The Richard Stockton College of New Jersey
Chemistry Program, School of Natural Sciences and Mathematics
PO Box 195, Pomoma, NJ
CHEM 3410: Physical Chemistry I — Fall 2008
October 20, 2008
Lecture 21: The Clapeyron Equation
References
1. Levine, Physical Chemistry, Sections 7.1–7.4
Key Concepts
•At equilibrium, the free energy is minimized. Therefore, the phase with the lowest chemical potential
will be stable.
•To understand the shape of the phase boundaries in single component systems, we derived the Clapey-
ron equation. This was possible because along the phase boundary, the chemical potential of both
phases are equal. If we have a boundary between an αphase and βphase:
dP
dT coexistence
=∆Sα→β
∆Vα→β
where ∆Sα→βand ∆Vα→βare the change in molar entropy and volume, respectively, in going from
the αto βphase.
•Since for a single component system, µ=G=H−T S, the Clapeyron equation can be rewritten in
terms of the enthalpy of the transformation:
dP
dT coexistence
=∆Hα→β
T∆Vα→β
•For a simple diagram with only solid, liquid, and gas phases we can determine the relative slopes of
the boundaries by considering the differences in molar entropy and volume for each transition. For
example, we would expect the liquid to gas boundary to have a small positive slope while the transition
from the solid to gas would have a larger (steeper) positive slope.
•The slope of the solid-liquid boundary depends on difference in molar volume of the solid and liquid
phases. For water, the liquid phase has a higher density (smaller molar volume) meaning the boundary
on the phase diagram will have large negative slope. Most other substances have a large positive slope
because the solid is denser than the liquid.
•For a condensed phase (solid or liquid) in equilibrium with a gas we derived the Clausius-Clapeyron
equation by treating the gas as ideal and assuming the molar volume of the gas is much larger than
that of the condensed phase. For liquid-vapor equilibrium:
dln P=dP
P=∆Hvap
RT 2dT
For solid-vapor equilibrium, the ∆Hvap would be replaced by ∆Hsublimation.
•If the enthalpy is independent of temperature over our temperature range of interest then we can relate
the vapor pressure of the gas at one temperature and pressure (T1, P1) to the vapor pressure at another
temperature (T2, P2).
ln P2
P1
=−∆Htrans
R1
T2
−1
T2
where ∆Htrans is either the enthalpy of sublimation or vaporization, depending on the phases involved.
Related Exercises in Levine
Exercises: 7.22, 7.24