The Phase Diagram of Carbon Dioxide

Drop an ice cube in a glass and it melts into a puddle of liquid water. Drop a piece of dry ice in the same glass and something odd happens: no puddle ever forms. The solid simply shrinks and vanishes into a cold white fog. This everyday contrast is a direct consequence of where carbon dioxide's triple point sits relative to atmospheric pressure. Carbon dioxide is the textbook counter-example to water in the study of one-component phase diagrams. Its triple point lies at 5.11 atm. This is well above the 1 atm we breathe at. That single fact rearranges the whole diagram. It also changes how CO2 behaves at ordinary pressure. This article builds the CO2 phase diagram from the phase rule. It locates its critical and triple points precisely. It also derives why solid CO2 sublimes at 1 atm instead of melting. It closes with the link to supercritical CO2, the workhorse solvent of decaffeination and green chemistry.

Key takeaways

CO2's triple point is at $T_3 = 216.6\text{ K}$ ( $-56.6\,^\circ\text{C}$ ) and $p_3 = 5.11\text{ atm}$ , above normal atmospheric pressure.
Because 1 atm lies below the triple-point pressure, the liquid phase cannot exist at 1 atm; solid CO2 sublimes directly to gas.
The solid–liquid coexistence line has a positive (steep, near-vertical) slope, unlike water's, because solid CO2 is denser than liquid CO2.
The critical point sits at $T_c = 304.2\text{ K}$ ( $31.0\,^\circ\text{C}$ ) and $p_c = 72.8\text{ atm}$ , easily reachable with modest engineering.
Liquid CO2 exists only above about 5.11 atm, which is why fire extinguishers and CO2 cartridges are always pressurized vessels.
Key idea: the position of the triple point relative to 1 atm — not any exotic chemistry — determines whether a substance melts or sublimes at ordinary pressure.

The core insight

Every feature of the CO2 phase diagram that surprises a first-time viewer — dry ice "melting" without a puddle, the near-vertical fusion line, the easy accessibility of the supercritical region — follows from one geometric fact: the triple point of CO2 lies at 5.11 atm, above 1 atm, so the horizontal line $p = 1\text{ atm}$ crosses only the solid–gas sublimation curve.

The physical picture

A one-component phase diagram plots pressure against temperature and divides the plane into regions where solid, liquid, or gas is the thermodynamically stable phase, separated by coexistence curves along which two phases share equal chemical potential. Three such curves — sublimation (solid–gas), fusion (solid–liquid), and vaporization (liquid–gas) — meet at the triple point, the unique $(T_3, p_3)$ where all three phases coexist. The vaporization curve terminates at the critical point $(T_c, p_c)$ , beyond which liquid and gas become indistinguishable. For water, the triple point (0.0098 °C, 0.00604 atm) sits far below 1 atm, so a horizontal line at 1 atm crosses both the fusion curve (near 0 °C) and the vaporization curve (100 °C): ice melts to liquid, and liquid boils to gas. Carbon dioxide's triple point, by contrast, sits at 5.11 atm — nearly five times atmospheric pressure. The 1 atm line therefore passes below the triple point and cuts only through the solid and gas regions, crossing the sublimation curve at $-78.5\,^\circ\text{C}$ . Liquid CO2 simply does not exist along that line. Warm solid CO2 at 1 atm and it converts straight to vapor, the phenomenon that gives "dry ice" its name and its fog-machine popularity. A second, more subtle feature is the slope of the fusion curve. For most substances (CO2 included), the solid is denser than the liquid, so by the Clapeyron equation the fusion curve has a steep positive slope — the melting point rises slightly with pressure. Water is the famous exception, with a negative-slope fusion curve, because ice is less dense than liquid water. CO2 behaves normally in this respect; its fusion line tilts only slightly off vertical, rising from the triple point toward higher temperature as pressure increases.

Figure

The CO2 phase diagram: three curves meeting at the triple point, well above 1 atm

Figure

Why dry ice sublimes: the 1 atm isobar intersects only the sublimation curve

The core relations

F = C - P + 2

where: $F$ = degrees of freedom, $C$ = number of components (here $C=1$ ), $P$ = number of phases present; at the triple point $P=3$ so $F=0$ , fixing $T_3$ and $p_3$ uniquely.

\frac{dp}{dT} = \frac{\Delta_{\text{trs}}H}{T\,\Delta_{\text{trs}}V}

where: this is the Clapeyron equation giving the exact slope of any two-phase coexistence curve, with $\Delta_{\text{trs}}H$ and $\Delta_{\text{trs}}V$ the molar enthalpy and volume change of the transition.

\ln\frac{p_2}{p_1} = -\frac{\Delta_{\text{sub}}H}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)

where: the Clausius–Clapeyron equation, an integrated approximation valid for solid/liquid–gas equilibria where the gas is ideal and $\Delta V \approx V_{\text{gas}}$ .

T_3 = 216.6\text{ K}, \quad p_3 = 5.11\text{ atm}; \qquad T_c = 304.2\text{ K}, \quad p_c = 72.8\text{ atm}

where: the four numbers that anchor the entire CO2 diagram; all measured quantities, not derived from theory.

Derivation AWhy 1 atm crosses only the sublimation curve

Goal. Show quantitatively that liquid CO2 cannot exist at $p = 1\text{ atm}$ , using only the triple-point data and the Clausius–Clapeyron equation, and find the temperature at which solid CO2 in equilibrium with vapor reaches 1 atm.

Step 1 — state the physical criterion. Liquid is stable at some pressure $p$ only if $p$ lies between the sublimation-curve pressure and the vaporization-curve pressure at temperatures above $T_3$ ; equivalently, liquid can exist only for $p \ge p_3 = 5.11\text{ atm}$ , since the fusion and vaporization curves both originate at the triple point and rise from it. Since $1\text{ atm} < 5.11\text{ atm}$ , liquid CO2 is thermodynamically forbidden at 1 atm at any temperature — this is the entire argument in one inequality.

Step 2 — locate where the sublimation curve crosses 1 atm. Treat the solid–gas equilibrium with Clausius–Clapeyron, using the triple point as the reference state $(T_1, p_1) = (216.6\text{ K}, 5.11\text{ atm})$ and the enthalpy of sublimation $\Delta_{\text{sub}}H \approx 26.1\text{ kJ mol}^{-1}$ (a standard tabulated value near the triple point). We solve for $T_2$ at $p_2 = 1\text{ atm}$ :

\ln\frac{p_2}{p_1} = -\frac{\Delta_{\text{sub}}H}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)

Step 3 — insert numbers.

\ln\frac{1}{5.11} = \ln(0.1957) = -1.631

-1.631 = -\frac{26100\text{ J mol}^{-1}}{8.314\text{ J K}^{-1}\text{mol}^{-1}}\left(\frac{1}{T_2}-\frac{1}{216.6\text{ K}}\right)

-1.631 = -3140\text{ K}\left(\frac{1}{T_2}-4.617\times10^{-3}\text{ K}^{-1}\right)

Step 4 — solve for $T_2$ .

\frac{1}{T_2} = 4.617\times10^{-3}\text{ K}^{-1} + \frac{1.631}{3140\text{ K}} = 4.617\times10^{-3}\text{ K}^{-1} + 5.195\times10^{-4}\text{ K}^{-1} = 5.137\times10^{-3}\text{ K}^{-1}

T_2 = 194.7\text{ K} = -78.5\,^\circ\text{C}

Step 5 — interpret. This matches the well-known sublimation temperature of dry ice at 1 atm, $-78.5\,^\circ\text{C}$ , confirming the model. Because this crossing occurs entirely within the solid–gas curve (we never needed the fusion or vaporization curve), the calculation independently verifies that solid CO2 goes directly to vapor at atmospheric pressure.

\boxed{T_{\text{sub}}(1\text{ atm}) = 194.7\text{ K} = -78.5\,^\circ\text{C}, \text{ and liquid CO}_2 \text{ cannot exist below } 5.11\text{ atm}}

Worked examples

Example 1 — Foundational: verifying the phase rule at the triple point and along a coexistence curve. Confirm the degrees of freedom at the CO2 triple point and along the sublimation curve, and state what each value means operationally. At the triple point, three phases coexist: $C = 1$ , $P = 3$ .

F = C - P + 2 = 1 - 3 + 2 = 0

$F = 0$ means the system is invariant: temperature and pressure are both fixed exactly at $216.6\text{ K}$ and $5.11\text{ atm}$ . You cannot independently choose either variable and still have three phases in equilibrium — nature picks the point for you, which is precisely why triple points serve as fixed calibration references. Along the sublimation curve away from the triple point, only two phases coexist (solid, gas): $C=1$ , $P=2$ .

F = C - P + 2 = 1 - 2 + 2 = 1

$F = 1$ means one intensive variable — say $T$ — can be chosen freely, and $p$ is then fixed by the coexistence curve (this is exactly the Clausius–Clapeyron relationship used above). Sanity check: this matches the general rule that on a one-component diagram, coexistence curves (lines) have $F=1$ , single-phase regions (areas) have $F=2$ , and the triple point (a single dot) has $F=0$ — dimensionality of the geometric feature equals $F$ .

Example 2 — Applied: pressure inside a sealed CO2 fire extinguisher at two temperatures. A carbon dioxide fire extinguisher contains liquid CO2 in equilibrium with its vapor at room temperature. Estimate the internal pressure at $25\,^\circ\text{C}$ (298 K) and at a hot summer temperature of $45\,^\circ\text{C}$ (318 K), using the Clausius–Clapeyron equation anchored at the normal boiling point of CO2 under its own vapor pressure — take the reference state as the critical point is too far away, so instead anchor at a known vapor-pressure point: $T_1 = 253.1\text{ K}$ (−20.1 °C), $p_1 = 19.7\text{ atm}$ (a tabulated vapor-pressure datum for liquid CO2), with $\Delta_{\text{vap}}H \approx 16.0\text{ kJ mol}^{-1}$ near this range. Step 1 — write Clausius–Clapeyron for the vaporization curve:

\ln\frac{p_2}{p_1} = -\frac{\Delta_{\text{vap}}H}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)

Step 2 — evaluate at $T_2 = 298\text{ K}$ :

\frac{1}{T_2}-\frac{1}{T_1} = \frac{1}{298\text{ K}} - \frac{1}{253.1\text{ K}} = 3.356\times10^{-3} - 3.951\times10^{-3} = -5.95\times10^{-4}\text{ K}^{-1}

\ln\frac{p_2}{p_1} = -\frac{16000\text{ J mol}^{-1}}{8.314\text{ J K}^{-1}\text{mol}^{-1}}\times(-5.95\times10^{-4}\text{ K}^{-1}) = 1.145

p_2 = 19.7\text{ atm}\times e^{1.145} = 19.7\times3.142 = 61.9\text{ atm}

Step 3 — evaluate at $T_2 = 318\text{ K}$ :

\frac{1}{318\text{ K}} - \frac{1}{253.1\text{ K}} = 3.145\times10^{-3} - 3.951\times10^{-3} = -8.06\times10^{-4}\text{ K}^{-1}

\ln\frac{p_2}{p_1} = -1924.0\times(-8.06\times10^{-4}) = 1.551

p_2 = 19.7\times e^{1.551} = 19.7\times4.716 = 92.9\text{ atm}

Sanity check: the extrapolated value at 318 K (92.9 atm) exceeds the critical pressure (72.8 atm) — but 318 K is still below $T_c = 304.2\text{ K}$ ... wait, 318 K > 304.2 K, so the extrapolation has actually run past the critical point, where liquid–vapor equilibrium no longer exists at all. This is a genuine physical warning built into the numbers: a real extinguisher left in a hot car (45 °C) is dangerously close to, or past, the critical temperature of CO2 (31.0 °C), at which point there is no distinct liquid phase left and pressure rises sharply — exactly why extinguishers carry temperature limits. The 298 K estimate, about 62 atm, is a realistic reference value close to the literature figure (~64 atm at 25 °C).

Limiting cases & common traps

Situation	What happens	Why
CO2 warmed at 1 atm	Solid → gas directly (sublimation), never liquid	1 atm is below $p_3 = 5.11$ atm
CO2 compressed above 5.11 atm and warmed past 216.6 K	Solid → liquid → gas is possible	pressure now exceeds the triple-point pressure
CO2 heated above $T_c = 304.2$ K at any pressure	No liquid–gas distinction; supercritical fluid	beyond the critical point, the coexistence curve simply ends
Water compared to CO2	Ice melts at 1 atm; fusion curve slopes negative	ice is less dense than water — opposite of CO2's solid

The chemist behind it

The modern quantitative treatment of CO2's coexistence curves rests on the Clausius–Clapeyron framework, but the specific high-precision measurements that pin down the triple point and critical point of CO2 trace back to 19th-century experimentalists studying "permanent gases" and liquefaction. Thomas Andrews's systematic 1869 study of carbon dioxide under pressure first demonstrated the critical point experimentally, showing that above 31 °C no amount of pressure could produce a visible liquid–gas meniscus — a discovery that made CO2 the archetypal substance for critical phenomena and, later, for the entire field of supercritical fluids.

Origins · 1 minThomas Andrews and how this was discovered

Thomas Andrews

1813–1885 · Irish physical chemist who first demonstrated the critical point of carbon dioxide experimentally in 1869, establishing CO2 as the reference substance for critical-phenomena studies

Ask the tutor: How does the phase diagram change if we compare CO2 to a substance with an even lower critical temperature, like methane?

Why is CO2's triple point above atmospheric pressure while water's is far below it?

The location of the triple point is an empirical property of each substance, set by its intermolecular forces and molecular size, not by any general rule. Water's weak, sharply directional hydrogen bonds allow ice and vapor to coexist at extremely low pressure, near 0.006 atm. CO2 molecules interact through weaker, less specific dispersion and quadrupole forces, and the resulting sublimation and fusion curves happen to cross at a much higher pressure, 5.11 atm. There is no deeper thermodynamic reason both triple points could not have landed on either side of 1 atm; it is simply how each substance's forces work out.

Can liquid CO2 ever exist naturally on Earth's surface?

Not at typical atmospheric conditions, since sea-level pressure (1 atm) is below the 5.11 atm triple-point pressure. However, natural CO2-rich fluid inclusions and industrial CO2 pipelines routinely operate above 5.11 atm, where liquid CO2 is perfectly stable. Deep ocean settings with sufficient hydrostatic pressure can also stabilize liquid or hydrate forms of CO2. On the diagram, all of these correspond to points above the horizontal 5.11 atm line, where the vertical temperature axis then determines solid, liquid, or supercritical behavior.

How does the CO2 diagram connect to supercritical fluid extraction?

Supercritical CO2, used in decaffeination and green extraction chemistry, exploits the region beyond the critical point at 304.2 K and 72.8 atm. Past this point, the liquid–vapor coexistence curve simply terminates, and a single fluid phase exists with liquid-like density and gas-like diffusivity. Because $T_c$ is only slightly above room temperature, supercritical CO2 is unusually easy to generate and use industrially compared to substances like water, whose critical point (647 K, 218 atm) is far more extreme. This accessibility is the direct reason CO2 dominates supercritical-fluid applications.

Does the phase rule tell us the numerical values of the triple point and critical point?

No. The phase rule only tells you the number of degrees of freedom available — it dictates that the triple point must be a single fixed $(T,p)$ pair and that the critical point terminates a curve, but it cannot predict what those coordinates are. The actual values, 216.6 K/5.11 atm and 304.2 K/72.8 atm for CO2, come entirely from experimental measurement (or, at a deeper level, from statistical thermodynamic models of intermolecular forces), which lies beyond the reach of the phase rule itself.

Why does the fusion curve of CO2 slope the "normal" way while water's is anomalous?

The Clapeyron equation ties the slope of any coexistence curve to $\Delta H$ divided by $T\Delta V$ . For CO2, freezing releases heat ( $\Delta_{\text{fus}}H > 0$ on melting) and the solid is denser than the liquid, so $\Delta V > 0$ on melting; both terms are positive, giving a positive, steep slope — melting point rises with pressure, as it does for most substances. Water is anomalous because ice is less dense than liquid water, making $\Delta V$ negative on melting, which flips the fusion curve to a negative slope. CO2 is thus the "textbook normal" case that highlights how unusual water really is.