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SOLUTIONSMANUAL TO ACCOMPANY ATKINS' PHYSICAL CHEMISTRY 657
�e number of H2 molecules in this volume is found using the perfect gas law
N = NApV
RT
= (6.0221 × 1023mol−1) × (1.01325 × 105 Pa) × (1.44 × 10−6 m3)
(8.3145 JK−1mol−1) × (273.15 K)
= 3.86... × 1019
�e area of a molecule is estimated from the mass density of the liquid in the
following way. Consider a volume V of the liquid which has mass density ρ;
the mass of the liquid is Vρ and this corresponds to Vρ/M moles, where M
is the molar mass. �e number of molecules in the volume is N = NAVρ/M,
therefore the volume occupied by one molecule is V/N = M/NAρ.
If the molecule is considered to be a sphere of radius R, then 43πR3 = M/NAρ,
from which it follows that R = (3M/4NAρπ)1/3. �erefore the area of the
‘silhouette’ of the sphere is A = πR2 = π(3M/4NAρπ)2/3. With the data given
A = π ( 3 × (2.0158 gmol−1)
4 × (6.0221 × 1023mol−1) × (0.708 g cm−3) × π
)
2/3
= 3.40... × 10−16 cm2 = 3.40... × 10−20 m2
�e surface area is therefore (3.86... × 1019) × (3.40... × 10−20 m2) = 1.3 m2 .
P19B.6 (a) As is shown in Example 19B.1 on page 833, a suitable plot to �t data to the
Langmuir isotherm is of p/V against p; such a plot has intercept 1/αV∞
and slope 1/V∞.�e table of data is given below and the plot is shown in
Fig. 19.4; it is clear from this that the data do not conform to the Langmuir
isotherm.
p/kPa V/cm3 (p/V)/(kPa cm−3) z y/(10−3 cm−3)
13.3 17.9 0.743 0.067 3.980
26.7 33.0 0.809 0.134 4.669
40.0 47.0 0.851 0.200 5.319
53.3 60.8 0.877 0.267 5.976
66.7 75.3 0.886 0.334 6.645
80.0 91.3 0.876 0.400 7.302
(b) In [19B.7–836] the BET isotherm is manipulated into a straight-line plot
z
(1 − z)V
= 1
cVmon
+ (c − 1)
cVmon
z z = p/p∗
�us a plot of z/(1 − z)V against z is expected to be a straight line with
slope (c − 1)/cVmon and intercept 1/cVmon; note that (slope)/(intercept)
= c − 1. For brevity the term z/(1 − z)V is denoted y. �e manipulated
data is shown in the table above and the plot is shown in Fig. 19.5.