Physical Chemistry Third Edition

9.7 The Model System with Potential Energy 421

Solution

a.

u(σ) 4 ε

[(

σ

σ

) 12

−

(σ

σ

) 6 ]

 4 ε(1−1) 0

b.At an extremum, the derivativedu/dris equal to zero:

du

dr

 4 ε

[

− 12 σ^12

1

r^13

+ 6 σ^6

1

r^7

]

 0

0  4 ε

(

6 σ^6

1

r^7

)(

−

2 σ^6

r^6

+ 1

)

This expression equals zero ifr→∞or if

r 21 /^6 σ 1. 12246 σ

The minimum is atrrmin 21 /^6 σ. There is a relative maximum atr→∞.

Exercise 9.21

a.Show that

uLJ(rmin)−ε (9.7-8)

b.The force in therdirection is given by−du/dr. Show that the force on one particle due to

another particle at distanceris

Fr 4 ε

(

12

σ^12

r^13

− 6

σ^6

r^7

)

(9.7-9)

c.Show thatFr0ifr 21 /^6 σ.

In a liquid or solid the average separation of the molecules from their nearest neigh-

bors is approximately equal to the intermolecular distance at the minimum inu. Energy

is required either to compress a liquid or solid or to pull its molecules away from each

other, which results in a nearly constant volume. In a gas at ordinary pressures, the

average nearest-neighbor distance is roughly 10 times as great as in a liquid, and the

average intermolecular forces are relatively small.

A simpler but less realistic representation of the pair potential function is thesquare

well potential

u(r)

⎧

⎨

⎩

∞ (0<r<d)

−u 0 (d<r<c)

0(r>c)

(9.7-10)

This representation is shown in Figure 9.16. If optimum values ofdandcare chosen,

this function can approximate the actual intermolecular potential for some purposes.

u

r = dr = c

r

0

2 u 0

`

Figure 9.16 The Square-Well Repre-
sentation of the Intermolecular Poten-
tial of a Pair of Atoms.