In thermodynamic equilibrium, a necessary condition for stability is that pressure, ${\displaystyle p}$, does not increase with molar volume, ${\displaystyle v=V/N}$; this is expressed mathematically as ${\displaystyle \partial _{v}p|_{T}<0}$, where ${\displaystyle T}$ is the temperature.^[1]

This basic stability requirement, and similar ones for other conjugate pairs of variables, is violated in analytic models of first order phase transitions. The most famous case is the van der Waals equation,^[2][3]

$${\displaystyle p=RT/(v-b)-a/v^{2))$$

${\displaystyle a,b,R}$

Figure 1 shows an isotherm drawn, for ${\displaystyle v>b}$, as a continuously differentiable solid black, dotted black, and dashed gray curve. The decreasing right hand part of the curve in Fig. 1 describes a gas, while its left decreasing part describes a liquid. These two parts are separated by a region between the local minimum and local maximum on the curve with positive slope that violates the stability criterion. This mathematical criterion expresses a physical condition which Epstein^[4] described as follows:

It is obvious that this middle part, dotted in our curves [dashed in Fig.1 here], can have no physical reality. In fact, let us imagine the fluid in a state corresponding to this part of the curve contained in a heat conducting vertical cylinder whose top is formed by a piston. The piston can slide up and down in the cylinder, and we put on it a load exactly balancing the pressure of the gas. If we take a little weight off the piston, there will no longer be equilibrium and it will begin to move upward. However, as it moves the volume of the gas increases and with it its pressure. The resultant force on the piston gets larger, retaining its upward direction. The piston will, therefore, continue to move and the gas to expand until it reaches the state represented by the maximum of the isotherm. Vice versa, if we add ever so little to the load of the balanced piston, the gas will collapse to the state corresponding to the minimum of the isotherm`

This situation is similar to a body exactly balanced at the top of a smooth surface that, with the slightest disturbance will depart from its equilibrium position and continue until it reaches a local minimum. As they are described such states are dynamically unstable, and consequently they are not observed. The gap ${\displaystyle v_{\rm {min))\leq v\leq v_{\rm {max))}$ is a precursor of the actual phase change from liquid to vapor. The points ${\displaystyle (p_{\rm {min)),v_{\rm {min)))}$ and ${\displaystyle (p_{\rm {max)),v_{\rm {max)))}$, where ${\displaystyle \partial _{v}p|_{T}=0}$, that delimit the largest possible liquid and smallest possible vapor states are called spinodal points. Their locus forms a spinodal curve which bounds a region where no homogeneous stable states can exist.

Experiments show that if the volume of a vessel containing a fixed amount of liquid is heated and expands at constant temperature, at a certain pressure, ${\displaystyle p_{s}(T)}$, vapor, (denoted by dots at points ${\displaystyle f}$ and ${\displaystyle g}$ in Fig. 1) bubbles nucleate so the fluid is no longer homogeneous, but rather it has become heterogeneous. It is now a mixture of two separate components, a boiling (saturated) liquid, ${\displaystyle v_{f}=V_{f}/N_{f))$, and a condensing (saturated) vapor, ${\displaystyle v_{g}=V_{g}/N_{g))$ that coexist at the same saturation temperature and pressure. As the heating continues the amount of vapor, ${\displaystyle N_{g))$ increases and that of the liquid, ${\displaystyle N_{f}=N-N_{g))$, decreases. All the while the pressure, ${\displaystyle p_{s))$ and temperature, ${\displaystyle T}$, remain constant and the volume ${\displaystyle V=V_{f}+V_{g))$ increases. In this situation the molar volume of the mixture is a weighted average of its components

$${\displaystyle v=V/N=(V_{f}/N_{f})(N-N_{g})/N+(V_{g}/N_{g})(N_{g}/N)=v_{f}(1-x)+v_{g}x}$$

${\displaystyle x=N_{g}/N}$

${\displaystyle 0\leq x\leq 1}$

${\displaystyle p(T)}$

${\displaystyle v_{f))$

${\displaystyle v_{g))$

The dotted parts of the curve in Fig. 1 are metastable states. For many years such states were an academic curiosity; Callen^[7] gave as an example,

water that has been cooled below 0°C at a pressure of 1 atm. A tap on a beaker of water in this condition precipitates a sudden dramatic crystallization of the system.

However, studies of boiling heat transfer have made clear that metastable states occur routinely as an integral part of this process. In it the heating surface temperature is higher than the saturation temperature, often significantly so, hence the adjacent liquid must be superheated.^[8] Further the advent of devices that operate with very high heat fluxes has created interest in the metastable states, and the thermodynamic properties associated with them, in particular the superheated liquid states.^[9] Moreover, the fact that they are predicted by the van der Waals equation, and cubic equations in general, is compelling evidence of its efficacy in describing phase transitions; Sommerfield^[10] described this as follows:

It is very remarkable that the theory due to van der Waals is in a position to predict, at least qualitatively, the existence of the unstable [called metastable here] states along the branches AA` or BB` [BC and FE in Fig. 1 here].

Equal area rule

The discontinuity in ${\displaystyle v}$, and other properties, e.g. internal energy, ${\displaystyle u}$, and entropy,${\displaystyle s}$, of the substance, is called a first order phase transition.^[11][12] In order to specify the unique experimentally observed pressure, ${\displaystyle p_{s}(T)}$, at which it occurs another thermodynamic condition is required, for from Fig.1 it could clearly occur for any pressure in the range ${\displaystyle p_{\rm {min))\leq p\leq p_{\rm {max))}$. Such a condition was first enunciated in a clever thermodynamic argument by Maxwell at a lecture he delivered to the British Chemical Society on Feb 18, 1875^[13] (Fig.1, including the letters B C D E F, is the curve he described):

     The portion of the curve from C to E represents points which are essentially unstable, and which cannot therefore be realized.

     Now let us suppose the medium to pass from B to F along the hypothetical curve B C D E F in a state always homogeneous, and to return along the straight line path F B in the form of a mixture of liquid and vapor. Since the temperature has been constant throughout, no heat can have been transformed into work. Now the heat transformed into work is represented by the excess of the area F D E over B C D. Hence the condition which determines the maximum pressure of the vapor at given temperature is that the line B F cuts off equal areas from the curve above and below.

The easiest way to understand Maxwell's argument is to consider the cycle he suggested on a temperature—molar entropy plane. Every introductory thermodynamics text^{[14] [15]} presents the fact that on such a plane the area under any curve is the heat transfer to the substance per mole, positive going from left to right and negative from right to left; moreover, in a cyclic process the net heat transfer to the substance is the area enclosed by the cycle's closed curve. Since the cycle he considered is composed of the two gray dashed isothermals at the same temperature, one proceeding from B to F (through C D and E), and the other directly back from F to B, the two lines are identical, just traversed in reverse; there is zero area enclosed, and hence ${\displaystyle q=0}$.

Further the same texts describe the area under these curves when plotted on a pressure—molar volume, see Fig. 1, as being the work done by the substance, positive going from left to right, and negative from right to left. Likewise the net work done in a cycle is the area enclosed by the closed curve. Since the first law of thermodynamics yields in the special case of a cycle ${\displaystyle w=q}$, for the cycle envisioned by Maxwell ${\displaystyle w=q=0}$; then since the area enclosed is I+II=0, see Fig.1, with I positive and II negative, the transition pressure must be such that the two areas are equal.

Written as a mathematical equation in terms of the work done in each process this is

$${\displaystyle \int _{v_{g))^{v_{f))\,p\,dv+\int _{v_{f))^{v_{g))\,p_{s}\,dv=-\int _{v_{f))^{v_{g))\,p\,dv+p_{s}(v_{g}-v_{f})=0\quad {\mbox{for))\quad T={\mbox{constant))}$$

${\displaystyle f}$

${\displaystyle g}$

$${\displaystyle p_{s}=p(v_{f},T)\qquad p_{s}=p(v_{g},T)}$$

${\displaystyle p_{s},T,v_{f},v_{g))$

${\displaystyle T}$

${\displaystyle p_{s}(T)}$

${\displaystyle v_{f}(T)}$

${\displaystyle v_{g}(T)}$

Gibbs criterion

At the end of his lecture, after complimenting van der Waals by referring to his work as "an exceedingly ingenious thesis", Maxwell finished it by saying:

I must not, however, omit to mention a most important American contribution to this part of thermodynamics by Prof. Willard Gibbs of Yale College U.S., who has given us a remarkably simple and thoroughly satisfactory method of representing the relations of the different states of matter by means of a model. By means of this model, problems which had long resisted the efforts of myself and others may be solved at once.

This remark proved prescient because in 1876-1878 Gibbs published his definitive work on thermodynamics ^[16] in which he showed that thermodynamic equilibrium of a heterogeneous substance requires that, in addition to mechanical equilibrium (the same pressure for each component) and thermal equilibrium (the same temperature for each component), there must also be material equilibrium (the same chemical potential for each component). In the present instance of one substance and two phases in addition to ${\displaystyle p_{f}=p_{g}=p_{s))$ and ${\displaystyle T_{f}=T_{g}=T}$, material equilibrium requires ${\displaystyle g_{f}=g_{g))$ (for the special case of one substance its chemical potential is the molar Gibbs function, ${\displaystyle \mu =g\equiv G/N}$ where ${\displaystyle g=u+pv-Ts}$). This condition can be deduced by a simple physical argument as follows: the energy required to vaporize a mole is from the second law at constant temperature ${\displaystyle q_{\rm {vap))=T(s_{g}-s_{f})}$, and from the first law at constant pressure ${\displaystyle q_{\rm {vap))=h_{g}-h_{f))$, then equating these two and rearranging produces the result (note that ${\displaystyle h=u+pv}$). The conditions of material equilibrium lead to the famous phase rule, ${\displaystyle D=n-r+2}$, where ${\displaystyle n}$ is the number of substances, ${\displaystyle r}$ the number of phases, and ${\displaystyle D}$ the number of independent intensive variables required to specify the state.^[17][18] In the case of one substance and two phases discussed here this gives, ${\displaystyle D=1}$, the experimentally observed number.

Now ${\displaystyle g(p,T)}$ is a thermodynamic potential function, its differential is^[19]

$${\displaystyle dg=\partial _{p}g|_{T}\,dp+\partial _{T}g|_{p}\,dT=v\,dp-s\,dT}$$

Integrating this at constant temperature produces

$${\displaystyle g(p,T)=g_{A}(T)+\int _{p_{A))^{p}\,v({\bar {p)),T)\,d{\bar {p))}$$

${\displaystyle g_{A))$

${\displaystyle T}$

${\displaystyle g}$

${\displaystyle p=p(v,T)}$

${\displaystyle v=v(p,T)}$

${\displaystyle v}$

$${\displaystyle pv^{3}-(pb+RT)v^{2}+av-ab=0,}$$

a cubic with either 1 or, in this case, 3 real roots. Thus there are three curves, as seen in Fig. 2, consisting of stable (shown solid black), metastable (shown dotted black), and unstable (shown dashed gray) states.

Actually, the figure was not produced by solving the cubic and integrating, rather ${\displaystyle g(v,T)}$ was obtained from its definition by first obtaining ${\displaystyle u(v,T)}$ and ${\displaystyle s(v,T)}$, which is easily done analytically for the van der Waals equation, and plotting it parametrically with ${\displaystyle p(v,T)}$, using ${\displaystyle v}$ as the parameter. Note that ${\displaystyle g(p,T)}$ is continuous with discontinuous partial derivatives, ${\displaystyle \partial _{p}g|_{T}=v}$ and ${\displaystyle \partial _{T}g|_{p}=-s}$, at the phase transition point. In the Ehrenfest classification, a first order phase transition refers to the discontinuity of the first partial derivatives of ${\displaystyle g}$ while a second order phase transition would involve discontinuities of the second partial derivatives.^[21]

Relationship between the Gibbs and Maxwell criteria

Evaluating the integral expression for ${\displaystyle g(p,T)}$ given previously between the saturated liquid and vapor states and applying the Gibbs criterion of material equilibrium to this phase change process requires writing it as

$${\displaystyle g_{g}-g_{f}=\int _{p_{s))^{p_{\rm {min))}\,v_{l}\,dp+\int _{p_{\rm {min))}^{p_{\rm {max))}\,v_{u}\,dp+\int _{p_{\rm {max))}^{p_{s))\,v_{v}\,dp=0}$$

Here the integral has been split into three parts using the three real roots of the cubic corresponding to the liquid, ${\displaystyle v_{l))$, unstable, ${\displaystyle v_{u))$, and vapor, ${\displaystyle v_{v))$, states respectively. These integrals can best be visualized by viewing Fig. 1 rotated so that ${\displaystyle v}$ is plotted on the ordinate of the curve as shown in the accompanying graph. In this view the function ${\displaystyle v(p,T)}$ clearly is multi-valued; this is the reason it requires three real functions to describe its behavior between ${\displaystyle p_{\rm {min))}$ and ${\displaystyle p_{\rm {max))}$. Now on splitting the middle integral into two

$${\displaystyle g_{g}-g_{f}=\int _{p_{s))^{p_{\rm {min))}\,v_{l}\,dp+\int _{p_{\rm {min))}^{p_{s))\,v_{u}\,dp+\int _{p_{s))^{p_{\rm {max))}\,v_{u}\,dp+\int _{p_{\rm {max))}^{p_{s))\,v_{v}\,dp=0}$$

${\displaystyle d(pv)=pdv+vdp}$

$${\displaystyle dg=v\,dp-s\,dT=-p\,dv+d(pv)-s\,dT}$$

${\displaystyle f}$

${\displaystyle g}$

$${\displaystyle g_{g}-g_{f}=-\int _{v_{f))^{v_{g))\,p(v,T)\,dv+p_{s}(v_{g}-v_{f})=0}$$

${\displaystyle g_{f}=g_{g))$

Application to the van der Waals equation

From the van der Waals equation applied to the saturated liquid, ${\displaystyle v_{f}<v_{\rm {min))}$, and vapor, ${\displaystyle v_{g}>v_{\rm {max))}$, states

$${\displaystyle p_{s}={\frac {RT_{s)){v_{f}-b))-{\frac {a}{v_{f}^{2))}\qquad p_{s}={\frac {RT_{s)){v_{g}-b))-{\frac {a}{v_{g}^{2))))$$

${\displaystyle p_{s},T_{s))$

${\displaystyle v_{f},v_{g))$

$${\displaystyle p_{s}=p^{*}{\frac {v^{*2}[v_{f}v_{g}-v^{*}(v_{f}+v_{g})]}{v_{f}^{2}v_{g}^{2))}\qquad T_{s}=T^{*}{\frac {v^{*}(v_{f}+v_{g})(v_{f}-v^{*})(v_{g}-v^{*})}{v_{f}^{2}v_{g}^{2))))$$

${\displaystyle p^{*}=a/b^{2))$

${\displaystyle v^{*}=b}$

${\displaystyle T^{*}=a/(Rb)}$

${\displaystyle p^{*}v^{*}/T^{*}=R}$

$${\displaystyle -T_{s}\ln \left({\frac {v_{g}-v^{*)){v_{f}-v^{*))}\right)+T^{*}{\frac {v^{*}(v_{g}-v_{f})}{v_{f}v_{g))}+p_{s}(v_{g}-v_{f})/R=0}$$

${\displaystyle T_{s))$

${\displaystyle p_{s))$

${\displaystyle p_{s))$

${\displaystyle T_{s))$

${\displaystyle \varrho =v^{*}/(v-v^{*})}$

${\displaystyle \infty }$

${\displaystyle v}$

${\displaystyle v^{*))$

${\displaystyle \infty }$

$${\displaystyle \ln \left({\frac {\varrho _{f)){\varrho _{g))}\right)={\frac {(\varrho _{f}-\varrho _{g})(\varrho _{g}+\varrho _{f}+2)}{\varrho _{f}+\varrho _{g}+2\varrho _{f}\varrho _{g))))$$

${\displaystyle (s_{g}-s_{f})/R}$

$${\displaystyle \ln \left({\frac {\varrho _{f)){\varrho _{g))}\right)={\frac {s_{g}-s_{f)){R))=\delta =2y}$$

${\displaystyle \varrho _{f}=e^{\delta }\varrho _{g))$

${\displaystyle \varrho _{f))$

${\displaystyle \varrho _{g))$

$${\displaystyle \varrho _{g}(y)=f(y)e^{-y}\quad {\mbox{and))\quad \varrho _{f}(y)=f(y)e^{y}\quad {\mbox{where))\quad f(y)={\frac {y\cosh y-\sinh y}{\sinh y\cosh y-y))}$$

${\displaystyle (s_{g}-s_{f})/(2R)=y}$

$${\displaystyle T_{s}(y)=T^{*}{\frac {2f(y)[\cosh y+f(y)]}{g(y)^{2))}\qquad p_{s}(y)=p^{*}{\frac {f(y)^{2}[1-f(y)^{2}]}{g(y)^{2))))$$

$${\displaystyle v_{f}(y)=v^{*}{\frac {1+f(y)e^{y)){f(y)e^{y))}\quad \quad \quad \quad \quad \quad v_{g}(y)=v^{*}{\frac {1+f(y)e^{-y)){f(y)e^{-y))))$$

${\displaystyle g(y)=[1+\varrho _{g}(y)][1+\varrho _{f}(y)]=1+2f(y)\cosh y+f(y)^{2))$

The values of all other property discontinuities across the saturation curve also follow from this solution.^[26]

These functions define the coexistence curve which is the locus of the saturated liquid and saturated vapor states of the van der Waals fluid. In Fig. 3 this curve is plotted in blue together with the spinodal curve in black, calculated from

$${\displaystyle T=T^{*}({\tilde {v))-1)^{2}/{\tilde {v))^{3}\quad \quad p=p^{*}({\tilde {v))-2)/{\tilde {v))^{3))$$

${\displaystyle {\tilde {v))=v/v^{*))$

${\displaystyle p_{r}=p/p_{c))$

${\displaystyle v_{r}=v/v_{c))$

${\displaystyle T_{r}=T/T_{c))$

${\displaystyle c}$

${\displaystyle \partial _{v}p|_{T}=0}$

${\displaystyle \partial _{v}^{2}p|_{T}=0}$

${\displaystyle p^{*}/p_{c}=27}$

${\displaystyle v^{*}/v_{c}=1/3}$

${\displaystyle T^{*}/T_{c}=27/8}$

${\displaystyle c}$

${\displaystyle v_{r))$

${\displaystyle v_{r))$

Over the parameter range ${\displaystyle 0\leq y<\infty }$, ${\displaystyle f(y)}$ decreases monotonically from ${\displaystyle f(0)=1}$ and approaches 0 as${\displaystyle f(y)\sim 2(y-1)e^{-y))$ in the limit ${\displaystyle y\rightarrow \infty }$. Therefore ${\displaystyle \varrho _{f}(0)=\varrho _{g}(0)=1/2}$ and in the limit ${\displaystyle y\rightarrow \infty }$, ${\displaystyle \varrho _{f}(y)\sim 2(y-1)}$ and ${\displaystyle \varrho _{g}(y)\sim 2(y-1)e^{-2y))$. The behavior of ${\displaystyle p_{s}(y)}$ and ${\displaystyle T_{s}(y)}$ follow from the equations. Both these properties also decrease monotonically from ${\displaystyle p_{s}(0)=p^{*}/27}$ and ${\displaystyle T_{s}(0)=8T^{*}/27}$, and approach 0 as ${\displaystyle p_{s}\sim p^{*}\varrho _{g}/\varrho _{f))$ and ${\displaystyle T_{s}\sim T^{*}/\varrho _{f))$ in the limit ${\displaystyle y\rightarrow \infty }$. Note from these that ${\displaystyle p_{s}\sim (p^{*}/T^{*})T_{s}\varrho _{g}\sim (p^{*}v^{*}/T^{*})T_{s}\rho _{g}=\rho _{g}RT_{s))$; the van der Waals saturated vapor is an ideal gas in this limit. To paraphrase Sommerfield, it is remarkable that the theory due to van der Waals is able to predict that when ${\displaystyle s_{g}-s_{f}\gg 2R}$ the saturated vapor behaves like an ideal gas; the saturated vapor of real gases behave exactly this way.

In addition for ${\displaystyle T_{r}<27/32=0.84375}$ the liquid spinodal point occurs at a negative pressure, and the isotherm ${\displaystyle T_{r}=0.8}$ is included in Fig. 4 to illustrate this point. This means that some part of those liquid metastable states are in tension, and the lower the temperature the greater the tensile stress. Although this seems counterintuitive it is known that under some circumstances liquids can support tension. Tien and Lienhard^[28] noted this and wrote:

The van der Waals equation predicts that at low temperatures liquids sustain enormous tension---a fact that has led some authors to take the equation lightly. In recent years measurements have been made that reveal this to be entirely correct.^[29] Liquids that are clean and free of dissolved gas can be subjected to tensions greater in magnitude than ${\displaystyle p_{c))$.

This is another interesting feature of the van der Waals theory.

Common tangent construction and lever rule

The Maxwell construction is related to the common tangent construction^[30][31] and the lever rule.^[32]