Physical Chemistry Third Edition

9.8 The Hard-Sphere Gas 429

Solution

a.

N 

N

V



6. 022 × 1023 mol−^1

0 .02445 m^3 mol−^1

 2. 463 × 1025 m−^3

〈v〉

(

8(8.3145 J K−^1 mol−^1 )(298.15 K)

π(0.0280 kg mol−^1 )

) 1 / 2

475 m s−^1

〈vrel〉

√

2 〈v〉671 m s−^1

λ

1

√

2 π(3. 7 × 10 −^10 m^3 )^2 (2. 463 × 1025 m−^3 )

 6. 7 × 10 −^8 m

b.

τcoll

λ

〈v〉



6. 7 × 10 −^8 m

475 m s−^1

 1. 4 × 10 −^10 s

z

1

1. 4 × 10 −^10 s

 7. 1 × 109 s−^1

Remember the approximate size of the collision rate and mean free path for an

ordinary gas. A gas molecule ordinarily undergoes several billion collisions per second

and moves several tens of nanometers between collisions.

Exercise 9.24

For helium gas at a molar volume of 24.45 L:

a.Find the length of a cube containing on the average one atom.

b.Find the mean free path.

c.Why is the mean free path so much larger than the length of the cube of part a?

To obtain a formula for thetotal rate of collisionsper unit volume we multiply

the mean molecular collision rate by the number of molecules per unit volume, but

we must correct for the fact that this would count each collision twice. For example,

the collision between molecule number 1 and molecule number 37 would be included

once for molecule 1 and once for molecule 37. We correct for this double counting by

dividing by 2:

Z

1

2

zN 

1

2

πd^2 〈vrel〉N^2 πd^2

√

4 kBT

πm

N^2 (9.8-22)

where the total collision rate per unit volume is denoted byZ.

Notice the following important physical facts: (1) the total collision rate per unit

volume is proportional to the square of the number density, (2) it is proportional to

the collision cross section, and (3) it is proportional to the mean speed and thus to the