SME
Physical Chemistry I: Thermodynamics & Kinetics · Problem type
Mean free path and collision frequency
Use this for “Argon atoms (39.95 g/mol) are 340 pm across. Find their mean free path and collision frequency at 300 K and 5000 Pa.”
Diagram
R = 8.314 J/(mol·K).
PV = nRT, which gives 101,325 × 0.02446 = 1 × 8.314 × 298.15 = 2,478.8191.
Each particle drawn stands for 0.05 mol: 20 particles.
This is an example. Type your own values to start a new problem.
d
Molecular diameter
example · pm
σ
Collision cross-section
calculated · m²
T
Temperature
example · K
p
Pressure
example · Pa
M
Molar mass
example · g/mol
λ
Mean free path
calculated · nm
⟨v⟩
Mean speed
calculated · m/s
z
Collision frequency
calculated · s⁻¹
Formulas
σ
=
π
×
(
d
×
10
-12
)
2
4.3008 ×
10
-19
=
π
×
(370 ×
10
-12
)
2
λ
=
1.381 ×
10
-23
×
T
√
2
×
σ
×
p
×
10
9
66.8
=
1.381 ×
10
-23
×
298.15
√
2
× (4.3008 ×
10
-19
) × 101,325
×
10
9
⟨v⟩
=
√(8
×
8.314
×
T
π ×
M
÷ 1,000
)
474.7
=
√(8
×
8.314
×
298.15
π × 28.01 ÷ 1,000
)
z
=
⟨v⟩
÷
(
λ
×
10
-9
)
7.1055 ×
10
9
=
474.7
66.8 ×
10
-9
Assumptions
•
Hard spheres of diameter d: two molecules collide when their centres come within d, so each one sweeps out a cross-section σ = πd².
•
An ideal gas holds p ÷ kBT molecules per cubic metre; the √2 allows for the other molecules moving too.
•
kB = 1.381 × 10⁻²³ J/K and R = 8.314 J/(mol·K); the diameter is in pm (10⁻¹² m) and λ is given in nm (10⁻⁹ m).
Step-by-step
WE KNOW
d = 370 pm, T = 298.15 K, p = 101,325 Pa, M = 28.01 g/mol
FIND
collision cross-section (σ), mean free path (λ), mean speed (⟨v⟩), collision frequency (z)
Step 1 · Find collision cross-section (σ)
Use σ = π × (d × 10⁻¹²)²
The
target
is
a
circle
of
radius
d
around
each
centre,
with
d
in
metres
(pm
×
10
-12
).
σ
=
π
×
(370 ×
10
-12
)
2
σ
=
(1.369
×
10
-19
)
×
π
σ
=
4.3008 ×
10
-19
m
2
Step 2 · Find mean free path (λ)
Use λ = 1.381 × 10⁻²³ × T ÷ (√2 × σ × p) × 10⁹
One
over
the
molecules
met
per
metre
of
path:
√
2
σ
times
p
kBT
of
them
per
cubic
metre.
×
10
9
gives
nm.
λ
=
1.381 ×
10
-23
×
298.15
√
2
× (4.3008 ×
10
-19
) × 101,325
×
10
9
λ
=
(1.381
×
10
-23
)
×
298.15
1.414 × (4.3008 ×
10
-19
) × 101,325
×
10
9
λ
=
(1.381
×
10
-23
)
×
298.15
(6.0813 ×
10
-19
) × 101,325
×
10
9
λ
=
(1.381
×
10
-23
)
×
298.15
6.1619 ×
10
-14
×
10
9
λ
=
4.1175 ×
10
-21
6.1619 ×
10
-14
×
10
9
λ
=
(6.682
×
10
-8
)
×
10
9
λ
=
66.81
nm
Step 3 · Find mean speed (⟨v⟩)
Use ⟨v⟩ = √(8 × 8.314 × T ÷ (π × M ÷ 1,000))
The
mean
speed
of
the
Maxwell
distribution,
√(8R
T
πM
),
with
M
in
kg/mol.
⟨v⟩
=
√(8
×
8.314
×
298.15
π × 28.01 ÷ 1,000
)
⟨v⟩
=
√(8
×
8.314
×
298.15
0.088
)
⟨v⟩
=
√(66.512
×
298.15
0.088
)
⟨v⟩
=
√(
19,830.5528
0.088
)
⟨v⟩
=
√
225
,300
⟨v⟩ = 474.72 m/s
Step 4 · Find collision frequency (z)
Use z = ⟨v⟩ ÷ (λ × 10⁻⁹)
In
one
second
a
molecule
runs
⟨v⟩
metres,
one
collision
every
λ
of
them.
z
=
474.72
66.81 ×
10
-9
z
=
7.1055 ×
10
9
s
-1
Check
Put the numbers back in. Both sides should match.
4.3008 ×
10
-19
=
π
×
(370 ×
10
-12
)
2
✓
66.81
=
1.381 ×
10
-23
×
298.15
√
2
× (4.3008 ×
10
-19
) × 101,325
×
10
9
✓
474.72
=
√(8
×
8.314
×
298.15
π × 28.01 ÷ 1,000
)
✓
7.1055 ×
10
9
=
474.72
66.81 ×
10
-9
✓
Topic check
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