SME
Chemical Engineering Thermodynamics · Problem type
Liquid fugacity with the Poynting factor
Use this for “Water at 25 °C (Pₛₐₜ = 3.17 kPa, VL = 18.07 cm³/mol) is held at 100 bar (10,000 kPa). What are the Poynting factor and its fugacity?”
This is an example. Type your own values to start a new problem.
ln
FP
=
VL (cm³/mol)
R
×
T (K)
×
(
P (kPa)
−
Pₛₐₜ (kPa)
)
f (kPa)
=
FP
×
Pₛₐₜ (kPa)
Formulas
FP
=
e
VL ×
10
-6
× (
P
− Pₛₐₜ) ÷ (8.314 ×
T
)
1.0756
=
e
18.07 ×
10
-6
× (10,000,000 − 3,170) ÷ (8.314 × 298.15)
f
=
FP
×
Pₛₐₜ
3,409.6379 = 1.0756 × 3,170
Formulas are worked in Pa. The step-by-step shows the conversions.
Assumptions
•
φₛₐₜ ≈ 1: the saturated vapour is nearly ideal, so its fugacity is Pₛₐₜ.
•
The liquid is nearly incompressible: VL stays the same from Pₛₐₜ up to P.
•
R = 8.314 J/(mol·K); VL is in cm³/mol, and cm³ to m³ is × 10⁻⁶.
Step-by-step
WE KNOW
P = 10,000 kPa, Pₛₐₜ = 3.17 kPa, VL = 18.07 cm³/mol, T = 298.15 K
FIND
Poynting factor (FP), liquid fugacity (f)
Step 1 · Convert to the formula’s units
The formulas don’t work directly in the units you chose, so convert first (to Pa).
P = 10,000 kPa = 10,000,000 Pa (1 kPa = 1000 Pa)
Pₛₐₜ = 3.17 kPa = 3,170 Pa (1 kPa = 1000 Pa)
Step 2 · Find Poynting factor (FP)
Use FP = eVL × 10⁻⁶ × (P − Pₛₐₜ) ÷ (8.314 × T)
Squeezing
the
liquid
from
Pₛₐₜ
up
to
P
adds
VL
dP
of
work
per
mole;
over
RT
that
is
ln
FP.
FP
=
e
18.07 ×
10
-6
× (10,000,000 − 3,170) ÷ (8.314 × 298.15)
FP
=
e
(18.07 ×
10
-6
) × 9,996,830 ÷ 2,478.8191
FP
=
e
180.64 ÷ 2,478.8191
FP
=
e
0.072873
FP = 1.0756
Step 3 · Find liquid fugacity (f)
Use f = FP × Pₛₐₜ
The saturated fugacity Pₛₐₜ, raised by the Poynting factor.
f
=
1.0756
×
3,170
f
=
3,409.6379
Pa
Step 4 · Convert the answers
Change each answer back to the unit you chose.
f = 3,409.6379 Pa = 3.4096 kPa (1 kPa = 1000 Pa)
Check
Put the numbers back in. Both sides should match.
1.0756
=
e
18.07 ×
10
-6
× (10,000,000 − 3,170) ÷ (8.314 × 298.15)
✓
3,409.638 = 1.0755955 × 3,170 ✓
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