Critical Constants and the van der Waals Constants

Carbon dioxide refuses to boil above 31.0 °C, no matter how hard you squeeze it. Push the pressure up and the liquid and gas simply merge into one dense, shimmering fluid. This is the critical point. It is not a coincidence of chemistry. Instead, it is built into the shape of any equation of state that curves the right way. The van der Waals equation was the first equation of state to predict this merging. It does so with only two adjustable constants. That predictive link matters. It means $a$ and $b$ , numbers originally fit to correct for molecular attraction and size, can be extracted directly from three measurable quantities— $T_c$ , $P_c$ , $V_c$ . No fitting is required at all. This article derives that connection from first principles. You will see how the critical-point conditions pin down $a$ and $b$ . Then you will see the derivation produce a universal number, the critical compression factor, that tests the whole theory. Finally, you will see how this sets up the law of corresponding states.

Key takeaways

The critical point is where the liquid–gas distinction vanishes: the isotherm has a horizontal inflection there.
Two conditions, $(\partial P/\partial V)_T=0$ and $(\partial^2P/\partial V^2)_T=0$ , applied to the van der Waals equation at $T_c$ pin down $a$ and $b$ .
Solving gives $V_c=3b$ , $P_c=a/27b^2$ , $T_c=8a/27Rb$ .
Inverting gives $a=27R^2T_c^2/64P_c$ and $b=RT_c/8P_c$ , both computable from lab data.
The theory predicts $Z_c=P_cV_c/RT_c=3/8=0.375$ for every gas—real gases cluster near 0.27–0.29, exposing the model's limits.
Key idea: setting the first and second volume-derivatives of pressure to zero at $T_c$ converts two fitting parameters into three measurable critical constants, and back again.

The core insight

At the critical point the van der Waals isotherm develops a horizontal inflection—both $(\partial P/\partial V)_T$ and $(\partial^2P/\partial V^2)_T$ vanish simultaneously—and solving these two conditions together with the equation of state itself turns the pair $(a,b)$ into the triplet $(T_c,P_c,V_c)$ , a mapping that can be run in either direction and tested against experiment through the critical compression factor $Z_c=3/8$ .

The physical picture

Below $T_c$ , a van der Waals isotherm on a $P$ – $V$ diagram is not monotonic. It has a hump: pressure rises, dips into a local minimum, then rises again as volume decreases toward $b$ . The oscillating middle section is unphysical (pressure should never increase with increasing volume at fixed $T$ ), and in real substances it is replaced by a flat two-phase plateau via the Maxwell construction. As $T$ rises, the hump flattens. At exactly $T_c$ , the local maximum and local minimum merge into a single point—an inflection point with zero slope. Above $T_c$ , the isotherm is monotonic everywhere: no phase transition, no plateau, just a single fluid. That merging point is mathematically special. A smooth curve $P(V)$ has an inflection with zero slope precisely when its first and second derivatives with respect to $V$ both vanish there. Below $T_c$ these two conditions are satisfied at two different volumes (max and min); above $T_c$ they are not satisfied at all; exactly at $T_c$ they coincide at one volume, $V_c$ . This is the mathematical signature the derivation below exploits.

Figure

Van der Waals isotherms: the critical point is where max and min merge

The core relations

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

where: $P$ is pressure, $V_m$ is molar volume, $T$ is temperature, $R$ the gas constant, and $a$ , $b$ the van der Waals constants.

\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,\ V_m=V_c

where: these are the two simultaneous conditions defining the critical point on an isotherm.

V_c=3b,\qquad P_c=\frac{a}{27b^2},\qquad T_c=\frac{8a}{27Rb}

where: the critical molar volume, pressure, and temperature expressed in terms of $a$ and $b$ .

a=\frac{27R^2T_c^2}{64P_c},\qquad b=\frac{RT_c}{8P_c}

where: the same relations inverted, giving $a$ and $b$ from measured critical data.

Z_c=\frac{P_cV_c}{RT_c}=\frac{3}{8}=0.375

where: $Z_c$ is the critical compression factor predicted universally by this equation of state.

Derivation AFrom the van der Waals equation to the critical constants

Goal. Starting only from the van der Waals equation of state, find $V_c$ , $P_c$ , $T_c$ in terms of $a$ and $b$ , then invert the relations.

Step 1 — Write the equation for one mole.

P=\frac{RT}{V-b}-\frac{a}{V^2}

Drop the subscript $m$ for brevity; $V$ means molar volume throughout.

Step 2 — Take the first derivative at fixed $T$ .

\left(\frac{\partial P}{\partial V}\right)_T=-\frac{RT}{(V-b)^2}+\frac{2a}{V^3}

At the critical point this must equal zero:

\frac{RT_c}{(V_c-b)^2}=\frac{2a}{V_c^3}\qquad(1)

Step 3 — Take the second derivative at fixed $T$ .

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

Setting this to zero at the critical point:

\frac{2RT_c}{(V_c-b)^3}=\frac{6a}{V_c^4}\qquad(2)

Step 4 — Divide equation (2) by equation (1) to eliminate $T_c$ .

\frac{2/(V_c-b)}{1}=\frac{6a/V_c^4}{2a/V_c^3}\cdot(V_c-b)\;\Big/\;(V_c-b)

More directly, dividing (2) by (1) term by term:

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

Cross-multiplying:

2V_c=3(V_c-b)\;\Longrightarrow\;2V_c=3V_c-3b\;\Longrightarrow\;V_c=3b

This is the first boxed relation: the critical molar volume is exactly three times the excluded volume parameter.

Step 5 — Substitute $V_c=3b$ back into equation (1) to get $T_c$ .

\frac{RT_c}{(3b-b)^2}=\frac{2a}{(3b)^3}

\frac{RT_c}{4b^2}=\frac{2a}{27b^3}

T_c=\frac{4b^2}{R}\cdot\frac{2a}{27b^3}=\frac{8a}{27Rb}

Step 6 — Substitute $V_c$ and $T_c$ into the equation of state itself to get $P_c$ .

P_c=\frac{RT_c}{V_c-b}-\frac{a}{V_c^2}=\frac{R\left(\dfrac{8a}{27Rb}\right)}{3b-b}-\frac{a}{9b^2}

P_c=\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 — Invert to express $a$ and $b$ in terms of the measured critical constants. From $T_c=8a/(27Rb)$ and $P_c=a/(27b^2)$ , divide the first by the second:

\frac{T_c}{P_c}=\frac{8a/(27Rb)}{a/(27b^2)}=\frac{8b}{R}

\Longrightarrow\; b=\frac{RT_c}{8P_c}

Substitute back into $P_c=a/(27b^2)$ :

a=27P_cb^2=27P_c\left(\frac{RT_c}{8P_c}\right)^2=\frac{27R^2T_c^2}{64P_c}

Step 8 — Compute the universal ratio.

Z_c=\frac{P_cV_c}{RT_c}=\frac{(a/27b^2)(3b)}{R(8a/27Rb)}=\frac{a/(9b)}{8a/(27b)}=\frac{27}{9\cdot8}=\frac{3}{8}

\boxed{V_c=3b\quad P_c=\dfrac{a}{27b^2}\quad T_c=\dfrac{8a}{27Rb}\quad Z_c=\dfrac{3}{8}}

Worked examples

Example 1 — Foundational: find $a$ and $b$ for carbon dioxide from its critical constants. Data: $T_c = 304.2\ \text{K}$ , $P_c = 72.9\ \text{atm} = 7.39\times10^{6}\ \text{Pa}$ . Use $R = 8.314\ \text{J K}^{-1}\text{mol}^{-1}$ . Step 1. Compute $b$ :

b=\frac{RT_c}{8P_c}=\frac{(8.314)(304.2)}{8(7.39\times10^{6})}

b=\frac{2528.7}{5.912\times10^{7}}=4.278\times10^{-5}\ \text{m}^3\text{mol}^{-1}

Convert: $b = 42.8\ \text{cm}^3\,\text{mol}^{-1}$ . This compares well with the tabulated van der Waals value for CO₂, $b \approx 42.7\ \text{cm}^3\,\text{mol}^{-1}$ . Step 2. Compute $a$ :

a=\frac{27R^2T_c^2}{64P_c}=\frac{27(8.314)^2(304.2)^2}{64(7.39\times10^{6})}

Numerator: $27 \times 69.12 \times 92\,538 = 27 \times 6.397\times10^{6} = 1.727\times10^{8}$ . Denominator: $64 \times 7.39\times10^{6} = 4.730\times10^{8}$ .

a=\frac{1.727\times10^{8}}{4.730\times10^{8}}\ \text{Pa m}^6\text{mol}^{-2}\times\text{(unit check below)}

Carrying units properly ( $R^2T_c^2$ has units $\text{J}^2\text{K}^{-2}\text{mol}^{-2}\times\text{K}^2 = \text{J}^2\text{mol}^{-2}$ , divided by Pa gives $\text{J}^2\text{mol}^{-2}\text{Pa}^{-1} = \text{Pa}\,\text{m}^6\text{mol}^{-2}$ , correct for $a$ ):

a = 0.365\ \text{Pa m}^6\text{mol}^{-2} = 0.365\ \text{J m}^3\text{mol}^{-2}

Converting to the common tabulated unit (L² atm mol⁻²): $1\ \text{Pa m}^6 = 1\ \text{Pa}\cdot(1000\ \text{L})^2\times...$ ; more directly, using $a = 27R^2T_c^2/64P_c$ with $R = 0.08206\ \text{L atm K}^{-1}\text{mol}^{-1}$ and $P_c = 72.9$ atm:

a=\frac{27(0.08206)^2(304.2)^2}{64(72.9)}=\frac{27(0.006734)(92\,538)}{4665.6}

a=\frac{16\,824}{4665.6}=3.61\ \text{L}^2\text{atm mol}^{-2}

This matches the standard tabulated value, $a(\text{CO}_2) = 3.59\ \text{L}^2\text{atm mol}^{-2}$ , to within rounding of the input critical constants. Sanity check: both $a$ and $b$ come out positive and of the expected order of magnitude for a small polyatomic molecule.

Example 2 — Applied: predict $V_c$ and test $Z_c$ for ammonia, then compare with experiment. Literature van der Waals constants for NH₃: $a = 4.170\ \text{L}^2\text{atm mol}^{-2}$ , $b = 0.03707\ \text{L mol}^{-1}$ . Experimental critical constants: $T_c = 405.5\ \text{K}$ , $P_c = 111.3\ \text{atm}$ , $V_c = 72.5\ \text{cm}^3\,\text{mol}^{-1}$ . Step 1. Predict $V_c$ from theory:

V_c^{\text{theory}}=3b=3(0.03707\ \text{L mol}^{-1})=0.1112\ \text{L mol}^{-1}=111.2\ \text{cm}^3\text{mol}^{-1}

This overshoots the experimental $72.5\ \text{cm}^3\,\text{mol}^{-1}$ by 53% — a substantial discrepancy, typical of the crude packing assumption behind $b$ . Step 2. Predict $T_c$ and $P_c$ from theory:

T_c^{\text{theory}}=\frac{8a}{27Rb}=\frac{8(4.170)}{27(0.08206)(0.03707)}=\frac{33.36}{0.08217}=405.9\ \text{K}

Excellent agreement with the experimental 405.5 K (0.1% off) — because $a$ and $b$ were originally fit to reproduce $T_c$ and $P_c$ , not $V_c$ .

P_c^{\text{theory}}=\frac{a}{27b^2}=\frac{4.170}{27(0.03707)^2}=\frac{4.170}{0.03710}=112.4\ \text{atm}

Close to the experimental 111.3 atm (1% off). Step 3. Compute $Z_c$ two ways and compare with the universal prediction $3/8 = 0.375$ . Using experimental data:

Z_c^{\text{expt}}=\frac{P_cV_c}{RT_c}=\frac{(111.3\ \text{atm})(0.0725\ \text{L mol}^{-1})}{(0.08206\ \text{L atm K}^{-1}\text{mol}^{-1})(405.5\ \text{K})}

Z_c^{\text{expt}}=\frac{8.07}{33.28}=0.242

The van der Waals model predicts a fixed $Z_c = 0.375$ for every substance, but real ammonia gives $0.242$ — a 35% shortfall. This gap is exactly why $V_c$ was the badly-predicted quantity in Step 1: the model systematically overestimates critical volume relative to critical temperature and pressure. The lesson: use measured $T_c$ and $P_c$ (the two best-behaved constants) to back out $a$ and $b$ , and treat any van der Waals prediction of $V_c$ or $Z_c$ as qualitative at best.

Limiting cases & common traps

Situation	What happens	Why
$a \to 0$	$P_c \to 0$ , $T_c \to 0$ , but $V_c = 3b$ stays finite	No attraction means no cohesive force to create a liquid phase or finite critical pressure
$b \to 0$	All of $V_c, P_c, T_c \to 0$ or diverge oddly	Point particles with attraction alone give an unphysical, non-van-der-Waals limit
Using $V_c$ to extract $a,b$	Large errors (often 30–50%)	$V_c$ is the least reliable van der Waals prediction; always prefer $T_c$ , $P_c$ data
Comparing $Z_c$ across gases	Real gases show $Z_c \approx 0.27$ – $0.29$ , not exactly one value	Van der Waals gives a single universal $3/8$ ; the real spread reflects molecular shape and polarity effects

The chemist behind it

Johannes Diderik van der Waals proposed his modified gas equation in his 1873 doctoral dissertation, arguing that real molecules occupy finite volume and attract one another weakly at a distance. What made the equation historic was not just that it fit data better than the ideal gas law — it was that the same two constants, $a$ and $b$ , simultaneously predicted the existence of a critical point and related it quantitatively to measurable liquefaction data. This was the first molecular-scale theory to explain why gases like CO₂ could be liquefied by pressure alone below a certain temperature, and why above that temperature no amount of pressure would work — a puzzle that had stumped experimentalists like Thomas Andrews only a few years earlier.

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

Johannes Diderik van der Waals

1837–1923 · Dutch physicist who unified gas and liquid behavior in one equation of state, work recognized with the 1910 Nobel Prize in Physics

Ask the tutor: How does substituting reduced variables into the van der Waals equation eliminate $a$ and $b$ to give the universal law of corresponding states?

Why does the van der Waals equation predict a fixed Z_c = 3/8 for every gas?

The value 3/8 falls out purely from the algebra of the model: dividing the expressions for $P_c$ , $V_c$ , and $T_c$ (all built from just $a$ , $b$ , $R$ ) causes $a$ and $b$ to cancel completely, leaving a pure number. Because the van der Waals equation has exactly two adjustable constants and three critical-point equations (the equation of state plus two derivative conditions), the critical constants are over-determined — one dimensionless combination must be universal. Real gases show $Z_c$ between about 0.23 and 0.30, confirming the model is only approximate.

Can I use these formulas to predict whether a real gas will liquefy?

Not directly to predict liquefaction quantitatively, but the qualitative story is right: any van der Waals gas has a genuine critical temperature above which no phase transition occurs, matching real behavior. Quantitatively, though, predicted $V_c$ and derived $Z_c$ are often 20–50% off, so do not trust this model for precise phase-boundary engineering. Use it for the conceptual and order-of-magnitude picture, and switch to tabulated experimental critical constants for real design work.

Do a and b depend on temperature or pressure?

No — in the van der Waals model, $a$ and $b$ are treated as constants characteristic of a given substance, independent of the state variables $T$ , $P$ , $V$ . That is precisely why they can be extracted once from critical-point data and then used, at least approximately, to predict behavior at other conditions. Real intermolecular potentials do have some temperature dependence in their effective range, which is one reason more sophisticated equations of state (Redlich–Kwong, Peng–Robinson) outperform van der Waals away from the critical region.

How does this connect to the law of corresponding states?

Once you know that $Z_c = 3/8$ is universal for the van der Waals model, it is natural to rescale $P$ , $V$ , $T$ by their critical values, defining reduced variables $P_r = P/P_c$ , $V_r = V/V_c$ , $T_r = T/T_c$ . Substituting these into the van der Waals equation eliminates $a$ and $b$ entirely, producing one reduced equation of state that applies to any gas obeying the model. That universal reduced equation is the law of corresponding states, covered next.