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SOLUTIONSMANUAL TO ACCOMPANY ATKINS' PHYSICAL CHEMISTRY 495
and hence
µ = (9ε0k
NA
× (2.091 × 10−2 m3mol−1 K))
1/2
= (9 × (8.8542 × 10−12 J−1 C2m−1) × (1.3806 × 10−23 JK−1)
6.0221 × 1023mol−1
×(2.09 × 10−2 m3mol−1 K))
1/2
= 6.17... × 10−30 Cm = 1.85 D
Similarly, identifying 43πNAα′ with the intercept gives
α′ = 3
4πNA
× (3.27 cm3mol−1)
= 3
4π × (6.0221 × 1023mol−1)
× (3.27 cm3mol−1) = 1.30 × 10−24 cm3
P14A.12 �e relationship between refractive index and relative permittivity is given by
[14A.13–592], nr = ε1/2r . In addition the relationship between relative permit-
tivity and the molar polarization is given by the Debye equation [14A.10–590],
(εr − 1)/(εr + 2) = ρPm/M where ρ is the mass density and M is the molar
mass.
�emass density ρ is given byM/Vm, whereVm is themolar volume. Assuming
perfect gas behaviour, Vm = RT/p, so ρ = pM/RT and hence the Debye
equation becomes
εr − 1
εr + 2
= pPm
RT
hence εr − 1 =
pPm
RT
(εr + 2)
If εr ≈ 1 then (εr + 2) ≈ 3 so the equation becomes
εr − 1 =
pPm
RT
× 3 hence εr = 1 +
3pPm
RT
Introducing nr = ε1/2r and using the series expansion (1 + x)1/2 ≈ 1 + 1
2 x for
small x gives
nr = (1 + 3pPm
RT
)
1/2
≈ 1 + 1
2 ×
3pPm
RT
= 1 + 3pPm
2RT
�is expression implies that a plot of nr against p should be a straight line of
slope 3Pm/2RT . Hence, by measuring nr as a function of pressure, the molar
polarization Pm can be determined.
�e molar polarization is given by [14A.11–590], Pm = (NA/3ε0)(α+ µ2/3kT).
However, assuming that the frequency of oscillation of the electric �eld is high,
as it will be for visible light, there will be no contribution from orientation po-
larization because themolecules cannot reorientate themselves quickly enough
to follow the �eld. �erefore the term µ2/3kT in the expression for Pm does