What this chapter is about
Kinetic theory explains the behaviour of gases by treating them as collections of a very large number of tiny molecules in constant random motion. Instead of describing a gas only through measurable quantities like pressure, volume and temperature, this chapter connects these bulk properties to the microscopic motion of individual molecules. You learn how pressure arises from countless molecular collisions with container walls, and how temperature is a measure of the average kinetic energy of molecules.
This chapter builds on the ideal gas laws you met earlier and gives them a molecular foundation. You will derive expressions that link pressure to molecular speeds, understand the distribution of speeds among molecules, and see why concepts like mean free path matter. The kinetic approach also helps explain why gases have specific heat capacities and how energy is shared among different degrees of freedom of a molecule.
After studying this chapter you should be able to estimate molecular speeds at room temperature, relate pressure and temperature to molecular kinetic energy, distinguish between monatomic and diatomic gases in terms of energy sharing, and apply the equipartition theorem to predict heat capacities.
Key ideas
- A gas consists of a large number of molecules moving randomly in all directions; intermolecular forces are negligible except during brief collisions.
- Pressure of a gas equals one-third the product of its density and the mean of the squared molecular speeds: P = (1/3) ρ v̄², where v̄² is the mean square speed.
- Temperature is directly proportional to the average translational kinetic energy of molecules: (1/2) m v̄² = (3/2) kB T, where kB is Boltzmann's constant.
- The root-mean-square (rms) speed of molecules is v_rms = √(3 kB T / m) = √(3 R T / M), with M being molar mass.
- The equipartition theorem states that each quadratic term (degree of freedom) in the energy expression of a molecule contributes (1/2) kB T to its average energy.
- Monatomic gases have 3 translational degrees of freedom; diatomic gases (at moderate temperatures) have 3 translational and 2 rotational, giving 5 degrees of freedom.
- Mean free path λ is the average distance a molecule travels between successive collisions: λ = 1 / (√2 π d² n), where d is molecular diameter and n is number density.
Formulas and facts to remember
- Ideal gas equation in molecular form: P V = N kB T, where N is the number of molecules and kB = 1.38 × 10⁻²³ J K⁻¹.
- Pressure from kinetic theory: P = (1/3) (N/V) m v̄² = (1/3) ρ v̄².
- Mean translational kinetic energy per molecule: ⟨KE⟩ = (3/2) kB T.
- Root-mean-square speed: v_rms = √(3 R T / M), with R = 8.314 J mol⁻¹ K⁻¹ and M in kg mol⁻¹.
- Equipartition: each degree of freedom contributes (1/2) kB T per molecule or (1/2) R T per mole.
- Molar heat capacity at constant volume: Cv = (f/2) R, where f is the number of degrees of freedom.
- For monatomic gas (f = 3): Cv = (3/2) R ≈ 12.5 J mol⁻¹ K⁻¹; γ = Cp/Cv = 5/3.
- For diatomic gas (f = 5 at moderate T): Cv = (5/2) R ≈ 20.8 J mol⁻¹ K⁻¹; γ = 7/5 = 1.4.
- Mean free path: λ = kB T / (√2 π d² P).
Worked examples
Example 1: Finding rms speed of nitrogen at room temperature
Problem: Calculate the rms speed of nitrogen molecules (N₂) at 300 K. Take molar mass M = 28 g mol⁻¹.
Solution:
Convert molar mass to SI: M = 28 × 10⁻³ kg mol⁻¹.
Use v_rms = √(3 R T / M).
v_rms = √(3 × 8.314 × 300 / 28 × 10⁻³)
Numerator = 3 × 8.314 × 300 = 7482.6 J mol⁻¹.
v_rms = √(7482.6 / 0.028) = √(2.673 × 10⁵) ≈ 517 m s⁻¹.
The nitrogen molecules in air around you move at roughly 500 m s⁻¹ on average (rms sense).
Example 2: Average kinetic energy of a helium atom
Problem: Find the average translational kinetic energy of one helium atom at 27 °C.
Solution:
Temperature T = 27 + 273 = 300 K.
Average KE = (3/2) kB T = (3/2) × 1.38 × 10⁻²³ × 300
= 6.21 × 10⁻²¹ J.
This tiny energy, multiplied by the enormous number of atoms in a sample, gives measurable thermal energy.
Example 3: Molar heat capacity of oxygen
Problem: Oxygen is diatomic. At room temperature, predict its Cv and Cp, then find γ.
Solution:
Diatomic molecule at moderate temperature has f = 5 (3 translational + 2 rotational).
Cv = (f/2) R = (5/2) × 8.314 = 20.8 J mol⁻¹ K⁻¹.
Cp = Cv + R = 20.8 + 8.314 = 29.1 J mol⁻¹ K⁻¹.
γ = Cp / Cv = 29.1 / 20.8 ≈ 1.4.
Experimental values for oxygen at room temperature match this closely.
Common mistakes
- Forgetting to convert molar mass from g mol⁻¹ to kg mol⁻¹ when using R in SI units → always express M in kg mol⁻¹ for consistency.
- Confusing average speed with rms speed; they differ by a numerical factor → use the correct formula for each quantity.
- Assuming diatomic molecules have only 3 degrees of freedom like monatomic ones → include 2 rotational degrees at room temperature for diatomic gases.
- Writing pressure as P = ρ v̄² instead of P = (1/3) ρ v̄² → remember the factor of one-third from averaging over three perpendicular directions.
- Using Celsius directly in kinetic theory formulas → always convert temperature to kelvin before substituting.
Quick revision
- Pressure arises from molecular collisions; P = (1/3) ρ v̄².
- Temperature measures average translational KE: (3/2) kB T per molecule.
- rms speed ∝ √T and ∝ 1/√M; lighter molecules move faster at the same temperature.
- Equipartition: each degree of freedom adds (1/2) kB T energy per molecule.
- Monatomic Cv = (3/2) R; diatomic Cv = (5/2) R at room temperature.
- Mean free path decreases when pressure rises or molecular size increases.