Virial Equation of State and the Second Virial Coefficient

By Aniket Bhardwaj · 30 October 2026 · Physical Chemistry: Real Gases

The ideal gas law fails at high pressure and low temperature. The van der Waals equation gives one model for the failure. The virial equation is more general: it writes the deviation as a power series, and each coefficient has a physical meaning. In IIT-JAM, GATE and CSIR-NET, questions ask for the sign of B, the Boyle temperature, or a value of Z from a given B. This page covers all three with computed examples.

The equation

Z = PV_m / RT = 1 + B(T)/V_m + C(T)/V_m² + …
Pressure form: Z = 1 + B′P + C′P² + …
To first order: B′ = B / RT, so Z ≈ 1 + BP/RT

Z = compressibility factor; Z = 1 for an ideal gas.
B = second virial coefficient, with units of molar volume (L/mol or cm³/mol). It depends only on temperature, and it measures the effect of pairs of molecules interacting.
C = third virial coefficient (groups of three molecules); it matters at higher density.

What the sign of B tells you

Sign of B	Dominant effect	Z compared with 1
B < 0 (low T)	Attraction between molecules	Z < 1
B = 0 (Boyle temperature)	Attraction and repulsion balance for pairs	Z ≈ 1 over a wide range of low pressure
B > 0 (high T)	Repulsion (molecular size)	Z > 1

The Boyle temperature T_B is the temperature at which B(T) = 0. Near this temperature the gas follows Boyle’s law closely at low pressures.

For a van der Waals gas, expanding the equation gives B(T) = b − a/RT. Setting B = 0 gives T_B = a/Rb. The constants a (attraction) and b (excluded volume) are specific to each gas, so use the values supplied in the question.

Worked example 1: Z and pressure from B

For a gas at 300 K, B = −120 cm³/mol (practice value). The molar volume is 1.000 L. Ignore C. Find Z, and the pressure compared with an ideal gas. (R = 0.082057 L atm K⁻¹ mol⁻¹)

B = −120 cm³ = −0.120 L/mol. Z = 1 + B/V_m = 1 + (−0.120 / 1.000) = 0.880.

RT = 0.082057 × 300 = 24.617 L atm/mol. Ideal pressure = RT/V_m = 24.617 atm.

Real pressure = Z × RT/V_m = 0.880 × 24.617 = 21.66 atm. The real pressure is lower because attractions pull the molecules together.

Worked example 2: Boyle temperature from B(T)

For a gas, B(T) = 60 − 24000/T in cm³/mol, with T in kelvin (practice expression). Find B at 300 K and the Boyle temperature.

At 300 K: B = 60 − 24000/300 = 60 − 80 = −20 cm³/mol (negative, so attraction dominates and Z < 1).

Boyle temperature: B = 0, so 60 = 24000/T, giving T_B = 24000/60 = 400 K.

Worked example 3: B and T_B for a van der Waals gas

A gas has a = 1.36 L² atm mol⁻² and b = 0.0318 L/mol. Find B at 300 K and the Boyle temperature.

RT = 24.617 L atm/mol. a/RT = 1.36 / 24.617 = 0.05525 L/mol.

B = b − a/RT = 0.0318 − 0.05525 = −0.02345 L/mol = −23.4 cm³/mol.

T_B = a / (Rb) = 1.36 / (0.082057 × 0.0318) = 1.36 / 0.0026094 = 521 K.

Since 300 K is below 521 K, B is negative, consistent with the result above.

Worked example 4: Z at a given pressure

Using the gas of Example 3 at 300 K and 10 atm, find Z using the pressure form Z ≈ 1 + BP/RT.

BP/RT = (−0.02345 × 10) / 24.617 = −0.2345 / 24.617 = −0.00953.

Z = 1 − 0.00953 = 0.9905.

The gas is only about 1% away from ideal at this pressure. This truncated form is reliable at low to moderate pressure only; at high pressure the C term must be added.

Common mistakes

Mixing units of B. If B is in cm³/mol and V_m in litres, convert (1 L = 1000 cm³) before dividing.
Using the P-form with the V-form coefficient. B′ = B/RT, not B.
Thinking B > 0 means attraction. A positive B means repulsion dominates.
Treating B as constant. B depends on temperature; C does too.
Confusing Boyle temperature with critical temperature. T_B is where B = 0; T_c is above which no liquid forms. For a van der Waals gas, T_B = a/Rb and T_c = 8a/27Rb, so T_B is larger.

Exam relevance

Question type	Where to look
Z from B and V_m	Z = 1 + B/V_m
Sign of B and meaning	Negative: attraction. Positive: repulsion
Boyle temperature	Solve B(T) = 0
B for a van der Waals gas	B = b − a/RT