1 (506)

Prévia do material em texto

498 14MOLECULAR INTERACTIONS
-Q
-Q
+2Q
-Q
-Q
+2Q
l
l
r
A
B
C
D
E
F
l
l
r
r 
2+l 2
r 
2+4l 2
Figure 14.8
term and r12 is the distance between them.
V = − 1
4πε0
(
A–D
³¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹·¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹µ
(−Q)(−Q)
r
+
A–E
³¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹·¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹µ
(−Q)(2Q)√
r2 + l 2
+
A–F
³¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹·¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹µ
(−Q)(−Q)√
r2 + 4l 2
+
B–D
³¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹·¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹µ
(2Q)(−Q)√
r2 + l 2
+ (2Q)(2Q)
r
´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶
B–E
+ (2Q)(−Q)√
r2 + l 2
´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶
B–F
+ (−Q)(−Q)√
r2 + 4l 2
´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶
C–D
+ (−Q)(2Q)√
r2 + l 2
´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶
C–E
+ (−Q)(−Q)
r
´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶
C–F
)
= 1
4πε0
(6Q
2
r
− 8Q2√
r2 + l 2
+ 2Q2√
r2 + 4l 2
)
= Q2
4πε0r
(6 − 8√
1 + (l/r)2
+ 2√
1 + 4(l/r)2
)
= Q2
4πε0r
(6 − 8(1 + x2)−1/2 + 2(1 + 4x2)−1/2)
where x = l/r. Assuming that r ≫ l , so that x ≪ 1, the series expansion
(1 + y)n = 1 + ny + 1
2n(n − 1)y
2 + ... can be used: it is su�cient to retain
the �rst three terms, (1 + y)−1/2 ≈ 1 − 1
2 y +
3
8 y
2.�e required expansions are
therefore (1 + x2)−1/2 ≈ 1 − 1
2 x
2 + 3
8 x
4 and (1 + 4x2)−1/2 ≈ 1 − 2x2 + 6x4
V = Q2
4πε0r
(6 − 8[1 − 1
2 x
2 + 3
8 x
4 + ...] + 2[1 − 2x2 + 6x4 + ...])
= Q2
4πε0r
(6 − 8 + 4x2 − 3x4 + 2 − 4x2 + 12x4) = Q2
4πε0r
× 9x4
= Q2
4πε0r
× 9( l
r
)
4
= 9Q2 l 4
4πε0r5
E14B.4(b) �e average energy of interaction between rotating polar molecules is given by
the Keesom interaction [14B.4–596].
⟨V⟩ = − C
r6
C = 2µ21µ22
3(4πε0)2kT