SME
Chemical Engineering Thermodynamics · Problem type
Dew pressure
Use this for “A vapor of 50% benzene and 50% toluene is at 90 °C, where P₁sat = 1021 mmHg and P₂sat = 406.7 mmHg. Find its dew pressure.”
Diagram
Pxy diagram of 1–2: the bubble curve runs from P₂ˢᵃᵗ = 406.7 to P₁ˢᵃᵗ = 1,021 mmHg, the dew curve under it; between them liquid and vapor coexist.
1
P
=
y₁
P₁
ˢᵃᵗ
+
1 − y₁
P₂
ˢᵃᵗ,
so
P
=
581.7
mmHg:
the
dew
point
of
vapor
y₁
=
0.5,
its
first
drop
of
liquid
x₁
=
0.285.
This is an example. Type your own values to start a new problem.
1
P (mmHg)
=
P₁ˢᵃᵗ (mmHg)
+
1
−
P₂ˢᵃᵗ (mmHg)
x₁
=
×
P (mmHg)
P₁ˢᵃᵗ (mmHg)
Formulas
1
P
=
y₁
P₁
ˢᵃᵗ
+
1 −
y₁
P₂
ˢᵃᵗ
1
581.7
=
0.5
1,021
+
1 − 0.5
406.7
x₁
=
y₁
×
P
P₁
ˢᵃᵗ
0.2849
=
0.5
×
581.7
1,021
Assumptions
•
Ideal liquid (Raoult’s law) and ideal vapor; component 1 is the more volatile one.
•
At the dew point the first drop of liquid forms; its mole fractions xᵢ = yᵢP ÷ Pᵢsat add to 1.
•
Both vapor pressures are at the vapor’s temperature.
Step-by-step
WE KNOW
P₁ˢᵃᵗ = 1,021 mmHg, P₂ˢᵃᵗ = 406.7 mmHg, y₁ = 0.5
FIND
dew pressure (P), mole fraction of 1 in the first drop (x₁)
Step 1 · Find dew pressure (P)
Use 1 ÷ P = y₁ ÷ P₁ˢᵃᵗ + (1 − y₁) ÷ P₂ˢᵃᵗ
The
first
drop’s
mole
fractions
yᵢP
Pᵢsat
add
to
1:
add
y₁
P₁
sat
and
y₂
P₂
sat,
then
take
the
reciprocal.
P
=
1
y₁
÷ P₁ˢᵃᵗ + (1 −
y₁
) ÷ P₂ˢᵃᵗ
P
=
1
0.5 ÷ 1,021 + (1 − 0.5) ÷ 406.7
P
=
1
0.5 ÷ 1,021 + 0.5 ÷ 406.7
P
=
1
0.00048972 + 0.0012294
P
=
1
0.0017191
P
=
581.7
mmHg
Step 2 · Find mole fraction of 1 in the first drop (x₁)
Use x₁ = y₁ × P ÷ P₁ˢᵃᵗ
Raoult’s
law
for
the
first
drop:
its
x₁
times
P₁sat
is
the
partial
pressure
y₁
P
.
x₁
=
0.5
×
581.7
1,021
x₁
=
290.85
1,021
x₁
=
0.2849
Check
Put the numbers back in. Both sides should match.
1
581.7
=
0.5
1,021
+
1 − 0.5
406.7
✓
0.2849
=
0.5
×
581.7
1,021
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
Related lessons
Vapor–liquid equilibrium
Main lesson
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