PVT behaviour, virial and cubic equations of state

Why it matters

Compressor sizing, vessel inventories, pipeline pressure drop, and every fugacity and VLE calculation need the molar volume of a real fluid at given T and P. The ideal-gas law fails at high pressure and near saturation, so engineers use the virial equation for moderate pressures and cubic equations of state (van der Waals, Redlich-Kwong, Soave, Peng-Robinson) for gases and liquids alike.

Key ideas

PVT surface and the critical point. A pure substance has a two-phase region on the P-V diagram bounded by the saturated-liquid and saturated-vapour curves, which meet at the critical point (Tc, Pc, Vc). At the critical point the critical isotherm has a horizontal inflection: (∂P/∂V)_T = 0 and (∂²P/∂V²)_T = 0. Above Tc no amount of compression produces a separate liquid phase.

Compressibility factor. Z = PV/(RT) with V the molar volume. Z = 1 for an ideal gas. Z < 1 means attractive forces dominate (the gas is more compressible than ideal); Z > 1 means repulsive (size) effects dominate, typical at very high pressure.

Virial equation. Z is expanded as a power series in density or in pressure. The coefficients B, C, … depend only on temperature for a pure gas and have a statistical-mechanics meaning: B reflects interactions between pairs of molecules, C among triplets. B is negative at ordinary temperatures (attraction) and becomes positive at high temperature. The two-term form Z = 1 + BP/(RT) is reliable for gases at low to moderate pressure (roughly where V exceeds about twice the critical volume); it cannot describe liquids. For mixtures, B_mix = ΣΣ yᵢyⱼBᵢⱼ, quadratic in mole fraction.

Cubic equations of state. They give P explicitly and are cubic in V. Below Tc, at a given P inside the two-phase region, they have three real roots: the smallest is the liquid volume, the largest the vapour volume, the middle one has no physical meaning. Above Tc there is one real root.

• van der Waals: a corrects for attraction, b for the volume of the molecules. Simple but quantitatively poor (it predicts Zc = 0.375, while real fluids have Zc of about 0.23 to 0.31).
• Redlich-Kwong: the attraction term depends on temperature as T^(−0.5).
• Soave-Redlich-Kwong (SRK) and Peng-Robinson (PR): the attraction parameter is a·α(Tr, ω) with acentric factor ω, which makes vapour pressures and VLE of hydrocarbons good. PR also gives better liquid densities than SRK, which is why it dominates in oil and gas simulation.

Corresponding states. Using reduced variables Tr = T/Tc and Pr = P/Pc, fluids with similar molecular shape have similar Z. The two-parameter principle (Tr, Pr) works for simple spherical molecules; the three-parameter form adds the acentric factor ω = −1 − log₁₀(Pr_sat at Tr = 0.7) to account for non-spherical molecules. Generalised correlations (Pitzer) build on this and are covered in the residual-properties topic.

Formulas

Z = P·V / (R·T) Z = 1 + B/V + C/V² + … (density form) Z = 1 + B·P/(R·T) (two-term virial, low to moderate P) P = R·T/(V − b) − a/V² (van der Waals) a = 27·R²·Tc² / (64·Pc), b = R·Tc / (8·Pc), Zc = 3/8 (van der Waals) P = R·T/(V − b) − a / (T^0.5·V·(V + b)) (Redlich-Kwong) P = R·T/(V − b) − a·α / (V² + 2·b·V − b²) (Peng-Robinson) Tr = T/Tc, Pr = P/Pc, ω = −1 − log₁₀(Pr_sat) at Tr = 0.7

Symbols: P pressure (Pa); V molar volume (m³/mol); T temperature (K); R = 8.314 J/(mol·K); B second virial coefficient (m³/mol), C third virial coefficient (m⁶/mol²); a attraction parameter (Pa·m⁶/mol² for van der Waals; RK and PR units follow their forms); b co-volume (m³/mol); Tc, Pc critical temperature (K) and pressure (Pa); α temperature function of Tr and ω; ω acentric factor (dimensionless).

Worked examples

Example 1 (standard): two-term virial equation. Given: a vapour at 473.15 K and 10 bar has B = −388 cm³/mol. Find Z and the molar volume, and compare with the ideal-gas value.

1. Z = 1 + B·P/(R·T) with B = −388 × 10⁻⁶ m³/mol, P = 10 × 10⁵ Pa.
2. B·P/(R·T) = (−388 × 10⁻⁶)(10⁶) / (8.314 × 473.15) = −388 / 3933.8 = −0.0986.
3. Z = 0.9014.
4. V = Z·R·T/P = 0.9014 × 3933.8 × 10⁻⁶ m³/mol = 3.546 × 10⁻³ m³/mol = 3546 cm³/mol.
5. Ideal gas: R·T/P = 3934 cm³/mol, about 11% high. Answer: Z = 0.901, V = 3546 cm³/mol

Example 2 (GATE level): van der Waals parameters from critical data. Given: CO₂ with Tc = 304.2 K, Pc = 73.83 bar. Find a and b, then the pressure at T = 320 K and V = 5.0 × 10⁻⁴ m³/mol.

1. a = 27·R²·Tc²/(64·Pc) = 27 × 8.314² × 304.2² / (64 × 73.83 × 10⁵) = 0.3655 Pa·m⁶/mol².
2. b = R·Tc/(8·Pc) = 8.314 × 304.2 / (8 × 73.83 × 10⁵) = 4.282 × 10⁻⁵ m³/mol.
3. Repulsive term: R·T/(V − b) = 8.314 × 320 / (5.0 × 10⁻⁴ − 0.4282 × 10⁻⁴) = 2660.5 / 4.572 × 10⁻⁴ = 5.819 × 10⁶ Pa.
4. Attractive term: a/V² = 0.3655 / (2.5 × 10⁻⁷) = 1.462 × 10⁶ Pa.
5. P = 5.819 − 1.462 = 4.357 × 10⁶ Pa = 43.6 bar.
6. Ideal gas would give R·T/V = 53.2 bar; Z = P·V/(R·T) = 0.819. Answer: a = 0.366 Pa·m⁶/mol², b = 4.28 × 10⁻⁵ m³/mol, P ≈ 43.6 bar (Z ≈ 0.82)

Common mistakes

• Mixing units: B in cm³/mol with P in Pa, or R in L·atm/(mol·K) with a in Pa·m⁶/mol².
• Using the two-term virial equation for a liquid or at high pressure, where it fails.
• Picking the wrong root of a cubic: the smallest root is the liquid, the largest the vapour; the middle root is meaningless.
• Forgetting that b must be smaller than V; if V − b is negative the data are inconsistent.
• Reading Z < 1 as "the gas is lighter"; it means the gas occupies less volume than an ideal gas at the same T and P.

For GATE CH

• NAT problems: Z or V from the two-term virial equation; P from van der Waals with given or derived a and b; a and b from Tc and Pc.
• Conceptual MCQs: conditions at the critical point, Zc of van der Waals (3/8), meaning of the roots of a cubic, sign of B and what it implies.
• Practise solving a cubic for V by iteration (start from the ideal-gas volume for vapour, from b for liquid) and checking units at every step.

Quick check

1. What two conditions define the critical point on a P-V isotherm?
2. What is Zc predicted by the van der Waals equation?
3. A gas has Z = 0.9. Is its molar volume larger or smaller than the ideal-gas value at the same T and P?
4. Which root of a cubic EOS corresponds to saturated vapour?
5. On what does the second virial coefficient of a pure gas depend?

Answers: 1. (∂P/∂V)_T = 0 and (∂²P/∂V²)_T = 0. 2. 0.375. 3. Smaller (10% smaller). 4. The largest root. 5. Temperature only.