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The volumetric coefficient of thermal expansion is the fractional volume increase per temperature increase at fixed pressure and composition:The mass density form requires fixed mass $m$. Its units are ${\mathrm{K}}^{-1}$; ${\alpha }_T$ is another notation for this coefficient. For small temperature changes $\theta$ at fixed pressure,For a finite change in a single thermodynamic phase, integrate the possibly varying coefficient: $\ln (V_2/V_1)={\int }_{T_1}^{T_2}\alpha (T,P,N)dT$. A constant-coefficient approximation gives an exponential, while the displayed linear formula additionally requires $|\alpha \theta |\ll 1$.

For an ideal gas at fixed particle number,Integrating $1/T$ recovers the exact fixed-pressure thermal expansion ratio $V_2/V_1=T_2/T_1$. Combining expansion with isothermal compressibility gives the local response of any smooth simple compressible system of fixed composition,The ratio has units $\mathrm{Pa}{\mathrm{K}}^{-1}$. For the one-mole ideal gas state in isothermal compressibility, it is $P_0/T_0=333.33\mathrm{Pa}{\mathrm{K}}^{-1}$: a $1\mathrm{K}$ increase at fixed volume raises the pressure by about $333.33\mathrm{Pa}$. Maxwell relations identify this same ratio as the entropy slope with respect to volume and give $C_P-C_V=TV{\alpha }^2/{\kappa }_T$.

Under the Boussinesq approximation, the small fractional mass density change produces buoyancy while mass density is otherwise treated as constant. A positive temperature excess $\theta$ gives reduced gravity $g^{\prime }=g{\alpha }_T\theta$ in the convention used in natural ventilation. The coefficient therefore converts a heat budget into a buoyancy budget. The same measurable expansion response supplies entropy changes and heat capacity differences through Maxwell relations, connecting fluid motion to equilibrium thermodynamics.

Compressibility measures how readily a sample's volume decreases when its pressure increases. For a simple compressible system with fixed particle number $N$ and composition, specify the constraint $X$:Here $P$ denotes pressure, also written $p$ elsewhere. The minus sign makes compressibility positive on a mechanically stable equilibrium branch. Its units are inverse pressure, $\mathrm{P}{\mathrm{a}}^{-1}$; its reciprocal $K_X=1/{\kappa }_X$ is the constrained bulk modulus, measured in pascal units. Holding temperature fixed gives isothermal compressibility; holding entropy fixed gives isentropic compressibility ${\kappa }_S$ along an isentropic process. The latter describes a reversible adiabatic compression, since an irreversible adiabatic process can increase entropy. ^[ref]

For an ideal gas, the ideal gas law $PV=N{k}_{B}T$ and the reversible ideal-gas adiabat give, at the same equilibrium state,Here $C_P$ and $C_V$ are the whole sample's heat capacity at constant pressure and heat capacity at constant volume, and $\gamma$ is the heat capacity ratio. Constant $\gamma$ is required for the finite-path formula $P{V}^{\gamma }=\mathrm{constant}$. Reversible adiabatic compression raises the temperature, making the pressure rise more steeply than in an isothermal compression. Thus keeping entropy fixed makes the gas harder to compress locally. For a classical monatomic gas at $100\mathrm{kPa}$, $\gamma =5/3$, so ${\kappa }_T=10^{-5}\mathrm{P}{\mathrm{a}}^{-1}$ and ${\kappa }_S=6\times 10^{-6}\mathrm{P}{\mathrm{a}}^{-1}$; the corresponding values of the bulk modulus are $100\mathrm{kPa}$ and $166.67\mathrm{kPa}$.

For a smooth, stable single thermodynamic phase at positive temperature, Maxwell relations also implywhere $\alpha$ is the coefficient of thermal expansion. The ideal gas result is one example of this broader link between mechanical and thermal responses. The equality ${\kappa }_T={\kappa }_S$ is possible when $\alpha =0$ and the other responses remain finite. These formulas compare equilibrium partial derivatives, rather than arbitrary fast or dissipative compression histories. ^[ref]

The compressibility factor is the ratio $p/(n{k}_{B}T)$, where $n$ is number density. A classical ideal gas has $Z=1$. The virial expansion writes $Z=1+B_{2}n+B_3{n}^2+\cdots$ for a dilute gas; this dimensionless ratio is distinct from isothermal compressibility, which is a response derivative with units of inverse pressure.

Here $n$ is average number density and $\mu$ the ground-state chemical potential. This commonly used condensate response has units of inverse energy; thermodynamic isothermal compressibility is instead $\kappa /n$. In a weakly interacting repulsive condensate, the normalized one-cell ground state $f_s$, with average $|f_s{|}^2=1$, gives $\mu =E_0(s)+UnI(s)+\cdots$, $I(s)=d^{-1}{\int }_0^d|f_s{|}^{4}dx$. Localization raises $I$ and reduces $\kappa$ at fixed $n,U,d$. In deep harmonic wells $I\propto s^{1/4}$, hence $\kappa \propto s^{-1/4}$. This is a ground-band, repulsive-superfluid statement with the held variables specified, not a theorem for arbitrary signs of the interaction or arbitrary excited branches.

For a nonrelativistic zero-temperature free electron model with fixed particle number, electron degeneracy pressure obeys $P\propto (N/V{)}^{5/3}$. Consequently its bulk modulus is $K=-V(\partial P/\partial V{)}_N=5P/3=2n{E}_F/3$. At temperatures small compared with the electron Fermi temperature, the leading isothermal compressibility is $1/K$. This electronic contribution alone omits ionic binding and interaction corrections in a real metal.

In the classical high-temperature limit, each independent harmonic oscillator contributes $k_B$ to the heat capacity at constant volume, where $k_B$ is the Boltzmann constant. A three-dimensional crystalline solid with $N$ chemical atoms has $3N$ vibrational normal modes in the bulk harmonic approximation. Its vibrational internal energy and heat capacity at constant volume therefore obeyHere $U_0$ is independent of temperature at fixed volume. Thus the Dulong-Petit law predicts $C_V\simeq 3N{k}_B$ for a real solid in this regime. Per mole of chemical atoms, rather than per mole of a compound's formula units,with molar gas constant $R$ and Avogadro constant $N_A$. A formula unit containing $r$ chemical atoms consequently has a classical vibrational prediction of $3rR$ per mole of formula units.

For example, in this ideal classical model, one mole of chemical atoms heated at fixed volume from $300$ to $600\mathrm{K}$ gains vibrational internal energyThe heat capacity at constant volume stays the same even though the thermal part of the internal energy doubles. These temperatures illustrate the stipulated classical model; they do not assert that every real solid is classical at $300\mathrm{K}$.

The proof outline is the equipartition theorem: one classical harmonic oscillator has a quadratic kinetic energy and a quadratic potential energy, each with mean $k_{B}T/2$. Their sum is $k_{B}T$, whose derivative with respect to temperature is $k_B$. Summing over $3N$ normal modes gives $3N{k}_B$. The classical result does not require equal angular frequencies: each harmonic normal mode contributes the same amount once it is classical. Overall rigid motions give only a finite-size correction to the bulk count. This independence from the individual angular frequencies makes the Dulong-Petit law useful as a quick estimate of a solid's heating requirement.

Experiments on freely expanding solids often measure the heat capacity at constant pressure $C_P$, whereas the harmonic prediction is for the heat capacity at constant volume $C_V$. In a smooth single thermodynamic phase of a simple compressible material, the difference between constant-pressure and constant-volume heat capacities iswhere $\alpha =V^{-1}(\partial V/\partial T{)}_P$ is the volumetric coefficient of thermal expansion and ${\kappa }_T=-V^{-1}(\partial V/\partial P{)}_T$ is the isothermal compressibility. Using measured $C_P$ as an approximation to $C_V$ requires this correction to be small compared with $C_V$. ^[ref]

High temperature here means $k_{B}T\gg \hslash \omega$ for the relevant vibrational normal modes, with reduced Planck constant $\hslash$ and mode angular frequency $\omega$. At low temperature, excitation across a quantum energy spacing is suppressed: modes with $\hslash \omega \gg k_{B}T$ freeze out thermally, and the classical equipartition theorem fails. The Einstein model of a solid makes this loss of vibrational heat capacity explicit. The Dulong-Petit law is therefore an ideal high-temperature vibrational limit, not a statement that all solids have the same heat capacity at every temperature. Other contributions and departures from the harmonic approximation can matter even at high temperature. ^[ref]

Historically, Petit and Dulong's 1819 experiments compared the specific heat capacities of several elemental solids and found an approximately constant product with their relative atomic weights. Their original empirical regularity came before the classical harmonic explanation; it suggested that counting chemical atoms reveals a simpler pattern than comparing equal masses. ^[ref]

In a smooth fixed-composition thermodynamic region, write $a=(\partial S/\partial T{)}_V$, $b=(\partial S/\partial V{)}_T$, $c=(\partial p/\partial T{)}_V$, and $d=(\partial p/\partial V{)}_T$, with $a,d\ne 0$. The chain rule gives $(\partial S/\partial T{)}_p=a-bc/d$ and $(\partial p/\partial V{)}_S=d-bc/a$. Multiplying these expressions by $d$ and $a$, respectively, gives the same Jacobian determinant $ad-bc$.

Using heat capacity at constant pressure $C_p=T(\partial S/\partial T{)}_p$, heat capacity at constant volume $C_V=Ta$, isothermal compressibility ${\kappa }_T=-1/(Vd)$ and isentropic compressibility ${\kappa }_S=-1/[V(\partial p/\partial V{)}_S]$ proves the displayed ratio. The derivation does not require division by thermal expansion, so it still applies when that derivative vanishes. For a stable single phase, the Maxwell relation $b=c$ also explains why $C_p\ge C_V$ and ${\kappa }_T\ge {\kappa }_S$.

For a fixed-composition simple compressible system, a Maxwell relation gives $C_p-C_V=T(\partial p/\partial T{)}_V(\partial V/\partial T{)}_p$. At fixed volume, the differential equation of state gives $(\partial p/\partial T{)}_V={\alpha }_V/{\kappa }_T$, proving the displayed result. Positive absolute temperature and positive isothermal compressibility imply a nonnegative difference. It vanishes when thermal expansion does, so strict inequality requires a nonzero expansion coefficient. This stability argument includes nonideal internal-energy changes, unlike a purely ideal-gas expansion-work explanation.

Isentropic compressibility is the compressibility at fixed entropy, particle number and composition. The conventional name adiabatic compressibility refers to a reversible adiabatic process; an irreversible adiabatic compression need not keep entropy fixed. For an ideal gas, it is ${\kappa }_S=1/(\gamma P)$, where $\gamma =C_P/C_V$ is the heat capacity ratio. It describes a steeper local pressure response than isothermal compressibility when $\gamma >1$.

Isothermal compressibility is the compressibility measured at fixed temperature, particle number and composition. A slowly compressed sealed piston in contact with a thermal reservoir can approach this constraint. For an ideal gas, differentiating $V=N{k}_{B}T/P$ at fixed temperature gives $(\partial V/\partial P{)}_{T,N}=-V/P$, hence ${\kappa }_T=1/P$. A small pressure increase $\Delta P$ therefore gives $\Delta V/V\simeq -\Delta P/P$. For a finite change, use the whole isothermal process curve $V_2/V_1=P_1/P_2$ instead.

Take one mole of classical monatomic gas, with stipulated molar gas constant $R=8.314\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$, initially at $T_0=300\mathrm{K}$ and $P_0=100\mathrm{kPa}$. The initial volume is $V_0=nR{T}_0/P_0=0.024942{\mathrm{m}}^3$. Compressing to $V_0/2$ givesThe isothermal compression releases heat to the thermal reservoir; the reversible adiabatic compression retains the input work as internal energy. Reversing either curve gives the corresponding expansion, with adiabatic expansion cooling the gas.

The equilibrium curvature of Gibbs free energy provides another measurement of isothermal compressibility:This follows from $V=(\partial G/\partial P{)}_{T,N}$, with $G$ denoting Gibbs free energy. Within a stable single thermodynamic phase, increasing pressure reduces volume and makes this curvature negative. At phase coexistence, changing the phase fraction can change the volume at constant temperature and pressure; the usual finite single-phase isothermal compressibility then need not exist. Near a critical point, it can diverge. ^[ref]

For an isotropically compressed cubic crystal, $d\ln a/dp=-\kappa /3$. Differentiating Bragg's law at fixed wavelength gives the displayed derivative. Constant isothermal compressibility gives $a(p)=a_0{e}^{-\kappa p/3}$, not an exactly linear lattice parameter. The integrated prediction presumes no intervening structural phase change.

Maxwell relations turn pressure, volume and temperature measurements into information about entropy. For example, the pressure rise when heating a rigid sealed vessel determines the entropy change when expanding the same sample isothermally:The quantities on the two sides describe different experiments, yet both are mixed partial derivatives of one Helmholtz free energy. This is the useful surprise: a mechanical measurement can reveal a thermal property. Similarly, the coefficient of thermal expansion gives $(\partial S/\partial P{)}_{T,N}=-\alpha V$. ^[ref]

For a simple compressible system, hold particle number $N$ and composition fixed, and allow only pressure-volume work. Write $U$ for internal energy, $H=U+PV$ for enthalpy, $F=U-TS$ for Helmholtz free energy, and $G=H-TS$ for Gibbs free energy. Their natural-variable differentials areThus the first partial derivatives, with the subscript recording the variables kept fixed, areThe symmetry of second derivatives produces all four standard Maxwell relations, including their signs:For a brief derivation of the $F$ relation, differentiate $F_T=-S$ with respect to $V$ and $F_V=-P$ with respect to $T$: $-S_V=F_{TV}=F_{VT}=-P_T$. Each other row follows in the same way. Assume continuous second partial derivatives of the relevant thermodynamic potential in the region considered. At a phase transition, the equilibrium potential may lose this smoothness; apply the identities within each smooth thermodynamic phase, and handle latent heat and phase coexistence separately. If particle numbers vary, the potential differentials acquire chemical potential terms, so the fixed-$N$ constraints must remain explicit. ^[ref]

For an ideal gas with fixed $N$, the ideal gas law gives $P=N{k}_{B}T/V$. At fixed volume, heating changes the pressure at rate $N{k}_B/V$. Consequently an isothermal expansion at positive temperature hasFor one mole with $R=8.314\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$ and $V_2=2V_1$, this is $\Delta S=R\ln 2\simeq 5.763\mathrm{J}{\mathrm{K}}^{-1}$. At $300\mathrm{K}$, a reversible isothermal expansion absorbs $Q=T\Delta S\simeq 1728.85\mathrm{J}$ and performs equal work, since its internal energy is unchanged. The entropy difference is a state function difference: the same equilibrium endpoints give the same $\Delta S$ in an insulated free expansion, although there $Q=0$ and the process is irreversible.

The same Maxwell relations let thermal measurements determine the difference between constant-pressure and constant-volume heat capacities. In a smooth single thermodynamic phase,using heat capacity at constant volume, heat capacity at constant pressure, coefficient of thermal expansion and isothermal compressibility. Along fixed pressure, $dV=\alpha VdT$, soThus measured expansion and compression responses determine a heat capacity difference without measuring both heat capacities separately. They do not determine the absolute heat capacities or absolute entropy by themselves; an independent thermal measurement supplies that missing information. For an ideal gas, $\alpha =1/T$ and ${\kappa }_T=1/P$ reduce the result to $C_P-C_V=PV/T=N{k}_B$, which is Mayer's relation.

When the necessary higher partial derivatives exist and commute, differentiating the Maxwell relations once more also givesIntegrating these slopes determines how a heat capacity changes with volume or pressure at fixed temperature. For an ideal gas, both right-hand sides vanish because the ideal gas law is linear in temperature at either fixed volume or fixed pressure. Consequently $C_V$ and $C_P$ depend only on temperature at fixed $N$, although they need not be constant in temperature. ^[ref]

Historically, James Clerk Maxwell's Theory of Heat, first published in 1871, presents these connections using the geometry of isothermal processes and reversible adiabatic processes. His fourth edition, page 167, explicitly relates expansion on heating at fixed pressure to heat emitted during an isothermal compression: the modern $G$ relation gives $\delta Q_{\mathrm{rev}}=TdS=-T\alpha VdP$. This explains why a substance with positive coefficient of thermal expansion emits heat when compressed isothermally, whereas one with negative coefficient absorbs it. ^{[ref] [ref]}

Mayer's relation states that the difference between heat capacity at constant pressure and heat capacity at constant volume for a fixed-composition ideal gas isHere both capacities are total heat capacities in $\mathrm{J}{\mathrm{K}}^{-1}$, $n$ is amount of substance, $N$ is the number of particles, $R$ is the molar gas constant and $k_B$ is the Boltzmann constant. In other normalizations the same statement isThe first is a difference of molar heat capacities, in $\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$; the second uses specific heat capacities, in $\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$. Writing $C_p-C_V=R$ with total capacities would omit the sample's amount of substance. The relation holds pointwise even when the capacities depend on temperature. ^[ref]

The proof outline is short: enthalpy is $H=U+pV=U+nRT$, and internal energy of an ideal gas is independent of volume at fixed temperature. Differentiating at fixed composition and amount of substance therefore gives $C_p=dH/dT=dU/dT+nR=C_V+nR$. It uses the ideal gas law, rather than assuming any particular number of molecular degrees of freedom. At fixed pressure, the additional heat supplies expansion work, $pdV=nRdT$. The two processes can have the same temperature endpoints and equal $\Delta U$ while requiring different heat inputs.

For one mole of the rigid-rotation diatomic ideal gas model, $C_V=5R/2$ and $C_p=7R/2$. Heating by $10\mathrm{K}$ with $R\simeq 8.31446\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$ gives $Q_V\simeq 207.86\mathrm{J}$ in a rigid sealed vessel and $Q_p\simeq 291.01\mathrm{J}$ under a frictionless piston at fixed pressure. Both gain $207.86\mathrm{J}$ of internal energy, while the piston delivers approximately $83.14\mathrm{J}$ of work. This is a calculated model budget. If the capacities vary with temperature, the corresponding finite statement is stillunder the fixed-pressure and fixed-volume heating assumptions in heat capacity. Their final volume values generally differ. Combining Mayer's relation with heat capacity ratio also gives $C_V=nR/(\gamma -1)$ and $C_p=\gamma nR/(\gamma -1)$ locally; constant values cannot be extrapolated across intervals in which $\gamma (T)$ changes.

For a smooth stable single thermodynamic phase of a general simple compressible system, the broader Maxwell relation identity iswhere $\alpha$ is the coefficient of thermal expansion and ${\kappa }_T$ the isothermal compressibility, with composition fixed throughout. Substituting the ideal gas law gives $nR$; a real gas need not do so. This distinction matters near a phase transition or when interactions appreciably change the equation of state. It explains why Mayer's relation is useful: thermal measurements under different constraints become linked to a mechanical response, while the difference retains a particularly simple value for an ideal gas.^[ref]

Use the integrated constant-compressibility model from c, with $\kappa p=0.0059\times 100=0.59$:ThereforeCompression reduces the spacing and moves the same allowed reflection to larger deflection, as the positive pressure derivative predicts. The input compressibility corresponds to a constant bulk modulus $1/\kappa \simeq 169.5$ GPa in this ideal model.

This is the stipulated unchanged cubic-phase, constant-compressibility extrapolation, rather than a reliable prediction for real Fe at 100 GPa. Real Fe exhibits a pressure-induced body-centered-cubic to hexagonal-close-packed transformation, and compressibility can change with pressure.^[ref] A real high-pressure diffraction interpretation would require its actual phase and equation of state. The source gives neither, so its numerical exercise is answered within its explicitly stated idealization.

The isothermal volume compressibility is $\kappa =-(1/V)(\partial V/\partial p{)}_T$. For isotropic cubic compression $V\propto a^3$ and $d_{hkl}=a/\sqrt{h^2+k^2+l^2}$, soDifferentiate Bragg's law $2d_{hkl}\sin \theta =\lambda$ at fixed incident wavelength and fixed reflection:This derivative uses radians. It remains locally valid with a pressure-dependent compressibility, provided the cubic isotropic phase is retained. For a constant stipulated $\kappa$, integration givesThus $a_0(1-\kappa p/3)$ is only the first-order approximation, not the exact consequence of a constant logarithmic compressibility. The finite-pressure calculation also assumes that the reflection still exists, the crystal remains in the same structure and $\sin {\theta }_0{e}^{\kappa p/3}\le 1$. These limitations matter for a large extrapolation.

At fixed particle number, expand the entropy differential in $(T,p)$. The definition of heat capacity at constant pressure and the Maxwell relation $(\partial S/\partial p{)}_T=-(\partial V/\partial T{)}_p$ giveSimilarly expansion in $(T,V)$, using heat capacity at constant volume and $(\partial S/\partial V{)}_T=(\partial p/\partial T{)}_V$, givesAt constant pressure, $dV=V\alpha dT$. Also $dV/V=\alpha dT-\kappa dp$, so $(\partial p/\partial T{)}_V=\alpha /\kappa$. Equating the two entropy expressions at constant pressure yieldsFor a mechanically stable system with positive isothermal compressibility, this also gives $C_p\ge C_V$.

An equation of state relates equilibrium state variables such as pressure, volume, temperature and material amount. At fixed composition, it reduces the number of independent thermodynamic variables. It does not by itself specify all caloric properties: information such as heat capacity or internal energy is also needed.

At a liquid-gas first-order transition, an equilibrium isotherm can contain a constant-pressure interval in which changing total volume changes the phase fractions. A metastable single-phase extrapolation can instead show a region with $(\partial p/\partial V{)}_T>0$; that region has negative isothermal compressibility and is mechanically unstable. Thermodynamic phase coexistence requires equal temperature, pressure and chemical potential, and a consistent free-energy description replaces the unstable region by coexistence. For an appropriate analytic fluid equation of state, a liquid-gas critical point occurs where $(\partial p/\partial V{)}_T=({\partial }^{2}p/\partial V^2{)}_T=0$.

These tests diagnose the phase behaviour represented by that equation of state. They do not guarantee detection of every possible transition involving an additional order parameter which the mechanical equation omits.

At every $T>0$, $V>b$,There is no constant-pressure coexistence interval, unstable loop or finite-temperature critical inflection. The isothermal compressibility is positive:Thus this equation of state predicts no liquid-gas phase transition in its physical domain. Its divergence at the excluded-volume boundary is a repulsive constraint, not a liquid-gas coexistence mechanism. This conclusion concerns the stated model and cannot rule out an ordering transition requiring additional physics or an additional thermodynamic variable. Extrapolation to absolute zero is also outside the classical gas model's reliable regime.

Pressure and its physical terms. Solving the molar Van der Waals equation for pressure givesThe excluded-volume parameter $b$ reduces the volume available for translational motion, raising the kinetic pressure relative to $RT/V$. The attraction parameter $a$ lowers the pressure because attractive interactions reduce the momentum transferred to the boundary. This mean-field model treats $a,b$ as temperature-independent material constants. For one mole, $V$ is the sample volume with the molar amount already fixed; $b$ has units of that volume and $a$ of pressure times volume squared.

volume dependence of entropy. The Helmholtz free energy has differential $dF=-SdT-PdV$. Equality of mixed derivatives yields the Maxwell relationFor $a=0$ this derivative is $R/(V-b)$, givingEquivalently, $S=R\ln (V-b)+f(T)$ with a common reference volume understood inside the logarithm. The equation of state fixes this volume dependence, not the otherwise unspecified function of temperature.

Excess entropy at fixed volume and temperature. The ideal gas has $(\partial S_{\mathrm{id}}/\partial V{)}_T=R/V$. ThereforeIntegrate from infinite volume, where the entropy difference is specified to vanish:It is negative because excluding configurations reduces the accessible state count. It tends to zero as $V\to \infty$ and has leading dilute limit $-Rb/V$. This is the entropy excess of a van der Waals gas.

Effect of attraction. Restoring a temperature-independent $a$ changes the pressure by $-a/V^2$, but not its derivative with respect to temperature. With the same infinite-volume entropy reference, it therefore does not change $\Delta S^{\mathrm{ex}}$ at fixed $V,T$. This can also be seen fromThe attraction affects internal energy and the equilibrium volume, so it can affect entropy indirectly when different state points are compared. If $a$ were temperature dependent, its derivative would contribute directly to the entropy; that is outside the stipulated constant-parameter model.

Coexistence isotherm. Below the Van der Waals critical point, the formal homogeneous isotherm has a local minimum and maximum. Its middle branch has $\partial P/\partial V>0$, hence negative isothermal compressibility, and is mechanically unstable. Portions of the negative-slope branches can be metastable rather than globally stable. The equilibrium isotherm replaces the loop between liquid volume $V_l$ and vapour volume $V_g$ by a constant saturation pressure $P_s$. The two volume are fixed byThe Maxwell equal-area construction expresses equality of the two phases' chemical potentials. Intermediate total volume are mixtures: changing the vapour fraction changes total volume at the same pressure, rather than following the unstable homogeneous curve. The spinodals are the turning points of the formal isotherm; they are distinct from the coexistence endpoints.

Dense-phase volume at low temperature. At a general subcritical temperature, the two coexistence equations above must be solved together. For this equation of state the area condition is explicitlyThis supplies an exact implicit determination of the liquid volume even when a simple closed form is unavailable.

For the asymptotically low-temperature limit, put $\tau =RTb/a\ll 1$. The liquid is dense, $V_l\simeq b$, while the vapour is dilute, $P_s{V}_g\simeq RT$ and $V_g\gg V_l$. The area condition then givesThus the vapour volume grows exponentially in $1/\tau$, and the saturation pressure is exponentially small. It is consequently consistent to neglect $P_s$ in the liquid equation to algebraic order in $\tau$:Choose the root near $b$, not the large-volume zero-pressure root:Expanding gives the leading requested dense-phase behavior,This is the low-temperature liquid volume of a van der Waals gas. Setting $P_s=0$ is an asymptotic approximation, not an exact coexistence condition at every temperature below $T_c$. The figure illustrates generic subcritical coexistence; its finite-temperature liquid volume is determined by the full equal-area equations. The van der Waals liquid also ceases to be a realistic description at sufficiently low physical temperature, where other phases or quantum effects can matter.

An equation of state relates equilibrium pressure, temperature and density. The ideal-gas form $p=n{k}_{B}T$ neglects intermolecular forces and excluded volume. A low-density real-gas expansion iswhere the virial coefficients encode interaction effects and generally depend on temperature.

A simple phenomenological example is the molar Van der Waals equationHere v is molar volume, b represents excluded volume and the attractive interaction lowers pressure through a. Its dilute expansion gives $pv/(RT)=1+[b-a/(RT)]/v+\cdots$. It recovers the ideal-gas limit but is not a quantitatively universal equation for all real gases.

Below its critical temperature, the raw isotherm contains a region with $(\partial p/\partial v{)}_T>0$, corresponding to negative isothermal compressibility and mechanical instability. Equilibrium phase separation replaces that segment by a coexistence pressure satisfying equal chemical potentials, equivalently the Maxwell equal-area construction. The critical point follows from $p_v=p_{vv}=0$:The distinction between an unstable homogeneous branch and a stable two-phase state is essential when interpreting a real-gas equation of state.