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A first-order phase transition has a discontinuity in a first derivative of the equilibrium free energy, such as entropy or the order parameter; a thermal first-order transition normally has latent heat. At a continuous phase transition the order parameter goes continuously to zero and there is no latent heat, although response functions and higher derivatives of the free energy can be singular.

For the symmetric scalar Landau free energy at zero conjugate field, minimizeIf $u>0$, the sixth-order term is unnecessary near $r=0$. For $r>0$ the minimum is $M=0$, whereas for $r<0$ there are two symmetry-related minima with $M^2=-r/u$ to leading order. The order parameter vanishes continuously as $r\uparrow 0$. The equilibrium free-energy density changes from zero to $-r^2/(4u)$, so its first temperature derivative is continuous and its second derivative has a jump. This is the ordinary mean-field continuous phase transition.

If $u<0$, keep $v>0$ for stability. A nonzero stationary order parameter, with $s=M^2$, obeys $r+us+v{s}^2=0$. For phase coexistence with $M=0$ its free energy must also vanish. Substituting $r=-us-v{s}^2$ into $V$ givesThe nonzero solution of $V=0$ is thereforeThe competing minima exchange stability at positive $r_t$, with a finite jump of the order parameter. This is a first-order phase transition. The metastability region is also visible: nonzero stationary points first appear at the spinodal $r=u^2/(4v)$, and the disordered local minimum loses stability at $r=0$.

For example, holding $u,v$ fixed and taking $r=a(T-T_0)$, the ordered phase has entropy lower than the disordered phase by $a{M}_t^2/2$ at coexistence. Heating through the transition absorbs latent heat per volume $T_{t}a{M}_t^2/2$. More generally the temperature dependence of all coefficients contributes to this entropy difference. With no inversion symmetry, a cubic term can also produce a first-order phase transition, as comparison of the competing minima of the Landau free energy shows.

The spinodal is the boundary where a metastable phase loses local stability. In a mean-field free-energy description the relevant curvature vanishes there, separating local minima from unstable configurations.

A Van der Waals isotherm is the pressure–particle volume curve at a fixed temperature. Put $x=v/v_c$, $t=T/T_c$, and $q=p/p_c$, using the Van der Waals critical point values below. Then the existing Van der Waals law of corresponding states becomesAll curves in the figure are calculations in this reduced model.

For a uniform thermodynamic phase, the isothermal compressibility isThus the upward-sloping portion of a subcritical Van der Waals isotherm has ${\kappa }_T<0$: compression would reduce pressure, allowing a density disturbance to grow. It is mechanically unstable. The spinodal points bound that portion; setting the derivative $\partial q/\partial x$ to zero gives their curve $t=(3x-1{)}^2/(4x^3)$. Negative slope alone guarantees only local mechanical stability: portions between the spinodal and the phase coexistence endpoints have metastability. Equilibrium replaces the entire interval between those endpoints, including these metastable portions, by a Maxwell construction plateau.^[ref]

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.