Real Gases & Equations of State

Why Ideal Gas Fails

• • Attractive forces: At moderate distances molecules attract one another (van der Waals, dipole-dipole, London dispersion). These attractions reduce the pressure below the ideal prediction because molecules approaching the walls are pulled back by interior molecules.
• • Finite molecular volume: Each molecule occupies a nonzero volume, reducing the effective volume available for translation. At high densities, the excluded volume becomes a significant fraction of the container volume.
• • Critical phenomena: Near the critical point, density fluctuations extend over macroscopic length scales and no simple equation of state can capture the true critical exponents. Nevertheless, classical equations of state provide a qualitatively correct description of the liquid-gas transition.

Step 1: Volume Correction (Excluded Volume)

Each molecule has a finite size characterised by an effective hard-sphere diameter $d$. When two molecules approach, the centre of one cannot come closer than a distance $d$ from the centre of the other. The excluded volume per pair of molecules is:

For $N$ molecules there are $N(N-1)/2 \approx N^2/2$ pairs, so the total excluded volume per mole is:

The effective free volume available per mole is therefore $V - nb$ rather than $V$. The parameter $b$ is the co-volume and is roughly four times the actual molecular volume per mole.

Step 2: Pressure Correction (Intermolecular Attraction)

A molecule about to strike the container wall experiences a net inward pull from the surrounding molecules. The magnitude of this internal pressure is proportional to the square of the number density $(n/V)^2$, because both the number of molecules at the wall and the number pulling them back scale as $n/V$:

Here $a > 0$ is a constant characterising the strength of the attractive interactions. The measured pressure $P$ is less than the kinetic pressure by this amount, so the kinetic pressure is $P + a(n/V)^2$.

Step 3: The Van der Waals Equation

Substituting both corrections into the ideal gas law $P_{\text{kinetic}} \cdot V_{\text{free}} = nRT$ gives:

For one mole ($n = 1$), with $V$ denoting the molar volume:

This is a cubic equation in $V$ at fixed $T$ and $P$. Below the critical temperature, it admits three real roots corresponding to liquid, unstable, and gas volumes.

Step 4: Critical Constants from Van der Waals Equation

At the critical point, the three roots of the cubic equation coalesce into one. This means both the first and second derivatives of $P$ with respect to $V$ vanish simultaneously. Writing the pressure explicitly:

Setting the first and second derivatives to zero at the critical point:

Dividing the first equation by the second eliminates $T_c$:

Substituting back gives the critical temperature and pressure:

The critical compressibility factor is a universal constant for van der Waals gases:

Step 5: Law of Corresponding States

Introducing reduced variables $P_r = P/P_c$, $V_r = V/V_c$, and $T_r = T/T_c$, the van der Waals equation takes a universal, parameter-free form. Substituting $P = P_r P_c$,$V = V_r V_c$, $T = T_r T_c$ and the critical constants:

This is the law of corresponding states: all van der Waals gases obey the same reduced equation of state. When plotted in reduced variables, the isotherms of all gases collapse onto a single set of universal curves. This principle extends approximately to real gases and is the theoretical basis for generalised correlation charts in engineering thermodynamics.

Step 1: The Virial Expansion

The compressibility factor is expanded as a power series in inverse molar volume:

Equivalently, as a power series in pressure:

The coefficients $B(T)$, $C(T)$, $D(T)$, ... are the second, third, fourth, ... virial coefficients. They depend only on temperature and encode successively higher-order molecular interactions: $B$ describes pair interactions, $C$ describes three-body interactions, and so on.

Step 2: Statistical Mechanical Foundation

From the grand canonical partition function, the pressure of a classical gas can be expressed through the cluster expansion. For a gas of $N$ identical particles interacting via pairwise additive potentials $u(r_{ij})$, the configuration integral is:

Introducing the Mayer f-function, $f(r) = e^{-u(r)/k_BT} - 1$, which vanishes at large separations, we expand the product of Boltzmann factors. Systematic grouping by cluster diagrams yields the virial expansion, where each virial coefficient corresponds to a sum of connected cluster integrals.

Step 3: Second Virial Coefficient from Pair Potential

The second virial coefficient is given exactly by the integral over the Mayer f-function. For a spherically symmetric pair potential $u(r)$:

This remarkable equation connects a macroscopic measurable quantity $B(T)$ directly to the microscopic pair potential $u(r)$. The integral has two contributions:

• • Repulsive core ($u \gg k_BT$): The exponential is nearly zero, so the integrand is $\approx -r^2$, giving a positive contribution to $B$(excluded volume).
• • Attractive well ($u < 0$): The exponential exceeds unity, giving a negative contribution to $B$ (cohesive effect).

At low temperatures the attractive well dominates and $B < 0$; at high temperatures the repulsive core dominates and $B > 0$.

Step 4: B(T) for the Lennard-Jones Potential

The Lennard-Jones 6-12 potential is the standard model for nonpolar molecules:

Substituting into the integral for $B(T)$ and changing variables to $x = r/\sigma$:

where $T^* = k_BT/\varepsilon$ is the reduced temperature and $b_0 = (2\pi/3)N_A\sigma^3$. This reduced form depends only on $T^*$, providing another manifestation of the principle of corresponding states. The integral must be evaluated numerically; see the Fortran simulation below.

Step 1: Z in Terms of Virial Coefficients

From the virial expansion truncated at the second term:

Converting to a pressure series using $V \approx RT/P$ at leading order:

The slope of $Z$ versus $P$ at low pressures is therefore $B(T)/(RT)$. When $B < 0$, $Z$ initially decreases below unity (attractive forces dominate); when $B > 0$, $Z$ increases above unity (repulsive forces dominate).

Step 2: The Boyle Temperature

The Boyle temperature $T_B$ is defined as the temperature at which the second virial coefficient vanishes:

At the Boyle temperature, the initial slope of $Z$ versus $P$ is zero, so$Z \approx 1$ for moderate pressures. The attractive and repulsive contributions to the second virial coefficient exactly cancel, and the gas behaves most nearly ideally.

Step 3: Boyle Temperature from Van der Waals Equation

To find $T_B$ for a van der Waals gas, we first extract its virial coefficients. Expanding the van der Waals equation at low density (large $V$):

Expanding $(1 - b/V)^{-1} = 1 + b/V + b^2/V^2 + \cdots$ and collecting powers of $1/V$:

The van der Waals second virial coefficient is therefore $B_{\text{vdW}}(T) = b - a/(RT)$. Setting this to zero:

Notice that $T_B/T_c = 27/8 = 3.375$. The Boyle temperature is always significantly above the critical temperature. For example, nitrogen has $T_c = 126\,\text{K}$ and$T_B \approx 327\,\text{K}$, consistent with this ratio.

Step 4: Physical Interpretation

Below $T_B$, attractive forces dominate at moderate pressures ($Z < 1$). Above $T_B$, repulsive forces dominate even at moderate pressures ($Z > 1$). Exactly at $T_B$, the gas follows Boyle's law ($PV = \text{const}$) over an extended pressure range. This is why the Boyle temperature is named after Robert Boyle — at this temperature, the gas obeys his law most faithfully. For the Lennard-Jones potential,$T_B^* = k_BT_B/\varepsilon \approx 3.42$, which is very close to the van der Waals prediction.

Step 1: Definition of Fugacity

The fugacity $f$ of a real gas is defined so that the chemical potential retains its ideal-gas form:

with the boundary condition that as $P \to 0$, $f \to P$ (the gas becomes ideal). The fugacity coefficient is defined as $\phi = f/P$, so $\phi \to 1$ for an ideal gas.

Step 2: Deriving the Fugacity Coefficient

Starting from the fundamental relation for the molar Gibbs energy at constant temperature:

For a real gas: $d\mu = V_m\,dP = \frac{ZRT}{P}\,dP$. For an ideal gas at the same temperature: $d\mu^{\text{ideal}} = \frac{RT}{P}\,dP$. The difference is:

Integrating from $P = 0$ (where both real and ideal chemical potentials coincide) to pressure $P$:

Since $\mu - \mu^{\text{ideal}} = RT\ln(f/P)$, we obtain the key result:

Step 3: Fugacity from the Virial Equation

Using the truncated virial equation $Z = 1 + B'P$ where $B' = B/(RT)$:

Therefore:

At the Boyle temperature where $B = 0$, $f = P$ exactly, confirming that the gas behaves ideally. Below $T_B$, $B < 0$ and $f < P$: the gas is easier to compress than an ideal gas because of attractive forces. Above $T_B$, $f > P$.

Step 4: Fugacity from Van der Waals Equation

For the van der Waals gas, we use an alternative form of the fugacity relation based on volume integration. Starting from $\mu - \mu^{\text{ideal}} = \int_0^P (V_m - V_m^{\text{ideal}})\,dP$and transforming to a volume integral:

This expression shows explicitly how the excluded volume ($b$ terms) and attractive interactions ($a$ term) each contribute to the departure from ideality. In the limit$a \to 0$, $b \to 0$, we recover $\phi = 1$ as expected.