Fugacity and Fugacity Coefficient of a Real Gas
For an ideal gas, chemical potential is simple: μ = μ° + RT ln(P/P°). For a real gas this formula fails, especially at high pressure. Instead of rewriting every equation, thermodynamics keeps the same form and replaces the pressure by a corrected quantity called the fugacity. In IIT-JAM, GATE, CSIR-NET and CUET-PG questions, this appears as a calculation of f, φ or ΔG for a real gas. This article shows the definitions and a clear calculation route using the second virial coefficient.
Definitions
f = φ P (φ = fugacity coefficient)
φ → 1 as P → 0 (every gas behaves ideally at very low pressure)
Here f is the fugacity (it has units of pressure), P is the actual pressure, P° is the standard pressure (1 bar in the current IUPAC convention), and φ is dimensionless. Fugacity can be thought of as the “effective pressure” that makes the ideal-gas form of μ true for a real gas.
Calculating φ from the virial coefficient
If the gas follows the truncated virial equation Z = 1 + BP/RT (valid at low to moderate pressure), then integration of (Z − 1)/P from 0 to P gives:
Here B is the second virial coefficient (units of m3/mol, but often given in cm3/mol), P is pressure in Pa and T is in K. Notice the meaning of the sign. A negative B (attractions dominate) gives φ < 1, so f < P. A positive B (repulsions dominate, as at high temperature) gives φ > 1, so f > P.
The same integral can also be written in terms of the compressibility factor: ln φ = ∫(Z − 1) dP / P. With the van der Waals constants, approximate expressions exist, but these need extra data and are not used here.
Worked examples
Example 1: fugacity at 10 bar. A gas has B = −150 cm3/mol at 300 K. Find φ, f and Z at 10 bar. (R = 8.314 J/mol K.)
Convert units: B = −150 × 10−6 m3/mol; P = 10 bar = 106 Pa.
BP = −150 × 10−6 × 106 = −150 J/mol.
RT = 8.314 × 300 = 2494.2 J/mol.
ln φ = −150 / 2494.2 = −0.06014.
φ = e−0.06014 = 0.9416.
f = 0.9416 × 10 = 9.42 bar.
Z = 1 + BP/RT = 1 − 0.0601 = 0.940. (Note that φ and Z are close at low pressure, but they are not the same quantity.)
Example 2: effect on the free energy of compression. Using the same gas at 300 K, find ΔG for the isothermal compression of 1 mol from 1 bar to 10 bar, first treating the gas as ideal, then as real.
Ideal: ΔG = RT ln(P2/P1) = 2494.2 × ln 10 = 2494.2 × 2.3026 = 5743 J.
Real: first find φ at 1 bar: ln φ1 = −150 × 10−6 × 105 / 2494.2 = −15 / 2494.2 = −0.006014, so φ1 = 0.9940 and f1 = 0.9940 bar.
f2 = 9.416 bar from Example 1.
ΔG = RT ln(f2/f1) = 2494.2 × ln(9.416 / 0.9940) = 2494.2 × ln 9.473 = 5608 J.
Cross-check using ln φ: ΔG = RT[ln 10 + ln φ2 − ln φ1] = 5743 + 2494.2 × (−0.06014 + 0.006014) = 5743 − 135 = 5608 J. The two routes agree.
The difference of 135 J (about 2.4 %) is the effect of non-ideality.
Example 3: what pressure gives a stated fugacity? For the same gas at 300 K, estimate the pressure at which f = 20.0 bar.
f = P eBP/RT. With B/RT = −150 × 10−6 / 2494.2 = −6.014 × 10−8 Pa−1, written in bar: −6.014 × 10−3 bar−1.
Try P = 21.5 bar: BP/RT = −0.1293, e−0.1293 = 0.8787, f = 18.89 bar. This is too low.
Try P = 22.8 bar: BP/RT = −0.1371, e−0.1371 = 0.8719, f = 19.88 bar. Close.
Try P = 22.9 bar: BP/RT = −0.1377, e−0.1377 = 0.8714, f = 19.95 bar.
So P ≈ 23 bar. Remember that the truncated equation is reliable only up to modest pressure; at this value the answer is only an estimate, and the question must tell you to use the truncated form.
Standard state of a real gas
The standard state of a real gas is the hypothetical ideal gas at 1 bar. It is called hypothetical because at 1 bar a real gas is not exactly ideal. The standard chemical potential μ° belongs to that imaginary state, and then the correction is placed entirely in f. This is the reason that φ is used in the equilibrium constant of high-pressure reactions such as ammonia synthesis: K is written with fugacities, K = Π(f/P°)ν, and K depends on T only.
Common mistakes
- Mixing units of B: cm3/mol must be converted to m3/mol when P is in Pa.
- Treating φ as equal to Z. They are related by an integral, not equal.
- Using f in place of φ (or the other way) in the equilibrium constant.
- Forgetting the sign of B. A negative B gives f < P at moderate pressure.
- Applying ln φ = BP/RT at very high pressure, where higher virial terms matter.
- Saying that fugacity applies only to gases. The idea also extends to liquids and solids, where it is called fugacity of the condensed phase.
Quick summary
| Quantity | Meaning | Ideal limit |
|---|---|---|
| f | Effective pressure (bar or Pa) | f = P |
| φ | f / P, dimensionless | φ = 1 |
| ln φ = BP/RT | First-order estimate from virial B | B = 0 gives φ = 1 |
| ΔG (isothermal) | RT ln(f2/f1) | RT ln(P2/P1) |
| B < 0 | Attraction dominates | φ < 1 |
| B > 0 | Repulsion dominates | φ > 1 |
The treatment of real gases can differ between textbooks and exams, so check the current official syllabus and the book recommended for your exam.
The ideal-gas tool gives the ideal value of pressure, volume or moles; compare it with the real-gas correction in the examples above.
Open the Ideal Gas Law Calculator →Previous-year CSIR-NET questions and timed quizzes to check what you just read.
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