Supercritical Fluids and the Critical Point

Turn up the pressure on carbon dioxide gas past 73 atm while holding it above 31°C, and something strange happens. The meniscus separating liquid from vapour simply vanishes. No boiling, no condensing — the two phases merge into a single fluid that is neither. This is the critical point, and the fluid beyond it is supercritical. Supercritical fluids matter far beyond a laboratory curiosity. Supercritical CO₂ decaffeinates coffee, extracts hop oils for beer, and cleans dry-cleaning fabrics without toxic solvents. Supercritical water oxidizes hazardous waste at industrial scale. Understanding why the liquid–vapour boundary terminates is central to reading any one-component phase diagram correctly. This article builds the thermodynamic picture of the critical point from the van der Waals equation. You will see why $\left(\partial p/\partial V_m\right)_T$ and its curvature vanish there. Then you will see what "supercritical" really means physically. Finally you will see how density, viscosity, and solvating power behave in this unusual regime.

Key takeaways

The critical point $(T_c, p_c, V_{m,c})$ is where liquid and vapour branches of the coexistence curve merge and the distinction between them disappears.
Mathematically it is defined by $\left(\partial p/\partial V_m\right)_T = 0$ and $\left(\partial^2 p/\partial V_m^2\right)_T = 0$ simultaneously.
Above $T_c$ , no amount of pressure liquefies a gas — only one fluid phase exists at any density.
Supercritical fluids combine liquid-like density (hence solvating power) with gas-like diffusivity and viscosity.
Near the critical point, compressibility diverges and density fluctuations scatter light strongly — critical opalescence.
Key idea: the critical point is the endpoint of the liquid–vapour coexistence line, where the free-energy surface loses the double-well structure that distinguishes two coexisting phases.

The core insight

The critical point exists because the free-energy difference driving phase separation — rooted in molecular attractions overcoming thermal motion — shrinks to zero as temperature rises; above $T_c$ thermal energy always dominates intermolecular cohesion enough that only one homogeneous fluid state is mechanically stable, however high the pressure.

The physical picture

Consider compressing a real gas at some temperature $T$ below $T_c$ . As pressure rises, the gas eventually reaches its vapour pressure and liquid droplets condense. On a $p$ – $V_m$ diagram this shows up as a horizontal (constant-pressure) segment: gas and liquid coexist, their relative amounts shifting as you compress further, until all the gas has liquefied. The horizontal segment sits between two points on the coexistence curve — one for the vapour molar volume, one for the liquid molar volume — and these two points are connected on the phase diagram by the liquid–vapour boundary. As $T$ is raised, that horizontal segment shrinks: the vapour and liquid molar volumes at coexistence draw closer together, because thermal expansion swells the liquid while compression shrinks the gas. At the critical temperature $T_c$ the segment collapses to a single point — the critical point — where liquid and vapour become identical in density, refractive index, and every other property. Above $T_c$ , isotherms are smooth and monotonic; no flat segment remains, so no compression, however strong, produces two coexisting phases. This is the origin of the empirical rule already met in the carbon dioxide phase diagram: the liquid–vapour boundary simply stops at $(T_c, p_c)$ , it does not cross into some new line. Physically, condensation occurs because intermolecular attractions can organize molecules into a denser phase when thermal agitation is weak enough relative to the attractive well depth. As $T$ increases, mean kinetic energy per molecule grows, and thermal motion increasingly overwhelms the cohesive energy that would otherwise sort molecules into a dilute gas phase and a dense liquid phase. At $T_c$ , attraction and thermal disorder are so evenly matched that the system cannot decide between two distinct densities — density fluctuations of all length scales become equally favorable, and the fluid is critically unstable in this specific mathematical sense, even though it is mechanically stable overall.

Figure

Collapse of the liquid–vapour coexistence region as T approaches Tc

Figure

Pressure–temperature diagram: the liquid–vapour line ends at the critical point

The core relations

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

where: these two conditions together locate the critical point as the inflection point of the critical isotherm.

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

where: $a$ measures attractive interactions, $b$ the excluded molar volume, in the van der Waals equation of state.

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

where: these van der Waals critical constants follow from solving the two derivative conditions simultaneously.

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

where: $Z_c$ is the universal van der Waals critical compressibility factor, a testable (if approximate) prediction.

\kappa_T = -\frac{1}{V_m}\left(\frac{\partial V_m}{\partial p}\right)_T \to \infty \quad \text{as } T \to T_c

where: $\kappa_T$ , the isothermal compressibility, diverges at the critical point, driving critical opalescence.

Derivation AVan der Waals critical constants from the double inflection condition

Goal. Find $(T_c, p_c, V_{m,c})$ in terms of $a$ and $b$ by imposing that the critical isotherm has a horizontal inflection point in $p(V_m)$ .

Step 1. Write the van der Waals equation of state per mole,

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

At the critical point the isotherm $p(V_m)$ at $T = T_c$ has neither a maximum nor a minimum nor ordinary curvature — the flat coexistence step and the two humps flanking it (visible at $T < T_c$ ) have merged into a single saddle-like inflection. This requires both the first and second derivatives with respect to $V_m$ to vanish at fixed $T$ .

Step 2. Differentiate once:

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

Step 3. Differentiate again:

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

Step 4. Divide the Step 3 equation by the Step 2 equation term-by-term to eliminate $T$ . From Step 2,

RT = \frac{2a(V_m - b)^2}{V_m^3}.

Substitute this into Step 3:

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

Simplify the left side:

\frac{4a}{V_m^3(V_m - b)} = \frac{6a}{V_m^4}.

Cross-multiply and cancel $a$ (nonzero) and one power of $V_m^3$ :

4V_m = 6(V_m - b) \;\Longrightarrow\; 2V_m = 6b \;\Longrightarrow\; V_{m,c} = 3b.

Step 5. Substitute $V_{m,c} = 3b$ back into the Step 2 result to get $T_c$ :

RT_c = \frac{2a(3b-b)^2}{(3b)^3} = \frac{2a(2b)^2}{27b^3} = \frac{8ab^2}{27b^3} = \frac{8a}{27b}.

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

Step 6. Substitute $V_{m,c}$ and $T_c$ into the original 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} = \frac{8a/27b}{2b} - \frac{a}{9b^2} = \frac{4a}{27b^2} - \frac{a}{9b^2}.

Common denominator $27b^2$ :

p_c = \frac{4a}{27b^2} - \frac{3a}{27b^2} = \frac{a}{27b^2}.

Step 7. Combine into the 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}} = \frac{3}{8}.

\boxed{V_{m,c} = 3b, \quad p_c = \dfrac{a}{27b^2}, \quad T_c = \dfrac{8a}{27Rb}, \quad Z_c = \dfrac{3}{8} = 0.375}

Real gases give $Z_c$ typically between 0.23 and 0.31, so the van der Waals value is qualitatively right but quantitatively approximate — a reminder that this two-parameter model captures the existence and topology of the critical point better than its precise numbers.

Worked examples

Example 1 — Locating carbon dioxide's critical point from van der Waals constants. Carbon dioxide has van der Waals parameters $a = 3.640\ \text{L}^2\,\text{atm}\,\text{mol}^{-2}$ and $b = 4.267\times10^{-2}\ \text{L mol}^{-1}$ (standard tabulated values). Use the boxed formulas to predict $T_c$ and $p_c$ , and compare with the experimental values $T_c = 304.2$ K, $p_c = 72.9$ atm. Molar volume at the critical point:

V_{m,c} = 3b = 3 \times 4.267\times10^{-2}\ \text{L mol}^{-1} = 0.1280\ \text{L mol}^{-1}.

Critical temperature:

T_c = \frac{8a}{27Rb} = \frac{8 \times 3.640}{27 \times 0.08206 \times 4.267\times10^{-2}}\ \text{K}.

Numerator: $8 \times 3.640 = 29.12$ . Denominator: $27 \times 0.08206 \times 0.04267 = 0.09455$ .

T_c = \frac{29.12}{0.09455}\ \text{K} = 308.0\ \text{K}.

Critical pressure:

p_c = \frac{a}{27b^2} = \frac{3.640}{27 \times (4.267\times10^{-2})^2}\ \text{atm}.

Denominator: $27 \times 1.8217\times10^{-3} = 4.919\times10^{-2}$ .

p_c = \frac{3.640}{4.919\times10^{-2}}\ \text{atm} = 74.0\ \text{atm}.

Sanity check. The van der Waals prediction ( $T_c = 308.0$ K, $p_c = 74.0$ atm) lands within about 1–2% of the experimental values (304.2 K, 72.9 atm) — the kind of agreement expected from a mean-field model with only two adjustable parameters. This confirms the CO₂ phase diagram fact that beyond roughly 31°C and 73 atm, CO₂ can exist only as one supercritical fluid phase, however high the pressure climbs.

Example 2 — Density and solvating regime of supercritical CO₂ used in decaffeination. Industrial decaffeination runs supercritical CO₂ near $T = 323$ K (50°C) and $p = 250$ atm, well above $T_c = 304.2$ K and $p_c = 72.9$ atm. Estimate the reduced conditions and use them to judge whether the fluid behaves more liquid-like or gas-like. Reduced temperature and pressure:

T_r = \frac{T}{T_c} = \frac{323\ \text{K}}{304.2\ \text{K}} = 1.062.

p_r = \frac{p}{p_c} = \frac{250\ \text{atm}}{72.9\ \text{atm}} = 3.43.

Because $T_r$ is only marginally above 1 while $p_r$ is far above 1, the fluid sits in the region of the supercritical phase diagram where compressing strongly increases density even though no phase transition occurs. Experimental data place the density of CO₂ under these conditions around

\rho \approx 0.80\text{--}0.85\ \text{g cm}^{-3},

comparable to liquid CO₂ densities (liquid CO₂ near its normal boiling range runs about 0.77–1.18 g cm⁻³ depending on conditions) and roughly 400–500 times the density of CO₂ gas at 1 atm and 298 K,

\rho_{\text{gas,1atm}} = \frac{pM}{RT} = \frac{(1\ \text{atm})(44.01\ \text{g mol}^{-1})}{(0.08206\ \text{L atm K}^{-1}\text{mol}^{-1})(298\ \text{K})} = 1.80\ \text{g L}^{-1} = 1.80\times10^{-3}\ \text{g cm}^{-3}.

Ratio of densities:

\frac{\rho_{\text{supercritical}}}{\rho_{\text{gas,1atm}}} \approx \frac{0.82}{1.80\times10^{-3}} \approx 460.

Sanity check and interpretation. A nearly 500-fold density enhancement over ambient gas — while viscosity stays close to a gas-like $10^{-4}$ to $10^{-5}$ Pa s rather than a liquid's $10^{-3}$ Pa s — is exactly why supercritical CO₂ dissolves nonpolar caffeine and hop oils like a liquid solvent, yet diffuses into the coffee bean matrix and can be flashed off cleanly like a gas. This dual character, tunable continuously by adjusting $p$ and $T$ without crossing any phase boundary, is the operational definition of a useful supercritical solvent.

Limiting cases & common traps

Situation	What actually happens
$T < T_c$ , compress the gas	Liquid condenses at fixed $p$ ; two phases coexist until liquefaction is complete
$T > T_c$ , compress the fluid	Density rises smoothly and continuously; never two coexisting phases
Exactly at $(T_c, p_c)$	Compressibility diverges; density fluctuations at all scales; critical opalescence
$T_r \gg 1$ far above $T_c$	Fluid behaves essentially as an ordinary dense gas, weak solvating power
$T_r \approx 1$ , $p_r$ large	Liquid-like density with gas-like transport — the useful supercritical solvent regime

The chemist behind it

The critical point was identified experimentally by Thomas Andrews in the 1860s, who studied carbon dioxide under carefully controlled pressure and temperature in sealed glass tubes. Andrews observed that above roughly 31°C, no pressure — however great — would produce a visible meniscus between liquid and gas; he coined the term "critical point" for this threshold. His painstaking p–V–T measurements on CO₂ gave the field its founding experimental phase diagram and directly inspired van der Waals's equation of state a few years later, which for the first time gave critical phenomena a molecular, mathematical explanation rooted in intermolecular attraction and finite molecular size.

Origins · 1 minThomas Andrews and how this was discovered

Thomas Andrews

1813–1885 · Irish chemist who discovered the critical point of carbon dioxide and coined the term "critical point"

Ask the tutor: How does the principle of corresponding states let us predict one substance's phase behavior from another's using only reduced variables?

What exactly disappears at the critical point?

The distinction between liquid and vapour disappears, not matter itself. Below $T_c$ , liquid and vapour have different densities, refractive indices, and other properties; a visible meniscus separates them. At $T_c$ , the vapour and liquid branches of the coexistence curve converge to identical values of these properties. Above $T_c$ , the system is a single homogeneous fluid phase at every pressure — there is no longer any density gap to create two distinguishable coexisting states.

Why does compressibility diverge at the critical point?

Isothermal compressibility $\kappa_T$ measures how much volume changes for a given pressure change. Near $T_c$ , the free-energy landscape flattens along the density direction, meaning huge volume (density) fluctuations cost almost no free energy. This flattening is exactly the vanishing of $\left(\partial p/\partial V_m\right)_T$ derived from the van der Waals equation, and it drives the strong, all-length-scale density fluctuations responsible for critical opalescence.

Is Zc = 3/8 exactly right for real substances?

No. The van der Waals prediction $Z_c = 0.375$ is a mean-field approximation from a simple two-parameter equation of state. Real substances typically show $Z_c$ between about 0.23 (water, strongly hydrogen-bonded) and 0.31 (simple nonpolar gases like argon). The van der Waals model correctly predicts that a universal, substance-independent $Z_c$ should exist — the principle of corresponding states — even though its numerical value needs refinement from more accurate equations of state.

How does the critical point connect to the phase rule?

By the phase rule, $F = C - P + 2$ , a one-component system with two coexisting phases ( $C=1$ , $P=2$ ) has one degree of freedom: fixing $T$ fixes the coexistence pressure. The critical point is where $P$ effectively drops from two phases to one identical phase; the coexistence line simply terminates there rather than continuing indefinitely, consistent with there being no way to maintain two distinguishable phases beyond $T_c$ .

Do all substances have a critical point?

Every pure substance with an ordinary liquid–vapour equilibrium has one, since the same argument — thermal energy eventually overwhelming intermolecular cohesion — applies generally. Critical temperatures vary enormously: helium's $T_c$ is only 5.2 K, water's is 647 K, and heavy hydrocarbons can exceed 700 K. Substances that decompose chemically before reaching their predicted $T_c$ never display a directly observable critical point, though the underlying equation of state still formally has one.