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SOLUTIONSMANUAL TO ACCOMPANY ATKINS' PHYSICAL CHEMISTRY 137
0.0 0.2 0.4 0.6 0.8 1.0
−4
−2
0
2
xA
(µ
A
−
µ∗ A
)/
RT
g = 0
g = 0.5
g = 2.0
g = 6.0
Figure 5.6
where R′ = R/g.�e �nal equation is of the form of the van ’t Ho� equa-
tion, but with a modi�ed measure of concentration, c/M, and a modi�ed
gas constant, R′.
�is equation is rearranged to give an expression for R′, and the units
of this quantity are then found by inserting the units of the quantities
involved
R′ = PM
cT
= (g cm−2) × (g mol−1)
(g cm−3) × (K)
= g cm K−1 mol−1
�e numerical value of R′ is found from its de�nition
R′ = R
g
= 8.3145 JK
−1mol−1
9.8067 ms−2
= 0.84784 kg m K−1 mol−1
�e units of R′ are found by using 1 J = 1 kgm2 s−2. �e �nal step is to
convert the numerical value of R′ from (kg m K−1 mol−1) to the required
(g cm K−1 mol−1)
(0.84784 kg m K−1 mol−1)× 1000 g
1 kg
× 100 cm
1 m
= 84784 g cm K−1 mol−1
(b) From now on the van ’t Ho� equation is writtenΠ = (c/M)RT where the
units ofΠ and c are as described in (a), andR is 84784 g cm K−1 mol−1. As
described in Section 5B.2(e) on page 162, if it is assumed that the osmotic
pressure is given by a virial-type equation [5B.18–163] where just the �rst
two terms are retained, a plot of Π/c against c should give a straight line
with intercept RT/M.�e data are plotted in Fig. 5.7.