Interações Moleculares

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496 14MOLECULAR INTERACTIONS
not contribute, and the equation reduces to Pm = NAα/3ε0.�is expression is
used in that for the slope
slope = 3Pm
2RT
= 3
2RT
NAα
3ε0
= α
2kTε0
It is therefore possible to determine α from the slope of the graph, and hence
α′.
14B Interactions betweenmolecules
Answer to discussion questions
D14B.2 �e arrangement of charges corresponding to the variousmultipoles are shown
in Fig. 14A.2 on page 587. According to [14B.4–596], the potential energy of
interaction between an n-pole and anm-pole goes as 1/rn+m+1. Two examples
of this relationship are derived in the text for the interaction between: (i) a point
charge (n = 0) and a dipole (m = 1), giving a 1/r0+1+1 = 1/r2 dependence; and
(ii) a dipole (n = 1) and a dipole (m = 1) giving a 1/r1+1+1 = 1/r3 dependence.
In both cases this form of the interaction arises by assuming that the distance
r is much greater than the separation between charges which form the dipole.
Consider the interaction between a point charge and a dipole, with the point
charge along the line of the dipole. �e point charge interacts with the two
charges which form the dipole: these interactions are opposite in sign but dif-
ferent in magnitude because the distances between the point charge and the
two charges which form the dipole are not the same. As the point chargemoves
further away, these interactions both decrease inmagnitude. However, because
the distance between these charges is becoming less signi�cant compared to
the distance of the point charge, the magnitude of the interaction with each
becomes more similar. As a result, the terms with opposite sign come closer
to cancelling one another out and the overall interaction therefore goes to zero
faster than does the simple interaction between point charges.
E�ectively, as r increases the two charges which form the dipole merge and
begin to cancel one another. �is is why the potential energy of interaction
falls o�more quickly than it does for the interaction between two point charges.
For quadrupoles and high n-poles the e�ect is greater, leading to an even faster
fall o� of the interaction, presumably because the e�ective cancellation of the
charges happensmore quicklywhen there aremore charges forming the n-pole.
D14B.4 �is is discussed in Section 14B.2 on page 598.
Solutions to exercises
E14B.1(b) �e interaction between a point charge and a point dipole orientated directly
away the charge is given by [14B.1–593], V = −µ1Q2/4πε0r2. In this case the