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Physics · Class 11 · Chapter 12

Kinetic Theory

Kinetic theory explains the pressure and temperature of a gas in terms of its molecules rushing about and hitting the walls. It is a short chapter, and the specific-heat results at the end connect it directly to thermodynamics.

In this chapter: molecular nature of matter, the ideal gas equation and gas laws, pressure of an ideal gas from kinetic theory, rms speed, kinetic interpretation of temperature, degrees of freedom and the law of equipartition of energy, specific heats of gases, solids and water, and mean free path.

Molecular nature of matter

Matter is made of molecules that are in constant motion. In a gas at ordinary pressure, the molecules are about ten times farther apart than their own size, so they interact only during collisions and otherwise move in straight lines. Kinetic theory treats a gas as a huge number of such molecules and derives its bulk properties from their motion.

Behaviour of gases

At low pressures and high temperatures, real gases approach ideal behaviour, described by

PV = μRT = NkBTμ = number of moles, N = number of molecules, R = 8.314 J mol−1 K−1, kB = R/NA = 1.38 × 10−23 J K−1, NA = 6.02 × 1023 mol−1.
  • Boyle's law: at constant temperature, PV is constant for a given mass of gas.
  • Charles' law: at constant pressure, V ∝ T.
  • Avogadro's hypothesis: equal volumes of all gases at the same temperature and pressure contain the same number of molecules. One mole of any ideal gas at STP (273 K, 1 atm) occupies V = RT/P = 8.314 × 273/(1.013 × 105) ≈ 22.4 L.
  • Dalton's law of partial pressures: the total pressure of a mixture of non-reacting ideal gases is the sum of their partial pressures.

Real gases deviate from ideal behaviour most at high pressure and low temperature, when the molecules are close together and intermolecular forces matter.

Pressure of an ideal gas

The assumptions: molecules are point masses in random motion, they collide elastically with each other and with the walls, and they exert no forces except during collisions. Consider a molecule of mass m with velocity components (vx, vy, vz) hitting a wall perpendicular to the x axis. After the elastic collision vx reverses, so the momentum given to the wall is 2mvx.

A molecule colliding elastically with a wallwww.iitmedicoguide.comwall of area Abefore: (vx, vy)after: (−vx, vy)lxyMomentum given to the wallin one collision = 2mvxAdding over all molecules:P = ⅓ n m (v²)avgn = molecules per unit volumevy is unchanged; only thecomponent normal to thewall reverseswww.iitmedicoguide.com
Only the velocity component normal to the wall reverses in an elastic collision. Summing the momentum transferred by all molecules per unit time and per unit area gives the pressure.

In time Δt, molecules within a distance vxΔt of the wall can hit an area A, and half of them are moving towards it. Adding their momentum transfers and using the fact that, for random motion, the average of vx2 is one third of the average of v2, gives

P = ⅓ n m v̄2 = ⅓ ρ vrms2PV = ⅔ En is the number of molecules per unit volume, v̄2 the mean of the squared speeds, and E the total translational kinetic energy. So pressure equals two thirds of the kinetic energy per unit volume.

Kinetic interpretation of temperature

Comparing PV = ⅔E with PV = NkBT gives E = (3/2)NkBT. The average translational kinetic energy of a molecule is

½ m v̄2 = (3/2) kBTvrms = √3kBT/m = √3RT/MM is the molar mass in kg mol−1.

So the absolute temperature of a gas is a measure of the average kinetic energy of its molecules. At the same temperature, all ideal gases have the same average kinetic energy per molecule, whatever their mass. Lighter molecules therefore move faster: vrms ∝ 1/√M.

Worked example: Find the rms speed of oxygen molecules at 300 K. By what factor is the rms speed of hydrogen molecules larger at the same temperature? (MO2 = 0.032 kg mol−1, R = 8.314 J mol−1 K−1)
Solution: vrms = √(3 × 8.314 × 300/0.032) = √(2.34 × 105) ≈ 484 m s−1. For hydrogen (M = 0.002 kg mol−1), the ratio is √(32/2) = √16 = 4.

Law of equipartition of energy

The number of independent ways in which a molecule can store energy is its number of degrees of freedom. A monatomic molecule (He, Ar) has 3 translational degrees of freedom. A rigid diatomic molecule (O2, N2) has 3 translational and 2 rotational degrees of freedom.

Law of equipartition of energy: in thermal equilibrium at temperature T, the total energy is shared equally among all energy modes, each quadratic term in the energy having an average value of ½kBT. A vibrational mode has both kinetic and potential energy terms, so it contributes kBT.

Degrees of freedom of a rigid diatomic moleculewww.iitmedicoguide.comx (bond)yzrotation about yrotation about zRigid diatomic molecule3 translational(motion along x, y, z)2 rotational(about y and z; rotationabout the bond is negligible)f = 5, so U = (5/2)RTper mole: Cv = 5R/2,γ = 7/5www.iitmedicoguide.com
A rigid diatomic molecule can move along three axes and rotate about the two axes perpendicular to its bond, giving five degrees of freedom and an average energy of (5/2)kBT.

Specific heat capacity

Using equipartition, the internal energy per mole and hence Cv = dU/dT can be found, with Cp = Cv + R.

GasDegrees of freedomU per moleCvCpγ
Monatomic3(3/2)RT(3/2)R(5/2)R5/3 ≈ 1.67
Diatomic, rigid5(5/2)RT(5/2)R(7/2)R7/5 = 1.4
Diatomic, with vibration5 + one vibrational mode(7/2)RT(7/2)R(9/2)R9/7 ≈ 1.29
Polyatomic, non-linear rigid6 (3 + 3)3RT3R4R4/3 ≈ 1.33

In general, for a gas with f degrees of freedom (counting a vibrational mode as two), Cv = (f/2)R and γ = 1 + 2/f.

For a solid, each atom oscillates in three dimensions, and each direction contributes kBT (kinetic plus potential). So U = 3RT per mole and C = 3R, close to the measured values for many solids at room temperature (the Dulong-Petit law). For water, treating each of the three atoms as a solid-like oscillator gives C ≈ 9R, about 75 J mol−1 K−1, which matches 4186 J kg−1 K−1 × 0.018 kg mol−1. At low temperatures, some modes freeze out and the classical predictions fail; quantum theory explains this.

Mean free path

The average distance a molecule travels between successive collisions is the mean free path:

l = 1√2 n π d2n = number density, d = diameter of a molecule. l increases as the gas becomes less dense.

For air at STP (n ≈ 2.7 × 1025 m−3, d ≈ 2 × 10−10 m), l ≈ 2 × 10−7 m, about a thousand times the size of a molecule. This is why the smell of a perfume takes time to spread across a room even though the molecules move at hundreds of metres per second: each one keeps changing direction.

Common mistakes: (1) Using the molar mass in g mol−1 instead of kg mol−1 in vrms = √(3RT/M). (2) Using Celsius temperature in kinetic theory formulas. (3) Counting 3 rotational degrees of freedom for a diatomic molecule; rotation about the bond axis does not count. (4) Forgetting that a vibrational mode adds kBT, not ½kBT. (5) Thinking heavier gas molecules have more kinetic energy at the same temperature; they have the same average kinetic energy but lower speed.

JEE and NEET focus

  • vrms and its dependence on T and M; ratio problems between two gases.
  • P = ⅓ρvrms2 and the relation between pressure and kinetic energy per unit volume.
  • Degrees of freedom, Cv, Cp and γ for different gases, and for a mixture of gases.
  • Average kinetic energy per molecule, (3/2)kBT, and total energy per mole (f/2)RT.
  • Mean free path and how it changes with pressure, temperature and molecular size.

Practice questions

The average translational kinetic energy of a gas molecule depends on:

  1. Its mass only
  2. The temperature only
  3. The pressure only
  4. The volume only
Show answer
B. It equals (3/2)kBT.

To double the rms speed of the molecules of a gas, its absolute temperature must be:

  1. Doubled
  2. Halved
  3. Made four times
  4. Made √2 times
Show answer
C. vrms ∝ √T.

The ratio of the rms speeds of hydrogen and oxygen molecules at the same temperature is:

  1. 1 : 4
  2. 4 : 1
  3. 1 : 16
  4. 16 : 1
Show answer
B. vrms ∝ 1/√M: √(32/2) = 4.

For a rigid diatomic gas, γ is:

  1. 5/3
  2. 7/5
  3. 9/7
  4. 4/3
Show answer
B. f = 5, so γ = 1 + 2/5.

The pressure of an ideal gas in terms of its total translational kinetic energy per unit volume E is:

  1. E/3
  2. 2E/3
  3. E
  4. 3E/2
Show answer
B. P = ⅓ρv̄2 = ⅔(½ρv̄2).

One mole of a monatomic gas is mixed with one mole of a rigid diatomic gas. The molar Cv of the mixture is:

  1. (3/2)R
  2. 2R
  3. (5/2)R
  4. 4R
Show answer
B. [(3/2)R + (5/2)R]/2 = 2R.

At constant temperature, the pressure of a gas is doubled. Its mean free path:

  1. Doubles
  2. Halves
  3. Stays the same
  4. Becomes four times
Show answer
B. n = P/kBT doubles, and l ∝ 1/n.

According to equipartition, each vibrational mode of a molecule contributes an average energy of:

  1. ½kBT
  2. kBT
  3. (3/2)kBT
  4. 2kBT
Show answer
B. It has both kinetic and potential energy terms.
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