Pressure from Molecular Collisions
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You should be comfortable with Newton’s second law, the definition of momentum, and the basic postulates of the kinetic molecular model of gases. Familiarity with the ideal gas law \(PV=NkT\) is assumed.
Why this matters
Pressure is not a mysterious property of a gas; it is the cumulative result of countless tiny impacts.
Understanding how macroscopic pressure arises from microscopic collisions bridges the gap between the visible world of pistons and tyres and the invisible world of molecular motion. This derivation provides the physical foundation for the ideal gas law, transforming it from an empirical rule into a consequence of mechanics and statistics. It explains why pressure depends on temperature and volume, and why different gases behave similarly under the same conditions, provided they are dilute and at moderate temperatures.
The Physical Picture
Imagine a box filled with a vast number of tiny, hard spheres moving in straight lines at constant speeds. These spheres represent gas molecules. They do not interact with each other except during brief, instantaneous collisions. When a molecule hits a wall of the box, it bounces off elastically, meaning it loses no kinetic energy in the process. However, its direction changes. Because the molecule’s velocity changes, its momentum changes. According to Newton’s second law, a change in momentum implies a force. The wall exerts a force on the molecule to reverse its component of velocity perpendicular to the wall, and by Newton’s third law, the molecule exerts an equal and opposite force on the wall.
Individually, the force from a single molecular collision is infinitesimally small and lasts for an incredibly short time. However, in a macroscopic sample of gas, there are on the order of \(10^{23}\) molecules. At room conditions each square metre of wall is struck roughly \(3 \times 10^{27}\) times per second. The total force on the wall is the sum of all these tiny impulses. Pressure is defined as this total force divided by the area of the wall. Therefore, pressure is a statistical average of the momentum transfer rate per unit area. This perspective reveals that pressure is a direct measure of the intensity of molecular bombardment. It depends on how many molecules are hitting the wall (related to density), how fast they are moving (related to temperature), and how much momentum each one carries (related to mass). This mechanical view allows us to derive the relationship between pressure, volume, and temperature from first principles, rather than relying solely on experimental observation.
Momentum Transfer in a Single Collision
Consider a single molecule of mass \(m\) moving with velocity components \(v_x\), \(v_y\), and \(v_z\). Let us focus on a wall perpendicular to the \(x\)-axis. When the molecule strikes this wall, the component of its velocity perpendicular to the wall, \(v_x\), reverses direction. The components parallel to the wall, \(v_y\) and \(v_z\), remain unchanged because the collision is elastic and the wall is smooth. The initial momentum in the \(x\)-direction is \(p_{x,i} = m v_x\). The final momentum in the \(x\)-direction is \(p_{x,f} = -m v_x\). The change in momentum of the molecule is \(\Delta p_x = p_{x,f} - p_{x,i} = -m v_x - m v_x = -2m v_x\).
By Newton’s third law, the change in momentum of the wall is equal and opposite, so the impulse delivered to the wall is \(+2m v_x\). This impulse occurs over a very short time interval \(\Delta t\). The average force exerted by this single molecule on the wall during the collision is \(F = \frac{\Delta p_{\text{wall}}}{\Delta t} = \frac{2m v_x}{\Delta t}\). However, this force is not continuous; it is a series of discrete spikes. To find the average force over a longer time period \(T\), we must consider how often this molecule hits the wall. If the box has length \(L\) in the \(x\)-direction, the molecule travels a distance \(2L\) between successive collisions with the same wall. The time between collisions is \(\Delta t_{\text{collision}} = \frac{2L}{v_x}\). Therefore, the number of collisions in time \(T\) is \(\frac{T}{\Delta t_{\text{collision}}} = \frac{T v_x}{2L}\). The total impulse delivered to the wall in time \(T\) is the impulse per collision multiplied by the number of collisions: \(J_{\text{total}} = (2m v_x) \left( \frac{T v_x}{2L} \right) = \frac{m v_x^2 T}{L}\). The average force exerted by this single molecule is \(F_{\text{avg}} = \frac{J_{\text{total}}}{T} = \frac{m v_x^2}{L}\). This result shows that the force contribution of a single molecule is proportional to the square of its velocity component perpendicular to the wall.
Summing Contributions from All Molecules
In a real gas, there are \(N\) molecules, each with potentially different velocities. Let the \(i\)-th molecule have an \(x\)-component of velocity \(v_{x,i}\). The total average force exerted on the wall by all molecules is the sum of the individual forces: \(F_{\text{total}} = \sum_{i=1}^{N} \frac{m v_{x,i}^2}{L}\). We can factor out the constants \(m\) and \(L\): \(F_{\text{total}} = \frac{m}{L} \sum_{i=1}^{N} v_{x,i}^2\). The sum \(\sum_{i=1}^{N} v_{x,i}^2\) is related to the mean-square speed in the \(x\)-direction. By definition, the mean-square \(x\)-velocity is \(\langle v_x^2 \rangle = \frac{1}{N} \sum_{i=1}^{N} v_{x,i}^2\). Therefore, \(\sum_{i=1}^{N} v_{x,i}^2 = N \langle v_x^2 \rangle\). Substituting this back into the force equation gives \(F_{\text{total}} = \frac{m}{L} N \langle v_x^2 \rangle\).
Pressure \(P\) is defined as force per unit area. The area of the wall perpendicular to the \(x\)-axis is \(A\). Thus, \(P = \frac{F_{\text{total}}}{A} = \frac{m N \langle v_x^2 \rangle}{L A}\). The volume of the box is \(V = L A\). Therefore, \(P = \frac{m N \langle v_x^2 \rangle}{V}\), or \(PV = m N \langle v_x^2 \rangle\). This equation relates the macroscopic variables \(P\) and \(V\) to the microscopic quantity \(\langle v_x^2 \rangle\). However, we usually know the total speed \(v\), not just the \(x\)-component. The total square speed is \(v^2 = v_x^2 + v_y^2 + v_z^2\). Taking the average over all molecules, \(\langle v^2 \rangle = \langle v_x^2 \rangle + \langle v_y^2 \rangle + \langle v_z^2 \rangle\). Due to the isotropy of the gas (no preferred direction), the average squared velocity components are equal: \(\langle v_x^2 \rangle = \langle v_y^2 \rangle = \langle v_z^2 \rangle\). Therefore, \(\langle v^2 \rangle = 3 \langle v_x^2 \rangle\), which implies \(\langle v_x^2 \rangle = \frac{1}{3} \langle v^2 \rangle\). Substituting this into the pressure equation yields the fundamental result of kinetic theory: \(PV = \frac{1}{3} N m \langle v^2 \rangle\). This shows that pressure is directly proportional to the number of molecules, their mass, and their mean-square speed.
Connecting to Temperature and the Ideal Gas Law
We now have a theoretical expression for pressure: \(PV = \frac{1}{3} N m \langle v^2 \rangle\). We also have the empirical ideal gas law: \(PV = N k T\), where \(k\) is the Boltzmann constant and \(T\) is the absolute temperature. By equating the two expressions for \(PV\), we get \(\frac{1}{3} N m \langle v^2 \rangle = N k T\). We can cancel \(N\) from both sides (assuming \(N\) is non-zero) to obtain \(\frac{1}{3} m \langle v^2 \rangle = k T\). Rearranging this gives \(\frac{1}{2} m \langle v^2 \rangle = \frac{3}{2} k T\). The left-hand side, \(\frac{1}{2} m \langle v^2 \rangle\), is the average translational kinetic energy of a single molecule. This is a profound result: it identifies temperature as a direct measure of the average kinetic energy of the molecules. Specifically, the average kinetic energy per molecule is \(\frac{3}{2} k T\). This explains why temperature is proportional to the intensity of molecular motion. It also explains why the ideal gas law holds for all ideal gases, regardless of their chemical identity: the relationship between \(P\), \(V\), and \(T\) depends only on the number of molecules and their average kinetic energy, not on their specific mass or size, provided they behave ideally.
We can also define the root-mean-square speed, \(v_{\text{rms}}\), as the square root of the mean-square speed: \(v_{\text{rms}} = \sqrt{\langle v^2 \rangle}\). From the equation \(\frac{1}{2} m v_{\text{rms}}^2 = \frac{3}{2} k T\), we can solve for \(v_{\text{rms}}\): \(v_{\text{rms}} = \sqrt{\frac{3 k T}{m}}\). If we work with moles instead of molecules, we use the molar mass \(M\) (mass per mole) and the gas constant \(R\). Since \(m = M/N_A\) and \(k = R/N_A\), we have \(v_{\text{rms}} = \sqrt{\frac{3 (R/N_A) T}{M/N_A}} = \sqrt{\frac{3 R T}{M}}\). This formula allows us to calculate the typical speed of gas molecules. For example, lighter molecules move faster than heavier molecules at the same temperature. This is why helium escapes from balloons faster than air, and why diffusion rates depend on molecular mass. The kinetic theory thus provides a complete microscopic explanation for the macroscopic behaviour of gases, linking the invisible world of atoms to the measurable properties of pressure and temperature.
The result
Kinetic theory of gases
Modelling a gas as a large number of point-like molecules undergoing elastic collisions with a container's walls, the rate of momentum transfer per unit wall area gives \(PV=\tfrac13Nm\langle v^2\rangle\), where \(N\) is the number of molecules, \(m\) their mass, and \(\langle v^2\rangle\) the mean-square speed. Comparing this to the empirically established ideal-gas-law (\(PV=NkT\), with Boltzmann constant \(k=R/N_A\)) identifies temperature directly with average molecular translational kinetic energy: \(\tfrac12m\langle v^2\rangle=\tfrac32kT\) per molecule.
A common misconception is that pressure is caused by the weight of the gas molecules pressing down on the bottom of the container. In reality, in a container at rest, pressure is isotropic; it acts equally on all walls, including the top and sides. The force on the bottom wall is due to molecular collisions, not gravity. Gravity only creates a slight pressure gradient in tall columns of gas, which is negligible in most laboratory settings. Another subtle point is that the derivation assumes elastic collisions. If collisions were inelastic, kinetic energy would be lost, and the gas would cool, violating the assumption of constant temperature. The validity of the ideal gas law relies on these assumptions holding true, which is why real gases deviate from ideality at high pressures or low temperatures where intermolecular forces and molecular volume become significant. C
Worked example
Problem. A 1.00 L flask holds 0.0400 mol of helium (\(M = 4.003\ \mathrm{g\,mol^{-1}}\)) whose root-mean-square speed is 1360 m s-1. Use \(PV = \tfrac13 Nm\langle v^2 \rangle\) to find the pressure, then check it against the ideal gas law.
- Total mass of gas: \(Nm = nM = 0.0400 \times 4.003 \times 10^{-3} = 1.601 \times 10^{-4}\ \mathrm{kg}\).
- Kinetic-theory pressure: \(P = \dfrac{Nm\langle v^2 \rangle}{3V} = \dfrac{1.601 \times 10^{-4} \times 1360^2}{3 \times 1.00 \times 10^{-3}} = 9.87 \times 10^{4}\ \mathrm{Pa}\).
- The temperature implied by the speed: \(T = \dfrac{M v_{\mathrm{rms}}^2}{3R} = \dfrac{4.003 \times 10^{-3} \times 1360^2}{3 \times 8.314} = 296.8\ \mathrm{K}\).
- Ideal gas law at that temperature: \(P = \dfrac{nRT}{V} = \dfrac{0.0400 \times 8.314 \times 296.8}{1.00 \times 10^{-3}} = 9.87 \times 10^{4}\ \mathrm{Pa}\).
- The two routes agree, as they must, since \(\tfrac12 m\langle v^2 \rangle = \tfrac32 kT\) is exactly what links them. The pressure is 98.7 kPa, a little below atmospheric.
Units check. \(\mathrm{kg} \times \mathrm{m^2\,s^{-2}} / \mathrm{m^3} = \mathrm{kg\,m^{-1}\,s^{-2}} = \mathrm{N\,m^{-2}} = \mathrm{Pa}\).
Three quick tests
- What is the relationship between the average kinetic energy of a gas molecule and the absolute temperature?
Answer
The average translational kinetic energy per molecule is directly proportional to the absolute temperature: \(\langle E_k \rangle = \frac{3}{2} k T\). - If the temperature of an ideal gas is doubled while the volume is held constant, what happens to the root-mean-square speed of the molecules?
Answer
The root-mean-square speed increases by a factor of \(\sqrt{2}\), since \(v_{\text{rms}} \propto \sqrt{T}\). - Why does helium diffuse faster than oxygen at the same temperature?
Answer
Helium has a lower molar mass than oxygen. Since \(v_{\text{rms}} = \sqrt{\frac{3RT}{M}}\), a smaller \(M\) results in a higher \(v_{\text{rms}}\), leading to faster diffusion.
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Lecture notes checked line by line by the editor; every number in the worked example was recomputed.