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Physics · Ch 12 — Kinetic Theory

Kinetic Theory of an Ideal Gas

12.4

Kinetic Theory of an Ideal Gas

The Kinetic Theory of an Ideal Gas

The kinetic theory of gases connects the macroscopic behaviour of a gas — its pressure, temperature, and volume — to the microscopic motion of its molecules. For an ideal gas, we make a set of simplifying assumptions that let us derive precise relationships.

The fundamental assumptions of the kinetic theory for an ideal gas are:

  1. A gas consists of a very large number of identical molecules, treated as point particles — their size is negligible compared to the average distance between them.
  2. The molecules are in constant, random motion, colliding elastically with each other and with the walls of the container. No kinetic energy is lost in these collisions.
  3. The molecules exert no forces on each other except during collisions. Between collisions, they move in straight lines at constant speeds.
  4. The duration of a collision is negligible compared to the time between collisions.
  5. The gas obeys Newton's laws of motion.

These assumptions define the ideal gas — a model that works well for real gases at low densities and high temperatures.

Pressure of an Ideal Gas

Consider a cubical container of side length LL, containing NN molecules of an ideal gas, each of mass mm. Let the total number of molecules be N=nNAN = n N_A, where nn is the number of moles and NAN_A is Avogadro's number.

Take a molecule with velocity components vxv_x, vyv_y, vzv_z. Its speed is v=vx2+vy2+vz2v = \sqrt{v_x^2 + v_y^2 + v_z^2}.

When this molecule collides with a wall perpendicular to the x-axis (say the wall at x=Lx = L), its x-component of velocity reverses from vxv_x to −vx-v_x, while vyv_y and vzv_z remain unchanged. The change in momentum of the molecule is:

Δp=m(−vx)−m(vx)=−2mvx\Delta p = m(-v_x) - m(v_x) = -2mv_x

The impulse imparted to the wall is therefore +2mvx+2mv_x per collision.

The time between successive collisions of this molecule with the same wall is the time needed to travel to the opposite wall and back: Δt=2Lvx\Delta t = \frac{2L}{v_x}.

The force exerted on the wall by this single molecule is the rate of change of momentum:

Fmolecule=2mvxΔt=2mvx2L/vx=mvx2LF_{\text{molecule}} = \frac{2mv_x}{\Delta t} = \frac{2mv_x}{2L/v_x} = \frac{mv_x^2}{L}

The total force on the wall from all NN molecules is the sum over all molecules:

F=mL∑i=1Nvxi2F = \frac{m}{L} \sum_{i=1}^{N} v_{xi}^2

Pressure is force per unit area. The area of the wall is L2L^2, so:

P=FL2=mL3∑i=1Nvxi2=mV∑i=1Nvxi2P = \frac{F}{L^2} = \frac{m}{L^3} \sum_{i=1}^{N} v_{xi}^2 = \frac{m}{V} \sum_{i=1}^{N} v_{xi}^2

where V=L3V = L^3 is the volume of the container.

Now, for any molecule, vi2=vxi2+vyi2+vzi2v_i^2 = v_{xi}^2 + v_{yi}^2 + v_{zi}^2. Because the motion is random and there is no preferred direction, the average values of the squared velocity components are equal:

vx2‾=vy2‾=vz2‾=13v2‾\overline{v_x^2} = \overline{v_y^2} = \overline{v_z^2} = \frac{1}{3}\overline{v^2}

where the bar denotes an average over all molecules. Therefore:

∑i=1Nvxi2=Nvx2‾=N3v2‾\sum_{i=1}^{N} v_{xi}^2 = N \overline{v_x^2} = \frac{N}{3} \overline{v^2}

Substituting this into the pressure expression:

P=mV⋅N3v2‾=13NmVv2‾P = \frac{m}{V} \cdot \frac{N}{3} \overline{v^2} = \frac{1}{3} \frac{Nm}{V} \overline{v^2}

P=13NmVv2‾P = \frac{1}{3} \frac{Nm}{V} \overline{v^2}

This is the fundamental equation for the pressure of an ideal gas from kinetic theory.

Relation Between Pressure and Kinetic Energy

The average translational kinetic energy of a single molecule is:

K‾=12mv2‾\overline{K} = \frac{1}{2} m \overline{v^2}

From the pressure equation:

P=13NmVv2‾=23NV(12mv2‾)=23NVK‾P = \frac{1}{3} \frac{Nm}{V} \overline{v^2} = \frac{2}{3} \frac{N}{V} \left( \frac{1}{2} m \overline{v^2} \right) = \frac{2}{3} \frac{N}{V} \overline{K}

So:

PV=23NK‾PV = \frac{2}{3} N \overline{K}

This is a key result — the product PVPV is directly proportional to the total translational kinetic energy of all molecules.

Root Mean Square Speed

The root mean square (rms) speed is defined as:

vrms=v2‾=v12+v22+⋯+vN2Nv_{\text{rms}} = \sqrt{\overline{v^2}} = \sqrt{\frac{v_1^2 + v_2^2 + \cdots + v_N^2}{N}}

From the ideal gas equation PV=nRT=NkBTPV = nRT = N k_B T (where kB=R/NAk_B = R/N_A is Boltzmann's constant), and using PV=13Nmv2‾PV = \frac{1}{3} N m \overline{v^2}, we get:

13Nmv2‾=NkBT\frac{1}{3} N m \overline{v^2} = N k_B T

v2‾=3kBTm\overline{v^2} = \frac{3 k_B T}{m}

Therefore:

vrms=3kBTm=3RTMv_{\text{rms}} = \sqrt{\frac{3 k_B T}{m}} = \sqrt{\frac{3 R T}{M}}

where M=mNAM = m N_A is the molar mass of the gas.

Watch out

The rms speed is not the same as the average speed. For a given temperature, vrmsv_{\text{rms}} is always slightly larger than the average speed vˉ\bar{v}. Do not confuse them in calculations.

Kinetic Interpretation of Temperature

From PV=23NK‾PV = \frac{2}{3} N \overline{K} and PV=NkBTPV = N k_B T, we obtain:

23NK‾=NkBT\frac{2}{3} N \overline{K} = N k_B T

K‾=32kBT\overline{K} = \frac{3}{2} k_B T

Important

The average translational kinetic energy of a molecule depends only on temperature, not on the mass of the molecule or the type of gas. At the same temperature, all ideal gas molecules have the same average translational kinetic energy.

This is the kinetic interpretation of temperature: temperature is a measure of the average translational kinetic energy of the molecules.

Mixture of Non-Reactive Gases: Dalton's Law from Kinetic Theory …