Fugacity and Fugacity Coefficient of Real Gases
In classical chemical thermodynamics, the chemical potential and Gibbs free energy of an ideal gas are expressed directly in terms of pressure. However, real gases deviate markedly from ideal behaviour at elevated pressures and lower temperatures because of finite molecular volumes and intermolecular attractive or repulsive forces. To preserve the elegant logarithmic mathematical forms of ideal gas equations while accurately describing non-ideal systems, G. N. Lewis introduced the concepts of fugacity and the fugacity coefficient. Advanced postgraduate entrance examinations, including CSIR-NET, GATE, and IIT-JAM, frequently test this topic through equation-of-state integrations, isothermal expansion work, and chemical potential shifts.
The Concept of Fugacity and the Fugacity Coefficient
For an ideal gas at constant temperature, the fundamental differential relation for molar Gibbs free energy is dG_m = V_m dP = (RT / P) dP = RT d(ln P). Integrating this from a standard reference pressure P° gives G_m(T, P) = G_m°(T) + RT ln(P / P°). For a real gas, molar volume V_m does not equal RT / P, meaning that replacing P directly inside the logarithm fails to yield the true free energy. Lewis defined fugacity, denoted by f, as an effective thermodynamic pressure such that the exact relation holds for any real gas:
G_m(T, P) = G_m°(T) + RT ln(f / P°)
Here, G_m is the molar Gibbs free energy in joules per mole (J mol⁻¹), R is the universal gas constant (8.314 J K⁻¹ mol⁻¹), T is the absolute temperature in kelvin (K), f is the fugacity, and P° is the standard state pressure (taken as 1 bar under current IUPAC conventions, or 1 atm in older thermodynamic literature). Because real gases approach ideal behaviour as pressure approaches zero, the boundary condition defining the absolute scale of fugacity is:
The dimensionless ratio of fugacity to the mechanical pressure is termed the fugacity coefficient, represented by the Greek letter φ (phi):
When φ = 1, the gas behaves ideally. When attractive forces dominate between molecules, the real gas is more compressible than an ideal gas, making the effective thermodynamic pressure lower than the measured pressure, giving φ < 1 and f < P. Conversely, when repulsive forces dominate at extremely high pressures, the gas resists compression, resulting in φ > 1 and f > P.
Evaluating Fugacity from Equations of State
The quantitative link between the fugacity coefficient and experimental PVT data is derived from the difference between real and ideal molar volumes, known as the residual volume α = V_m - (RT / P). Subtracting the ideal differential dG_m(ideal) = (RT / P) dP from the real differential dG_m(real) = V_m dP at constant temperature gives:
Integrating both sides from P = 0 (where f / P = 1 and ln(f / P) = 0) to the system pressure P yields the fundamental integral:
where Z = P V_m / (RT) is the dimensionless compressibility factor of the gas. If the equation of state expresses Z as an explicit function of pressure, such as a truncated virial equation Z = 1 + B'P or an empirical polynomial, the integral can be solved directly by analytical calculus.
Worked Numerical Examples
Question: A real gas at 300.0 K obeys the linear equation of state Z = 1 - 0.00250 P, where P is the pressure in bar. Calculate the fugacity coefficient φ and the fugacity f of this gas at an operating pressure of 20.0 bar.
Step 1: Identify the integrand for the fugacity coefficient relation.
From the given compressibility factor expression:
Z - 1 = -0.00250 P
Dividing by P yields:
(Z - 1) / P = -0.00250 bar⁻¹
Step 2: Evaluate the integral from 0 to 20.0 bar.
ln φ = ∫020.0 (-0.00250) dP
ln φ = -0.00250 × [P]020.0
ln φ = -0.00250 × 20.0 = -0.05000
Step 3: Compute the fugacity coefficient φ.
φ = exp(-0.05000) = 0.95123
Step 4: Calculate the fugacity f.
f = φ × P = 0.95123 × 20.0 bar = 19.025 bar
Step 5: Verification by secondary route.
Using the Taylor expansion exp(-x) ≈ 1 - x + x²/2 for small x:
φ ≈ 1 - 0.05000 + (0.00250 / 2) = 1 - 0.05000 + 0.00125 = 0.95125.
Multiplying by 20.0 bar: f = 20.0 × 0.95125 = 19.025 bar. Both calculation routes match exactly.
Answer: φ = 0.9512, f = 19.02 bar
Question: At 300.0 K, a moderately compressed gas obeys the equation of state P(V_m - b) = RT, where the excluded volume parameter b is equal to 0.0400 L mol⁻¹. Taking R = 0.08314 L bar K⁻¹ mol⁻¹, determine the fugacity coefficient φ and the fugacity f of the gas at a pressure of 50.0 bar.
Step 1: Express the compressibility factor Z in terms of P, b, and RT.
Rearranging the equation of state gives:
V_m = (RT / P) + b
Multiply both sides by P / (RT):
Z = (P V_m) / (RT) = 1 + (b P) / (RT)
Therefore:
(Z - 1) / P = b / (RT)
Step 2: Calculate the numerical constant b / (RT).
RT = 0.08314 L bar K⁻¹ mol⁻¹ × 300.0 K = 24.942 L bar mol⁻¹
b / (RT) = (0.0400 L mol⁻¹) / (24.942 L bar mol⁻¹) = 0.0016037 bar⁻¹
Step 3: Integrate from 0 to 50.0 bar to obtain ln φ.
ln φ = ∫050.0 [b / (RT)] dP = [b / (RT)] × P
ln φ = 0.0016037 bar⁻¹ × 50.0 bar = 0.080186
Step 4: Compute φ and f.
φ = exp(0.080186) = 1.08349
f = φ × P = 1.08349 × 50.0 bar = 54.174 bar
Step 5: Verification by secondary route.
ln(f / P) = (b P) / (RT) = (0.0400 × 50.0) / 24.942 = 2.000 / 24.942 = 0.080186.
f = 50.0 × exp(0.080186) = 50.0 × 1.08349 = 54.17 bar. The numerical results are completely consistent.
Answer: φ = 1.0835, f = 54.17 bar
Question: Calculate the change in molar Gibbs free energy (ΔG_m) when 1.00 mole of pure nitrogen gas at 298.15 K is compressed isothermally from an initial pressure of 1.00 bar (where it behaves ideally with φ₁ = 1.000) to a final pressure of 100.0 bar, where its fugacity coefficient is φ₂ = 0.970. Use R = 8.314 J K⁻¹ mol⁻¹.
Step 1: Compute the initial and final fugacities.
Initial state: f₁ = φ₁ × P₁ = 1.000 × 1.00 bar = 1.00 bar
Final state: f₂ = φ₂ × P₂ = 0.970 × 100.0 bar = 97.0 bar
Step 2: Formulate the Gibbs free energy change using fugacity.
ΔG_m = RT ln(f₂ / f₁)
RT product = 8.314 J K⁻¹ mol⁻¹ × 298.15 K = 2478.82 J mol⁻¹
Step 3: Evaluate the natural logarithm and compute ΔG_m.
Ratio f₂ / f₁ = 97.0 / 1.00 = 97.0
ln(97.0) = 4.57471
ΔG_m = 2478.82 J mol⁻¹ × 4.57471 = 11339.9 J mol⁻¹ = 11.34 kJ mol⁻¹
Step 4: Verification by separating ideal and excess contributions.
ΔG_ideal = RT ln(P₂ / P₁) = 2478.82 × ln(100.0 / 1.00) = 2478.82 × 4.60517 = 11415.4 J mol⁻¹.
ΔG_excess = RT ln(φ₂ / φ₁) = 2478.82 × ln(0.970 / 1.000) = 2478.82 × (-0.03046) = -75.5 J mol⁻¹.
ΔG_m = ΔG_ideal + ΔG_excess = 11415.4 J mol⁻¹ - 75.5 J mol⁻¹ = 11339.9 J mol⁻¹ = 11.34 kJ mol⁻¹.
Both computational routes yield identical results.
Answer: ΔG_m = 11.34 kJ mol⁻¹
Exam details
- Source: practice problem written for this article (not a past-paper question)
- Relevant exams: CSIR-NET, GATE, IIT-JAM, CUET-PG
- Topic: Real Gases, Fugacity, Fugacity Coefficient, and Chemical Potential
- Difficulty: Advanced / Graduate level
Common Student Mistakes
1. Confusing fugacity with the fugacity coefficient: Fugacity f carries physical units of pressure (bar, atm, or Pa), whereas the fugacity coefficient φ is a dimensionless ratio (φ = f / P). Writing units on φ or omitting units on f is a frequent point deduction in descriptive papers.
2. Forgetting the lower integration limit: When evaluating ln φ = ∫ [(Z - 1) / P] dP, the lower limit is strictly P = 0, not P = 1. At P = 0, all real gases approach ideal behaviour, which guarantees that Z → 1 and ln φ → 0.
3. Assuming φ is always less than 1: At moderate pressures where attractive forces dominate, Z < 1 and φ < 1. However, at very high pressures where repulsive hard-sphere interactions dominate, Z > 1 and φ becomes substantially greater than 1.
4. Inconsistent units for the gas constant R: If pressure is substituted in bar and volume in litres, R must be 0.08314 L bar K⁻¹ mol⁻¹. If pressure is in pascals and volume in cubic metres, R must be 8.314 J K⁻¹ mol⁻¹. Mixing bar with 8.314 leads to an error factor of 10⁵.
Exam Relevance and Typical Question Patterns
| Exam Name | Question Style | Focus Areas Tested |
|---|---|---|
| CSIR-NET Chemical Sciences | Multiple choice (Part B and Part C) | Analytical integration of ln φ for van der Waals and virial gases; chemical potential shifts |
| GATE Chemistry (CY) | Numerical Answer Type (NAT) and MCQ | Direct calculation of φ and f at given T and P; isotherm work differences between real and ideal gases |
| IIT-JAM Chemistry | Multiple choice and numerical section | Qualitative meaning of φ > 1 vs φ < 1; compressibility factor connection; basic truncated virial problems |
| CUET-PG Chemistry | Multiple choice | Definitions of fugacity, standard state conventions, and limit behaviours as P approaches zero |
Previous-year CSIR-NET questions and timed quizzes to check what you just read.
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