The Kinetic Molecular Theory describes gas molecules as particles moving continuously in random directions. As these molecules travel through a gas, they undergo frequent collisions with one another. These collisions are responsible for many important phenomena, including diffusion, effusion, transport of energy, and the rates of chemical reactions.
The frequency with which molecules collide depends on three important factors:
If the gas contains more molecules per unit volume, collisions occur more frequently. Likewise, larger molecules present a larger target and therefore collide more often than smaller molecules. Finally, faster-moving molecules encounter one another more frequently than slower-moving molecules.
To model molecular collisions, molecules are often treated as hard spheres. When the centers of two molecules come within one molecular diameter of one another, a collision occurs. The effective target area for a collision is called the collisional cross section, denoted by \(\sigma\).
The collision frequency, \(z\), is the average number of molecular collisions experienced by a molecule per unit time. According to the Kinetic Molecular Theory, the collision frequency is given by
\[ z = \sqrt{2}\, \sigma \left( \frac{N}{V} \right) v \]
where
| Selected Collisional Cross Sections | |
|---|---|
| Molecule | \(\sigma\) (nm2) |
| He | 0.21 |
| Ne | 0.24 |
| N2 | 0.43 |
| CO2 | 0.52 |
| C2H4 | 0.64 |
The collisional cross section, \(\sigma\), is a measure of the effective target area presented by a molecule during a collision. Larger values of \(\sigma\) generally correspond to larger molecules and lead to higher collision frequencies and shorter mean free paths.
This expression shows that collision frequency increases when the gas becomes more dense, when molecules become larger, or when molecular speeds increase.
Because increasing temperature generally increases molecular speeds, hotter gases tend to exhibit larger collision frequencies. Increasing pressure also increases collision frequency because the molecules are packed more closely together.
The mean free path, \(\lambda\), is the average distance a molecule travels between collisions. Rather than describing how often collisions occur, it describes how far a molecule can travel before encountering another molecule.
The mean free path is related to the collision frequency by
\[ \lambda = \frac{v}{z} \]
Substituting the expression for collision frequency yields
\[ \lambda = \frac{1} {\sqrt{2}\, \sigma \left( \frac{N}{V} \right)} \]
This equation reveals that the mean free path depends primarily on the number density and molecular size. Large molecules and dense gases have short mean free paths because collisions occur frequently. Small molecules and dilute gases have longer mean free paths because collisions are less frequent.
Under ordinary atmospheric conditions, the mean free path of a gas molecule is typically on the order of tens of nanometers. Under high-vacuum conditions, however, the mean free path can become several meters or even kilometers long.
Mean free path is an important quantity in many areas of chemistry and physics. Molecular beam experiments, vacuum systems, and gas-phase reaction studies all depend on controlling the frequency of molecular collisions.
If the mean free path is much larger than the dimensions of an apparatus, molecules can travel through the system with very few collisions. Conversely, if the mean free path is very short, molecules undergo many collisions and rapidly exchange energy and momentum with one another.
Big picture: Collision frequency and mean free path describe two complementary aspects of molecular motion. Collision frequency tells us how often molecules collide, while mean free path tells us how far they travel between collisions. Together, these quantities provide a microscopic picture of transport and reactivity in gases.
Calculate the collision frequency of nitrogen molecules, N2, at \(25^\circ\mathrm{C}\) and \(1.00\ \mathrm{atm}\).
The collision frequency is given by
\[ z = \sqrt{2}\, \sigma \left( \frac{N}{V} \right) v \]
The required quantities are:
\[ \sigma = 0.43\ \mathrm{nm^2} \]
\[ \sigma = 0.43\times10^{-18}\ \mathrm{m^2} \]
The number density can be obtained from the Ideal Gas Law:
\[ \frac{N}{V} = \frac{P}{RT}N_A \]
\[ \frac{N}{V} = \frac{ (1.01325\times10^5\ \mathrm{Pa}) } { (8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}) (298.15\ \mathrm{K}) } (6.022\times10^{23}\ \mathrm{mol^{-1}}) \]
\[ \frac{N}{V} = 2.46\times10^{25}\ \mathrm{m^{-3}} \]
From the previous module, the average speed of N2 at \(298\ \mathrm{K}\) is
\[ v_{ave} = 474\ \mathrm{m\,s^{-1}} \]
Substituting these values into the collision-frequency expression:
\[ z = \sqrt{2} \left( 0.43\times10^{-18}\ \mathrm{m^2} \right) \left( 2.46\times10^{25}\ \mathrm{m^{-3}} \right) \left( 474\ \mathrm{m\,s^{-1}} \right) \]
\[ z = 7.08\times10^{9}\ \mathrm{s^{-1}} \]
Therefore, a nitrogen molecule undergoes approximately
\[ z = 7.1\times10^{9} \ \text{collisions per second} \]
Physical interpretation: Although molecules travel hundreds of meters per second, they collide billions of times each second under ordinary atmospheric conditions. This frequent collisional mixing is why gas samples rapidly become thermalized and why the Maxwell-Boltzmann distribution is such a successful description of molecular speeds.
A time-of-flight mass spectrometer has a \(1.00\ \mathrm{m}\) flight tube. What is the maximum pressure that will allow N2 molecules at \(25^\circ\mathrm{C}\) to have a mean free path of at least \(1.00\ \mathrm{m}\)?
The mean free path is
\[ \lambda = \frac{1} {\sqrt{2}\sigma(N/V)} \]
Using the ideal gas relationship \(N/V = P/(k_BT)\), this becomes
\[ \lambda = \frac{k_BT} {\sqrt{2}\sigma P} \]
Solving for pressure:
\[ P = \frac{k_BT} {\sqrt{2}\sigma\lambda} \]
Substitute \(T=298.15\ \mathrm{K}\), \(\sigma=0.43\times10^{-18}\ \mathrm{m^2}\), and \(\lambda=1.00\ \mathrm{m}\):
\[ P = \frac{ (1.380649\times10^{-23}\ \mathrm{J\,K^{-1}}) (298.15\ \mathrm{K}) } { \sqrt{2} (0.43\times10^{-18}\ \mathrm{m^2}) (1.00\ \mathrm{m}) } \]
\[ P = 6.77\times10^{-3}\ \mathrm{Pa} \]
\[ P = 5.08\times10^{-5}\ \mathrm{Torr} \]
Therefore, the pressure must be no greater than about \(6.77\times10^{-3}\ \mathrm{Pa}\), or \(5.08\times10^{-5}\ \mathrm{Torr}\).
Comparison: At \(1.00\ \mathrm{atm}\), the mean free path of N2 is only about
\[ \lambda = 6.68\times10^{-8}\ \mathrm{m} = 66.8\ \mathrm{nm} \]
This is much shorter than a \(1.00\ \mathrm{m}\) flight tube, so a time-of-flight instrument must operate under vacuum conditions to allow molecules to travel through the tube without frequent collisions.
In the time-of-flight experiment, it is important that molecules NOT undergo collisions, which may significantly degrade signal integrity. This is why These instruments must operate at very low pressures.
Choose the correct answer for the randomly generated problem.
Big picture: Collision frequency and mean free path provide a molecular picture of gas behavior. Collision frequency describes how often molecules collide, while mean free path describes how far they travel between collisions. Together, these quantities help explain diffusion, effusion, reaction rates, and the operation of many experimental instruments.