1. Physics
  2. Kinetic Theory of Gases
  3. Mean Free Path and Molecular Collisions

Kinetic Theory of Gases · JEE & NEET Physics

Mean Free Path and Molecular Collisions: notes and previous year questions

How far a molecule flies between collisions, the collision tube, λ = kT/(√2πd²P), collision frequency, vacuum and transport.

Mean Free Path and Molecular Collisions in short

  • Molecules zigzag: the mean free path λ is the average distance between collisions.
  • A molecule sweeps a collision tube of radius d, so its collision cross-section is πd².
  • With still targets λ = 1/(nπd²); because all molecules move, λ = 1/(√2 πnd²).
  • For an ideal gas λ = kT/(√2πd²P): longer at low pressure and high temperature, shorter for big molecules.

1Fast but slow

Perfume molecules fly at about 500 m/s and would cross a 5 m room in a hundredth of a second in a straight line. But each one hits another molecule, changes direction, hits another, billions of times a second. By this zigzag alone a smell would take days to cross a still room; air currents do most of the carrying.

  • Mean free path λ\lambda: the average distance a molecule travels between two collisions.
  • Collision frequency ν\nu: the average number of collisions a molecule makes each second.
  • Mean free time τ\tau: the average time between collisions.
  • Molecular diameter dd: molecules are treated as hard balls, about 2–5 Å across (1 Å = 10−1010^{-10} m).

2The collision tube

Let one molecule fly while the others stand still. Two balls of diameter dd touch when their centres are dd apart, so it hits every molecule whose centre is within dd of its path: it sweeps a tube of radius dd (not d/2d/2), with cross-section πd2\pi d^2.

In time tt the tube has volume πd2vˉt\pi d^2 \bar v t and holds nπd2vˉtn\pi d^2 \bar v t molecules (nn per m³), each a collision. The distance divided by the number of collisions is:

λ=1nπd2\lambda = \frac{1}{n\pi d^2}Simple model with the targets standing still. λ ∝ 1/n and λ ∝ 1/d².

3Everyone moves

The other molecules move too. Two molecules moving at random are, on average, at right angles, so their relative speed is 2\sqrt{2} times their speed: a molecule meets 2\sqrt{2} times as many targets over the same path.

λ=12 πnd2\lambda = \frac{1}{\sqrt{2}\,\pi n d^2}
λ=kT2 πd2P\lambda = \frac{kT}{\sqrt{2}\,\pi d^2 P}Using n = P/kT for an ideal gas.
  • λ∝T/P\lambda \propto T/P: longer when hotter at the same pressure, or at lower pressure.
  • λ∝1/d2\lambda \propto 1/d^2: bigger molecules, shorter path. λ∝1/n\lambda \propto 1/n: more crowded, shorter path.
  • At a fixed nn (a closed rigid box) heating does not change λ\lambda; the molecules just meet their targets sooner.
  • Pressure doubled and kelvin temperature ×4: λ\lambda doubles.

4Numbers for air

With d=2d = 2 Å and n=2.7×1025n = 2.7 \times 10^{25} per m³: λ=1/(2π×2.7×1025×4×10−20)≈2.08×10−7\lambda = 1/(\sqrt{2}\pi \times 2.7 \times 10^{25} \times 4 \times 10^{-20}) \approx 2.08 \times 10^{-7} m. For λ=1\lambda = 1 m at 300 K with d=3d = 3 Å: P=kT/(2πd2λ)≈0.01P = kT/(\sqrt{2}\pi d^2 \lambda) \approx 0.01 Pa (about 10−710^{-7} atm).

5Collision frequency

Follow one molecule: its flights are all different lengths, but total distance ÷ number of collisions settles down to λ\lambda. Covering vˉ\bar v metres each second, it collides vˉ/λ\bar v/\lambda times a second.

ν=vˉλ=2 πnd2vˉ\nu = \frac{\bar v}{\lambda} = \sqrt{2}\,\pi n d^2 \bar v
τ=1ν=λvˉ\tau = \frac{1}{\nu} = \frac{\lambda}{\bar v}
ν=4πd2PkTkTπm\nu = \frac{4\pi d^2 P}{kT}\sqrt{\frac{kT}{\pi m}}Using n = P/kT and v̄ = √(8kT/πm).
Z=12 nνZ = \tfrac{1}{2}\,n\nuCollisions per m³ per second; the ½ because each collision involves two molecules.

6Vacuum

Since λ∝1/P\lambda \propto 1/P, pumping a chamber from 1 atm (about 100 nm) to 10−310^{-3} atm makes the path about 0.1 mm, and at 10−610^{-6} atm about 10 cm, as big as the chamber: molecules fly from wall to wall, hardly meeting.

PressureMean free path (air, roughly)Where
10⁵ Pa (1 atm)70–100 nmthe air around us
10³ Pa10 μmrough vacuum
10⁻¹ Pa10 cmvacuum tubes, thin-film coating
10⁻⁵ Pa1 kmelectron microscopes
10⁻¹⁰ Pa10⁵ kmspace-like vacuum

About 100 km above the ground the air is roughly a million times thinner, and λ\lambda is tens of centimetres. When λ\lambda is much larger than the container (a large Knudsen number λ/L\lambda/L), the gas no longer flows like a fluid; the molecules move one by one.

7Collisions at work

D≈13vˉλ∝T3/2PD \approx \tfrac{1}{3}\bar v\lambda \propto \frac{T^{3/2}}{P}Diffusion: longer flights (low pressure) and faster molecules spread a gas faster.
η≈13ρvˉλ∝mTd2\eta \approx \tfrac{1}{3}\rho\bar v\lambda \propto \frac{\sqrt{mT}}{d^2}Viscosity of a gas.

Since ρ∝n\rho \propto n and λ∝1/n\lambda \propto 1/n, the viscosity of a gas does not depend on its pressure (Maxwell's surprising result), and it grows with temperature, the opposite of a liquid. Thermal conductivity likewise grows as T\sqrt{T}.

Summary

Key ideas

  • Molecules zigzag: the mean free path λ is the average distance between collisions.
  • A molecule sweeps a collision tube of radius d, so its collision cross-section is πd².
  • With still targets λ = 1/(nπd²); because all molecules move, λ = 1/(√2 πnd²).
  • For an ideal gas λ = kT/(√2πd²P): longer at low pressure and high temperature, shorter for big molecules.
  • In a closed rigid box heating does not change λ.
  • Collision frequency ν = v̄/λ and mean free time τ = 1/ν.
  • In air λ ≈ 10⁻⁷ m, a few hundred molecular diameters, and ν ≈ 5 × 10⁹ per second.
  • Put d in metres and n per cubic metre.
  • Pumping a vacuum makes λ grow in proportion to 1/P.
  • Collisions set diffusion and viscosity; gas viscosity is independent of pressure and rises with temperature.

Every equation

Still targets
λ=1/(nπd2)\lambda = 1/(n\pi d^2)
Mean free path
λ=1/(2πnd2)\lambda = 1/(\sqrt{2}\pi n d^2)
With P and T
λ=kT/(2πd2P)\lambda = kT/(\sqrt{2}\pi d^2 P)
Number density
n=N/V=P/kT=NAP/RTn = N/V = P/kT = N_AP/RT
Collision frequency
ν=vˉ/λ=2πnd2vˉ\nu = \bar v/\lambda = \sqrt{2}\pi n d^2 \bar v
Frequency with P
ν=4πd2PkTkT/πm\nu = \tfrac{4\pi d^2 P}{kT}\sqrt{kT/\pi m}
Mean free time
τ=1/ν=λ/vˉ\tau = 1/\nu = \lambda/\bar v
Collisions per volume
Z=12nνZ = \tfrac{1}{2}n\nu
Scaling
λ∝T/P\lambda \propto T/P
Knudsen number
Kn=λ/LKn = \lambda/L
Diffusion
D≈13vˉλD \approx \tfrac{1}{3}\bar v\lambda
Viscosity
η≈13ρvˉλ\eta \approx \tfrac{1}{3}\rho\bar v\lambda

Previous year questions with solutions

Real JEE and NEET questions on mean free path and molecular collisions. Try each one before you open the solution.

Q1JEE Main 2026Numerical answer

An insulated cylinder of volume 60cm360{\mathrm{cm}}^{3} is filled with a gas at 27∘C{27}^{\circ }C and 2 atmospheric pressure. Then the gas is compressed making the final volume as 20cm320{\mathrm{cm}}^{3} while allowing the temperature to rise to 77∘C{77}^{\circ }C. The final pressure is ____\_\_\_\_ atmospheric pressure.

Show answer and solution

Answer: 7

The amount of gas is fixed, so p1V1T1=p2V2T2\dfrac{p_{1}V_{1}}{T_{1}} = \dfrac{p_{2}V_{2}}{T_{2}}, with the temperatures in kelvin: 27 ∘C=300 K27\ {}^{\circ}\mathrm{C} = 300\ \mathrm{K} and 77 ∘C=350 K77\ {}^{\circ}\mathrm{C} = 350\ \mathrm{K}. The volumes can stay in cm3\mathrm{cm^{3}} and the pressures in atmospheres, since only ratios appear.

p2=p1×V1V2×T2T1=2×6020×350300=7 atmp_{2} = p_{1} \times \dfrac{V_{1}}{V_{2}} \times \dfrac{T_{2}}{T_{1}} = 2 \times \dfrac{60}{20} \times \dfrac{350}{300} = 7\ \mathrm{atm}

The trap is forgetting the heating: squeezing to a third of the volume alone gives 6 atm6\ \mathrm{atm}, and the rise to 350 K350\ \mathrm{K} adds the rest. Putting the Celsius values into the ratio, 7727\dfrac{77}{27}, gives about 17 atm17\ \mathrm{atm} — far too much.

Q2NEET 2026One correct option

A flask contains argon and chlorine in the ratio of 2:12:1 by mass. The temperature of the mixture is 27∘C{27}^{\circ }C. The ratio of root mean square speed of the molecules of the two gases (VrmsArVrmsCl)(\frac{V_{\mathrm{rms}}^{\mathrm{Ar}}}{V_{\mathrm{rms}}^{\mathrm{Cl}}}) is:

(Atomic mass of argon =40.0u=40.0u and molecular mass of chlorine =70.0u=70.0u )

  1. A72\frac{\sqrt{7}}{2}
  2. B74\frac{7}{4}
  3. C72\frac{7}{2}
  4. D27\frac{2}{\sqrt{7}}
Show answer and solution

Answer: Option A

The two gases share one temperature, so in vrms=3RTMv_{\mathrm{rms}} = \sqrt{\dfrac{3RT}{M}} only the molar mass differs, and vrms∝1Mv_{\mathrm{rms}} \propto \dfrac{1}{\sqrt{M}}:

vrmsArvrmsCl=MClMAr=7040=74=72\dfrac{v_{\mathrm{rms}}^{\mathrm{Ar}}}{v_{\mathrm{rms}}^{\mathrm{Cl}}} = \sqrt{\dfrac{M_{\mathrm{Cl}}}{M_{\mathrm{Ar}}}} = \sqrt{\dfrac{70}{40}} = \sqrt{\dfrac{7}{4}} = \dfrac{\sqrt{7}}{2}

The 2:12 : 1 ratio of masses in the flask plays no part: how much of each gas there is does not change how fast its molecules move.

The trap is D, 27\dfrac{2}{\sqrt{7}}, the ratio upside down — it would make the heavier chlorine molecules the faster ones. B, 74\dfrac{7}{4}, forgets the square root, and C, 72\dfrac{7}{2}, lets the 2:12 : 1 mass ratio in as well.

Q3JEE Main 2026One correct option

An air bubble of volume 2.9cm32.9{\mathrm{cm}}^{3} rises from the bottom of a swimming pool of 5 m deep. At the bottom of the pool water temperature is 17∘C{17}^{\circ }C. The volume of the bubble when it reaches the surface, where the water temperature is 27∘C{27}^{\circ }C, is ____\_\_\_\_ cm3{\mathrm{cm}}^{3}.

( g=10m/s2g=10 m/s^{2}, density of water =103kg/m3={10}^{3} \mathrm{kg}/m^{3}, and 1 atm pressure is 105Pa{10}^{5} \mathrm{Pa} )

  1. A2.0
  2. B4.2
  3. C3.0
  4. D4.5
Show answer and solution

Answer: Option D

The air in the bubble is a fixed amount of gas, so p1V1T1=p2V2T2\dfrac{p_{1}V_{1}}{T_{1}} = \dfrac{p_{2}V_{2}}{T_{2}}. At the bottom it is pressed by the air above the pool and the water: p1=p0+ρgh=105+103×10×5=1.5×105 Pap_{1} = p_{0} + \rho g h = 10^{5} + 10^{3} \times 10 \times 5 = 1.5 \times 10^{5}\ \mathrm{Pa}. At the surface, p2=105 Pap_{2} = 10^{5}\ \mathrm{Pa}. In kelvin, T1=290 KT_{1} = 290\ \mathrm{K} and T2=300 KT_{2} = 300\ \mathrm{K}.

V2=V1×p1p2×T2T1=2.9×1.5×300290=4.5 cm3V_{2} = V_{1} \times \dfrac{p_{1}}{p_{2}} \times \dfrac{T_{2}}{T_{1}} = 2.9 \times 1.5 \times \dfrac{300}{290} = 4.5\ \mathrm{cm^{3}}

The trap is C, 3.0 cm33.0\ \mathrm{cm^{3}}, which allows for the warming but forgets that the pressure falls as the bubble rises. B, 4.24.2, turns the temperature ratio upside down, and A, 2.02.0, the pressure ratio. A rising bubble must grow: less pressure and more warmth both make it expand.

Practice questions, easy to hard

Three questions from the mean free path and molecular collisions practice ladder: one easy, one medium, one hard.

Q4Numerical answer

In p1V1T1=p2V2T2\dfrac{p_{1}V_{1}}{T_{1}} = \dfrac{p_{2}V_{2}}{T_{2}} only ratios appear, so pressures may stay in atmospheres and volumes in litres or cm3\mathrm{cm^{3}} — as long as both sides use the same unit. Temperatures are different: they must be in kelvin, because the laws say pp and VV are proportional to the absolute temperature.

A rigid steel cylinder holds gas at 2.0 atm2.0\ \mathrm{atm} and 27 ∘C27\ {}^{\circ}\mathrm{C}. It is left in the sun until the gas is at 177 ∘C177\ {}^{\circ}\mathrm{C}. What is the pressure now, in atm?

Show answer and solution

Answer: 3 atm

The volume is fixed, so pT\dfrac{p}{T} is constant. In kelvin, T1=300 KT_{1} = 300\ \mathrm{K} and T2=177+273=450 KT_{2} = 177 + 273 = 450\ \mathrm{K}, so

p2=2.0×450300=3.0 atmp_{2} = 2.0 \times \dfrac{450}{300} = 3.0\ \mathrm{atm}

The trap is using the Celsius values: 2.0×17727≈13 atm2.0 \times \dfrac{177}{27} \approx 13\ \mathrm{atm}. A rise from 300300 to 450 K450\ \mathrm{K} is only half as much again, and so is the pressure.

Q5One correct option

The molar heat capacity at constant volume, CVC_{V}, is the heat that warms 1 mol1\ \mathrm{mol} of gas by 1 K1\ \mathrm{K} with its volume held fixed. Then all the heat goes into internal energy, so from U=f2nRTU = \dfrac{f}{2}nRT,

CV=f2RC_{V} = \dfrac{f}{2}R

At constant pressure the gas expands as it warms and pushes back its surroundings, doing work nRΔTnR\Delta T as it does. That takes extra heat, so the molar heat capacity at constant pressure is larger by RR:

CP=CV+RC_{P} = C_{V} + R (Mayer's relation)

To warm nn moles by ΔT\Delta T takes nCVΔTnC_{V}\Delta T at constant volume and nCPΔTnC_{P}\Delta T at constant pressure. Either way the internal energy rises by nCVΔTnC_{V}\Delta T, because UU depends on the temperature alone.

How much heat warms 2 mol2\ \mathrm{mol} of helium by 10 K10\ \mathrm{K} at constant pressure? Take R=8.3 J mol−1 K−1R = 8.3\ \mathrm{J\,mol^{-1}\,K^{-1}}.

  1. A415 J415\ \mathrm{J}
  2. B249 J249\ \mathrm{J}
  3. C166 J166\ \mathrm{J}
  4. D581 J581\ \mathrm{J}
Show answer and solution

Answer: Option A

Helium is monatomic, f=3f = 3, so CV=32RC_{V} = \dfrac{3}{2}R and CP=52RC_{P} = \dfrac{5}{2}R. The heat is nCPΔT=2×52×8.3×10=415 JnC_{P}\Delta T = 2 \times \dfrac{5}{2} \times 8.3 \times 10 = 415\ \mathrm{J}: 249 J249\ \mathrm{J} of it raises the internal energy and 166 J166\ \mathrm{J} goes into pushing back the surroundings.

The trap is B, 249 J249\ \mathrm{J}, which uses CVC_{V} — right at constant volume, but here the gas also expands. C, 166 J166\ \mathrm{J}, is only the extra part, nRΔTnR\Delta T. D, 581 J581\ \mathrm{J}, uses CP=72RC_{P} = \dfrac{7}{2}R, which belongs to a rigid diatomic gas.

Q6Numerical answer

For one gas, vrms=3RTMv_{\mathrm{rms}} = \sqrt{\dfrac{3RT}{M}} grows as the square root of the absolute temperature.

The rms speed of the molecules of a gas is 400 m/s400\ \mathrm{m/s} at 27 ∘C27\ {}^{\circ}\mathrm{C}. What is it at 327 ∘C327\ {}^{\circ}\mathrm{C}? Give your answer to the nearest whole m/s.

Show answer and solution

Answer: 565.7 m/s

In kelvin the gas goes from 300 K300\ \mathrm{K} to 600 K600\ \mathrm{K}, which doubles TT. So vrms=400×2≈566 m/sv_{\mathrm{rms}} = 400 \times \sqrt{2} \approx 566\ \mathrm{m/s}.

The trap is the Celsius ratio: 400×32727≈1392 m/s400 \times \sqrt{\dfrac{327}{27}} \approx 1392\ \mathrm{m/s}. Another is forgetting the square root and doubling the speed to 800 m/s800\ \mathrm{m/s}.