SME
Chemical Engineering Thermodynamics
Equations of state
Use this for “Propane (Tc = 369.8 K, Pc = 42.48 bar, ω = 0.152) is at 400 K and 10 bar. Find its compressibility factor Z.”
Diagram
Virial
isotherm
of
the
gas
at
T
=
400
K:
Z
=
1
+
BP
RT
,
so
V
=
RT
P
+
B,
with
B
=
(Z
−
1)
RT
P
=
−0.208
L/mol.
The ideal gas is dashed.
At
P
=
10
bar,
Z
=
0.9376
puts
the
state
at
V
=
ZRT
P
=
3.118
L/mol,
left
of
the
ideal
gas’s
3.326
L/mol.
This is an example. Type your own values to start a new problem.
Tᵣ
=
T (K)
Tc (K)
Pᵣ
=
P (bar)
Pc (bar)
Z
=
1
+
(
B⁰
+
ω
×
B¹
)
×
Tᵣ and Pᵣ are T and P over the critical Tc and Pc; B⁰ and B¹ are Pitzer’s two virial terms, worked from Tᵣ.
Formulas
Tᵣ
=
T
Tc
1.082
=
400
369.8
Pᵣ
=
P
Pc
0.2354
=
10
42.48
B
0
=
0.083
−
0.422
Tᵣ
1.6
−0.2892
=
0.083
−
0.422
1.082
1.6
B
1
=
0.139
−
0.172
Tᵣ
4.2
0.01531
=
0.139
−
0.172
1.082
4.2
Z
=
1
+
(
B
0
+
ω
×
B
1
)
×
Pᵣ
Tᵣ
0.9376
=
1
+
(−0.2892
+
0.152
×
0.01531)
×
0.2354
1.082
Assumptions
•
Low to moderate pressure: the two-term virial form holds roughly where the reduced volume Vᵣ is above 2.
•
A nonpolar or slightly polar gas; read Tc, Pc and the acentric factor ω from a table.
•
Z = PV ÷ RT compares the gas with an ideal one; R = 8.314 J/(mol·K) draws the isotherm.
Step-by-step
WE KNOW
T = 400 K, P = 10 bar, Tc = 369.8 K, Pc = 42.48 bar, ω = 0.152
FIND
reduced temperature (Tᵣ), reduced pressure (Pᵣ), first virial term (B⁰), second virial term (B¹), compressibility factor (Z)
Step 1 · Find reduced temperature (Tᵣ)
Use Tᵣ = T ÷ Tc
Reduced temperature: the temperature over the critical temperature.
Tᵣ
=
400
369.8
Tᵣ = 1.08167
Step 2 · Find reduced pressure (Pᵣ)
Use Pᵣ = P ÷ Pc
Reduced pressure: the pressure over the critical pressure.
Pᵣ
=
10
42.48
Pᵣ = 0.2354
Step 3 · Find first virial term (B⁰)
Use B⁰ = 0.083 − 0.422 ÷ Tᵣ1.6
Pitzer’s first term: raise Tᵣ to 1.6, divide 0.422 by it, take that from 0.083.
B
0
=
0.083
−
0.422
1.08167
1.6
B
0
=
0.083
−
0.422
1.1338
B
0
=
0.083
−
0.3722
B
0
=
−0.2892
Step 4 · Find second virial term (B¹)
Use B¹ = 0.139 − 0.172 ÷ Tᵣ4.2
Pitzer’s second term: raise Tᵣ to 4.2, divide 0.172 by it, take that from 0.139.
B
1
=
0.139
−
0.172
1.08167
4.2
B
1
=
0.139
−
0.172
1.3906
B
1
=
0.139
−
0.12369
B
1
=
0.01531
Step 5 · Find compressibility factor (Z)
Use Z = 1 + (B⁰ + ω × B¹) × Pᵣ ÷ Tᵣ
Add
ω
times
B
1
to
B
0
,
multiply
by
Pᵣ
Tᵣ
,
then
add
1.
Z
=
1
+
(−0.2892
+
0.152
×
0.01531)
×
0.2354
1.08167
Z
=
1
+
((−0.2892)
+
0.002327)
×
0.2354
1.0817
Z
=
1
+
(−0.2869)
×
0.2354
1.0817
Z
=
1
+
−0.06754
1.0817
Z
=
1
+
(−0.06244)
Z
=
0.9376
Check
Put the numbers back in. Both sides should match.
1.08167
=
400
369.8
✓
0.2354
=
10
42.48
✓
−0.2892
=
0.083
−
0.422
1.08167
1.6
✓
0.01531
=
0.139
−
0.172
1.08167
4.2
✓
0.9376
=
1
+
(−0.2892
+
0.152
×
0.01531)
×
0.2354
1.08167
✓
Topic check
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