Kinetic Theory · Lesson 5 of 6
Mean Free Path and Molecular Transport
“Derive the collision length scale, distinguish flight time from collision duration and explain slow spreading despite large thermal speeds.”
• Derive the stationary-target swept-volume estimate using collision cross-section πd². • Apply the relative-motion correction and explain the assumptions behind ℓ =1/(√2πnd²). • Relate mean free path, mean flight time and collision frequency with consistent speed estimates. • Independently calculate the source’s air, water-vapour and nitrogen examples. • Connect molecular collisions to diffusion, viscosity, conduction and low-density limits.
Fast molecules do not imply fast straight-line spreading
A room-temperature molecule may travel hundreds of metres per second, yet gas from a small leak takes time to spread across a room. Between one side and the other it is repeatedly deflected by collisions; its long total path is not the same as its net displacement. Smoke patterns can also remain visible even though air molecules move rapidly. Visible smoke additionally involves suspended particles and air currents, so it is not a direct trace of one air molecule’s ballistic journey.
The mean free path ℓ is the average distance a molecule travels between successive molecular collisions. The mean flight time τ is the average time between those encounters. Collision duration is the much shorter time spent in a close interaction; it is not the same as τ. Molecular size, intermolecular spacing and ℓ are three distinct lengths, and confusing them hides the reason a dilute gas has long free flights.
The collision cylinder has radius d, not d/2
Model identical molecules as hard spheres of diameter d, hence radius d/2. They collide when their centres come within d =d/2+d/2 of one another. To count encounters, one may imagine the moving molecule’s centre as a point and enlarge every stationary target to radius d. This converts a two-sphere collision question into a swept-cylinder geometry problem.
First temporarily hold all targets at rest and let the chosen molecule move at a representative mean speed 〈v〉. In time Δt it travels length 〈v〉Δt and sweeps collision volume πd²〈v〉Δt. Multiplying by target number density n gives the expected encounter count. Dividing by Δt gives the stationary-target collision frequency z₀ =n πd²〈v〉.
Moving targets increase the encounter rate
All molecules actually move. A target approaching the chosen molecule can meet it sooner than a stationary target, while one moving away can meet it later. The collision count therefore depends on their relative speed g, not simply the chosen molecule’s speed. For identical molecules in the classical equilibrium hard-sphere model, a more complete statistical treatment gives mean relative speed 〈g〉 =√2〈v〉. This factor is stated here, as its full velocity-distribution derivation is beyond the chapter.
Check the dimensions: n πd² has unit m⁻¹, so its reciprocal is a length. Multiplying it by a speed produces s⁻¹, as required for collision frequency. The typical molecular speed cancels from ℓ at fixed n and d, but not from the collision frequency or flight time. Faster particles cover the free-flight length more quickly.
Density and size control the free-flight length
Doubling n doubles the number of possible collision targets per volume and halves ℓ. Doubling d quadruples collision cross-section and reduces ℓ by four. For a fixed n in this simple model, changing temperature alone does not alter ℓ if d is treated as constant. At fixed pressure, however, n =P/(kBT) decreases as T rises, so ℓ increases with T. The constraint being held fixed matters.
Problem
Use this approximate gas model with n =2.7×10²⁵ m⁻³, molecular diameter d =2.0×10⁻¹⁰ m and mean speed 〈v〉 =485 m s⁻¹. Find both the initial stationary-target estimate and the relative-motion-corrected mean free path and flight time.
- 1.Stationary targets give ℓ₀ =1/(n πd²) =1/[(2.7×10²⁵)π(2.0×10⁻¹⁰)²] ≈2.95×10⁻⁷ m.
- 2.The associated time is τ₀ =ℓ₀/485 ≈6.08×10⁻¹⁰ s.
- 3.For moving molecules, divide both by √2: ℓ ≈2.08×10⁻⁷ m and τ ≈4.30×10⁻¹⁰ s.
- 4.The corrected collision frequency is z =1/τ ≈2.33×10⁹ s⁻¹. The corrected free path is about 1040 molecular diameters.
- 5.The values near 2.9×10⁻⁷ m and 6.1×10⁻¹⁰ s correspond to the stationary-target approximation. They should not be labelled as evaluations of the corrected formula containing √2.
The water-vapour worked estimate
Problem
Estimate ℓ for water vapour at 373 K and the same 1 atm pressure as the reference gas. Follow the stated effective-diameter model d =2.0×10⁻¹⁰ m and reference number density 2.7×10²⁵ m⁻³ at 273 K.
- 1.At the same pressure, nT is constant for an ideal gas. Thus nvapour = (2.7×10²⁵)(273/373) ≈1.976×10²⁵ m⁻³.
- 2.Use the corrected formula: ℓ =1/[√2π(1.976×10²⁵)(2.0×10⁻¹⁰)²] ≈2.85×10⁻⁷ m, or roughly 3×10⁻⁷ m.
- 3.Without the √2 correction the estimate would instead be about 4.03×10⁻⁷ m. Keeping the two approximations separate explains the numerical difference.
- 4.Compared with the rough vapour spacing of about 4×10⁻⁹ m, the corrected mean free path is about 70 times larger. A molecule passes many average spacings before its direction is randomised by a close encounter.
- 5.The imposed effective collision diameter is a model input. It is not the same measurement as the roughly 2 Å radius estimated from liquid-volume geometry. Use the diameter specified for this collision model rather than silently replacing it with that radius.
A cylinder problem: include diameter, frequency and encounter duration
Problem
Estimate mean free path, collision frequency and the comparison of close-encounter duration with free-flight time for nitrogen at 2.0 atm and 17 °C. Use molecular radius 1.0 Å, M =0.028 kg mol⁻¹, R =8.314 J mol⁻¹ K⁻¹, kB =1.38×10⁻²³ J K⁻¹, 1 atm =1.013×10⁵ Pa and T≈290 K.
- 1.The given 1.0 Å is a radius, so d =2.0×10⁻¹⁰ m. Absolute pressure is 2.026×10⁵ Pa.
- 2.Number density is n =P/(kBT) =2.026×10⁵/[(1.38×10⁻²³)(290)] ≈5.06×10²⁵ m⁻³.
- 3.ℓ =1/(√2πnd²) ≈1.11×10⁻⁷ m, of order 10⁻⁷ m.
- 4.From the temperature relation, vᵣₘₛ =√(3 RT/M) ≈508 m s⁻¹. Using this as a characteristic speed gives a collision-frequency estimate z≈vᵣₘₛ/ℓ≈4.6×10⁹ s⁻¹ and a flight-time estimate τ≈ℓ/vᵣₘₛ≈2.2×10⁻¹⁰ s.
- 5.The exact mean flight-frequency relation uses mean speed 〈v〉 rather than vᵣₘₛ. Here the rms value gives a characteristic estimate, not an exact statistical coefficient.
- 6.A rough real-encounter duration scale is d/vᵣₘₛ≈3.9×10⁻¹³ s, roughly 2×10⁻³ of the flight time. The interaction interval is therefore hundreds of times shorter than a free flight; ideal hard-sphere collisions are treated as instantaneous.
- 7.This calculation answers both parts of the comparison: a molecule spends most of its time flying freely, even though it has billions of encounters each second.
Collisions connect microscopic motion to transport
A molecule’s successive free-flight segments point in different directions. Many segments can cancel in net displacement, so large accumulated travel distance need not imply equally large progress toward the far side of a room. This is the qualitative reason thermal speed cannot be used as the speed of diffusion. A concentration difference allows a net transfer of particles from a more crowded region to a less crowded one even though individual motions remain random.
The same molecular traffic transports other quantities. Molecules moving between regions with different bulk flow velocities transfer momentum, contributing to gas viscosity. Molecules travelling from warmer to cooler regions transport energy, contributing to thermal conduction. Kinetic descriptions can relate those measurable transport properties to molecular mass, size, speed and collision length. Such relations historically helped infer molecular dimensions. Detailed viscosity and conductivity formulas are beyond this chapter’s scope.
In a highly evacuated vessel, the small number density gives a large gas–gas mean free path. It may exceed the vessel or tube length, so molecules can reach a wall before meeting another gas molecule. This does not eliminate wall encounters; it changes which boundaries limit free motion. The expression also predicts why increasing pressure generally shortens free paths and makes straight long-range molecular travel less likely.
Quick Quiz
Quiz
If each identical molecule has diameter d, the collision cross-section is
The centre-to-centre collision distance is the sum of two radii, d. The swept cylinder therefore has radius d.
Relative molecular motion changes the stationary-target mean free path ℓ₀ to
The collision rate is larger by √2 in the stated identical-particle equilibrium model, reducing mean path by √2.
At fixed number density and unchanged effective diameter, raising temperature does what to ℓ in the simple model?
ℓ =1/(√2πnd²) contains n and d, not the molecular speed. At fixed pressure the density changes, giving a different conclusion.
A gas has ℓ =2.0×10⁻⁷ m and mean speed 500 m s⁻¹. Its mean flight time is
τ =ℓ/〈v〉 =2.0×10⁻⁷/500 =4.0×10⁻¹⁰ s.
Why can a gas spread slowly despite high molecular speeds?
Many short flights in changing directions do not add into one long directed path across the room.
Practice Problems
- Derive the stationary-target mean free path from the swept collision cylinder. Explain why its radius is d even though each molecule has radius d/2, and state how relative motion changes the result.
- A gas has number density 2.5×10²⁵ m⁻³, effective molecular diameter 3.0×10⁻¹⁰ m and mean speed 500 m s⁻¹. Find corrected ℓ, mean flight time τ and collision frequency z.
- In the identical hard-sphere model, determine the factor change in ℓ when(a) number density doubles,(b) diameter doubles,(c) pressure halves at fixed T and(d) temperature doubles at fixed pressure. Explain each constraint.
- For n =2.7×10²⁵ m⁻³ and d =2.0×10⁻¹⁰ m, compute ℓ₀ and corrected ℓ. State which one must be used when quoting the formula with √2.
- At 1.0 atm and 300 K, estimate nitrogen number density and corrected ℓ using radius 1.0 Å, kB =1.38×10⁻²³ J K⁻¹ and 1 atm =1.013×10⁵ Pa. Use vᵣₘₛ from M =0.028 kg mol⁻¹ and R =8.314 J mol⁻¹ K⁻¹ to estimate flight time, clearly labelling the speed approximation.
- Explain the difference between mean flight time and collision duration. For ℓ =1.0×10⁻⁷ m, d =2.0×10⁻¹⁰ m and characteristic speed 500 m s⁻¹, estimate both times and their ratio.
- Explain how the same random molecular motion can contribute to particle diffusion, momentum transfer between flowing gas layers and energy transfer between regions of different temperature.
- A gas–gas mean free path becomes much longer than a vacuum tube’s length. Explain what that says about intermolecular collisions, what still happens at the walls and why “no collisions at all” would be an incorrect conclusion.
Key Takeaways
• Mean free path is the average distance between molecular encounters, not the size or spacing of molecules. • The collision cross-section for identical spheres of diameter d is πd². • Relative motion gives ℓ =1/(√2πnd²) in the stated equilibrium hard-sphere model. • Mean flight time is ℓ/〈v〉; collision duration is much shorter, and using rms speed instead gives a characteristic estimate. • Randomly redirected flights explain slow diffusion and connect molecular motion to gas transport properties.