SME
Physical Chemistry I: Thermodynamics & Kinetics
Phase equilibria
Use this for “Water’s vapor pressure is 1 atm at 373.15 K, and ΔHᵥₐₚ = 40.7 kJ/mol. Find its vapor pressure at 323.15 K.”
Diagram
Water: the vapor curve runs from the triple point (273.2 K, 0.00604 atm) to the critical point (647.1 K, 217.7 atm), through (373.15 K, 1 atm) and (323.15 K, 0.1314 atm); pressure on a log scale.
Between
neighbouring
points
ln
P
is
a
straight
line
in
1
T
:
Clausius–Clapeyron,
ln(
P₂
P₁
)
=
−(
ΔHᵥₐₚ
R
)(
1
T₂
−
1
T₁
),
with
ΔHᵥₐₚ
constant
on
each
piece.
The melting line leans left: dP/dT = −133 atm/K, so raising the pressure lowers the melting point.
This is an example. Type your own values to start a new problem.
ln
P₂ (atm)
P₁ (atm)
=
−(
ΔHᵥₐₚ (kJ/mol)
÷
R)
×
1
T₂ (K)
−
1
T₁ (K)
Formulas
ln(
P₂
P₁
)
=
−(ΔHᵥₐₚ
×
1,000
8.314
)
×
(
1
T₂
−
1
T₁
)
ln(
0.1314
1
)
=
−(40.7
×
1,000
8.314
)
×
(
1
323.15
−
1
373.15
)
Assumptions
•
The vapor is an ideal gas and the liquid’s volume is tiny next to it, so Clapeyron’s equation becomes d ln P ÷ dT = ΔHᵥₐₚ ÷ RT².
•
ΔHᵥₐₚ stays the same between the two temperatures; it is in kJ/mol, so it is multiplied by 1000 for J/mol, with R = 8.314 J/(mol·K).
•
The diagram is water’s; your two points are marked on it when they fall inside it. Temperatures are in kelvins.
Step-by-step
WE KNOW
ΔHᵥₐₚ = 40.7 kJ/mol, T₁ = 373.15 K, P₁ = 1 atm, T₂ = 323.15 K
FIND
vapor pressure at T₂ (P₂)
Step 1 · Find vapor pressure at T₂ (P₂)
Use ln(P₂ ÷ P₁) = −(ΔHᵥₐₚ × 1,000 ÷ 8.314) × (1 ÷ T₂ − 1 ÷ T₁)
Work
out
the
exponent,
ΔHᵥₐₚ
R
times
the
change
in
1
T
with
its
sign
flipped,
then
raise
e
to
it
and
scale
P₁
.
P₂
=
P₁
×
e^(−(ΔHᵥₐₚ
×
1,000
8.314
)
×
(
1
T₂
−
1
T₁
))
P₂
=
1
×
e^(−(40.7
×
1,000
8.314
)
×
(
1
323.15
−
1
373.15
))
P₂
=
e^(−(40.7
×
1,000
8.314
)
×
(0.00309454
−
0.00267989))
P₂
=
e^(−(40.7
×
1,000
8.314
)
×
0.00041465)
P₂
=
e^(−(
40,700
8.314
)
×
0.00041465)
P₂
=
e
−4,895.36 × 0.00041465
P₂
=
e
−2.02986
P₂
=
0.1314
atm
→
Colder,
so
fewer
molecules
have
the
energy
to
escape:
the
vapor
pressure
is
lower.
Check
Put the numbers back in. Both sides should match.
ln(
0.1314
1
)
=
−(40.7
×
1,000
8.314
)
×
(
1
323.15
−
1
373.15
)
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
More problem types
Slope of a melting line (Clapeyron)
Use this for “Benzene melts at 278.7 K with ΔHfus = 9870 J/mol, and it grows by 10.3 cm³/mol as it melts. Find the slope dP/dT of its melting line.”
Vapor pressure of an ideal mixture (Raoult)
Use this for “Hexane (151 torr) and heptane (45.8 torr) are mixed at 25 °C with xₕₑₓₐₙₑ = 0.350. Find the total vapor pressure and the vapor’s mole fraction of hexane.”
Degrees of freedom (phase rule)
Use this for “A mixture of ethanol and water boils, with liquid and vapor present. How many intensive variables can be changed on their own?”