Page 234 - Physical chemistry understanding our chemical world

QUANTITATIVE EFFECTS OF PRESSURE AND TEMPERATURE CHANGE 201

In this example, it is simpler to insert values into Equation (5.5) and Notice how the ratio
to rearrange later. Inserting values gives
within the bracket on
the left-hand side of the
−1
6 × p O −50 000 J mol 1 1 Clausius–Clapeyron
ln = −1 × −
p O 8.314 J K −1 mol T 2 373 K equation permits us
to dispense with abso-
lute pressures.
We can omit the units of the two pressures on the left-hand side
because Equation (5.5) is written as a ratio, so the units cancel: we
require only a relative change in pressure.

1 1
ln 6.0 =−6104 K × −
T 2 373 K
where ln 6.0 has a value of −1.79. Next, we rearrange slightly by dividing both sides by
6104 K, to yield:

−1.79 1 1
= −
6104 K T 2 373 K
so
1 −4 −1 1
=−2.98 × 10 K +
T 2 373 K
and
1 −3 −1
= 2.38 × 10 K
T 2
We obtain the temperature at which water boils by taking the reciprocal of both side.
◦
O
T 2 , is 420 K, or 147 C at a pressure of 6 × p , which is much higher than the normal
◦
boiling temperature of 100 C.
SAQ 5.4 A mountaineer climbs Mount Everest and wishes to make a
strong cup of tea. He boils his kettle, but the final drink tastes lousy
because the water boiled at too low a temperature, itself because the
pressure at the top of the mountain is only 0.4 × p . Again taking the
O
enthalpy of boiling the water to be 50 kJ mol −1 and the normal boiling
temperature of water to be 373 K, calculate the temperature of the water
as it boils at the top of the mountain.

The form of the Clausius–Clapeyron equation in Equation (5.5) is called the inte-
grated form. If pressures are known for more than two temperatures, an alternative
form may be employed:

H O 1
(boil)
ln p =− × + constant (5.6)
R T