Kinetic Theory explains the behaviour of gases by treating them as a huge number of tiny molecules in constant random motion, linking the large-scale quantities pressure, volume and temperature to molecular motion. It builds the ideal gas equation and gas laws, derives pressure and root-mean-square speed from molecular collisions, shows that temperature is simply a measure of average kinetic energy, and uses the law of equipartition to explain specific heats and the mean free path. It is a high-yield NEET chapter that ties thermodynamics, gas laws and molecular physics together.
Table of Contents
- Key Concepts - Molecular nature of matter, gas laws, kinetic theory postulates, pressure, temperature, RMS speed, degrees of freedom, mean free path
- Weightage in Board & Entrance Exams
- Important Definitions
- Solved Examples
- Important Questions for Board Exams
- Quick Revision Points
Key Concepts
1. Molecular Nature of Matter
All matter is made up of tiny particles - atoms and molecules - that are in continuous motion. This idea, first proposed by John Dalton, explains why matter exists as solids, liquids, and gases depending on how strongly the molecules are held together.
In a solid, molecules are tightly packed and only vibrate. In a liquid, they are loosely bound and can slide past each other. In a gas, molecules are far apart, move freely at high speed, and the intermolecular forces are almost negligible.
Key idea: Gases are the simplest state to model because the molecules barely interact except during collisions - this is exactly why kinetic theory works best for gases.
2. Behaviour of Gases and the Gas Laws
Real gases at low pressure and high temperature behave almost like an ideal gas. The experimental gas laws describe how pressure (P), volume (V), and temperature (T) of a fixed amount of gas are related.
- Boyle’s Law (constant T): PV = constant, so P ∝ 1/V.
- Charles’s Law (constant P): V/T = constant, so V ∝ T.
- Gay-Lussac’s Law (constant V): P/T = constant, so P ∝ T.
- Avogadro’s Law: equal volumes of all gases at the same T and P contain equal numbers of molecules.
[DIAGRAM: P–V curve for Boyle’s law (a rectangular hyperbola) alongside the straight-line V–T graph for Charles’s law passing through −273.15 °C.]
3. Ideal Gas Equation
Combining the three gas laws with Avogadro’s law gives the ideal gas equation, the master equation of this chapter.
PV = nRT
- n = number of moles, R = universal gas constant = 8.314 J mol⁻¹ K⁻¹
- In terms of number of molecules N: PV = NkT, where k is the Boltzmann constant.
- Boltzmann constant k = R/Nₐ = 1.38 × 10⁻²³ J K⁻¹
An ideal gas is one that obeys PV = nRT at all pressures and temperatures. No real gas is perfectly ideal, but most gases come close at low density.
4. Postulates of Kinetic Theory of Gases
Kinetic theory rests on a set of simplifying assumptions about gas molecules. These postulates let us derive gas behaviour from pure mechanics.
- A gas consists of a very large number of identical molecules in random motion.
- The size of a molecule is negligible compared to the average distance between molecules.
- Molecules exert no force on each other except during collisions.
- All collisions (molecule–molecule and molecule–wall) are perfectly elastic, so kinetic energy is conserved.
- The time of a collision is negligible compared to the time between collisions.
- Between collisions, molecules move in straight lines obeying Newton’s laws.
5. Pressure of an Ideal Gas
Pressure arises because molecules continuously strike the walls of the container and transfer momentum. Applying Newton’s laws to these collisions gives the central result of kinetic theory.
P = (1/3)(mN/V) v̄² = (1/3)ρv̄²
- m = mass of one molecule, N = number of molecules, V = volume, ρ = density.
- v̄² is the mean of the squares of molecular speeds (mean-square speed).
This can also be written as PV = (1/3)Nm v̄², directly connecting a macroscopic quantity (pressure) to microscopic motion.
6. Kinetic Interpretation of Temperature
Comparing P = (1/3)(Nm/V)v̄² with PV = NkT gives a remarkable result: temperature is a direct measure of the average kinetic energy of molecules.
(1/2)m v̄² = (3/2)kT
- Average translational KE per molecule = (3/2)kT.
- Average KE per mole = (3/2)RT.
- This energy depends only on temperature, not on the nature, mass, or pressure of the gas.
Key idea: At absolute zero (T = 0 K), molecular translational motion would theoretically cease.
7. RMS Speed of Gas Molecules
The root-mean-square (RMS) speed is the square root of the mean-square speed - a useful single number for the “typical” molecular speed.
v_rms = √(v̄²) = √(3kT/m) = √(3RT/M)
- M = molar mass of the gas.
- v_rms ∝ √T - hotter gas means faster molecules.
- v_rms ∝ 1/√M - lighter molecules (like H₂) move faster than heavier ones (like O₂) at the same temperature.
Two other useful speeds: average speed v_avg = √(8RT/πM) and most probable speed v_mp = √(2RT/M). Their ratio is v_mp : v_avg : v_rms = 1 : 1.128 : 1.224.
8. Degrees of Freedom
The degrees of freedom (f) of a molecule is the number of independent ways it can store energy - that is, the number of independent coordinates needed to describe its motion.
| Type of gas | Translational | Rotational | Total f (room temp) |
|---|---|---|---|
| Monatomic (He, Ar) | 3 | 0 | 3 |
| Diatomic (O₂, N₂) | 3 | 2 | 5 |
| Triatomic (linear, CO₂) | 3 | 2 | 5 |
| Triatomic (non-linear, H₂O) | 3 | 3 | 6 |
At high temperatures, diatomic and polyatomic molecules also gain vibrational degrees of freedom, each contributing 2 to f.
9. Law of Equipartition of Energy
The law of equipartition of energy states that in thermal equilibrium, the total energy is shared equally among all degrees of freedom, and each degree of freedom contributes an average energy of (1/2)kT per molecule.
- Each translational and each rotational degree of freedom → (1/2)kT per molecule.
- Each vibrational mode → 2 × (1/2)kT = kT (it has both kinetic and potential energy terms).
So total internal energy per mole = U = (f/2)RT, where f is the number of degrees of freedom.
10. Specific Heat Capacity of Gases
Using U = (f/2)RT, the molar specific heats of an ideal gas follow directly from the degrees of freedom.
- At constant volume: C_v = (f/2)R
- At constant pressure: C_p = C_v + R = (f/2 + 1)R (Mayer’s relation)
- Ratio: γ = C_p/C_v = 1 + 2/f
| Gas type | f | C_v | C_p | γ |
|---|---|---|---|---|
| Monatomic | 3 | (3/2)R | (5/2)R | 1.67 |
| Diatomic | 5 | (5/2)R | (7/2)R | 1.40 |
| Polyatomic (non-linear) | 6 | 3R | 4R | 1.33 |
11. Mean Free Path
The mean free path (λ) is the average distance a molecule travels between two successive collisions. The more crowded or larger the molecules, the shorter this distance.
λ = 1/(√2 · π d² n)
- d = diameter of a molecule, n = number of molecules per unit volume.
- λ ∝ 1/n, so λ ∝ T/P (using n = P/kT) - at higher temperature λ increases, at higher pressure λ decreases.
- For air at NTP, λ ≈ 10⁻⁷ m.
12. Avogadro’s Number and Avogadro’s Law
Avogadro’s number (Nₐ = 6.022 × 10²³ mol⁻¹) is the number of molecules in one mole of any substance.
Avogadro’s law: equal volumes of all gases under the same conditions of temperature and pressure contain an equal number of molecules. At STP, one mole of any ideal gas occupies 22.4 litres.
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Weightage in Board & Entrance Exams
| Exam | Typical Weightage | Most-Tested Areas |
|---|---|---|
| CBSE Board (Class 11) | 4–5 marks | Postulates, RMS speed, equipartition, specific heats |
| JEE Main / Advanced | 1–2 questions | Pressure derivation, RMS speed, γ, mean free path |
| NEET | 1–2 questions | Kinetic energy–temperature relation, RMS speed, degrees of freedom |
[TABLE: Question-type split - VSA (1 mark): definitions, value of R/k/Nₐ; SA (2–3 marks): RMS speed numericals, C_p − C_v = R, degrees of freedom; LA (5 marks): derivation of pressure of an ideal gas, kinetic interpretation of temperature.]
Important Definitions
| Term | Definition |
|---|---|
| Ideal gas | A gas that obeys PV = nRT at all temperatures and pressures |
| Pressure (kinetic) | Force per unit area from molecular collisions: P = (1/3)ρv̄² |
| Mean-square speed | Average of the squares of molecular speeds, v̄² |
| RMS speed | Square root of mean-square speed: v_rms = √(3RT/M) |
| Boltzmann constant | k = R/Nₐ = 1.38 × 10⁻²³ J K⁻¹ |
| Degrees of freedom | Number of independent ways a molecule can store energy |
| Equipartition of energy | Each degree of freedom carries average energy (1/2)kT per molecule |
| Mean free path | Average distance between successive molecular collisions: λ = 1/(√2 π d² n) |
| Avogadro’s number | Number of molecules in one mole: Nₐ = 6.022 × 10²³ mol⁻¹ |
| Specific heat ratio | γ = C_p/C_v = 1 + 2/f |
Solved Examples
Example 1
Calculate the RMS speed of oxygen molecules at 300 K. (M = 32 g/mol = 0.032 kg/mol, R = 8.314 J mol⁻¹ K⁻¹)
Answer: v_rms = √(3RT/M) = √(3 × 8.314 × 300 / 0.032) = √(233 831) ≈ 484 m/s.
Example 2
At what temperature will the RMS speed of hydrogen molecules be double its value at 300 K?
Answer: Since v_rms ∝ √T, doubling the speed requires T to become 4 times. T = 4 × 300 = 1200 K.
Example 3
Find the average translational kinetic energy of a gas molecule at 27 °C. (k = 1.38 × 10⁻²³ J K⁻¹)
Answer: T = 27 + 273 = 300 K. KE = (3/2)kT = (3/2)(1.38 × 10⁻²³)(300) = 6.21 × 10⁻²¹ J.
Example 4
For a diatomic gas, find C_v, C_p, and γ. (R = 8.314 J mol⁻¹ K⁻¹)
Answer: f = 5, so C_v = (5/2)R = 20.8 J mol⁻¹ K⁻¹, C_p = (7/2)R = 29.1 J mol⁻¹ K⁻¹, and γ = C_p/C_v = 1.4.
Example 5
The RMS speed of a gas at 300 K is 500 m/s. What is its RMS speed at 1200 K?
Answer: v_rms ∝ √T, so v₂ = v₁√(T₂/T₁) = 500 × √(1200/300) = 500 × 2 = 1000 m/s.
Example 6
Calculate the number of molecules in 2 g of hydrogen gas. (Nₐ = 6.022 × 10²³ mol⁻¹, molar mass of H₂ = 2 g/mol)
Answer: Moles n = 2/2 = 1 mol. Number of molecules = n × Nₐ = 1 × 6.022 × 10²³ = 6.022 × 10²³ molecules.
Important Questions for Board Exams
1-Mark Questions (VSA)
- State the value and SI unit of Boltzmann’s constant.
- On what factor does the average kinetic energy of a gas molecule depend?
- Why does a lighter gas diffuse faster than a heavier gas at the same temperature?
- What are the degrees of freedom of a monatomic gas molecule?
- Define mean free path.
2–3-Mark Questions (SA)
- State the postulates of the kinetic theory of gases.
- Derive the relation C_p − C_v = R for an ideal gas.
- Show that the average kinetic energy of a molecule is (3/2)kT and is independent of the nature of the gas.
- Explain how mean free path depends on temperature and pressure.
5-Mark Questions (LA)
- Derive an expression for the pressure exerted by an ideal gas on the walls of its container using kinetic theory.
- State the law of equipartition of energy and use it to find C_v, C_p, and γ for monatomic and diatomic gases.
- Define RMS speed and derive v_rms = √(3RT/M). Hence compare the RMS speeds of two gases of different molar masses at the same temperature.
Quick Revision Points
- Ideal gas equation: PV = nRT = NkT; k = R/Nₐ = 1.38 × 10⁻²³ J K⁻¹
- Pressure of ideal gas: P = (1/3)ρv̄² = (1/3)(Nm/V)v̄²
- Temperature: (1/2)m v̄² = (3/2)kT → average KE depends only on T
- RMS speed: v_rms = √(3RT/M); v_rms ∝ √T and ∝ 1/√M
- Speed ratio: v_mp : v_avg : v_rms = 1 : 1.128 : 1.224
- Degrees of freedom: monatomic 3, diatomic 5, non-linear triatomic 6
- Equipartition: each degree of freedom → (1/2)kT per molecule; U = (f/2)RT
- Specific heats: C_v = (f/2)R, C_p = C_v + R, γ = 1 + 2/f
- γ values: monatomic 1.67, diatomic 1.40, polyatomic 1.33
- Mean free path: λ = 1/(√2 π d² n); λ ∝ T/P; ≈ 10⁻⁷ m for air at NTP
- Avogadro’s number Nₐ = 6.022 × 10²³ mol⁻¹; 1 mole of gas at STP = 22.4 L
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Ideal Gas Equation
One rule links pressure, volume, temperature and amount of gas.
R = 8.314 J mol⁻1K⁻1; k_B = R/N_A = 1.38×10⁻23 J K⁻1; T in kelvin always
- n = moles, N = number of molecules; n = N/N_A
- Use nRT for moles, Nk_BT for molecules
- Real gases behave ideally at low P, high T
The Gas Laws
Each historic gas law is PV = nRT with one quantity frozen.
Density form: P = ρRT/M (ρ = density, M = molar mass)
- Boyle: T fixed, PV = constant
- Charles: P fixed, V ∝ T (kelvin)
- Avogadro: equal V at same P,T ⇒ equal molecules
Postulates of Kinetic Theory
A gas is a huge swarm of point molecules in random motion.
Molecule ≈10⁻10 m wide, gaps ≈10⁻9 m — gas is mostly empty space
- Molecules are point masses; their volume is negligible
- Energy is purely kinetic, zero potential between collisions
- Collisions perfectly elastic: KE and momentum conserved
Kinetic Pressure
Pressure is the steady drumbeat of molecules hitting the walls.
v̄2 = mean square speed; the ⅓ comes from x,y,z sharing the motion
- ρ = density of the gas
- Random motion ⇒ equal pressure on all walls
- Watch the ⅓ — a common slip is writing ½
RMS Speed
The root-mean-square speed is the useful ‘typical’ molecular speed.
Use M in kg/mol with SI R (32 g/mol = 0.032 kg/mol)
- v_rms ∝ √T — quadruple T to double the speed
- v_rms ∝ 1/√M — lighter gases move faster (H2 > O2)
- At fixed T, v_rms depends only on M, not on pressure
Three Molecular Speeds
Most-probable, mean and rms speeds always rank in the same order.
Ratio v_p : v̄ : v_rms ≈ 1.41 : 1.60 : 1.73
- Most probable is the lowest, rms the highest
- All scale as √(T/M) for the same gas
- Order holds because 2 < 8/π < 3
KE & Temperature
Temperature is just a measure of average molecular kinetic energy.
Per molecule; per mole translational KE = (3/2)RT
- Average translational KE depends only on T, not gas type or P
- He and O2 at 300 K have equal mean KE but different speeds
- Monatomic internal energy U = (3/2)nRT = (3/2)Nk_BT
Degrees of Freedom & Specific Heats
Energy shares equally among all the ways a molecule can move.
Mayer’s relation C_P − C_V = R holds for all ideal gases
- Monatomic f=3: C_V=3R/2, γ = 5/3 ≈ 1.67
- Diatomic f=5: C_V=5R/2, γ = 7/5 = 1.4
- Polyatomic f=6: C_V=3R, γ = 4/3 ≈ 1.33
Mean Free Path
Average straight-line distance a molecule travels between collisions.
Air at STP: λ ≈ 10⁻7 m, hundreds of times the molecular size
- λ ∝ 1/n and λ ∝ 1/d2 — denser or bigger ⇒ more collisions
- λ ∝ T at constant P; λ ∝ 1/P at constant T
- Collision frequency = v̄ / λ
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Frequently Asked Questions
It is a model that explains gas behaviour by assuming a gas is made of a very large number of tiny molecules in constant random motion that collide elastically with each other and the walls. It connects macroscopic properties like pressure and temperature to the microscopic motion of these molecules.
Pressure is P = (1/3) (mN/V) times the mean square speed, which can be written as P = (1/3) times density times mean square speed. The factor of one third comes from the molecular motion being shared equally among the x, y and z directions.
The average translational kinetic energy of a molecule is (3/2) k_B T, where k_B is the Boltzmann constant and T is the absolute temperature in kelvin. This means temperature is a direct measure of the average kinetic energy and depends only on T, not on the type of gas or the pressure.
Yes, Kinetic Theory is part of the NEET Physics syllabus and usually carries around 1 to 2 questions. It is high yield because the formulas overlap with Thermodynamics and the Gas Laws, so the concepts get tested indirectly as well.
All three describe molecular speeds for the same gas and always rank as most probable speed, then average speed, then rms speed, in increasing order. Most probable speed is the square root of (2RT/M), average speed is the square root of (8RT/pi M), and rms speed is the square root of (3RT/M), giving an approximate ratio of 1.41 to 1.60 to 1.73.