SME
Physical Chemistry I: Thermodynamics & Kinetics · Problem type
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.”
Diagram
Water near its triple point (273.2 K, 0.00604 atm), both axes to linear scales: the vapor and solid–gas lines hug P ≈ 0 here.
The melting line leans left: dP/dT = −133.2 atm/K, so raising the pressure lowers the melting point.
This is an example. Type your own values to start a new problem.
dP/dT (atm/K)
=
ΔHfus (J/mol)
÷
(
T (K)
×
(
ΔV (cm³/mol)
)
)
Formulas
dP/dT
=
ΔHfus
T
× ΔV ×
10
-6
÷
101,325
−133.2
=
6,010
273.15 × (−1.63) ×
10
-6
÷
101,325
Assumptions
•
Along the line the solid and liquid stay at equilibrium, so dP/dT = ΔH ÷ (TΔV), Clapeyron’s equation.
•
ΔV is the liquid’s molar volume minus the solid’s: negative for water, because ice floats.
•
ΔV in cm³/mol is × 10⁻⁶ for m³/mol, which gives Pa/K; ÷ 101,325 turns Pa into atm.
•
ΔHfus and ΔV stay the same over the small stretch of line.
Step-by-step
WE KNOW
ΔHfus = 6,010 J/mol, T = 273.15 K, ΔV = −1.63 cm³/mol
FIND
slope of the melting line (dP/dT)
Step 1 · Find slope of the melting line (dP/dT)
Use dP/dT = ΔHfus ÷ (T × ΔV × 10⁻⁶) ÷ 101,325
Clapeyron:
divide
ΔH
by
T
times
ΔV
in
m
3
/mol
for
Pa/K,
then
divide
by
101,325
Pa
per
atm.
dP/dT
=
6,010
273.15 × (−1.63) ×
10
-6
÷
101,325
dP/dT
=
6,010
(−445.2345) ×
10
-6
÷
101,325
dP/dT
=
6,010
−0.0004452
÷
101,325
dP/dT
=
−1.35 ×
10
7
101,325
dP/dT = −133.2 atm/K → dP/dT < 0: the liquid is denser, so squeezing lowers the melting point (ice under pressure).
Check
Put the numbers back in. Both sides should match.
−133.2
=
6,010
273.15 × (−1.63) ×
10
-6
÷
101,325
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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