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126 5 SIMPLEMIXTURES
0.0 0.2 0.4 0.6 0.8 1.0
0
10
20
30
xA
p J
/(
kP
a)
A
B
Figure 5.4
Henry’s law [5A.24–152], applies to the solute at low concentrations pJ = KJxJ.
�e constant is estimated from the partial pressure at the smallest mole frac-
tion; these data are selected from the table.
KA =
pA
xA
= 1.399 kPa
0.0898
= 15.6 kPa KB =
pB
xB
= 4.209 kPa
0.0895
= 47.0 kPa
P5A.6 �e mass of haemoglobin in 100 cm3 of blood is [(100 cm3)/(1000 cm3)] ×
(150 g cm−3) = 15 g. In the lungs the haemoglobin is 97% saturated and so this
mass binds a volume 0.97 × (15 g) × (1.34 cm3 g−1) of O2. In the capillaries
the saturation drops to 75%, so the volume of O2 released is
(0.97 − 0.75) × (15 g) × (1.34 cm3 g−1) = 4.4 cm3
5B The properties of solutions
Answers to discussion question
D5B.2 �e excess volume is de�ned as VE = ∆mixV − ∆mixV ideal. �e volume of
mixing of an ideal solution is zero, which can be understood at amolecular level
as a result of the A and B molecules �tting together in just the same way that A
or B molecules �t with one another. A thermodynamic explanation is that for
an ideal system the partial molar volume is not a function of composition. It
follows that the excess volume is equal to the volume of mixing VE = ∆mixV ;
it is also useful to recall that ∆mixV = Vmixed − Vseparated.
Melons are large relative to oranges. When melons pack together it is conceiv-
able that oranges could �t into the interstices between the melons. If this were
the case there would initially be no increase in volume as oranges are added to
melons – meaning that the volume of the mixed oranges and melons is smaller