426
10 Heat and Matter
where x is the total distance travelled and N is the number of collisions.
In terms of frequency f , average time of collision T and the mean molecular
velocity v,
λ = vT =
v
f
=
1000
2 × 10 10 = 5 × 10
−8 m
10.2 (a)
v 2
=
3RT
M
=
3 × 8.31 × 300
28 × 10 −3 = 516.8 m/s
(b) R = kn 0 = 1.38 × 10
−23
× 6.02 × 10
23
= 8.3 J/mol K
N =
PV
RT
=
2 × 1.013 × 10 5 × 1
8.3 × 300
= 81.365 mol/m
3
n = number/m
3
=
6.02 × 10 23 × 81.365
28 × 10 −3
= 2.15 × 10
25
/m
3
λ =
1
σ n
=
1
0.43 × 10 −18 × 2.15 × 10 25 = 1.082 × 10
−7 m
f =
v
λ
=
516.8
1.082 × 10 −7 = 4.78 × 10
9
/s
10.3 (a) Collision cross-section between two molecules each of radius r is equivalent to collision of one molecule of radius 2r with another point size
molecule. Therefore the cross -section will be
σ = π(2r )
2
= 4πr
2
(1)
Consider a rectangular box of face area l m 2 and length λ metres. Then
the volume of the box V = λ m 3 . If n is the number of molecules per
unit volume then λ is such that the total projected area arising from n
molecules will just fill up an area of 1 m 2 .
∴
nλσ = 1
or
λ =
1
nσ
=
V
4πr 2 N
(2)
where we have used (1) and set n = N /V .
Equation (2) in based on the assumption that the target molecules are
stationary. In practice, the molecule hits moving targets. This leads to an
increase of collision frequency by a factor of
√
2 and therefore a decrease
in the cross-section by a factor
√
2. The corrected expression for mean
free path is then
λ =
V
4π
√
2r 2 N
(3)
10 Heat and Matter
where x is the total distance travelled and N is the number of collisions.
In terms of frequency f , average time of collision T and the mean molecular
velocity v,
λ = vT =
v
f
=
1000
2 × 10 10 = 5 × 10
−8 m
10.2 (a)
v 2
=
3RT
M
=
3 × 8.31 × 300
28 × 10 −3 = 516.8 m/s
(b) R = kn 0 = 1.38 × 10
−23
× 6.02 × 10
23
= 8.3 J/mol K
N =
PV
RT
=
2 × 1.013 × 10 5 × 1
8.3 × 300
= 81.365 mol/m
3
n = number/m
3
=
6.02 × 10 23 × 81.365
28 × 10 −3
= 2.15 × 10
25
/m
3
λ =
1
σ n
=
1
0.43 × 10 −18 × 2.15 × 10 25 = 1.082 × 10
−7 m
f =
v
λ
=
516.8
1.082 × 10 −7 = 4.78 × 10
9
/s
10.3 (a) Collision cross-section between two molecules each of radius r is equivalent to collision of one molecule of radius 2r with another point size
molecule. Therefore the cross -section will be
σ = π(2r )
2
= 4πr
2
(1)
Consider a rectangular box of face area l m 2 and length λ metres. Then
the volume of the box V = λ m 3 . If n is the number of molecules per
unit volume then λ is such that the total projected area arising from n
molecules will just fill up an area of 1 m 2 .
∴
nλσ = 1
or
λ =
1
nσ
=
V
4πr 2 N
(2)
where we have used (1) and set n = N /V .
Equation (2) in based on the assumption that the target molecules are
stationary. In practice, the molecule hits moving targets. This leads to an
increase of collision frequency by a factor of
√
2 and therefore a decrease
in the cross-section by a factor
√
2. The corrected expression for mean
free path is then
λ =
V
4π
√
2r 2 N
(3)