Maxwell construction

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):

This is rendition of a figure in the James Clerk-Maxwell paper in Nature explaining what we now call the Maxwell construction for the van der Waals gas

     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

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

This equation together with the equation of state written for each of the states ${\displaystyle f}$ and ${\displaystyle g}$

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

are three equation for the four variables, ${\displaystyle p_{s},T,v_{f},v_{g}}$, so given any one of them, say ${\displaystyle T}$, the other three are determined. In other words, there is a unique value of ${\displaystyle p_{s}(T)}$, as well as ${\displaystyle v_{f}(T)}$ and ${\displaystyle v_{g}(T)}$, at which the phase transition can occur.

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 ${\displaystyle q_{\rm {vap}}=T(s_{g}-s_{f})}$, 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

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

here ${\displaystyle g_{A}}$ is a constant of integration, but the constant is different for each isotherm hence it is written as a function of ${\displaystyle T}$.[20] In order to evaluate ${\displaystyle g}$ one must invert ${\displaystyle p=p(v,T)}$ to obtain ${\displaystyle v=v(p,T)}$. However, it is the nature of the phase transition phenomenon that this inversion is not unique; for example the van der Waals quation written for ${\displaystyle v}$ is,

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

Figure 2: The Gibbs function on the same isotherm shown in Fig. 1. The letters denote the same points here that they do in that figure.

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]

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 roots 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

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

The first two integrals here are area I while the second two are the negative of area II. The two areas add to zero hence their magnitudes are equal according to this Gibbs criterion. That this must be the same as the Maxwell construction seems obvious, and it can be easily shown analytically since ${\displaystyle d(pv)=pdv+vdp}$. As a result,

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

and then integrating for constant temperature from state ${\displaystyle f}$ to ${\displaystyle g}$ with the Gibbs condition produces

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

which is Maxwell's result. This equal area rule can also be derived by making use of the Helmholtz free energy.[22] In any event the two conditions are the same; the Maxwell construction derives from the Gibbs condition of material equilibrium. Although ${\displaystyle g_{f}=g_{g}}$ is more fundamental, the Maxwell construction often requires less effort to implement.

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

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

These two equations specify 4 variables so they can be solved for ${\displaystyle p_{s},T_{s}}$ in terms of ${\displaystyle v_{f},v_{g}}$. This results in

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

where ${\displaystyle p^{*}=a/b^{2}}$, ${\displaystyle v^{*}=b}$, and ${\displaystyle T^{*}=a/(Rb)}$ are a characteristic pressure, molar volume, and temperature defined by the constants (note that ${\displaystyle p^{*}v^{*}/T^{*}=R}$). Applying the Maxwell construction to the van der Waals equation gives

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

These three equations can be solved numerically. This has been done given a value for either ${\displaystyle T_{s}}$ or ${\displaystyle p_{s}}$, and tabular results presented;[23][24] however, the equations also admit an analytic parametric solution that, according to Lenkner,[25] was obtained by Gibbs. Lenkner himself devised a simple, elegant, method to obtain this solution, by eliminating ${\displaystyle p_{s}}$ and ${\displaystyle T_{s}}$ from the equations, and writing them in terms of a stretched dimensionless number density, ${\displaystyle \varrho =v^{*}/(v-v^{*})}$, that varies between ${\displaystyle \infty }$ and 0 as ${\displaystyle v}$ varies from ${\displaystyle v^{*}}$ to ${\displaystyle \infty }$; this produces

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

Although transcendental, this equation has a simple analytic, parametric solution obtained by writing the left side of the equation, which is just ${\displaystyle (s_{g}-s_{f})/R}$, as

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

Then ${\displaystyle \varrho _{f}=e^{\delta }\varrho _{g}}$ and when used to eliminate ${\displaystyle \varrho _{f}}$ from the right hand side a linear equation for ${\displaystyle \varrho _{g}}$ is obtained, whose solution is

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

Accordingly, the fundamental variable that specifies all the others in this phase transition process is ${\displaystyle (s_{g}-s_{f})/(2R)=y}$. This solution to the saturation problem is easily extended to encompass all its variables

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

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

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

Fig. 3: The analytical vdW spinodal (dashed dot blue) and coexistence (dashed dot black) curves on the ${\displaystyle p}$-${\displaystyle T}$ plane.

Figure 4: The isotherm of Fig.1, ${\displaystyle T_{r}=7/8}$, with the critical isotherm, ${\displaystyle T_{r}=1}$, a supercritical isotherm, and the spinodal and coexistence curves.

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

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

where ${\displaystyle {\tilde {v}}=v/v^{*}}$ is a parameter. The variables used in making these plots are the reduced (dimensionless) variables, ${\displaystyle p_{r}=p/p_{c}}$, ${\displaystyle v_{r}=v/v_{c}}$, and ${\displaystyle T_{r}=T/T_{c}}$ where the ${\displaystyle c}$ subscripted quantities are the critical point values. They are defined by, ${\displaystyle \partial _{v}p|_{T}=0}$, and ${\displaystyle \partial _{v}^{2}p|_{T}=0}$ at the critical point,[27] and are measurable quantities. The relations ${\displaystyle p^{*}/p_{c}=27}$, ${\displaystyle v^{*}/v_{c}=1/3}$, ${\displaystyle T^{*}/T_{c}=27/8}$ are used to convert the star quantities in the solution to the ${\displaystyle c}$ quantities used in the figures. The curve agrees completely with the numerical results referenced earlier. In the region inside the spinodal curve there are two states at each point, one stable and one metastable, either superheated liquid to the right of the blue curve, or subcooled vapor to the left, while outside the spinodal curve there is one stable state at each point. In Fig. 4 the region under the (dot dash black) spinodal curve contains no stable states while between the (dot dash red) coexistence and spinodal curves there is one metastable state at every point, and outside the coexistence curve there is one stable state at each point. The two blue and two green circles denote the saturated liquid and vapor states on their respective isotherms. There are also observed heterogeneous states everywhere under the red coexistence curve that satisfy the lever rule; however, they are not homogeneous states of the van der Waals equation, so their existence, indicated by horizontal lines connecting the saturation points on each subcritical isotherm, is not displayed. Also the abscissa in this figure is logarithmic, not linear, in order to show more of the vapor region at large ${\displaystyle v_{r}}$ without excessively compressing the liquid and unstable regions at small ${\displaystyle v_{r}}$; however, this device distorts areas, so the two areas I and II in Fig.1 would not appear equal here.

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.

The van der Waals gas equation (using reduced variables) can be expanded [30] to

${\displaystyle v_{R}^{3}-\left({\frac {8T_{R}}{p_{R}}}+1\right){\frac {v_{R}^{2}}{3}}+{\frac {9v_{R}}{3p_{R}}}-{\frac {1}{p_{R}}}=0}$

which is of the form

${\displaystyle v_{r}^{3}+a_{1}v_{r}^{2}+a_{2}v_{r}+a_{3}=0}$

To solve this Cubic function one defines several precursor terms :

${\displaystyle a_{1}(T_{R},p_{R})=-{\frac {{\frac {8T_{R}}{p_{R}}}+1}{3}}}$

${\displaystyle a_{2}(p_{R})={\frac {3}{p_{R}}}}$

and

${\displaystyle a_{3}(p_{R})=-{\frac {1}{p_{R}}}}$

so that the following become defined

${\displaystyle \eta =a_{2}-{\frac {a_{1}^{2}}{3}}}$

and

${\displaystyle q=-\left(+{\frac {2}{27}}a_{1}^{3}-{\frac {a_{1}a_{2}}{3}}+a_{3}\right)}$

The first root's precursor form is:

${\displaystyle z_{1}={\sqrt {\frac {4\eta }{-3}}}\cos \left[{\frac {1}{3}}\cos ^{-1}\left(-{\frac {3q}{2\eta }}{\sqrt {\frac {-3}{\eta }}}\right)\right]}$

leading to

${\displaystyle v_{r_{1}}=z_{1}-{\frac {a_{1}}{3}}}$

and

${\displaystyle v_{r_{2}}={\frac {-z_{1}+{\sqrt {z_{1}^{2}-4(\eta +z_{1}^{2})}}}{2}}-{\frac {a_{1}}{3}}}$

and finally

${\displaystyle v_{r_{3}}={\frac {-z_{1}-{\sqrt {z_{1}^{2}-4(\eta +z_{1}^{2})}}}{2}}-{\frac {a_{1}}{3}}}$

These last four equations depend on two variables, the temperature which is chosen when one determines which isotherm one is working on, and the pressure. One starts with an arbitrary (but reasonable) chosen value and adjusts its values as one solves the equation (below) ultimately obtaining ${\displaystyle p=p_{V}}$ or ${\displaystyle p=p_{\mathrm {eq} }}$ through the Maxwell construction (see below) at that temperature. With these two variables in hand, one can re-substitute the pressure value obtained into the root equations (above) to obtain the three roots.

The Maxwell construction requires solution of the equation (obtained by making the areas under the two loops equal and opposite in value from each other):

${\displaystyle {\frac {8T_{r}}{3}}\ln \left({\frac {3v_{r_{\ell }}-1}{3v_{r_{g}}-1}}\right)-8T_{r}\left({\frac {v_{r_{\ell }}}{3v_{r_{\ell }}-1}}-{\frac {v_{r_{g}}}{3v_{r_{g}}-1}}\right)+{\frac {6}{v_{r_{\ell }}}}-{\frac {6}{v_{r_{g}}}}=0}$

with a fixed reduced temperature and the solution depending then on the chosen variable reduced pressure, which becomes the reduced vapor pressure. Unfortunately, this equation can not be solved analytically, and requires numerical evaluation. The subscripts in this equation are ${\displaystyle \ell \equiv smallest}$ and ${\displaystyle g\equiv largest}$ have been changed to make clear which two roots of the cubic are to be employed; these roots themselves depend on the equations which precede them (above), and contain the reduced pressure and temperature which are treated as above.

pseudo 3D p-v-T diagram for van der Waals fluid showing some tie lines and isotherms

Maxwell construction for van der Waals fluid

van der Waals coexistence locus of liquid-gas equilibrium i.e., vapor pressure versus molar volumes

By equating the pressures corresponding to the two volumes where there is a discontinuity in the p-v isotherm, one obtains an expression for the temperature independent of pressure, i.e.,

$${\displaystyle T={\frac {{\frac {3}{v_{g}^{2}}}-{\frac {3}{v_{\ell }^{2}}}}{\left({\frac {8}{3v_{g}-1}}-{\frac {8}{3v_{\ell }-1}}\right)}}}$$

which allows elimination of the temperature from the prior equation. Its solution is complicated, but finally becomes

$${\displaystyle v_{g}={\frac {f(d)e^{d}+1}{3}}}$$

and

$${\displaystyle v_{\ell }={\frac {f(d)e^{-d}+1}{3}}}$$

where

$${\displaystyle f(d)={\frac {1}{2}}{\frac {4de^{2d}-e^{4d}+1}{d(e^{3d}+e^{d})-e^{3d}+e^{d}}}.}$$

The pseudo-3D ${\displaystyle p_{r}-v_{r}-T_{r}}$ diagram for the van der Waals fluid is shown in the accompanying Figure. For details see lectures from University of Connecticut, articles 88, 93, 95 and especially 96.[31]

The Maxwell construction is rarely derived from the condition that the Gibbs free energies of the gas and the liquid must be equal when they coexist. However, it can be shown that this condition is fulfilled. Essentially the same applies to any other thermodynamic system, where ${\displaystyle P}$ and ${\displaystyle V}$ are replaced by a different pair of conjugate variables, e.g. magnetic field and magnetization or chemical potential and number of particles.

The Maxwell construction is related to the common tangent construction[32][33] and the lever rule.[34]

• Conjugate variables (thermodynamics)

• Maxwell, J. C. (1875). "On the Dynamical Evidence of the Molecular Constitution of Bodies". Nature. 11 (279): 357–359. Bibcode:1875Natur..11..357C. doi:10.1038/011357a0.

• Reichl, L. E. (2009). A Modern Course in Statistical Physics (3rd ed.). New York, NY USA: Wiley-VCH. ISBN 9783527407828.