Physical Chemistry , 1st ed.

have to be pumped down to in order that a nitrogen molecule has a reason-

able chance of not colliding with another nitrogen molecule going from one

side to the other (that is, the mean free path is 1.00 m)? Assume a tempera-

ture of 22.0°C.

Solution

If the radius of the molecule is 1.60 Å, then the diameter dis 3.20 Å. The

question is essentially asking what pressure is necessary for a mean free path

of 1.00 m. Using equation 19.39:

1.00 m 

Again, in order for the units to work out properly, several conversion factors

must be applied (see Example 19.5):

1.00 m

1

1

L

0



0

ba

J

r



(0.

1

1

L

m)^3



1.00

1

1

b

0

a

0

r



(0.

1

1)^3



Solving for p:

p1.27  10 ^7 bar

This pressure, about one ten-millionth of an atmosphere, is easily obtainable

in the laboratory (using oil diffusion pumps, for example).

Now that we know roughly how far a gas particle typically travels between
collisions, we can determine an average collision frequency,which tells us about
how many times a gas particle comes in contact with another gas particle each
second. Collision frequency is a useful concept to apply to gas-phase chemical
reactions. We will start with a simple estimate from classical mechanics. Using
the definition

average speed 

di

t

s

i

t

m

an

e

ce



we can use the mean free path as our distance and one of the definitions of
average speed defined earlier in this chapter. A frequency is usually defined as
the reciprocal of time. In this case, “frequency” is interpreted as the number of
collisions per second, so it will have units of s^1. Using the average speed v,we
apply the definition of average velocity from equation 19.36 and the mean free
path from equation 19.38 to get an average collision frequency z:

z

v

 (19.40)

If we recognize that the fraction N/Vis the density of gas (in units of num-
ber of gas particles per m^3 , or 1/m^3 ), then we can substitute for Vand Nin
equation 19.40 to get

zd

(^2
8) kT

m
Nd^2
8 kT

V m


8

k

m

T



1/2





N

V

d^2



(1.381  10 ^23 )(295.15)



3.14159(3.20  10 ^10 )^2 p

(1.381  10 ^23 J/K)(295.15 K)



(3.20  10 ^10 m)^2 p

(1.381 ^10 ^23 J/K)(295.15 K)

(3.20  10 ^10 m)^2 p

19.4 Collisions of Gas Particles 669