1 (491)

Prévia do material em texto

SOLUTIONSMANUAL TO ACCOMPANY ATKINS' PHYSICAL CHEMISTRY 139
this the molar mass is calculated
intercept = RT/M hence M = (84784 g cm K−1 mol−1) × (298.15 K)
102 × 2.307 g cm−2/g cm−3
which gives M = 1.1 × 105 gmol−1 .
(c) Over the full range of concentrations the plot of Π/c against c is non-
linear, so the solvent is characterised a ‘good’. �is may be attributed to
both solvent and polymer being non-polar.
(d) �e virial-style equation with three coe�cients is
Π = c
M
RT [1 + B c
M
+ C ( c
M
)
2
] hence Π
c
= RT
M
[1 + B c
M
+ C ( c
M
)
2
]
It is convenient to take the factor ofM into the virial coe�cients to give
Π
c
= RT
M
(1 + B′c + C′c2)
where B′ = B/M and likewise for C′. Using this, the data Π/c are �tted
to a quadratic in c.
If the �tted function is required to have the same intercept at c = 0 as in
part (a), the �tted function to the �rst 8 data points (chosen as these gave
the best �t) is
(Π/c)/( g cm−2/g cm−3) = 55563(c/(g cm−3))2+3784.2(c/(g cm−3))+230.7
It follows that
RT
M
B′ = 3784.2 g−1 cm4
From part (a) RT/M = 230.7 g cm−2/g cm−3, therefore
B′ = (3784.2 g−1 cm4)/(230.7 g cm−2/g cm−3) = 16.4 g−1 cm3
By a similar line of argument, C′ = 241 g−2 cm6 .
(e) �e proposed virial equation with g = 1
4 is developed into a straight-line
plot as follows
Π
c
= RT
M
(1 + B′c + 1
4B
′2c2)
= RT
M
(1 + 1
2B
′c)2
hence (Π
c
)
1/2
= (RT
M
)
1/2
(1 + 1
2B
′c) (5.5)
A plot of (Π/c)1/2 against c should be a straight line; such a plot is shown
in Fig. 5.9.