SME
Chemical Engineering Thermodynamics
Vapor–liquid equilibrium
Use this for “A liquid of 50% benzene and 50% toluene is at 90 °C. Find its bubble pressure and the first vapor’s composition.”
Diagram
Pxy diagram of 1–2 at 90 °C: the bubble curve runs from P₂ˢᵃᵗ = 406.7 to P₁ˢᵃᵗ = 1,021 mmHg, the dew curve under it; between them liquid and vapor coexist.
P = x₁P₁ˢᵃᵗ + (1 − x₁)P₂ˢᵃᵗ = 0.5 × 1,021 + 0.5 × 406.7 = 713.9 mmHg.
y₁
=
x₁P₁
ˢᵃᵗ
P
=
0.7151:
the
vapor
is
richer
in
1,
the
more
volatile.
This is an example. Type your own values to start a new problem.
log₁₀(
P₁ˢᵃᵗ (mmHg)
)
=
A₁
−
B₁ (°C)
T (°C)
+
C₁ (°C)
log₁₀(
P₂ˢᵃᵗ (mmHg)
)
=
A₂
−
B₂ (°C)
T (°C)
+
C₂ (°C)
P (mmHg)
=
x₁
×
P₁ˢᵃᵗ (mmHg)
+
(1
−
x₁
)
×
P₂ˢᵃᵗ (mmHg)
y₁
Vapor mole fraction of 1
calculated
Formulas
log₁₀(P₁ˢᵃᵗ)
=
A₁
−
B₁
T
+
C₁
log₁₀(1,021)
=
6.90565
−
1,211.033
90 + 220.79
log₁₀(P₂ˢᵃᵗ)
=
A₂
−
B₂
T
+
C₂
log₁₀(406.7)
=
6.95464
−
1,344.8
90 + 219.482
P
=
x₁
×
P₁ˢᵃᵗ
+
(1
−
x₁
)
×
P₂ˢᵃᵗ
713.9 = 0.5 × 1,021 + (1 − 0.5) × 406.7
y₁
=
x₁
×
P₁
ˢᵃᵗ
P
0.7151
=
0.5
×
1,021
713.9
P
=
x₁
×
10
A₁
−
B₁
÷ (
T
+
C₁
)
+
(1
−
x₁
)
×
10
A₂
−
B₂
÷ (
T
+
C₂
)
713.9
=
0.5 ×
10
6.90565 − 1,211.033 ÷ (90 + 220.79)
+
(1
−
0.5)
×
10
6.95464 − 1,344.8 ÷ (90 + 219.482)
Assumptions
•
Ideal liquid (Raoult’s law) and ideal vapor; component 1 is the more volatile one.
•
Antoine’s equation in mmHg and °C, with constants from a table whose range covers T.
•
Type P instead of T for the bubble temperature: it is found by trial.
Step-by-step
WE KNOW
A₁ = 6.9057, B₁ = 1,211.033 °C, C₁ = 220.79 °C, A₂ = 6.9546, B₂ = 1,344.8 °C, C₂ = 219.482 °C, T = 90 °C, x₁ = 0.5
FIND
vapor pressure of 1 (P₁ˢᵃᵗ), vapor pressure of 2 (P₂ˢᵃᵗ), bubble pressure (P), vapor mole fraction of 1 (y₁)
Step 1 · Find vapor pressure of 1 (P₁ˢᵃᵗ)
Use log₁₀(P₁ˢᵃᵗ) = A₁ − B₁ ÷ (T + C₁)
Antoine’s equation for 1: work out the exponent, then raise 10 to it.
P₁ˢᵃᵗ
=
10
A₁
−
B₁
÷ (
T
+
C₁
)
P₁ˢᵃᵗ
=
10
6.90565 − 1,211.033 ÷ (90 + 220.79)
P₁ˢᵃᵗ
=
10
6.9057 − 1,211.033 ÷ 310.79
P₁ˢᵃᵗ
=
10
6.9057 − 3.8966
P₁ˢᵃᵗ
=
10
3.009
P₁ˢᵃᵗ = 1,021 mmHg
Step 2 · Find vapor pressure of 2 (P₂ˢᵃᵗ)
Use log₁₀(P₂ˢᵃᵗ) = A₂ − B₂ ÷ (T + C₂)
Antoine’s equation for 2: work out the exponent, then raise 10 to it.
P₂ˢᵃᵗ
=
10
A₂
−
B₂
÷ (
T
+
C₂
)
P₂ˢᵃᵗ
=
10
6.95464 − 1,344.8 ÷ (90 + 219.482)
P₂ˢᵃᵗ
=
10
6.9546 − 1,344.8 ÷ 309.482
P₂ˢᵃᵗ
=
10
6.9546 − 4.3453
P₂ˢᵃᵗ
=
10
2.6093
P₂ˢᵃᵗ = 406.7 mmHg
Step 3 · Find bubble pressure (P)
Use P = x₁ × P₁ˢᵃᵗ + (1 − x₁) × P₂ˢᵃᵗ
Each liquid adds its mole fraction times its own vapor pressure (Raoult’s law); add the two.
P
=
0.5
×
1,021
+
(1
−
0.5)
×
406.7
P
=
0.5
×
1,021
+
0.5
×
406.7
P
=
510.5
+
203.35
P
=
713.9
mmHg
Step 4 · Find vapor mole fraction of 1 (y₁)
Use y₁ = x₁ × P₁ˢᵃᵗ ÷ P
The partial pressure of 1 over the total is its mole fraction in the vapor (Dalton’s law).
y₁
=
0.5
×
1,021
713.9
y₁
=
510.5
713.9
y₁
=
0.7151
Check
Put the numbers back in. Both sides should match.
log₁₀(1,021)
=
6.90565
−
1,211.033
90 + 220.79
✓
log₁₀(406.7)
=
6.95464
−
1,344.8
90 + 219.482
✓
713.9 = 0.5 × 1,021 + (1 − 0.5) × 406.7 ✓
0.7151
=
0.5
×
1,021
713.9
✓
713.9
=
0.5 ×
10
6.90565 − 1,211.033 ÷ (90 + 220.79)
+
(1
−
0.5)
×
10
6.95464 − 1,344.8 ÷ (90 + 219.482)
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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