Page 264 - Physical chemistry understanding our chemical world

PHASE EQUILIBRIA INVOLVING VAPOUR PRESSURE 231

In practice, we force water vapour (steam) at high pressure through the clove pulp to
obtain a significant partial pressure of eugenol (V).

Justification Box 5.5

When considering the theory behind steam distillation, we start with the ideal-gas
equation (Equation (1.13)), pV = nRT . We will consider two components: oil and
water. For the oil, we say p (oil) V = n (oil) RT , and for the water p (water) V = n (water) RT .
Dividing the two equations by R and V (which are both constant) yields
p (oil) = n (oil) × T for the oil

p (water) = n (water) × T for the water

We then divide each pressure by the respective number of moles n i , to obtain

p (oil) ÷ n (oil) = T for the oil
p (water) ÷ n (water) = T for the water

The temperature of the two materials will be T , which is the same for each as they are
in thermal equilibrium. We therefore equate the two expressions, saying

p (oil) ÷ n (oil) = p (water) ÷ n (water)
Dividing both sides by p (water) and multiplying both sides by n (oil) yields Equation (5.23):

p (oil) n (oil)
=
p (water) n (water)
so we see how the percentage of each constituent in the vapour depends only on its
vapour pressure at the distillation temperature.
To extract a relatively involatile oil such as eugenol (V) without charring requires a
◦
high pressure of steam, although the steam will not be hotter than 100 C, so we generate
a mixture of vapours at a temperature lower than that of the less volatile component.