Physical Chemistry Third Edition

428 9 Gas Kinetic Theory: The Molecular Theory of Dilute Gases at Equilibrium

The relative speed is the rate of change ofrand is larger than the speed of each particle

by a factor of

√

2:

vrel

∣

∣

∣

∣

dr

dt

∣

∣

∣

∣

√

2

∣

∣

∣

∣

dx

dt

∣

∣

∣

∣ (9.8-16)

If both particles are moving at the mean speed, we identify their relative speed as the

mean relative speed, denoted by〈vrel〉:

〈vrel〉

√

2 〈v〉

√

2

√

8 kBT

πm



√

16 kBT

πm

(9.8-17)

Our derivation is crude, but Eq. (9.8-17) is the correct formula for the mean relative

speed.

√If we assume that particle 1 is approaching the other particles at a mean speed of

2 〈v〉instead of〈v〉the mean free path is shorter by a factor of 1/

√

2:

λ

1

√

2 πd^2 N

(9.8-18)

We will use this equation rather than Eq. (9.8-13). Themean collision timeτcollis given

by the familiar relation: timedistance/speed:

τcoll

λ

〈v〉



√

πm

8 kBT

1

√

2 πd^2 N

(9.8-19)

Note that the mean speed enters in this formula, not the mean relative speed. However,

incorporation of the

√

2 factor in the formula for the mean free path allows us to write

τcoll

√

πm

16 kBT

1

πd^2 N



1

〈vrel〉πd^2 N

(9.8-20)

Themean molecular collision ratezis the reciprocal of the mean collision time:

z

1

τcoll

πd^2 N

√

16 kBT

πm

πd^2 N〈vrel〉 (9.8-21)

Notice how reasonable this equation is. The rate of collisions of a molecule is pro-

portional to the collision cross section, to the number density of molecules, and to the

mean relative speed. Under ordinary conditions, a gas molecule undergoes billions of

collisions per second.

EXAMPLE9.18

For nitrogen gas at 298 K and a molar volume of 24.45 L (approximately corresponding to

1.00 atm pressure):

a.Find the mean free path.

b.Find the mean collision time and the mean molecular collision rate.