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144 5 SIMPLEMIXTURES
0.0 0.2 0.4 0.6 0.8 1.0
40
45
50
xA
θ/
○ C
Figure 5.13
Solutions to problems
P5C.2 If it is assumed that Raoult’s law applies, [5A.22–151], the partial vapour pres-
sures of DE and DP are
pDE = xDEp∗DE pDP = xDPp∗DP
where xJ are the mole fractions and p∗J are the vapour pressures over the pure
liquids.�e total pressure is taken to be ptot = pDE + pDP.
(a) �e total pressure is given by
ptot = pDE + pDP = xDEp∗DE + xDPp∗DP
= 0.60 × (22.9 kPa) + 0.40 × (17.1 kPa) = 20.6 kPa
(b) �e mole fraction in the vapour, yJ, is related to the total pressure by pJ =
yJptot, so it follows that
yJ =
pJ
ptot
=
xJp∗J
ptot
�erefore
yDE =
xDEp∗DE
ptot
= 0.60 × (22.9 kPa)
0.60 × (22.9 kPa) + 0.40 × (17.1 kPa)
= 0.67
yDP =
xDPp∗DP
ptot
= 0.40 × (17.1 kPa)
0.60 × (22.9 kPa) + 0.40 × (17.1 kPa)
= 0.33
P5C.4 (a) �e temperature–composition phase diagram is a plot of the boiling point
against (1) composition of the liquid, xB, the mole fraction of butan-1-ol
and (2) composition of the vapour, yB. �e horizontal axis is labelled
zB, which is interpreted as xB or yB according to which set of data are
being plotted. In addition to the data in the table, the boiling point of
pure chlorobenzene is added. �e resulting phase diagram is shown in
Fig. 5.14; the data points are just connected by best-�t polynomials.