JEE Main Physics · Chapter 9

Kinetic Theory of Gases

Ideal gas, pressure from molecular motion, molecular speeds, degrees of freedom, equipartition and mean free path.

NCERT Class 11JEE Main: 1 Q per shift 9 min read

1. Chapter at a Glance

The kinetic theory of gases explains pressure and temperature from the motion of molecules. It links the microscopic world (speeds of molecules) with the gas laws we measure. The chapter covers the ideal gas equation, the pressure formula, the meaning of temperature, molecular speeds, degrees of freedom, the law of equipartition of energy, specific heats and mean free path. JEE Main asks about one question, typically on rms speed, degrees of freedom or γ of a mixture.

What JEE asks most
  • rms, average and most probable speeds and how they depend on T and M.
  • Degrees of freedom and equipartition of energy.
  • Cv, Cp and γ for gases and gas mixtures.
  • Mean free path and its dependence on pressure and temperature.

2. Ideal Gas and Its Assumptions

  • Ideal gas equation: PV = nRT = NkT, where R = 8.314 J mol⁻¹ K⁻¹, k = R/NA = 1.38 × 10⁻²³ J K⁻¹.
  • Assumptions: molecules are point-like, in random motion, collide elastically, exert no forces except during collisions, and the collision time is negligible.
  • Real gases behave ideally at low pressure and high temperature.
  • Dalton's law: total pressure of a mixture = sum of partial pressures.

3. Pressure and Kinetic Interpretation of Temperature

P = ⅓ (N m/V) vrms² = ⅓ ρ vrms²    P = ⅔ E (E = translational KE per unit volume)
  • Average translational KE per molecule = (3/2)kT — it depends only on temperature, not on the type of gas.
  • Total translational KE of one mole = (3/2)RT.
  • Absolute zero is the temperature at which (classically) molecular motion would stop.

4. Molecular Speeds

SpeedFormulaRelative size
Most probablevmp = √(2RT/M)smallest (√2)
Averagevavg = √(8RT/πM)middle (√(8/π) ≈ 1.60)
Root mean squarevrms = √(3RT/M) = √(3P/ρ)largest (√3)
T₁ (lower)T₂ > T₁ N(v)v
Maxwell distribution: at higher temperature the peak (vmp) moves to higher speed and the curve spreads out; the area (total number of molecules) stays the same.

5. Degrees of Freedom and Equipartition

Each degree of freedom (independent way of storing energy) gets an average energy ½kT per molecule (½RT per mole).

GasDegrees of freedom fCvCpγ = 1 + 2/f
Monoatomic (He, Ar)33R/25R/25/3 ≈ 1.67
Diatomic, rigid (O₂, N₂)55R/27R/27/5 = 1.4
Diatomic with vibration77R/29R/29/7
Non-linear polyatomic (rigid)63R4R4/3

Internal energy of n moles: U = (f/2)nRT. For a mixture, Cv,mix = (n₁Cv1 + n₂Cv2)/(n₁ + n₂).

6. Mean Free Path

λ = 1/(√2 π d² n) = kT/(√2 π d² P)

n is the number of molecules per unit volume and d the molecular diameter. λ increases when pressure falls or temperature rises (at constant pressure). At constant volume, λ does not change with temperature.

7. Formula Master Table

QuantityFormula
Ideal gasPV = nRT = NkT
PressureP = ⅓ρvrms²
KE per molecule(3/2)kT
rms speed√(3RT/M)
Speed ratiovmp : vavg : vrms = √2 : √(8/π) : √3
Energy per dof½kT
γ1 + 2/f
Mean free pathkT/(√2πd²P)

8. Mind Map

Kinetic Theory
Ideal gas
  • PV = nRT
  • Assumptions
  • Dalton's law
Pressure & T
  • P = ⅓ρv²
  • KE = (3/2)kT
Speeds
  • vmp < vavg < vrms
  • Maxwell curve
Energy
  • Degrees of freedom
  • Equipartition
  • γ, mean free path

9. Tips and Tricks

Shortcuts
  • vrms ∝ √(T/M): to double the speed, make T four times.
  • At the same temperature, all gases have the same average KE per molecule, but lighter molecules move faster.
  • Always use M in kg mol⁻¹ (O₂: 0.032) in speed formulas.
  • For mixtures, add Cv weighted by moles, then Cp = Cv + R.
Mistakes students make
  • Using M in g mol⁻¹, which makes speeds √1000 too large.
  • Thinking rms speed changes with pressure at constant temperature — it does not.
  • Giving diatomic molecules 3 degrees of freedom — rigid diatomic has 5.

10. Important One-liners

  • Temperature is a measure of the average kinetic energy of molecules.
  • vrms is independent of pressure at constant temperature.
  • Equal volumes of gases at the same T and P contain equal numbers of molecules (Avogadro).
  • Monoatomic gases have γ = 5/3.
  • Mean free path is inversely proportional to pressure.
  • Gases deviate from ideal behaviour at high pressure and low temperature.

11. Solved Numericals: Average to Advanced

Five average-level and five advanced-level problems. Click to reveal each answer and its full working. Take R = 8.314 J mol⁻¹ K⁻¹ and k = 1.38 × 10⁻²³ J K⁻¹.

Average level

N1Average

Find the rms speed of oxygen molecules at 300 K (M = 32 g mol⁻¹).

Show answer & solution

Concept: vrms = √(3RT/M) with M in kg mol⁻¹.

  1. M = 0.032 kg mol⁻¹.
  2. vrms = √(3 × 8.314 × 300/0.032) = √(233 831).
  3. ≈ 484 m s⁻¹.

Answer: about 484 m s⁻¹

N2Average

At what temperature will the rms speed of a gas be double its value at 27 °C?

Show answer & solution

Concept: vrms ∝ √T.

  1. T₁ = 300 K; doubling speed needs T₂ = 4T₁.
  2. T₂ = 1200 K = 927 °C.

Answer: 1200 K (927 °C)

N3Average

Find the average translational kinetic energy of a gas molecule at 300 K.

Show answer & solution

Concept: KE = (3/2)kT, the same for every gas.

  1. KE = 1.5 × 1.38 × 10⁻²³ × 300.
  2. = 6.21 × 10⁻²¹ J (about 0.039 eV).

Answer: 6.21 × 10⁻²¹ J

N4Average

Find the total translational kinetic energy of one mole of any ideal gas at 300 K.

Show answer & solution

Concept: per mole, translational KE = (3/2)RT.

  1. E = 1.5 × 8.314 × 300.
  2. ≈ 3741 J.

Answer: about 3.74 kJ

N5Average

How many gas molecules are in 1 litre of air at 1.0 × 10⁵ Pa and 300 K?

Show answer & solution

Concept: PV = NkT ⇒ N = PV/kT.

  1. V = 10⁻³ m³, so PV = 100 J.
  2. N = 100/(1.38 × 10⁻²³ × 300) = 100/(4.14 × 10⁻²¹).
  3. ≈ 2.4 × 10²² molecules.

Answer: about 2.4 × 10²²

Advanced level

N6Advanced

A mixture contains 2 mol of helium and 3 mol of oxygen (treat O₂ as rigid). Find Cv, Cp and γ of the mixture.

Show answer & solution

Concept: Cv,mix = Σn Cv/Σn; Cp = Cv + R.

  1. He: Cv = 1.5R; O₂: Cv = 2.5R.
  2. Cv,mix = (2 × 1.5R + 3 × 2.5R)/5 = 10.5R/5 = 2.1R.
  3. Cp,mix = 3.1R.
  4. γ = 3.1/2.1 ≈ 1.48.

Answer: 2.1R; 3.1R; about 1.48

N7Advanced

Estimate the mean free path of air molecules (diameter 3 × 10⁻¹⁰ m) at 300 K and 1.0 × 10⁵ Pa.

Show answer & solution

Concept: λ = kT/(√2 π d² P).

  1. Numerator kT = 1.38 × 10⁻²³ × 300 = 4.14 × 10⁻²¹ J.
  2. Denominator = 1.414 × 3.1416 × 9 × 10⁻²⁰ × 10⁵ ≈ 4.0 × 10⁻¹⁴.
  3. λ ≈ 1.0 × 10⁻⁷ m (about 300 molecular diameters).

Answer: about 1.0 × 10⁻⁷ m

N8Advanced

(a) Find the ratio of rms speeds of hydrogen and oxygen molecules at the same temperature. (b) At what temperature will oxygen molecules have the same rms speed as hydrogen molecules at 300 K?

Show answer & solution

Concept: vrms ∝ √(T/M); M(H₂) = 2, M(O₂) = 32 g mol⁻¹.

  1. (a) vH/vO = √(32/2) = √16 = 4.
  2. (b) Need TO/32 = 300/2.
  3. TO = 16 × 300 = 4800 K.

Answer: (a) 4 : 1; (b) 4800 K

N9Advanced

Calculate vmp, vavg and vrms for nitrogen (M = 28 g mol⁻¹) at 300 K and verify their ratio.

Show answer & solution

Concept: use the three formulas with M = 0.028 kg mol⁻¹; RT/M = 8.314 × 300/0.028 ≈ 89 079.

  1. vmp = √(2 × 89 079) ≈ 422 m s⁻¹.
  2. vavg = √(8 × 89 079/π) ≈ 476 m s⁻¹.
  3. vrms = √(3 × 89 079) ≈ 517 m s⁻¹.
  4. Ratio 422 : 476 : 517 ≈ 1 : 1.13 : 1.22 = √2 : √(8/π) : √3 ✓.

Answer: about 422, 476 and 517 m s⁻¹

N10Advanced

An insulated container of helium moves at 100 m s⁻¹ and is suddenly stopped. If all the ordered kinetic energy becomes internal energy of the gas, find the rise in temperature (M = 4 g mol⁻¹).

Show answer & solution

Concept: ½Mv² per mole = ΔU = n(f/2)RΔT with f = 3.

  1. Per mole: ½ × 0.004 × 100² = 20 J.
  2. ΔU per mole = (3/2) × 8.314 × ΔT.
  3. ΔT = 20/(12.47) ≈ 1.6 K (equivalently Mv²/(fR)).

Answer: about 1.6 K

12. PYQ-pattern Questions with Solutions

These questions are original and modelled on concepts frequently tested in JEE Main.

Q1Easy

The number of degrees of freedom of a rigid diatomic molecule is:

  1. 3
  2. 5
  3. 6
  4. 7
Show solution

3 translational + 2 rotational = 5.

Answer: (B)

Q2Easy

The average energy associated with each degree of freedom of a molecule at temperature T is:

  1. kT
  2. ½kT
  3. (3/2)kT
  4. 2kT
Show solution

Law of equipartition: ½kT per degree of freedom.

Answer: (B)

Q3Medium

If the pressure of a gas is doubled at constant temperature, its rms speed:

  1. Doubles
  2. Halves
  3. Remains the same
  4. Becomes √2 times
Show solution

vrms = √(3RT/M) depends only on T and M. (Density doubles along with pressure, keeping P/ρ constant.)

Answer: (C)

Q4Hard

The rms speed of oxygen molecules at temperature T is v. If the temperature is made 4T and all molecules dissociate into atoms, the rms speed becomes:

  1. 2v
  2. 2√2 v
  3. 4v
  4. √2 v
Show solution

v ∝ √(T/M): T becomes 4T and M becomes M/2, so the factor is √(4 × 2) = 2√2.

Answer: (B)

Q5Numerical

Find the ratio of rms speeds of H₂ and O₂ molecules at the same temperature.

Show solution

√(32/2) = 4.

Answer: 4

Q6Medium

The pressure exerted by a gas of density ρ and rms speed v is:

  1. ρv²
  2. ⅓ρv²
  3. ½ρv²
  4. ⅔ρv²
Show solution

From kinetic theory, P = ⅓ρvrms².

Answer: (B)

13. Quick Revision Summary

Remember before the exam
  • PV = nRT = NkT; P = ⅓ρvrms².
  • KE per molecule (3/2)kT; per mole (3/2)RT.
  • vrms = √(3RT/M) > vavg > vmp.
  • ½kT per degree of freedom; γ = 1 + 2/f.
  • λ = kT/(√2πd²P).
All Physics chaptersJEE Main Mock Tests