Van der Waals equation

A mean-field fluid model with excluded-volume and attractive-interaction parameters, described by the Van der Waals equation. Its pressure law alone does not fix every thermal property: the caloric closure of the van der Waals equation supplies the additional internal-energy assumption.

Isothermal free-energy change of a Van der Waals gas ( $Δ F = - RT ln [(V_{2} - b) / (V_{1} - b)] + a (1/ V_{1} - 1/ V_{2})$ )

For the one-mole Van der Waals equation with constant parameters, integrate

d F = - p d V

along an isotherm to obtain the displayed difference. The attractive contribution has a positive change on expansion, whereas the translational term lowers free energy. Thermodynamic consistency independently gives

Δ U = a (1/ V_{1} - 1/ V_{2})

and

Δ S = R ln [(V_{2} - b) / (V_{1} - b)]

. These homogeneous-branch expressions are not a substitute for equilibrium convexification through a subcritical coexistence region.

Caloric closure of the van der Waals equation

The thermodynamic identity

(\partial U / \partial V)_{T} = T (\partial P / \partial T)_{V} - P

fixes the volume-dependent term

- a N^{2} / V

but leaves an arbitrary function of temperature. Classical monatomic mean-field modelling adds

3 N k_{B} T /2

, giving

U = 3 N k_{B} T /2 - a N^{2} / V + C

Internal energy of a van der Waals gas ( $U (V, T) = h (T) - a / V$ )

For fixed amount of substance, thermodynamic consistency gives

(\frac{\partial U}{\partial V})_{T} = T (\frac{\partial P}{\partial T})_{V} - P .

Indeed, the first law of thermodynamics gives

U_{V} ∣_{T} = T S_{V} ∣_{T} - P

, and the Helmholtz free energy Maxwell relation gives

S_{V} ∣_{T} = P_{T} ∣_{V}

. For the fixed-amount convention

P = RT / (V - b) - a / V^{2}

with temperature-independent

a, b

, substitution gives

U_{V} ∣_{T} = a / V^{2}

. Integration yields

U = h (T) - a / V

. If the dilute-limit reference is

U \to c T

V \to \infty

for every

T

, then

h (T) = c T

The same formula may be interpreted molarly if

V, U, a, b

all use that convention. Mixing molar and total quantities changes the amount factors. The attraction controls the volume correction to internal energy, while excluded volume alone does not contribute to this derivative when

a, b

are independent of temperature.

Entropy excess of a van der Waals gas ( $Δ S^{ex} = R ln (1 - b / V)$ )

For one mole with temperature-independent parameters

a, b

, the equation of state is

P = RT / (V - b) - a / V^{2}

. A Maxwell relation gives

(\partial S / \partial V)_{T} = R / (V - b)

, so the volume dependence is

R ln (V - b)

. Matching the ideal gas entropy at infinite volume fixes the excess at the same

V, T

R ln (1 - b / V)

. The attraction contributes

- a / V

to the Helmholtz free energy but no excess entropy when

a

is independent of temperature. A temperature-dependent attraction parameter would change this conclusion.

Van der Waals molar equation ( $p = RT / (v_{m} - b_{m}) - a_{m} / v_{m}^{2}$ )

The molar volume

v_{m} = V / n_{mol}

is the volume per amount of substance, not the total container volume or the volume per particle. In the molar convention of the Van der Waals equation, take constant positive parameters

[a_{m}] = Pa m^{6} mo l^{- 2}, [b_{m}] = m^{3} mo l^{- 1}, [R] = Pa m^{3} mo l^{- 1} K^{- 1} .

Then

p = \frac{RT}{v _{m} - b _{m}} - \frac{a _{m}}{v _{m}^{2}}, v_{m} > b_{m},

or, for any fixed amount of substance,

(p + \frac{a _{m} n _{mol}^{2}}{V ^{2}}) (V - n_{mol} b_{m}) = n_{mol} RT .

Both terms in the first equation are pressures. The total excluded volume is

n_{mol} b_{m}

: using

V - b_{m}

would subtract quantities with different units. Changing the sample size at fixed molar volume and temperature changes

V

and

n_{mol}

proportionally, while leaving pressure unchanged. This is a useful unit and extensivity check before fitting a real gas.

For the molar Van der Waals critical point, differentiate with respect to molar volume at fixed temperature:

(\frac{\partial p}{\partial v _{m}})_{T} = - \frac{RT}{( v _{m} - b _{m} ) ^{2}} + \frac{2 a _{m}}{v _{m}^{3}}, (\frac{\partial ^{2} p}{\partial v _{m}^{2}})_{T} = \frac{2 RT}{( v _{m} - b _{m} ) ^{3}} - \frac{6 a _{m}}{v _{m}^{4}} .

(4)

Setting both to zero gives

R T_{c} = 2 a_{m} (v_{m, c} - b_{m})^{2} / v_{m, c}^{3}

and

R T_{c} = 3 a_{m} (v_{m, c} - b_{m})^{3} / v_{m, c}^{4}

. Equating them and dividing by the nonzero factors yields

2 v_{m, c} = 3 (v_{m, c} - b_{m})

, so

v_{m, c} = 3 b_{m}, T_{c} = \frac{8 a _{m}}{27 R b _{m}}, p_{c} = \frac{a _{m}}{27 b _{m}^{2}} .

(5)

Here

v_{m, c}

has units

m^{3} mo l^{- 1}

T_{c}

has units

K

, and

p_{c}

has units

Pa

. This is the molar version of the particle derivation below:

v_{m, c} = N_{A} v_{c}

, with the same physical

T_{c}

and

p_{c}

. If one writes

v_{c} = 3 b

T_{c} = 8 a / (27 R b)

and

p_{c} = a / (27 b^{2})

without subscripts, then all of

v_{c}, b, a

in those formulas must be molar quantities.

Now define the dimensionless reduced variables explicitly as

x = v_{m} / v_{m, c}

t = T / T_{c}

and

q = p / p_{c}

. Substituting the three critical values cancels

a_{m}, b_{m}, R

and gives

q = \frac{8 t}{3 x - 1} - \frac{3}{x ^{2}} .

(6)

Because

v_{m} / v_{m, c} = v / v_{c}

, these are exactly the reduced variables in the Van der Waals isotherm discussion. The Van der Waals law of corresponding states is therefore independent of whether one starts with particle or molar units. It predicts a critical ratio

p_{c} v_{m, c} / (R T_{c}) = 3/8

, providing a third check on any proposed parameter fit.

Molar Van der Waals worked example

Choose a hypothetical fluid with molar constants

a_{m} = 0.360 Pa m^{6} mo l^{- 2}

and

b_{m} = 4.00 \times 1 0^{- 5} m^{3} mo l^{- 1}

, using

R = 8.314462618 J mo l^{- 1} K^{- 1}

. These stipulated constants are not a fit to carbon dioxide. The molar critical formulas give

v_{m, c} = 120.0 c m^{3} mo l^{- 1}, T_{c} ≃ 320.7263 K, p_{c} = 8.333333 MPa .

T = 360 K

and

v_{m} = 240 c m^{3} mo l^{- 1}

, first convert

v_{m}

2.40 \times 1 0^{- 4} m^{3} mo l^{- 1}

. The repulsive-volume term and attractive correction are

\frac{RT}{v _{m} - b _{m}} = 14.966033 MPa, \frac{a _{m}}{v _{m}^{2}} = 6.250000 MPa,

p = 8.716033 MPa

. The ideal gas law would instead give

RT / v_{m} = 12.471694 MPa

. The excluded volume raises pressure relative to that ideal value, but attraction more than offsets the increase here. Neither correction alone describes this example.

Figure 1.
Molar Van der Waals equation: competing corrections and a calculated critical point
. Original calculations for the stipulated hypothetical fluid. Left: the ideal gas law, the excluded-volume term and their attraction-corrected result at $360 K$ . Right: molar Van der Waals isotherms; the dotted subcritical curve is a homogeneous continuation, while the green Maxwell construction plateau is equilibrium phase coexistence. None of the curves are measured carbon dioxide data. **Details**.

For

n_{mol} = 0.500 mol

, the same state occupies

V = n_{mol} v_{m} = 120 c m^{3}

. The total-volume formula gives

n_{mol} b_{m} = 20 c m^{3}

and the same

8.716033 MPa

. Halving the sample has halved both its physical volume and its excluded volume, while retaining its pressure. Conversely, if the two critical measurements

T_{c}, p_{c}

are used to determine model constants, inversion gives

b_{m} = \frac{R T _{c}}{8 p _{c}}, a_{m} = \frac{27 R ^{2} T _{c}^{2}}{64 p _{c}} .

An independent critical molar volume then tests the prediction

v_{m, c} = 3 R T_{c} / (8 p_{c})

; agreement with two fitted numbers does not ensure agreement with the third.

t = 0.9

, the Van der Waals liquid-gas coexistence example below becomes

T ≃ 288.6537 K

p_{*} ≃ 5.391653 MPa

v_{m, l} ≃ 72.40823 c m^{3} mo l^{- 1}

and

v_{m, g} ≃ 281.8611 c m^{3} mo l^{- 1}

. For a mean molar volume of

120 c m^{3} mo l^{- 1}

, the lever rule gives gas mole fraction

f_{g} ≃ 0.227219

. Its volume fraction is instead

f_{g} v_{m, g} / v_{m} ≃ 0.533703

: less than one quarter of the molecules occupy more than half the volume. The flat equilibrium pressure corresponds to changing these proportions. It does not describe a homogeneous sample sliding along the unstable loop.

Van der Waals law of corresponding states

The Van der Waals equation becomes independent of its material parameters when pressure, temperature and molecular volume are divided by their critical values:

\overset{p}{ˉ} = 8 \overset{ˉ}{T} / (3 \overset{v}{ˉ} - 1) - 3/ \overset{v}{ˉ}^{2}

. This explains the common reduced isotherms within this model, without claiming exact agreement for all real fluids.

Van der Waals isotherm ( $p (v, T)$ )

A Van der Waals isotherm is the pressure–particle volume curve at a fixed temperature. Put

x = v / v_{c}

t = T / T_{c}

, and

q = p / p_{c}

, using the Van der Waals critical point values below. Then the existing Van der Waals law of corresponding states becomes

q (x, t) = \frac{8 t}{3 x - 1} - \frac{3}{x ^{2}}, x > \frac{1}{3} .

All curves in the figure are calculations in this reduced model.

Figure 1.
Van der Waals isotherms below, at and above the Van der Waals critical point
. The dashed subcritical homogeneous continuation contains an unstable red segment. The green Maxwell construction plateau describes equilibrium phase coexistence; the two shaded areas are equal. Curves are model calculations, not carbon dioxide measurements. **Details**.

For a uniform thermodynamic phase, the isothermal compressibility is

κ_{T} = - \frac{1}{v} (\frac{\partial v}{\partial p})_{T} = - \frac{1}{v ( \partial p / \partial v ) _{T}} .

Thus the upward-sloping portion of a subcritical Van der Waals isotherm has

κ_{T} < 0

: compression would reduce pressure, allowing a density disturbance to grow. It is mechanically unstable. The spinodal points bound that portion; setting the derivative

\partial q / \partial x

to zero gives their curve

t = (3 x - 1)^{2} / (4 x^{3})

. Negative slope alone guarantees only local mechanical stability: portions between the spinodal and the phase coexistence endpoints have metastability. Equilibrium replaces the entire interval between those endpoints, including these metastable portions, by a Maxwell construction plateau.^[ref]

Van der Waals critical point ( $(v_{c}, T_{c}, p_{c})$ )

The liquid-gas critical point of the Van der Waals equation occurs where its two subcritical stationary points merge into a horizontal inflection point. The conditions are

(\partial p / \partial v)_{T} = (\partial^{2} p / \partial v^{2})_{T} = 0

. The short calculation is

- \frac{k _{B} T _{c}}{( v _{c} - b ) ^{2}} + \frac{2 a}{v _{c}^{3}} = 0, \frac{2 k _{B} T _{c}}{( v _{c} - b ) ^{3}} - \frac{6 a}{v _{c}^{4}} = 0.

Dividing these relations gives

2/ (v_{c} - b) = 3/ v_{c}

, hence

v_{c} = 3 b, T_{c} = \frac{8 a}{27 b k _{B}}, p_{c} = \frac{a}{27 b ^{2}} .

Substitution gives the Van der Waals law of corresponding states and the parameter-independent ratio

p_{c} v_{c} / (k_{B} T_{c}) = 3/8

. That ratio belongs to this equation of state; reducing a real fluid's variables does not enforce it.^[ref]

An exact rearrangement of the reduced critical Van der Waals isotherm makes its flatness visible:

q (x, 1) - 1 = - \frac{3 ( x - 1 ) ^{3}}{x ^{2} ( 3 x - 1 )} .

The first nonzero departure near

x = 1

- 3 (x - 1)^{3} /2

. On the critical particle volume approached from

t > 1

(\partial q / \partial x)_{x = 1} = - 6 (t - 1)

and

κ_{T} = \frac{1}{6 p _{c} ( t - 1 )} .

(4)

The divergent isothermal compressibility and cubic flattening explain why the critical region is mechanically soft. These are mean-field theory predictions; thermal fluctuations limit their use for real fluids very close to a thermodynamic critical point.

Critical-isochore compressibility of a van der Waals gas ( $κ_{T} = 4 b / [3 R (T - T_{c})]$ )

For molar equation

p = RT / (v - b) - a / v^{2}

, the critical values are

v_{c} = 3 b

and

T_{c} = 8 a / (27 R b)

. At

v = v_{c}

(\partial p / \partial v)_{T} = - R (T - T_{c}) / (4 b^{2})

. Thus

κ_{T} = - v_{c}^{- 1} (\partial v / \partial p)_{T} = 4 b / [3 R (T - T_{c})]

on the stable branch above

T_{c}

. The mean-field compressibility exponent is one; the continuation below the critical temperature is unstable rather than an equilibrium negative compressibility.

Van der Waals liquid-gas coexistence

Below

T_{c}

, the equilibrium liquid and gas have the same temperature, pressure

p_{*}

, and per-particle chemical potential. For macroscopic bulk thermodynamic phases, neglect interface contributions. The Maxwell equal-area construction determines

v_{l}

and

v_{g}

p (v_{l}, T) = p (v_{g}, T) = p_{*}, \int_{v_{l}}^{v_{g}} [p (v, T) - p_{*}] d v = 0.

It expresses equality of chemical potentials, since

d μ = v d p

at fixed temperature. Equivalently, the equilibrium Helmholtz free energy per particle follows a common tangent line between the two endpoint states, lying below the homogeneous continuation between them. The construction uses that mathematical continuation to compare endpoints; it is not a physical compression path through negative isothermal compressibility states.^[ref]

Here is a reproducible calculated example at

t = 0.9

. With

q_{*} = p_{*} / p_{c}

, the homogeneous intersections solve the cubic equation

3 q_{*} x^{3} - (q_{*} + 8 t) x^{2} + 9 x - 3 = 0.

The Maxwell construction gives

q_{*} ≃ 0.646998352, (x_{l}, x_{m}, x_{g}) ≃ (0.603401903, 1.090526633, 2.348842376) .

All three roots of a polynomial satisfy

x > 1/3

. Only the smallest and largest are the phase coexistence endpoints; the middle one has negative isothermal compressibility. Equal pressures at three admissible values of volume therefore do not imply three stable thermodynamic phases. An analytic area check is

\frac{8 t}{3} ln \frac{3 x _{g} - 1}{3 x _{l} - 1} + 3 (\frac{1}{x _{g}} - \frac{1}{x _{l}}) - q_{*} (x_{g} - x_{l}) = 0.

(4)

At full numerical precision, the residual is below

1 0^{- 11}

in reduced units; independent numerical integration gives the same cancellation. At

t = 0.9

, the spinodal points are instead

x ≃ 0.718597189

and

1.528504964

Figure 1.
Phase coexistence and spinodal curves meet at the Van der Waals critical point
. The horizontal coordinate is reduced number density $n / n_{c} = 1/ x$ , not molar volume. The green region is equilibrium liquid–gas coexistence, bounded by the binodal. Hatching marks unstable homogeneous states, and the strips between solid and dashed boundaries mark homogeneous metastability. This is a model phase diagram, not measured data. **Details**.

The equilibrium plateau changes the phase fraction rather than the endpoint densities. For a gas particle fraction

f_{g} = N_{g} / N

and mean particle volume

\overset{v}{ˉ} = V / N

, the lever rule reads

\overset{v}{ˉ} = (1 - f_{g}) v_{l} + f_{g} v_{g}, f_{g} = \frac{v ˉ - v _{l}}{v _{g} - v _{l}} .

(5)

Thus

\overset{v}{ˉ} = v_{c}

t = 0.9

gives

f_{g} ≃ 0.227219

: this is a particle fraction, not the volume fraction occupied by gas. Differentiating equality of chemical potentials along this model's phase coexistence curve yields the same Clapeyron equation explained below,

d p_{*} / d T = ℓ / [T (v_{g} - v_{l})]

, with latent heat

ℓ

per particle. Both endpoint differences vanish as the Van der Waals critical point is approached; the equation's critical limit must be taken as a ratio.

The Maxwell construction presumes a pure bulk fluid at thermodynamic equilibrium. Its endpoint equality does not specify how rapidly a real compression reaches equilibrium. A homogeneous metastable sample can persist while nucleation is delayed; the spinodal marks loss of local mechanical stability in this mean-field theory, rather than the equilibrium onset of condensation. Surface tension, finite sample size and impurities can affect the observed transition. These qualifications matter when comparing a mathematical Van der Waals isotherm with a visible liquid meniscus.

Low-temperature liquid volume of a van der Waals gas ( $V_{l} \sim b + RT b^{2} / a$ )

For one mole, liquid–vapour phase coexistence requires equal pressure and chemical potential. At

τ = RT b / a ≪ 1

, the saturation pressure is exponentially small. Neglecting it in the dense-phase equation of state gives

RT V_{l}^{2} - a V_{l} + ab = 0

. The root near the excluded volume is

V_{l} ≃ \frac{a}{2 RT} (1 - 1 - 4 RT b / a) = \frac{2 b}{1 + 1 - 4 τ} = b (1 + τ + 2 τ^{2} + \dots) .

This is an asymptotic low-temperature result, not an exact coexistence volume at every subcritical temperature. Away from that limit, solve equal pressure together with the Maxwell equal-area construction.

Andrews's carbon dioxide experiment