SME
Chemical Engineering Thermodynamics
Fugacity
Use this for “Propane at 400 K and 10 bar has Tᵣ = 1.082 and B⁰ + ωB¹ = −0.2868 (Pc = 42.48 bar). What are its fugacity coefficient and fugacity?”
Chart
P = 10 bar
f = 9.395 bar
Tᵣ = 1.0817
Pc = 42.48 bar
B̂ = −0.2869
This is an example. Type your own values to start a new problem.
ln φ
=
×
φ
=
e
f (bar)
=
φ
×
P (bar)
Pc
Critical pressure
example
Formulas
Pᵣ
=
P
Pc
0.2354
=
10
42.48
ln
φ
=
B̂
×
Pᵣ
Tᵣ
−0.06243
=
−0.2869
×
0.2354
1.0817
φ
=
e
ln
φ
0.9395
=
e
−0.06243
f
=
φ
×
P
9.3948 = 0.9395 × 10
Assumptions
•
Two-term virial form at low to moderate pressure: it holds roughly where the reduced volume Vᵣ exceeds 2.
•
B̂ = B⁰ + ωB¹ is the reduced second virial coefficient, set by Tᵣ and the acentric factor ω.
•
Fugacity f is the corrected pressure of a real gas: φ = f ÷ P, and φ = 1 is the ideal gas.
Step-by-step
WE KNOW
B̂ = −0.2869, Tᵣ = 1.0817, Pc = 42.48 bar, P = 10 bar
FIND
reduced pressure (Pᵣ), log of φ (ln φ), fugacity coefficient (φ), fugacity (f)
Step 1 · Find reduced pressure (Pᵣ)
Use Pᵣ = P ÷ Pc
Divide the pressure by the critical pressure.
Pᵣ
=
10
42.48
Pᵣ = 0.2354049
Step 2 · Find log of φ (ln φ)
Use ln φ = B̂ × Pᵣ ÷ Tᵣ
With
Z
=
1
+
B̂Pᵣ
Tᵣ
,
∫
Z − 1
P
dP
from
0
to
P
gives
ln
φ
=
B̂Pᵣ
Tᵣ
.
ln
φ
=
−0.28686199
×
0.2354049
1.0816658
ln
φ
=
−
0.06753
1.0817
ln
φ
=
−0.06243
Step 3 · Find fugacity coefficient (φ)
Use φ = eln φ
Undo the natural log.
φ
=
e
−0.06243
φ
=
0.93948
Step 4 · Find fugacity (f)
Use f = φ × P
The
fugacity
is
φ
times
the
pressure.
f
=
0.93948
×
10
f
=
9.3948
bar
Check
Put the numbers back in. Both sides should match.
0.2354049
=
10
42.48
✓
−0.06243
=
−0.28686199
×
0.2354049
1.0816658
✓
0.93948
=
e
−0.06243
✓
9.3948 = 0.93948 × 10 ✓
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Van der Waals pressure and volume
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Liquid fugacity with the Poynting factor
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