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198 PHASE EQUILIBRIA

phase boundary, which describes melting and solidification. For these reasons, the
Clapeyron equation is most effective when dp is relatively small, i.e. 2–10 atm
at most.
The worst deviations from the Clapeyron equation occur when one of the phases
is a gas. This occurs because the volume of a gas depends strongly on temperature,
whereas the volume of a liquid or solid does not. Accordingly, the value of V m is
not independent of temperature when the equilibrium involves a gas.

Why does deflating the tyres on a car improve its
road-holding on ice?

The Clapeyron equation, continued

We saw from the Clapeyron equation, Equation (5.1), how the decrease in freezing
temperature dT is proportional to the applied pressure dp, so one of the easiest ways
of avoiding the lethal conversion of solid ice forming liquid water is to apply a smaller
pressure – which will decrease dT in direct proportion.
The pressure change dp is caused by the additional weight of,
for example, a car, lorry or ice skater, travelling over the surface
The pressure beneath
of the ice. We recall our definition of ‘pressure’ as ‘force ÷ area’.
the blades of an ice- There is rarely a straightforward way of decreasing the weight of
skater’s shoe is enor-
a person or car exerting the force, so the best way to decrease
mous – maybe as much
as 100 atm when the the pressure is to apply the same force but over a larger area.
skater twists and turns An elementary example will suffice: cutting with a sharp knife
at speed. is easier than with a blunt one, because the active area along
the knife-edge is greater when the knife is blunt, thus causing p
to decrease.
In a similar way, if we deflate slightly the tyres on a car, we see the tyre bulge
a little, causing it to ‘sag’, with more of the tyre in contact with the road surface.
So, although the weight of the car does not alter, deflating the tyre increases the area
over which its weight (i.e. force) is exerted, with the result that we proportionately
decrease the pressure.
In summary, we see that letting out some air from a car tyre decreases the value of
dp, with the result that the change in melting temperature dT of the ice, as calculated
with the Clapeyron equation (Equation (5.1)), also decreases, thereby making driving
on ice much safer.

SAQ 5.3 A man is determined not to slip on the ice, so instead of wearing
2
skates of area 10 cm he now wears snow shoes, with the underside of
each sole having an extremely large area to spread his 100 kg mass
2
(equating to 1000 N). If the area of each snow shoe is 0.5m , what is the
depression of the freezing temperature of the ice caused by his walking
over it?
Use the thermodynamic data for water given in Worked Example 5.1.