Page 259 - Physical chemistry understanding our chemical world

226 PHASE EQUILIBRIA

depends on the liquid’s composition, according to Raoult’s law:

O
x
p (benzene) = p (benzene) (benzene) (5.21)
where x (benzene) is the mole fraction of the benzene in the liquid.
Raoult’s law states If we assume that liquid benzene and petrol have the same den-
that (at constant tem- sities (which is entirely reasonable), then petrol containing 3 per
perature) the partial
cent of benzene represents a mole fraction x (benzene) = 0.03; the
pressure of component mole fraction of the petrol in the liquid mixture is therefore 0.97
i in the vapour residing
at equilibrium above (or 97 per cent). The vapour above the petrol mixture will also
a liquid is proportional be a mixture, containing some of each hydrocarbon in the petrol.
to the mole fraction We call the pressure due to the benzene component its partial pres-
x i of component in sure p (benzene) . The constant of proportionality in Equation (5.21) is
the liquid. p O , which represents the pressure of gaseous benzene above
(benzene)
pure (i.e. unmixed) liquid benzene.

Calculations with Raoult’s law

If we know the mole fraction of a liquid i (via Equation (5.11))
If a two-component and the vapour pressures of the pure liquids p , then we can ascer-
O
system of A and B i
forms an ideal mixture, tain the total vapour pressure of the gaseous mixture hovering at
then we can calcu- equilibrium above the liquid.
The intensity of the benzene smell is proportional to the amount
late x A if we know x B
because x A + x B = 1, so of benzene in the vapour, p (benzene) . According to Equation (5.21),
x B = (1 − x A ). p (benzene) is a simple function of how much benzene resides within
the liquid petrol mixture. Figure 5.21 shows a graph of the partial
pressures of benzene and octane above a mixture of the two liq-
uids. (For convenience, we assume here that the mixture comprises only these two
components.)
The extreme mole fractions, 0 and 1, at either end of the graph relate to pure petrol
(x = 0) and pure benzene (x = 1) respectively. The mole fractions between these
values represent mixtures of the two. The solid, bold line represents the total mole
fraction while the dashed lines represent the vapour pressures of the two constituent
vapours. It is clear that the sum of the two dashed lines equals the bold line, and
represents another way of saying Dalton’s law: the total vapour pressure above a
mixture of liquids is the sum of the individual vapour pressures.

Benzene is more vola- Worked Example 5.8 The two liquids benzene and bromobenzene
tile than bromobenzene are mixed intimately at 298 K. At equilibrium, the pressures of the
because its vapour gases above beakers of the pure liquids are 100.1 kPa and 60.4 kPa
pressure is higher. respectively. What is the vapour pressure above the mixture if 3 mol
of benzene are mixed with 4 mol of bromobenzene?