Van der Waals Equation & the Critical Point

Pick a gas and drag the temperature down. The isotherm is smooth, then it kinks, then it flattens — and a flat stretch on a pressure–volume curve is a substance boiling.

Worked example — Van der Waals Equation & the Critical Point

Pressure	51.447 bar
What the ideal law says	71.267 bar
Compressibility Z	0.7219
Critical temperature	304.0 K
Critical pressure	74.044 bar
Critical compressibility	0.3750
Boiling pressure here	70.211 bar

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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Lesson

The theory — Van der Waals Equation & the Critical Point

The van der Waals equation is the ideal gas law with two things put back: (P + a/V_m²)(V_m − b) = RT. The molecules take up room, so the space they can move in is V_m − b rather than V_m. They attract each other, so one approaching the wall is pulled back by the others and strikes it less hard, which subtracts a term going as the square of the density. Two constants, fitted once per substance, and the ideal gas law is what is left when both are zero.

What each symbol means

a: the strength of the attraction between molecules, in Pa·m⁶·mol⁻²; it lowers the pressure the gas exerts
b: the volume one mole of the molecules themselves occupy, in m³·mol⁻¹; it raises the pressure by leaving less room
Z: PV_m/RT, the compressibility factor: exactly 1 for an ideal gas at every state, so its distance from 1 is the whole error
T_c: the temperature above which no pressure produces a liquid

Assumes: That the molecules are hard spheres of fixed size with an attraction that depends only on how densely they are packed. Both are caricatures. The attraction between real molecules depends on their orientation as well as their spacing, and nothing is a hard sphere, which is why a and b drift with temperature when they are fitted honestly instead of being taken from the critical point.
Breaks when: On the liquid side, badly. The equation gets the existence of a liquid right and its properties wrong: for CO₂ at 273.15 K it puts the boiling pressure at 47.572 bar where the real vapour pressure is 34.85 bar, 37% high. It is a theory of why a liquid must appear, not a way to predict one.

Right to a tenth of a percent, then wrong by a factor of three 🖖

Nitrogen in an ordinary room, 293 K and 24 litres per mole: the ideal gas law gives 1.015 bar where the full equation gives 1.014. Z, which is PV_m/RT and is exactly 1 for an ideal gas, comes to 0.9993. One part in fifteen hundred.

Now squeeze. Carbon dioxide at 300 K in 0.35 L/mol: the ideal law says 71.267 bar and the answer is 51.447 bar, so it is 38.5% high. At CO₂'s own critical point it is worse — Z there is 0.3750, and a Z of 0.3750 means the ideal law overstates the pressure by a factor of 2.667.

Both corrections are in the equation and they pull opposite ways. The molecules attract each other, which lowers the pressure on the wall; the molecules take up room, which raises it. Cold helium is the only gas in the menu where the second wins: its Z is 1.0220, on the wrong side of 1.

Nobody put a liquid into this equation 🖖

Van der Waals fitted two numbers to how gases fail to be ideal. Below a certain temperature the curve those two numbers draw does something no substance can do: over a stretch of the isotherm, pressure rises as the volume grows. Let such a state expand and it would push harder, expand further, push harder still.

The equation is saying that a single phase cannot exist there. Maxwell's construction replaces the impossible stretch with a flat line, placed so the two lobes it cuts off have equal area — which is the condition that the two phases have the same chemical potential.

Press Boiling. The flat line sits at 47.572 bar, running from 0.0770 L/mol to 0.3033 L/mol. One pressure, one temperature, two volumes four times apart. That is a liquid and its vapour, and it arrived out of an equation written about gases.

The one prediction fitting cannot fake, and it is 25% out 🖖

The tabulated a and b are usually worked backwards from a measured critical point, so an equation reproducing that critical point proves nothing. The agreement is arithmetic, not evidence.

One quantity escapes. Whatever a and b are, the closed forms give V_c = 3b, P_c = a/27b² and T_c = 8a/27Rb, and the combination P_cV_c/RT_c collapses to 3/8 with no a and no b left in it. Every gas in the menu reads 0.3750. No choice of constants can move it.

Real gases do not read 0.375. They cluster near 0.27 to 0.30, and water comes in lower still. So the equation makes exactly one falsifiable claim at the critical point, and misses by about a quarter. That it misses by roughly the same amount for everything is the interesting part: the error is in the model, not in the substances.

Problem solved in full

Finding a critical point with no thermometer 6 steps
Carbon dioxide has van der Waals constants a = 0.3640 Pa·m⁶·mol⁻² and b = 4.267 × 10⁻⁵ m³·mol⁻¹. Using nothing else, find the temperature, pressure and molar volume of its critical point, and the compressibility factor there.

Open this problem in the tool
1. The critical point is where the kink in the isotherm is about to appear, and "about to appear" is a statement about derivatives: the two turning points that bracket the impossible stretch have just merged into one. At a merged pair both the first and the second derivative vanish, so dP/dV_m = 0 and d²P/dV_m² = 0 at the same point.
  
  $\frac{d P}{d V _{m}}_{c} = 0, \frac{d ^{2} P}{d V _{m}^{2}}_{c} = 0$
2. Differentiating P = RT/(V_m − b) − a/V_m² twice gives −RT/(V_m − b)² + 2a/V_m³ = 0 and 2RT/(V_m − b)³ − 6a/V_m⁴ = 0. Dividing the second by the first eliminates both R and T in one step and leaves 2/(V_m − b) = 3/V_m, so V_m = 3b. The critical volume is three times the molecules' own volume and depends on nothing else.
  
  $- \frac{R T}{( V _{m} - b ) ^{2}} + \frac{2 a}{V _{m}^{3}} = 0, \frac{2 R T}{( V _{m} - b ) ^{3}} - \frac{6 a}{V _{m}^{4}} = 0 \Rightarrow \frac{2}{V _{m} - b} = \frac{3}{V _{m}} \Rightarrow V_{c} = 3 b$
3. Substituting V_c = 3b back into the first derivative gives RT_c/(2b)² = 2a/(3b)³, so T_c = 8a/(27Rb). With the numbers: 8 × 0.3640 ÷ (27 × 8.3145 × 4.267 × 10⁻⁵) = 2.912 ÷ 0.009578 = 304.0 K. The Critical temperature row says 304.0 K.
  
  $T_{c} = \frac{8 a}{27 R b} = \frac{8 \cdot 0 . 3640}{27 \cdot 8 . 3145 \cdot 4 . 267 \times 1 0 ^{- 5}} = 304.0 K$
4. And putting both into the equation itself gives P_c = a/(27b²) = 0.3640 ÷ (27 × 1.8207 × 10⁻⁹) = 74.0 bar. The Critical pressure row agrees, and the Critical temperature row reads 304.0 K against a measured 304.13 K.
  
  $P_{c} = \frac{a}{27 b ^{2}} = \frac{0 . 3640}{27 \cdot 1 . 8207 \times 1 0 ^{- 9}} = 7.404 \times 1 0^{6} Pa = 74.0 bar$
5. That agreement is worth less than it looks. The a and b in the table were themselves obtained from the measured critical point, so reproducing it is arithmetic running backwards. What cannot be faked is the combination: Z_c = P_cV_c/RT_c = (a/27b²)(3b) ÷ R(8a/27Rb). Every a cancels, every b cancels, and R cancels, leaving 3/8 = 0.3750.
  
  $Z_{c} = \frac{P _{c} V _{c}}{R T _{c}} = \frac{( \frac{a}{27 b ^{2}} ) ( 3 b )}{R ( \frac{8 a}{27 R b} )} = \frac{3}{8} = 0.3750$
6. So the equation says every substance has the same critical compressibility. Click through the six gases and the Critical compressibility row does not move: carbon dioxide, helium and water all read 0.3750. Measured values sit near 0.27 to 0.30. The prediction is universal and it is about a quarter too high, for everything at once.
  
  $Z_{c, measured} \approx 0.27 to 0.30 for every one of them$
Answer

T_c = 304.0 K, P_c = 74.0 bar, V_c = 0.1280 L/mol and Z_c = 0.3750. The last of those is the only one that was a prediction rather than a restatement, and it is the only one that is wrong. An equation that reproduces what it was fitted to has told you nothing; an equation that makes the same error for carbon dioxide, water and helium has told you where it is broken.

$V_{c} = 3 b, P_{c} = \frac{a}{27 b ^{2}}, T_{c} = \frac{8 a}{27 R b} ⟹ \frac{P _{c} V _{c}}{R T _{c}} = \frac{3}{8} always$

References (3)

The thesis the equation came from, whose title is already the claim — that the gas and the liquid are one substance and not two: J. D. van der Waals, Over de continuïteit van den gas- en vloeistoftoestand. Doctoral thesis, Leiden, 1873.
Where the equal-area rule comes from, in a lecture about why a substance would split into two phases at all: J. Clerk-Maxwell, "On the Dynamical Evidence of the Molecular Constitution of Bodies." Nature 11, 1875, pp. 357–359 and 374–377. DOI 10.1038/011357a0
A modern textbook treatment of real gases, the compressibility factor and the corresponding-states principle: P. Atkins and J. de Paula, Atkins' Physical Chemistry, 11th edition, Oxford University Press, 2018, chapter 1 (The properties of gases).

By the 42Math team · Reviewed against published sources. Corrections welcome.