Kinetic theory of gases for NEET

Kinetic theory is one of the shortest chapters in NEET physics, and one of the most predictable. The questions are mostly ratio problems: what happens to the rms speed, the kinetic energy or the mean free path when you change the temperature, the pressure or the gas. Add degrees of freedom and γ, and you have covered almost everything that gets asked.

The chapter also explains the numbers you used in thermodynamics: why C_v for a monatomic gas is 3R/2 and why γ for air is about 1.4.

Assumptions of the kinetic theory

An ideal gas is modelled as follows:

• It is made of a very large number of identical molecules in random motion.
• The molecules are point masses: their own volume is negligible compared with the volume of the container.
• They exert no forces on each other except during collisions.
• Collisions with each other and with the walls are perfectly elastic and last a negligible time.
• Between collisions, molecules move in straight lines with constant speed.

Real gases behave most like this model at low pressure and high temperature, where molecules are far apart and fast.

The ideal gas equation

PV = nRT = NkT

• n is the number of moles, N the number of molecules, and N = nN_A.
• R = 8.314 J mol⁻¹ K⁻¹, N_A = 6.022 × 10²³ mol⁻¹, and Boltzmann's constant k = R/N_A = 1.38 × 10⁻²³ J K⁻¹.
• In terms of density: P = ρRT/M, where M is the molar mass in kg mol⁻¹.

Avogadro's hypothesis follows directly: equal volumes of all gases at the same temperature and pressure contain the same number of molecules. One mole of an ideal gas occupies about 22.4 L at 273 K and 1 atm.

Pressure and the kinetic meaning of temperature

A molecule of mass m hitting a wall with velocity component v_x reverses it, giving a momentum change of 2mv_x. Adding up all such hits per second per unit area, and using the fact that motion is equally likely in x, y and z, gives

P = (1/3) n m v̄² = (1/3) ρ v_rms²

where n here is the number of molecules per unit volume and v̄² is the mean of the squared speeds.

Rearranging, PV = (2/3) × (total translational KE). Comparing with PV = NkT:

Average translational KE per molecule = (3/2) kT

This is the kinetic meaning of temperature. At the same temperature, the average translational KE of a molecule is the same for every gas, whether it is light hydrogen or heavy oxygen. The lighter molecule simply moves faster.

Molecular speeds

v_rms = √(3RT/M) = √(3kT/m) = √(3P/ρ)

• v_rms ∝ √T at fixed M, and ∝ 1/√M at fixed T.
• At constant temperature, changing the pressure or volume does not change v_rms. Pressure and density change together.

Older books also give two other averages: the mean speed √(8RT/πM) and the most probable speed √(2RT/M). The three are in the ratio v_rms : v_mean : v_mp ≈ 1.73 : 1.60 : 1.41. The syllabus names the rms speed, which is where questions concentrate.

Degrees of freedom and equipartition

A degree of freedom is an independent way a molecule can store energy.

• A monatomic molecule can only translate: 3 degrees of freedom.
• A rigid diatomic molecule translates (3) and rotates about two axes (2): 5 in total. Rotation about the bond axis does not count.
• At high temperatures, vibration adds 2 more (one kinetic, one potential term): 7.
• A rigid non-linear polyatomic molecule has 3 translational and 3 rotational: 6.

Law of equipartition: in thermal equilibrium, each quadratic energy term (each degree of freedom) gets an average of ½kT per molecule. So for f degrees of freedom:

• Internal energy of one mole: U = (f/2) RT
• C_v = (f/2) R, C_p = C_v + R = (f/2 + 1) R
• γ = 1 + 2/f

Mixtures. For n₁ moles of one gas and n₂ of another, add the internal energies: C_v(mix) = (n₁C_v1 + n₂C_v2)/(n₁ + n₂), and C_p(mix) = C_v(mix) + R.

Mean free path

The average distance a molecule travels between collisions is

λ = 1/(√2 π d² n) = kT/(√2 π d² P)

where d is the molecular diameter and n the number per unit volume.

• More molecules per unit volume means more collisions and a shorter free path.
• At constant temperature, doubling the pressure halves λ.
• At constant pressure, raising the temperature lengthens λ (the gas spreads out).

Worked numericals

Example 1: rms speed of nitrogen

Find v_rms of nitrogen (M = 28 g mol⁻¹) at 300 K. Take R = 8.31 J mol⁻¹ K⁻¹.

• 3RT = 3 × 8.31 × 300 = 7,479.
• 3RT/M = 7,479/0.028 ≈ 2.67 × 10⁵.
• v_rms = √(2.67 × 10⁵) ≈ 517 m s⁻¹.

Example 2: matching speeds of two gases

At what temperature will oxygen molecules (M = 32) have the same rms speed as hydrogen molecules (M = 2) at 300 K?

• Equal v_rms means equal T/M.
• T/32 = 300/2, so T = 4,800 K.

Example 3: speed from pressure and density

A gas has density 1.2 kg m⁻³ at a pressure of 1.0 × 10⁵ Pa. Find v_rms.

• v_rms = √(3P/ρ) = √(3 × 10⁵ / 1.2) = √(2.5 × 10⁵) = 500 m s⁻¹.

No temperature or molar mass needed. Questions like this test whether you know the √(3P/ρ) form.

Example 4: average kinetic energy

Find the average translational KE of a gas molecule at 27 °C, in joules and in electronvolts.

• T = 300 K. KE = (3/2)kT = 1.5 × 1.38 × 10⁻²³ × 300 = 6.21 × 10⁻²¹ J.
• In eV: 6.21 × 10⁻²¹ / 1.6 × 10⁻¹⁹ ≈ 0.039 eV.

Example 5: γ of a mixture

One mole of helium is mixed with one mole of oxygen. Find γ for the mixture.

• C_v(mix) = (1 × 3R/2 + 1 × 5R/2)/2 = 2R.
• C_p(mix) = 2R + R = 3R.
• γ = 3R/2R = 1.5.

Practice MCQs

1. At what temperature is the rms speed of a gas twice its value at 300 K? (a) 600 K (b) 900 K (c) 1,200 K (d) 2,400 K
2. γ for helium is: (a) 1.29 (b) 1.33 (c) 1.40 (d) 1.67
3. The ratio of the rms speed of oxygen (M = 32) to that of hydrogen (M = 2) at the same temperature is: (a) 1 : 4 (b) 4 : 1 (c) 1 : 16 (d) 1 : 8
4. A gas is at 27 °C. The temperature at which the average KE of its molecules doubles is: (a) 54 °C (b) 108 °C (c) 327 °C (d) 600 °C
5. The internal energy of 2 moles of a monatomic ideal gas at temperature T is: (a) (3/2)RT (b) 3RT (c) 5RT (d) 6RT
6. The volume of a gas is halved at constant temperature. Its rms speed: (a) doubles (b) halves (c) rises by √2 (d) does not change
7. At constant temperature, the pressure of a gas is doubled. The mean free path: (a) doubles (b) halves (c) stays the same (d) becomes four times
8. C_v for a rigid diatomic ideal gas is: (a) 3R/2 (b) 5R/2 (c) 7R/2 (d) 3R

Answers

1. (c) v ∝ √T, so T must be 4 × 300 = 1,200 K.
2. (d) Monatomic, f = 3, γ = 1 + 2/3 = 5/3.
3. (a) v ∝ 1/√M: √(2/32) = 1/4.
4. (c) KE ∝ T, so T = 600 K = 327 °C. Doubling the Celsius value is the trap.
5. (b) U = n(3/2)RT = 2 × 1.5RT = 3RT.
6. (d) v_rms depends only on T and M.
7. (b) λ ∝ T/P.
8. (b) f = 5, C_v = (5/2)R.

What to do next

• Write the four core relations on one card: PV = nRT, P = (1/3)ρv_rms², KE = (3/2)kT, v_rms = √(3RT/M).
• Rebuild the degrees-of-freedom table from γ = 1 + 2/f without looking.
• Solve ten ratio questions that change T, M, P or V, and say in words why each quantity changes or stays fixed.
• Revise the first law and the process table in thermodynamics for NEET, since γ feeds straight into adiabatic questions.

Next in the syllabus: oscillations and waves.