20.2.1 Mean Free Path

A molecule in a gas does not travel in a straight line because it is constantly colliding with other molecules. Instead, as FIGURE 20.3 shows, a molecule follows a zig-zag path in which short linear segments are connected by “kinks” where collisions change the molecule’s speed and direction.

The figure illustrates the path followed by a molecule through a gas.

FIGURE 20.3 Full Alternative Text

A question we could ask is: What is the average distance between collisions? If a molecule has $N_{coll}$ collisions as it travels distance L, the average distance between collisions, which is called the mean free path $λ$ (lowercase Greek lambda), is

λ = \frac{L}{N_{coll}}

(20.1)

FIGURE 20.4a shows two molecules approaching each other. We will assume that the molecules are spherical and of radius r. In that case, the molecules will collide if the distance between their centers is less than 2r. They will miss if the distance is greater than 2r.

The figure labeled (“a”) illustrates two spherical molecules, each of radius r, approaching each other.

The figure labeled (b) illustrates a bent cylinder of radius 2 r with three target molecules.

Full Alternative Text

In FIGURE 20.4b we’ve drawn a cylinder of radius 2r centered on the trajectory of a “sample” molecule. The sample molecule collides with any “target” molecule whose center is located within the cylinder, causing the cylinder to bend at that point. Hence the number of collisions $N_{coll}$ is equal to the number of molecules in a cylindrical volume of length L.

The volume of a cylinder is $V_{cyl} = A L = π {(2 r)}^{2} L$ . If the number density of the gas is $N / V$ particles per $m^{3}$ , then the number of collisions along a trajectory of length L is

N_{coll} = \frac{N}{V} V_{cyl} = \frac{N}{V} π {(2 r)}^{2} L = 4 π \frac{N}{V} r^{2} L

(20.2)

Thus the mean free path between collisions is

λ = \frac{L}{N_{coll}} = \frac{1}{4 π (N / V) r^{2}}

We made a tacit assumption in this derivation that the target molecules are at rest. While the general idea behind our analysis is correct, a more detailed calculation with all the molecules moving introduces an extra factor of $\sqrt{2}$ , giving

λ = \frac{1}{4 \sqrt{2} π (N / V) r^{2}} (mean free path)

(20.3)

Laboratory measurements are necessary to determine atomic and molecular radii, but a reasonable rule of thumb is to assume that atoms in a monatomic gas have $r \approx 0.5 \times 10^{- 10} m$ and diatomic molecules have $r \approx 1.0 \times 10^{- 10} m$ .

EXAMPLE 20.1 The mean free path at room temperature

What is the mean free path of a nitrogen molecule at 1.0 atm pressure and room temperature $(20 ° C) ?$

SOLVE Nitrogen is a diatomic molecule, so $r \approx 1.0 \times 10^{- 10} m$ . We can use the ideal-gas law in the form $p V = N k_{B} T$ to determine the number density:

\frac{N}{V} = \frac{p}{k_{B} T} = \frac{101, 300 Pa}{(1.38 \times 10^{- 23} J/K)(293 K)} = 2.5 \times 10^{25} m^{- 3}

Thus the mean free path is

\begin{matrix} λ & = & \frac{1}{4 \sqrt{2} π (N / V) r^{2}} \\ = & \frac{1}{4 \sqrt{2} π (2.5 \times 10^{25} m^{- 3}) (1.0 \times 10^{- 10} {m)}^{2}} \\ = & 2.3 \times 10^{- 7} m = 230 nm \end{matrix}

REVIEW You learned in Example 18.6 that the average separation between gas molecules at STP is $\approx 4 nm$ . It seems that any given molecule can slip between its neighbors, which are spread out in three dimensions, and travel—on average—about 60 times the average spacing before it collides with another molecule.

STOP TO THINK 20.1

The table shows the properties of four gases, each having the same number of molecules. Rank in order, from largest to smallest, the mean free paths $λ_{A}$ to $λ_{D}$ of molecules in these gases.

Gas	A	B	C	D
Volume	V	2V	V	V
Atomic mass	m	m	2m	m
Atomic radius	r	r	r	2r

20.2.1 Mean Free Path

FIGURE 20.3 A single molecule follows a zig-zag path through a gas.

FIGURE 20.4 A “sample” molecule collides with “target” molecules.

EXAMPLE 20.1 The mean free path at room temperature