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194 PHASE EQUILIBRIA

We move from the qualitative argument that T (melt) decreases as p increases, and
next look for a quantitative measure of the changes in melt temperature with pressure.
We will employ the Clapeyron equation:

dp H O
= (5.1)
dT T V

where T is the normal melting temperature, dT is the change in
In fact, it does not the melting temperature caused by changing the applied pressure
matter whether H by an amount of dp (in SI units of pascals), where 1 Pa is the
relates to the direction 2
pressure exerted by a force of 1 N over an area of 1 m . H is
of change of solid →
the enthalpy change associated with the melting of water and V
liquid or of liquid →
solid, provided that V is the change in volume on melting. Strictly, both H and V are
relates to the same molar quantities, and are often written as H m and V m , although
direction of change. the ‘m’ is frequently omitted.
−1
3
The molar change in volume V m has SI units of m mol .
We should note how these volumes are molar volumes, so they
refer to 1 mol of material, explaining why V is always very
The molar change in −6 −5 3 −1
small. The value of V m is usually about 10 to 10 m mol
volume V m has units 3 −1
3
of m mol −1 .Values in magnitude, equating to 1 to 10 cm mol respectively.
We recall from Chapter 1 how the symbol means ‘final state
typically lie in the
range 10 −5 –10 −6 m 3 minus initial state’, so a positive value of V m during melting
mol −1 . (which is V m (liquid) − V m (solid) ) tells us that the liquid has a slightly
larger volume than the solid from which it came. V m (melt) is
positive in the overwhelming majority of cases, but for water
−1
3
V m (melt) =−1.6 × 10 −6 m mol . This minus sign is extremely
The minus sign of V m
reflects the way water unusual: it means that ice is less dense than water. This explains
expands on freez- why an iceberg floats in water, yet most solids sink when immersed
ing. This expansion in their respective liquid phases.
explains why a car The enthalpy H O is the energy required to melt 1 mol of
radiator cracks in cold (melt)
material at constant pressure. We need to be careful when obtain-
weather (if it contains ing data from tables, because many books cite the enthalpy of
no ‘de-icer’): the water
freezes and, in expand- fusion, which is the energy released during the opposite process of
ing, exerts a huge a solidification. We do not need to worry, though, because we know
pressure on the metal. from Hess’s law that H O =− H O . The molar enthalpy
(melt) (fusion)
−1
of melting water is +6.0kJ mol .
Worked Example 5.1 Consider a car weighing 1000 kg (about
2200 lbs) parked on a sheet of ice at 273.15 K. Take the area under
Care: following Hess’s
2
2
wheels in contact with the ice as 100 cm i.e. 10 −2 m . What is the
law, we say: O
H O =− H O . new melting temperature of the ice – call it T (final) ?Take H (melt) =
(melt) (fusion) −1 −6 3 −1
6.0kJ mol and water V m (melt) =−1.6 × 10 m mol .