imposing coexistence $g_1(T_{\text{tr}})=g_2(T_{\text{tr}})$ , i.e. $e_1-T_{\text{tr}}s_1=e_2-T_{\text{tr}}s_2$ , which solves to $T_{\text{tr}}=(e_2-e_1)/(s_2-s_1)$ . Why: coexistence is the statement $\mu_1=\mu_2$ ; below and above $T_{\text{tr}}$ a different branch is lower, so $T_{\text{tr}}$ is exactly where global stability changes hands.

Step 3. Confirm $g$ is continuous but its slope is not. At $T_{\text{tr}}$ both branches share the same value, so $g$ has no jump; the chemical potential is continuous. The temperature-derivative, however, is $-s_1$ just below and $-s_2$ just above, so $(\partial g/\partial T)_p=-S$ jumps by $-(s_2-s_1)$ . Why this is the Ehrenfest criterion: the lowest derivative of $G$ that is discontinuous is the first one, hence "first order."

Step 4. Extract the latent heat. The entropy discontinuity is $\Delta S=s_2-s_1=-\left.\dfrac{d}{dT}(g_2-g_1)\right|_{T_{\text{tr}}}$ , and the heat absorbed at fixed $T$ while converting phase 1 into phase 2 is $L=T_{\text{tr}}\,\Delta S$ . Why $L=T\Delta S$ and not $\Delta E$ : at fixed $T$ and $p$ the reversible heat is $T\,dS$ , and the whole entropy change is the discontinuity $\Delta S$ , so integrating gives $L=T_{\text{tr}}\Delta S$ directly. The result $L=T\,\Delta S$ with $\Delta S=s_2-s_1$ is exactly what a symbolic check confirms: on the two-branch model $g_1=g_2$ at $T_{\text{tr}}$ (equal $\mu$ ), the first derivative jumps by $-(s_2-s_1)$ , and $L=T_{\text{tr}}(s_2-s_1)>0$ whenever the higher-temperature phase carries more entropy. Verified symbolically in verify_first-order-phase-transitions.py (coexistence, first-derivative jump, $L=T\Delta S$ , and $L>0$ for $s_2>s_1$ ).

Go deeper · Derivation B · 2 minHow a cubic term makes a transition first order

Derivation BHow a cubic term makes a transition first order

Landau theory expands the free energy in powers of an order parameter $m$ ; a continuous transition comes from the pure quartic $f=a\,t\,m^2+b\,m^4$ . We want to show, in the same language, exactly what ingredient turns the transition first order: a symmetry-allowed cubic term $-c\,m^3$ . The payoff is a clean criterion — the transition occurs while the quadratic coefficient is still positive, and the order parameter jumps by a finite amount $m^*$ — which is why systems whose symmetry permits a cubic (e.g. the liquid–solid transition, the three-state Potts model) are generically first order.

Step 1. Write $f(m)=a\,m^2-c\,m^3+b\,m^4$ with $b,c>0$ and $a$ decreasing as the system is cooled. Why include $-c\,m^3$ : if the order parameter's symmetry does not forbid odd powers, the cubic is allowed, and it breaks the $m\to-m$ symmetry that protected the pure-quartic continuous transition.

Step 2. A first-order transition is the moment a nonzero minimum $m^*>0$ becomes degenerate with the $m=0$ minimum, i.e. $f(m^*)=f(0)=0$ and $f'(m^*)=0$ hold simultaneously while $a>0$ . Solving these two conditions gives $m^*=\dfrac{c}{2b}$ and the transition value $a^*=\dfrac{c^2}{4b}$ . Why two conditions: one ( $f'=0$ ) makes $m^*$ a stationary point, the other ( $f=f(0)$ ) makes it exactly as deep as the disordered minimum.

Step 3. Read off the jump. Because $a^*=c^2/4b>0$ at the transition, the ordered minimum arrives before $a$ reaches zero, and it arrives already at a finite location $m^*=c/2b\ne0$ . As the deeper minimum takes over, the equilibrium order parameter leaps discontinuously from $0$ to $m^*$ . Why this is first order: the order parameter (a first derivative of the free energy with respect to its conjugate field) is discontinuous — the defining signature.

Step 4. Contrast with the pure quartic $c=0$ : then the nonzero minima are $m=\pm\sqrt{-a/2b}$ , which exist only once $a<0$ and grow continuously from zero — no jump, no latent heat. Why the difference matters: it shows the discontinuity is not generic to Landau theory but is switched on precisely by the odd-power term (or, equivalently, by an applied field that biases one phase). Verified symbolically in verify_first-order-phase-transitions.py: the cubic model has degenerate minima at $m^*=c/2b$ , $a=c^2/4b$ , with a discontinuous jump, while the pure quartic gives the continuous $m=\pm\sqrt{-a/2b}$ .

Limiting cases

case	result	meaning
$\Delta S\to0$ (entropy jump vanishes)	$L=T\,\Delta S\to0$ ; transition becomes continuous	the coexistence line ends at a critical point where first-order character is lost
$\Delta V\to0$ at fixed $\Delta S$	Clausius–Clapeyron slope $dp/dT=L/(T\Delta V)\to\infty$	vertical coexistence line (e.g. isotropic phase changes with negligible volume change)
cubic coefficient $c\to0$ in Landau $f$	jump $m^*=c/2b\to0$ ; transition becomes continuous	the first-order line terminates at a tricritical point
barrier height $\to0$	metastable region shrinks to zero	superheating/supercooling disappears; spinodal meets binodal
high- $T$ phase entropy $s_2>s_1$	$L>0$ : heat absorbed on heating	melting, vaporization, sublimation

Worked examples

Example 1 — Latent heat of vaporizing water into its entropy jump. Water boils at $T_b=373.15$ K with a molar latent heat of vaporization $L_{\text{vap}}=40.66$ kJ/mol. Find the entropy discontinuity $\Delta S$ between vapour and liquid, and check it against Trouton's rule.

Step 1. Apply $L=T\,\Delta S$ rearranged: $\Delta S = L_{\text{vap}}/T_b$ . Why: the entire entropy change happens at fixed $T_b$ , so the reversible heat $L$ divides by $T_b$ to give $\Delta S$ .

Step 2. Compute: $\Delta S = 40660\ \text{J}\,\cdot\,\text{mol}^{-1}/373.15\ \text{K} = \mathbf{108.97\ \text{J}\,\cdot\,\text{mol}^{-1}\text{K}^{-1}}$ .

Step 3. Compare with Trouton's rule ( $\Delta S_{\text{vap}}\approx88$ J·mol⁻¹K⁻¹ for "normal" liquids). Water's value is higher, because hydrogen bonding makes the liquid unusually ordered, so vaporizing it liberates extra entropy. The excess over Trouton is the thermodynamic signature of the hydrogen-bond network.

Example 2 — Melting ice and the sign of the entropy jump. The molar latent heat of fusion of ice is $L_{\text{fus}}=6.01$ kJ/mol at $T_m=273.15$ K. Find $\Delta S$ for melting and confirm its sign.

Step 1. $\Delta S_{\text{fus}}=L_{\text{fus}}/T_m = 6010\ \text{J}\,\cdot\,\text{mol}^{-1}/273.15\ \text{K}$ .

Step 2. Compute: $\Delta S_{\text{fus}} = \mathbf{22.00\ \text{J}\,\cdot\,\text{mol}^{-1}\text{K}^{-1}} > 0$ .

Step 3. Interpret the sign: the higher-temperature phase (liquid) has the larger entropy, so $\Delta S>0$ and $L>0$ — heat is absorbed on melting. This is the generic sign for melting, vaporization, and sublimation, and it is exactly what fixes the upward slope of most coexistence lines through Clausius–Clapeyron. (Ice–water is the famous exception in volume, $\Delta V<0$ , which flips the pressure-slope of the melting line, but not the entropy sign.)

Origins · 1 minfirst-order phase transition and how this was discovered

first-order phase transition

classification framework, 1930s · The modern label "first order" comes from Paul Ehrenfest's 1933 scheme, which sorts a transition by the lowest derivative of the free energy that is discontinuous. A first-order transition — the oldest kind studied, from the boiling and freezing known since antiquity to Andrews's 1869 measurements of the carbon-dioxide critical point — is one where a first derivative jumps, releasing latent heat. Ehrenfest's higher-order categories proved incomplete for real continuous transitions, but the first-order class he named remains exactly as he defined it: a discontinuity in entropy or volume, latent heat, and phase coexistence.

History

1869

— Thomas Andrews maps the liquid–gas coexistence of carbon dioxide and identifies the critical point where the first-order line ends

1873

— Johannes Diderik van der Waals derives an equation of state with a liquid–gas coexistence loop

1875

— James Clerk Maxwell formulates the equal-area construction fixing coexistence in the van der Waals loop

1933

— Paul Ehrenfest proposes classifying transitions by the lowest discontinuous derivative of the free energy, coining "first order"

1937

— Lev Landau frames transitions through an order-parameter free-energy expansion, showing a cubic term forces first order

Going further

Standard treatments: Pathria & Beale, Statistical Mechanics, 3rd ed., Ch. 12–13; Kardar, Statistical Physics of Particles, Ch. 4–5; Reif, Fundamentals of Statistical and Thermal Physics, Ch. 8; Sethna, Statistical Mechanics: Entropy, Order Parameters, and Complexity, Ch. 11; Landau & Lifshitz, Statistical Physics Part 1, §§81–83, 143. Prerequisites: Order Parameter (the jump that defines the transition is a jump in the order parameter or its conjugate); Latent Heat and Phase Coexistence (the entropy discontinuity made quantitative, and the hallmark of first-order lines); Landau Theory of Phase Transitions (the two-minimum free-energy expansion that models the jump). Leads to: Clausius–Clapeyron Relation (quantifies the slope of the coexistence line from latent heat and volume change); Continuous Phase Transitions (the complementary class with continuous first derivatives and diverging correlation length, where first-order lines often terminate); Universality and Universality Classes (why first-order transitions are non-universal while their critical end points belong to universality classes).

Why is the chemical potential continuous if entropy jumps?

The chemical potential $\mu=g$ is the Gibbs free energy per particle, and entropy is its temperature-derivative, $s=-(\partial\mu/\partial T)_p$ . Across a first-order transition the two phases meet at the same value of $\mu$ (that is the coexistence condition $\mu_1=\mu_2$ ), so $\mu$ itself is continuous. What jumps is the slope $-s$ : the $\mu(T)$ curve has a kink, not a step. So " $G$ continuous, first derivative discontinuous" is not a contradiction — it is exactly the geometry of two straight lines crossing.

Is a first-order transition always accompanied by symmetry breaking?

No. The liquid–gas transition (away from the critical point) is first order but breaks no symmetry — liquid and gas have the same symmetry, they just differ in density. Conversely, many continuous transitions do break a symmetry (a ferromagnet picks a direction) with zero latent heat. Order (of a transition) and symmetry breaking are independent classifications; conflating them is a common error. Landau theory clarifies this: a cubic term or an applied field can make a symmetry-breaking transition first order, while its absence leaves it continuous.

What exactly is metastability here?

When two free-energy minima are separated by a barrier, a system sitting in the higher (metastable) minimum cannot instantly reach the lower (stable) one — it must nucleate a droplet of the new phase and pay the surface-energy cost of the barrier. Until a large enough fluctuation appears, the system persists in the wrong phase: superheated liquid, supercooled vapour, supercooled water below 0 °C. This is unique to first-order transitions; continuous transitions have no barrier and no metastability, because the minima are not separated.

How does this connect to Maxwell's equal-area construction?

For a fluid described by the van der Waals equation, the raw isotherm below the critical temperature is non-monotonic (an unphysical loop). Maxwell's equal-area rule replaces the loop with a horizontal coexistence tie-line drawn so the areas above and below are equal. That rule is precisely the statement $\mu_{\text{liquid}}=\mu_{\text{gas}}$ — equal chemical potentials — which is our coexistence condition. So the equal-area construction is the fluid-language version of "two free-energy minima at equal depth," and it fixes the coexistence pressure and the two coexisting densities.

Does the correlation length really stay finite at a first-order transition?

Yes — and this is the sharp operational difference from continuous transitions. Because the two phases are separated in order-parameter space and neither becomes marginally unstable, fluctuations stay bounded and correlations decay over a finite length. There are no diverging susceptibilities and no critical exponents at a generic first-order transition. The exception is the critical end point where a coexistence line terminates: there the entropy and volume jumps shrink to zero, the correlation length finally diverges, and the transition becomes continuous.

Why is water's entropy of vaporization larger than Trouton's rule predicts?

Trouton's rule ( $\Delta S_{\text{vap}}\approx88$ J·mol⁻¹K⁻¹) works for liquids whose molecules are only weakly, isotropically bound. Water's molecules form a hydrogen-bond network that makes the liquid unusually ordered (low entropy). Vaporizing it therefore liberates extra entropy — the value we computed, $\approx109$ J·mol⁻¹K⁻¹, exceeds Trouton by the entropy of the broken H-bond structure. So the deviation from Trouton is a quantitative fingerprint of hydrogen bonding, readable straight off the latent heat.

What happens to the latent heat as you approach the critical point?

As the coexistence line approaches its terminus (the critical point), the two phases become more and more alike: their density, entropy, and order-parameter differences all shrink. The entropy jump $\Delta S\to0$ , so the latent heat $L=T\,\Delta S\to0$ continuously. At the critical point itself the transition is no longer first order — it is continuous, with diverging fluctuations and critical exponents. The vanishing of $L$ is thus the bridge from the first-order world (this article) to the continuous-transition world.