Isotherms and Critical Behavior of Real Gases

Push carbon dioxide gas hard enough at room temperature and something strange happens. The pressure climbs, then suddenly stalls flat while the gas condenses into liquid before your eyes. Above 31.0 °C, no matter how hard you compress it, that boundary vanishes and CO₂ just becomes a dense fluid with no visible transition at all. This behavior sits at the heart of real-gas thermodynamics. It explains why refrigerants liquefy, why supercritical CO₂ extracts caffeine from coffee beans, and why every real gas deviates from the ideal law. Understanding it means understanding the shape of pressure-volume isotherms. This article builds the van der Waals isotherm from its equation of state. You will see why isotherms below a critical temperature loop unphysically. Then you will see how the Maxwell construction fixes that loop into a flat coexistence line. Finally you will locate the critical point itself, where liquid and gas become indistinguishable.

Key takeaways

A $pV_m$ isotherm plots pressure against molar volume at fixed temperature; ideal gases give simple hyperbolas, real gases do not.
Below the critical temperature $T_c$ , the van der Waals equation predicts an unphysical wiggle with $\partial p/\partial V_m > 0$ in places.
The Maxwell equal-area construction replaces that wiggle with a flat horizontal segment representing genuine liquid-gas coexistence.
The critical point is where the flat coexistence region shrinks to a single inflection point: $(\partial p/\partial V_m)_T = 0$ and $(\partial^2 p/\partial V_m^2)_T = 0$ .
Above $T_c$ , no phase boundary exists — the substance is a single supercritical fluid at any pressure.
Key idea: the loop-shaped isotherm predicted by a mean-field equation of state is not wrong physics but incomplete physics; the Maxwell construction restores thermodynamic consistency by enforcing equal chemical potential for coexisting phases.

The core insight

A real gas isotherm below $T_c$ must contain a flat, horizontal segment where liquid and vapor coexist at a single equilibrium pressure — the van der Waals loop that replaces this segment is a mathematical artifact of the equation's smooth polynomial form, corrected by demanding equal Gibbs energy (equal area) on either side.

The physical picture

Consider tracking the pressure of a fixed amount of gas as you slowly compress it at constant temperature, plotting $p$ against molar volume $V_m$ . For an ideal gas this traces a simple hyperbola, $p = RT/V_m$ , dropping smoothly to zero volume with infinite pressure. Real substances behave nothing like this once you cool them enough. At high temperature, far above the substance's critical temperature $T_c$ , the isotherm is still smooth and monotonic — pressure rises steadily as volume shrinks, resembling the ideal gas but bowed by intermolecular forces. Cool the gas below $T_c$ , though, and compressing it reveals a plateau: over a range of volume, pressure stays constant. Physically, the sample is separating into two coexisting phases, liquid and vapor, each with its own well-defined molar volume ( $V_\text{liq}$ and $V_\text{gas}$ ), while the overall volume of the mixed system slides between these two extremes as more material condenses. Only once every last molecule has liquefied does pressure begin to climb steeply again, reflecting the near-incompressibility of the liquid. The van der Waals equation of state,

p = \frac{RT}{V_m - b} - \frac{a}{V_m^2},

is a cubic polynomial in $V_m$ at fixed $T$ and $p$ . Below $T_c$ this cubic has three real roots, producing a wave that dips down and rises back up — a region where $p$ decreases as $V_m$ decreases, i.e. $(\partial p/\partial V_m)_T > 0$ . That would mean compressing the fluid lowers its pressure, which is mechanically unstable and never observed. Nature replaces this unphysical loop with the flat coexistence line at the same temperature; the van der Waals loop is retained mathematically only as a bridge connecting the stable liquid and vapor branches. As $T$ rises toward $T_c$ , the flat plateau shrinks. Exactly at $T_c$ it contracts to a single point — the critical point $(T_c, p_c, V_{m,c})$ — where the liquid and gas branches merge and become indistinguishable. Above $T_c$ , the substance is called a supercritical fluid: it has no discrete gas-to-liquid transition at any pressure.

Figure

Van der Waals isotherms at three temperatures, showing the loop, the Maxwell-corrected plateau, and the critical point

The core relations

p = \frac{RT}{V_m - b} - \frac{a}{V_m^2}

where: $a$ measures attractive intermolecular forces (Pa m⁶ mol⁻²), $b$ is the excluded molar volume (m³ mol⁻¹).

\left(\frac{\partial p}{\partial V_m}\right)_T = 0 \qquad\text{and}\qquad \left(\frac{\partial^2 p}{\partial V_m^2}\right)_T = 0 \quad \text{at } T = T_c

where: these two conditions together define the inflection point that is the critical point.

V_{m,c} = 3b, \qquad p_c = \frac{a}{27b^2}, \qquad T_c = \frac{8a}{27Rb}

where: the critical constants, expressed purely in terms of the two van der Waals parameters.

Z_c = \frac{p_c V_{m,c}}{RT_c} = \frac{3}{8} = 0.375

where: the universal critical compressibility factor predicted by van der Waals theory (real gases typically show 0.23–0.31).

\int_{V_\text{liq}}^{V_\text{gas}} \big(p_\text{vdW}(V_m) - p_\text{eq}\big)\, dV_m = 0

where: the Maxwell equal-area condition; $p_\text{eq}$ is the flat coexistence pressure that makes the two enclosed loop areas equal.

The core derivation

Derivation ALocating the critical point on the van der Waals surface

Goal. Find $V_{m,c}$ , $p_c$ , $T_c$ from $a$ and $b$ by demanding that the isotherm's inflection point (where the loop degenerates to a single point) coincides with a horizontal tangent.

Step 1. Write the two critical conditions. At the critical point the cubic isotherm has a triple root, meaning both the slope and curvature of $p(V_m)$ vanish simultaneously:

\left(\frac{\partial p}{\partial V_m}\right)_T = -\frac{RT}{(V_m-b)^2} + \frac{2a}{V_m^3} = 0

\left(\frac{\partial^2 p}{\partial V_m^2}\right)_T = \frac{2RT}{(V_m-b)^3} - \frac{6a}{V_m^4} = 0

Step 2. Divide the two equations to eliminate $T$ . Dividing the second equation by the first:

\frac{2RT/(V_m-b)^3}{RT/(V_m-b)^2} = \frac{6a/V_m^4}{2a/V_m^3}

\frac{2}{V_m - b} = \frac{3}{V_m}

Cross-multiplying gives $2V_m = 3V_m - 3b$ , so

\boxed{V_{m,c} = 3b}

Step 3. Substitute back to find $T_c$ . From the first-derivative condition,

RT_c = \frac{2a(V_{m,c}-b)^2}{V_{m,c}^3} = \frac{2a(2b)^2}{27b^3} = \frac{8ab^2}{27b^3} = \frac{8a}{27b}

\boxed{T_c = \frac{8a}{27Rb}}

Step 4. Substitute $V_{m,c}$ and $T_c$ into the equation of state to get $p_c$ .

p_c = \frac{RT_c}{V_{m,c}-b} - \frac{a}{V_{m,c}^2} = \frac{R\left(\dfrac{8a}{27Rb}\right)}{3b-b} - \frac{a}{9b^2}

p_c = \frac{8a}{27b \cdot 2b} - \frac{a}{9b^2} = \frac{4a}{27b^2} - \frac{3a}{27b^2}

\boxed{p_c = \frac{a}{27b^2}}

Step 5. Form the critical compressibility factor.

Z_c = \frac{p_c V_{m,c}}{RT_c} = \frac{\left(\dfrac{a}{27b^2}\right)(3b)}{R\left(\dfrac{8a}{27Rb}\right)} = \frac{\dfrac{3a}{27b}}{\dfrac{8a}{27b}}

\boxed{Z_c = \frac{3}{8} = 0.375}

This single universal number — independent of $a$ , $b$ , or the identity of the gas — is the celebrated (and only approximately correct) prediction that seeds the law of corresponding states.

Worked examples

Example 1 — Critical constants of carbon dioxide from its van der Waals parameters. CO₂ has van der Waals constants $a = 0.3640\ \text{Pa m}^6\text{mol}^{-2}$ and $b = 4.267\times10^{-5}\ \text{m}^3\text{mol}^{-1}$ (standard tabulated values). Molar volume at the critical point:

V_{m,c} = 3b = 3 \times 4.267\times10^{-5}\ \text{m}^3\text{mol}^{-1} = 1.280\times10^{-4}\ \text{m}^3\text{mol}^{-1}

Critical temperature:

T_c = \frac{8a}{27Rb} = \frac{8(0.3640)}{27(8.314)(4.267\times10^{-5})}

T_c = \frac{2.912}{27 \times 8.314 \times 4.267\times10^{-5}} = \frac{2.912}{9.577\times10^{-3}} = 304.1\ \text{K}

Critical pressure:

p_c = \frac{a}{27b^2} = \frac{0.3640}{27\,(4.267\times10^{-5})^2} = \frac{0.3640}{27 \times 1.821\times10^{-9}}

p_c = \frac{0.3640}{4.914\times10^{-8}} = 7.41\times10^{6}\ \text{Pa} = 73.1\ \text{atm}

Sanity check: the experimental critical temperature of CO₂ is 304.13 K (30.98 °C) and the experimental critical pressure is 72.8 atm — our van der Waals estimates (304.1 K, 73.1 atm) agree to within 0.4%, an excellent showing for a two-parameter mean-field model. Predicted $Z_c = 0.375$ versus the true experimental value $Z_c \approx 0.274$ for CO₂ — the van der Waals model captures the location of the critical point well but is less accurate about compressibility right at that point, because it neglects the diverging fluctuations that occur there.

Example 2 — Locating the Maxwell coexistence pressure at a sub-critical temperature. Consider a hypothetical van der Waals fluid with $a = 0.50\ \text{Pa m}^6\text{mol}^{-2}$ and $b = 4.0\times10^{-5}\ \text{m}^3\text{mol}^{-1}$ , held at $T = 0.90\,T_c$ . First find $T_c$ :

T_c = \frac{8a}{27Rb} = \frac{8(0.50)}{27(8.314)(4.0\times10^{-5})} = \frac{4.00}{8.979\times10^{-3}} = 445.5\ \text{K}

So $T = 0.90 \times 445.5\ \text{K} = 401.0\ \text{K}$ , safely below $T_c$ , guaranteeing a loop. The equal-area (Maxwell) condition requires finding a pressure $p_\text{eq}$ and two volumes $V_\text{liq} < V_\text{gas}$ satisfying simultaneously: 1. Both points lie on the isotherm: $p_\text{eq} = RT/(V_m - b) - a/V_m^2$ evaluated at each root. 2. The areas under the loop above and below the line $p = p_\text{eq}$ are equal — equivalent to equal Gibbs energy, $\mu_\text{liq} = \mu_\text{gas}$ . This is normally solved numerically (Newton-Raphson on the two nonlinear conditions), but we can estimate the order of magnitude. Using the reduced-variable form (foreshadowed by the law of corresponding states), at $T_r = 0.90$ the reduced coexistence pressure for a van der Waals fluid is known from tabulated solutions to be approximately $p_r \approx 0.68$ . Since

p_c = \frac{a}{27b^2} = \frac{0.50}{27(4.0\times10^{-5})^2} = \frac{0.50}{4.32\times10^{-8}} = 1.157\times10^{7}\ \text{Pa}

the coexistence pressure is

p_\text{eq} = p_r \, p_c \approx 0.68 \times 1.157\times10^7\ \text{Pa} \approx 7.9\times10^{6}\ \text{Pa} \approx 78\ \text{atm}

Sanity check: this coexistence pressure sits below $p_c$ (as it must — the coexistence curve only reaches $p_c$ exactly at $T_c$ ) and reduces smoothly toward $p_c$ as $T_r \to 1$ , consistent with the plateau shrinking to the critical point.

Limiting cases & common traps

Situation	What happens	Common trap
$T \gg T_c$	Isotherm nearly ideal-gas hyperbola	Forgetting real gases still deviate at very high $p$ even here
$T$ slightly below $T_c$	Loop is shallow, plateau narrow	Assuming the loop itself is physically traversed — it is not
$T \ll T_c$	Deep loop, wide plateau, distinct liquid/gas volumes	Ignoring that the liquid branch is nearly incompressible (steep rise)
$T = T_c$ exactly	Inflection point, $Z_c = 3/8$ for vdW	Treating $Z_c = 3/8$ as universally exact — real gases give 0.23–0.31
Middle of the vdW loop	$(\partial p/\partial V_m)_T > 0$	Believing this region is metastable superheated/supercooled fluid; it is mechanically unstable, not just metastable

The chemist behind it

Johannes Diderik van der Waals proposed his equation of state in his 1873 doctoral dissertation, correcting the ideal gas law for finite molecular size and intermolecular attraction with just two empirical constants, $a$ and $b$ . What made the equation historic was not merely fitting real $pV$ data — it was that the same equation, for the first time, predicted a critical point and gas-liquid coexistence from first principles, unifying liquid and gaseous states as limits of one continuous equation of state. James Clerk Maxwell recognized shortly afterward that thermodynamic consistency (equal chemical potential of coexisting phases) demanded the equal-area rule that now bears his name, converting van der Waals' loop into a physically faithful isotherm.

Origins · 1 minJohannes Diderik van der Waals and how this was discovered

Johannes Diderik van der Waals

1837–1923 · Dutch physicist whose 1873 equation of state first unified the liquid and gas phases and earned the 1910 Nobel Prize in Physics

Ask the tutor: Want to see how the reduced form of the van der Waals equation eliminates $a$ and $b$ to give the universal law of corresponding states?

Why does the van der Waals equation predict an S-shaped loop instead of a flat line directly

The van der Waals equation is a smooth, single continuous cubic polynomial in molar volume. Smooth polynomials cannot spontaneously produce a flat, non-analytic segment on their own — that would require a discontinuity in slope. The equation instead threads a continuous wiggle through the region where two phases should coexist. Recovering the physically correct flat line requires an extra thermodynamic argument, the Maxwell construction, layered on top of the smooth equation of state; the equation of state alone is not thermodynamically sufficient at coexistence.

Is the critical compressibility factor of 3/8 accurate for real gases

Not precisely. The van der Waals prediction $Z_c = 0.375$ is a fixed universal number because it depends only on the algebraic structure of the equation, not on any specific gas. Real gases show $Z_c$ values clustered between about 0.23 (water) and 0.31 (helium, neon), reflecting the more complex intermolecular potentials that a two-parameter equation cannot capture. The value 3/8 is best treated as a rough benchmark and motivation for the law of corresponding states, not as an accurate universal constant.

What actually happens physically inside the flat coexistence plateau

As you compress the sample along the flat segment, the pressure stays fixed at $p_\text{eq}$ while the fraction of liquid grows and the fraction of vapor shrinks. Both phases individually keep constant molar volumes, $V_\text{liq}$ and $V_\text{gas}$ , throughout this process. Only the overall (lever-rule-weighted) volume of the two-phase mixture changes. Once the last bubble of vapor collapses, the system becomes homogeneous liquid, and further compression requires a steep pressure rise because liquids resist compression strongly.

What happens to the isotherm exactly at the critical point

At $T_c$ , the two roots that bound the coexistence plateau ( $V_\text{liq}$ and $V_\text{gas}$ ) merge into a single value, $V_{m,c}$ . The isotherm develops a horizontal inflection point there rather than a loop or plateau. Above $T_c$ , no coexistence plateau exists at any pressure; the substance transitions smoothly from gas-like to liquid-like density without a distinguishable phase transition, becoming what is called a supercritical fluid.

How does this topic connect to the law of corresponding states

Because the critical constants $T_c$ , $p_c$ , $V_{m,c}$ can be expressed purely through $a$ and $b$ , one can rewrite the van der Waals equation in reduced variables $T_r = T/T_c$ , $p_r = p/p_c$ , $V_r = V_m/V_{m,c}$ . The result eliminates $a$ and $b$ entirely, leaving a single universal equation valid for any van der Waals-like substance. That universality is the law of corresponding states, letting chemists predict one gas's behavior from another's using only reduced coordinates.