Page 228 - Physical chemistry understanding our chemical world

QUANTITATIVE EFFECTS OF PRESSURE AND TEMPERATURE CHANGE 195

Strategy. (1) Calculate the pressure exerted and hence the pressure change. (2) Insert
values into the Clapeyron equation (Equation (5.1)).

(1) The pressure exerted by the car is given by the equation
‘force ÷ area’. The force is simply the car weight expressed At sea level, a mass
in newtons (N): force = 10 000 N, so we calculate the of 1 kg has a weight
2
pressure exerted by the wheels as 10 000 N ÷ 10 −2 m , (i.e. exerts a force) of
6
which is 10 Pa. We see how a car exerts the astonishing approximately 10 N.
6
pressure beneath its wheels of 10 Pa (about 10 bar).
(2) Before inserting values into the Clapeyron equation, we
rearrange it slightly, first by multiplying both sides by
O
T V , then dividing both sides by H ,to give

dp T V
dT =
H O Notice how the free-
zing temperature of
6
−6
3
10 Pa × 273.15 K × (−1.6 × 10 ) m mol −1
dT = −1 water decreases when
3
6.0 × 10 Jmol a pressure is applied.
dT =−0.07 K This decrease is directly
attributable to the
minus sign of V.
Next, we recall that the symbol ‘ ’ means ‘final state − initial state’.
Accordingly, we say
T = T (final) − T (initial)

where the temperatures relate to the melting of ice. The normal melting temperature of
ice T (initial) is 273.15 K. The final temperature T (final) of the ice with the car resting on it
is obtained by rearranging, and saying

T + T (initial) =−0.07 K + 273.15 K
T (final) = 273.08 K

The new melting temperature of the ice T (final) is 273.08 K. Note how we performed
this calculation with the car parked and immobile on the ice. When driving rather than
parked, the pressure exerted beneath its wheels is actually considerably greater. Since
Equation (5.1) suggests that dp ∝ dT , the change in freezing temperature dT will be
proportionately larger (perhaps as much as −3 K), so there will be a layer of water on
◦
the surface of the ice even if the ambient temperature is −3 C. Drive
with care! Care:the word ‘normal’
here is code: it means
‘at p = p ’.
O
SAQ 5.2 Paraffin wax has a normal melting temperature
T (melt) of 320 K. The temperature of equilibrium is raised