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SOLUTIONSMANUAL TO ACCOMPANY ATKINS' PHYSICAL CHEMISTRY 487
where 1 D = 3.3356 × 10−30 Cm is used from inside the front cover.
µy =∑
J
QJ yJ
= −2 × (1.6022 × 10−19 C) × (143 × 10−12 m) × sin 30○
× 1 D
3.3356 × 10−30 Cm
= −6.86... D
�e magnitude of the resultant is
µ = (µ2x + µ2y)
1/2 = [(−27.4... D)2 + (−6.86... D)2] = 28 D
and, from the diagram on the right of Fig. 14.2, the direction is given by
θ = tan−1 (6.86... D
27.4... D
) = 14○
E14A.4(b) �e relationship between the induced dipole moment µ∗ and the electric �eld
strength E is given by [14A.5a–587], µ∗ = αE , where α is the polarizability.
�e polarizability volume α′ is related to the polarizability α by [14A.6–587],
α′ = α/4πε0. Combining these equations, rearranging for E , and using 1 V =
1 JC−1 gives
E = µ∗
4πε0α′
= (2.5 × 10−6 D) × [(3.3356 × 10−30 Cm)/(1 D)]
4π × (8.8542 × 10−12 J−1 C2m−1) × (1.05 × 10−29 m3)
= 7.1 × 103 Vm−1
E14A.5(b) �e molar polarization Pm is de�ned by [14A.11–590], Pm = (NA/3ε0)(α +
µ2/3kT), where α is the polarizability of the molecule and µ is its dipole mo-
ment.�is equation is written as
Pm = NAα
3ε0
+ NAµ
2
9ε0k
1
T
which implies that a graph of Pm against 1/T should be a straight line with
slope NAµ2/9ε0k and intercept NAα/3ε0. However, as there are only two data
points it is convenient to calculate the required quantities directly from the
data. Writing the molar polarization at the two temperatures as Pm(T1) and
Pm(T2) and considering Pm(T2) − Pm(T1) gives
Pm(T2) − Pm(T1) =
NAµ2
9ε0k
( 1
T2
− 1
T1
)