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⚛️ NEET Physics❓ Question

Mean free path: where does the √2 come from, and how does it change with temperature?

I made a small list of cases but I think two of them are wrong.

Aarav Kulkarni11th Sample

7 Oct 2026 · 1 min read

I'm making a checklist of all kinetic theory formulas (obviously) and mean free path is the one I'm unsure of.

NCERT gives

ℓ=12 nπd2\ell = \frac{1}{\sqrt{2}\,n\pi d^2}

where nn is number of molecules per unit volume and dd is the diameter. I understand the 1/(nπd2)1/(n\pi d^2) part: a molecule sweeps a cylinder of cross-section πd2\pi d^2 and the number of molecules in the cylinder is nπd2ℓn \pi d^2 \ell, so setting that to 1 gives ℓ\ell. But the book just says 'taking relative motion into account' and a 2\sqrt{2} appears. I'd like to know why 2\sqrt2 and not some other number.

Second, my list says:

  • doubling the temperature at constant volume: ℓ\ell doubles?
  • doubling the pressure at constant temperature: ℓ\ell halves.
  • doubling the diameter: ℓ\ell becomes 1/4.

I'm fairly sure the first one is wrong but I can't tell why. Also what is a typical value of ℓ\ell for air at room conditions? Would help me get a sense of scale.

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6 answers & replies

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  • [+]Arjun Nair12thSample· 12d ago

    The 2\sqrt2: in the simple derivation you assume the target molecules are standing still and only your molecule moves with the average speed vˉ\bar v. But all the molecules are moving. What matters for collisions is the relative speed. For two molecules moving in random directions, the average relative speed works out to 2 vˉ\sqrt2\,\bar v. So the sweeping rate goes up by 2\sqrt2, there are 2\sqrt2 times more collisions per second and the path between collisions shrinks by 2\sqrt2. That's why it sits in the denominator.

  • [+]Rohan Sharma11thSample· 12d ago

    temperature up = molecules faster = they go farther before hitting something. so mean free path increases with T always, your first point is right imo. like a fast bike in traffic covers more distance before the next signal, no? 😅 anyway that's how I picture it

    • [+]Rahul DubeyDropperSample· 12d ago

      Careful, Rohan. Speed being higher doesn't change how many molecules are in the way. ℓ\ell depends only on nn and dd, not on speed. So:

      • constant volume, change T: nn is the same, so ℓ\ell does not change. Aarav, your first point is wrong.
      • constant pressure, raise T: gas expands, n=P/kBTn = P/k_BT falls, so ℓ=kBT2πd2P∝T\ell = \frac{k_BT}{\sqrt2\pi d^2P} \propto T. Here it does increase.

      Same temperature, double the pressure: nn doubles, ℓ\ell halves. Your second point is fine. Diameter doubling gives ℓ→ℓ/4\ell \to \ell/4, also fine.

      • [+]Aarav Kulkarni11thSample· 12d agoAuthor

        this clears it. checklist fixed: ℓ∝1/n\ell \propto 1/n, ∝1/d2\propto 1/d^2, independent of speed. thank you Rahul bhaiya!

  • [+]Lakshmi TripathiDropperSample· 12d ago

    For scale, NCERT works it out for air at room conditions and gets roughly 10−710^{-7} m. A molecule's diameter is about 2×10−102\times10^{-10} m, so the mean free path is several hundred times the molecular diameter. So in the gas, molecules are far apart compared to their size. In a good vacuum it grows to metres. I liked that comparison.

  • [+]Ishaan Bose11thSample· 12d ago

    chemistry guy chiming in: this is the only part of physics that makes me feel smart. I think of a crowded railway platform. More people (n) and fatter people (d) = you bump into someone sooner. Doesn't matter how fast you walk, the crowd density is the same. 🙂

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