Ten small planes are flying at a speed of 150 km/h in total darkness in an air space that is 20 × 20 × 1.5 km$^3$ in volume. You are in one of the planes, flying at random within this space with no way of knowing where the other planes are. On the average about how long a time will elapse between near collision with your plane. Assume for this rough computation that a saftey region around the plane can be approximated by a sphere of radius 10m.

Ten small planes are flying at a speed of 150 km/h in total darkness in an air space that is 20 × 20 × 1.5 km³

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Concept understanding — Molecular Volume Fraction

Molecular Volume Fraction

Kinetic theory begins with a bold assumption: gas molecules are so small compared to the space they roam in that we can treat them as point particles with essentially zero volume. The molecular volume fraction is the number that checks whether that assumption is actually reasonable — it compares the volume genuinely occupied by the molecules themselves to the total volume the gas fills.

The Intuition

Picture a large empty hall with a handful of tennis balls rolling around freely. The balls take up hardly any of the hall's floor space — most of what you see is empty air. That is exactly the picture kinetic theory paints of an ordinary gas: the molecules are the tennis balls, and the container is the hall. The molecular volume fraction simply asks: what fraction of the hall's volume is actually tennis ball?

The Precise Statement

If a gas has $N$ molecules, each modelled as a sphere of radius $r$ (so each molecule has volume $\frac{4}{3}\pi r^{3}$ ), and the gas occupies a total volume $V$ , the molecular volume fraction is:

f = \frac{N \times \frac{4}{3}\pi r^{3}}{V}

This is a pure number (no units) — a ratio of two volumes.

A Worked Example (STP Oxygen)

Take one mole of oxygen gas at STP, where $N = N_A = 6.02 \times 10^{23}$ molecules and the molar volume is $V = 22.4\ \text{L} = 2.24 \times 10^{-2}\ \text{m}^3$ . Taking the molecular diameter to be about $3\ \text{Å} = 3 \times 10^{-10}\ \text{m}$ (radius $r = 1.5 \times 10^{-10}\ \text{m}$ ):

\text{Volume of one molecule} = \frac{4}{3}\pi r^{3} \approx 1.4 \times 10^{-29}\ \text{m}^3

\text{Total molecular volume} = N_A \times 1.4 \times 10^{-29} \approx 8.5 \times 10^{-6}\ \text{m}^3

f = \frac{8.5 \times 10^{-6}}{2.24 \times 10^{-2}} \approx 4 \times 10^{-4}

So the molecules themselves occupy only about 0.04% of the space the gas fills — the remaining 99.96% is empty. This tiny number is exactly why kinetic theory can safely ignore the size of individual molecules for an ordinary gas.

Note

This is also why the mean free path of a gas molecule (the average distance it travels between collisions) is far larger than the molecule's own diameter — there is enormous empty space to cross before the next collision.

Why It Matters: Ideal vs. Real Gas …

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Treating the ten planes as molecules in kinetic theory, a near-collision happens when two 10 m safety spheres touch -- i.e. when plane centres come within $d=20\ \text{m}$ . Using the kinetic-theory mean free path with this collision diameter gives a mean time between near-collisions of about $\boxed{8\times10^{5}\ \text{s}\approx225\ \text{hours}\ (\approx9\ \text{days})}$ .

Modelling the problem as kinetic theory

Ten planes fly at random within a fixed airspace, just like molecules of a gas bouncing around a container. A "near collision" with your plane is exactly analogous to a molecular collision, if we give each plane a collision radius equal to its safety sphere.

Step-by-step

1. Airspace volume and number density.

V = 20\times20\times1.5\ \text{km}^3 = 6\times10^{11}\ \text{m}^3, \qquad n = \frac{N}{V} = \frac{10}{6\times10^{11}} \approx 1.667\times10^{-11}\ \text{m}^{-3}

2. Effective collision diameter.

A near-collision occurs when the two planes' 10 m safety spheres touch -- that is, when their centres come within $d = r_1+r_2 = 10+10 = 20\ \text{m}$ of each other (both planes carry a safety sphere, so the danger radius is the sum of the two, not just one). The collision cross-section is:

\sigma = \pi d^2 = \pi(20)^2 \approx 1{,}257\ \text{m}^2

3. Mean free path (the $\sqrt2$ accounts for the relative motion of the other planes):

l = \frac{1}{\sqrt2\, n\,\sigma} = \frac{1}{\sqrt2\,(1.667\times10^{-11})(1{,}257)} \approx 3.38\times10^{7}\ \text{m}

4. Speed of your plane. …

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Modelling the problem as kinetic theory

Step-by-step

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