Class 11 Physics · NCERT Chapter 12
~27 min readKinetic Theory Class 11 Physics Notes
Complete, exam-ready notes on the kinetic theory of gases: the assumptions and model, the ideal gas equation and its laws, the pressure exerted by an ideal gas, the kinetic interpretation of temperature, root-mean-square speed, degrees of freedom, the law of equipartition of energy and specific heats of gases, and the Maxwell speed distribution — every NCERT topic with MCQs, mark-wise questions and solved numericals for CBSE, JEE and NEET.
What is the kinetic interpretation of temperature?
The average kinetic energy of a gas molecule is directly proportional to its absolute temperature: (½mv²)_avg = (3/2)k_B T. Temperature is thus a measure of the average translational kinetic energy of the molecules — at higher temperature molecules move faster on average.
Ideal Gas and Its Assumptions
Kinetic theory assumptions
The kinetic theory models a gas as a large number of identical molecules. The assumptions are: (1) molecules are in random motion, (2) the size of molecules is negligible compared with the distance between them, (3) collisions are perfectly elastic and the time of collision is negligible, and (4) the only interactions occur during collisions (no intermolecular forces otherwise).
Ideal gas
An ideal gas obeys the ideal gas equation PV = nRT exactly. Its molecules occupy no volume and exert no intermolecular forces, so its internal energy depends only on temperature. Real gases approach ideal behaviour at low pressure and high temperature.
The ideal gas equation combines Boyle's law (P ∝ 1/V at constant T), Charles' law (V ∝ T at constant P) and Avogadro's law (V ∝ n at constant P and T). R = 8.31 J mol⁻¹ K⁻¹ is the universal gas constant, and k_B = R/N_A ≈ 1.38 × 10⁻²³ J K⁻¹ is the Boltzmann constant.
- Boyle's law: PV = constant at constant temperature.
- Charles' law: V/T = constant at constant pressure.
- Avogadro's law: equal volumes of gases at the same P and T contain equal numbers of molecules.
Real gases become ideal
MCQ Questions
1 Mark Questions
- 1State the main assumptions of the kinetic theory of gases.
- 2Write the ideal gas equation and define each symbol.
- 3State Boyle's law.
2 Mark Questions
- 1What is an ideal gas? Under what conditions does a real gas approach ideal behaviour?
- 2State Boyle's law, Charles' law and Avogadro's law and show how they combine into the ideal gas equation.
- 3Find the number of moles in 44 g of CO₂. (Molar mass = 44 g mol⁻¹.)
3 Mark Questions
- 1State the postulates of the kinetic theory of gases and explain why the ideal gas equation holds at low pressure and high temperature.
- 2A gas occupies 2 m³ at a pressure of 10⁵ Pa and a temperature of 300 K. Find the number of moles and the number of molecules. (R = 8.31 J mol⁻¹ K⁻¹, N_A = 6.02 × 10²³.)
Pressure Exerted by an Ideal Gas
Pressure of a gas
The pressure of a gas arises from the collisions of its molecules with the walls of the container. Each collision transfers momentum 2mv to a wall (for a molecule moving perpendicular to it), and the rate of momentum transfer per unit area is the pressure.
Using the average of the square of the molecular speed v², the pressure is P = (1/3)ρv²_avg, where ρ = Nm/V is the density. This derivation connects the macroscopic quantity pressure with the microscopic motion of the molecules.
- Pressure depends on the mean-square speed of the molecules, not their individual speeds.
- The same formula holds in three dimensions because each of the three velocity components contributes equally on average.
- Doubling the mean-square speed doubles the pressure at constant volume.
P = ⅓ρv² is rigorous for an ideal gas
MCQ Questions
1 Mark Questions
- 1Explain how pressure arises in the kinetic theory of gases.
- 2Write the expression P = (1/3)ρv²_avg and identify the symbols.
- 3On which molecular quantity does gas pressure depend?
2 Mark Questions
- 1Derive the relation P = (1/3)ρv²_avg from the kinetic theory model.
- 2The density of a gas is 1.5 kg m⁻³ and its molecules have a mean-square speed of 10⁶ m² s⁻². Find the pressure it exerts.
- 3Show that the pressure of a gas is directly proportional to its density at constant mean-square speed.
3 Mark Questions
- 1Use the kinetic theory to derive an expression for the pressure exerted by an ideal gas in terms of the mass, number density and mean-square speed of the molecules.
- 2A gas has a density of 2 kg m⁻³ and exerts a pressure of 2 × 10⁵ Pa. Find the root-mean-square speed of its molecules.
Kinetic Interpretation of Temperature
Temperature and molecular energy
Combining the ideal gas equation with P = (1/3)ρv²_avg shows that the average translational kinetic energy per molecule is (1/2)mv²_avg = (3/2)k_B T. Thus the temperature of a gas is a direct measure of the average kinetic energy of its molecules.
At a given temperature all gases, regardless of molecular mass, have the same average translational kinetic energy per molecule (3/2)k_B T. Lighter molecules move faster to have the same energy — at the same temperature, hydrogen molecules are much faster than oxygen molecules.
- Temperature is proportional to the mean translational kinetic energy, not the total energy.
- At absolute zero all translational motion ceases in the classical model.
- Two gases at the same temperature have equal average translational KE per molecule.
Same T, same average KE
MCQ Questions
1 Mark Questions
- 1State the kinetic interpretation of temperature.
- 2Write the expression for the average kinetic energy of a gas molecule.
- 3Two gases at the same temperature: what is equal for their molecules?
2 Mark Questions
- 1Show that the average kinetic energy per molecule of a gas equals (3/2)k_B T.
- 2Find the average translational kinetic energy of a molecule at 300 K. (k_B = 1.38 × 10⁻²³ J K⁻¹.)
- 3Why is the average translational kinetic energy of all gases the same at a given temperature?
3 Mark Questions
- 1Derive the kinetic interpretation of temperature, starting from P = (1/3)ρv²_avg and the ideal gas equation.
- 2Compute the total translational kinetic energy of 2 moles of an ideal gas at 400 K. (R = 8.31 J mol⁻¹ K⁻¹.)
Root-Mean-Square Speed
RMS speed
The root-mean-square speed is the square root of the mean of the squares of the molecular speeds: v_rms = √(v²_avg). From the kinetic interpretation, v_rms = √(3k_B T/m) = √(3RT/M), where M is the molar mass.
The RMS speed is proportional to the square root of temperature and inversely proportional to the square root of molar mass. At 300 K, the RMS speed of nitrogen molecules (M = 28 g/mol) is about 517 m/s. For a given gas, raising the temperature increases the RMS speed, and for a given temperature, lighter gases have higher RMS speeds.
- v_rms ∝ √T at fixed mass.
- v_rms ∝ 1/√M at fixed temperature.
- The RMS speed is slightly larger than the mean speed and the most probable speed.
Remember the ordering of speeds
MCQ Questions
1 Mark Questions
- 1Define the root-mean-square speed of gas molecules.
- 2Write the expression for v_rms in terms of the molar mass M.
- 3How does v_rms depend on temperature?
2 Mark Questions
- 1Find the RMS speed of oxygen molecules at 300 K. (M = 32 × 10⁻³ kg mol⁻¹, R = 8.31 J mol⁻¹ K⁻¹.)
- 2Compare the RMS speeds of hydrogen and oxygen at the same temperature.
- 3If the temperature of a gas is raised from 300 K to 1200 K, by what factor does its RMS speed increase?
3 Mark Questions
- 1Derive an expression for the root-mean-square speed of gas molecules in terms of T and molar mass.
- 2At what temperature will the RMS speed of oxygen molecules equal that at which hydrogen has an RMS speed of 2000 m/s at 300 K? Derive your answer.
Degrees of Freedom and Specific Heats
Degrees of freedom
The number of degrees of freedom of a gas molecule is the number of independent ways it can possess energy. A monatomic molecule has 3 translational degrees of freedom. A diatomic molecule has 3 translational, 2 rotational and (at high temperature) 2 vibrational degrees of freedom.
Law of equipartition of energy
The law of equipartition of energy states that in thermal equilibrium, energy is shared equally among all degrees of freedom, each contributing (1/2)k_B T per molecule (or (1/2)RT per mole). Hence a monatomic gas has internal energy (3/2)nRT and C_V = (3/2)R.
For a monatomic gas C_V = 3R/2 and γ = 5/3. For a diatomic gas (translational + rotational) C_V = 5R/2 and γ = 7/5. At very high temperatures the vibrational degrees are also excited and C_V rises.
- Monatomic (He, Ar): f = 3, C_V = 3R/2, γ = 5/3.
- Diatomic (N₂, O₂) without vibration: f = 5, C_V = 5R/2, γ = 7/5.
- Each degree of freedom contributes (1/2)R per mole to C_V.
Which degrees are active?
MCQ Questions
1 Mark Questions
- 1Define the degrees of freedom of a gas molecule.
- 2State the law of equipartition of energy.
- 3Write the value of C_V for a monatomic gas.
2 Mark Questions
- 1Find the number of degrees of freedom of (a) monatomic and (b) diatomic gas molecules.
- 2State the law of equipartition of energy and hence write C_V and C_P for a monatomic gas.
- 3For a diatomic gas, using f = 5, find C_V, C_P and γ.
3 Mark Questions
- 1State the law of equipartition of energy and use it to obtain the specific heats and γ for a monatomic and a diatomic gas.
- 2Compute the internal energy of 3 moles of a monatomic gas at 400 K. (R = 8.31 J mol⁻¹ K⁻¹.)
The Maxwell Speed Distribution
Maxwell distribution
In a gas, molecules do not all move at one speed; their speeds follow the Maxwell speed distribution. The fraction of molecules with speed between v and v + dv depends on temperature and molecular mass. The distribution has a maximum at the most probable speed v_mp = √(2k_B T/m).
The three speeds are related by v_mp : v_mean : v_rms = √2 : √(8/π) : √3 ≈ 1 : 1.128 : 1.225. As temperature rises, the distribution broadens and shifts to higher speeds, and the most probable, mean and RMS speeds all increase.
- The most probable speed is the speed possessed by the largest number of molecules.
- The distribution spreads out with increasing temperature.
- At the same temperature, lighter gases have their distribution peak at higher speeds.
The three characteristic speeds
MCQ Questions
1 Mark Questions
- 1What is the Maxwell speed distribution?
- 2Define the most probable speed.
- 3Write the expressions for the three characteristic speeds of a gas.
2 Mark Questions
- 1Find the most probable speed of nitrogen molecules at 300 K. (M = 28 × 10⁻³ kg mol⁻¹, R = 8.31 J mol⁻¹ K⁻¹.)
- 2Explain why the Maxwell distribution broadens as the temperature increases.
- 3State the ordering v_mp < v_mean < v_rms and explain it briefly.
3 Mark Questions
- 1Describe the Maxwell speed distribution and explain the physical meaning of the most probable, mean and RMS speeds.
- 2Compute the most probable, mean and RMS speeds of argon atoms at 300 K. (M = 40 × 10⁻³ kg mol⁻¹, R = 8.31 J mol⁻¹ K⁻¹.)
NCERT Exercise Questions
From the textbook
Exercises (NCERT Q12.1–Q12.10)
- 12.1) Estimate the fraction of molecular volume to the actual volume occupied by oxygen gas at STP. Take the diameter of an oxygen molecule to be 3 Å.
- 12.2) Molar volume is the volume occupied by 1 mol of any (ideal) gas at standard temperature and pressure (STP: 1 atmospheric pressure, 0 °C). Show that it is 22.4 litres.
- 12.3) Figure 12.8 shows plot of PV/T versus P for 1.00 × 10⁻³ kg of oxygen gas at two different temperatures. (a) What does the dotted plot signify? (b) Which is true: T₁ > T₂ or T₁ < T₂? (c) What is the value of PV/T where the curves meet on the y-axis? (d) If we obtained similar plots for 1.00 × 10⁻³ kg of hydrogen, would we get the same value of PV/T at the point where the curves meet on the y-axis? If not, what mass of hydrogen yields the same value of PV/T (for low pressure high temperature region of the plot)? (Molecular mass of H₂ = 2.02 u, of O₂ = 32.0 u, R = 8.31 J mol⁻¹ K⁻¹.)
- 12.4) An oxygen cylinder of volume 30 litre has an initial gauge pressure of 15 atm and a temperature of 27 °C. After some oxygen is withdrawn from the cylinder, the gauge pressure drops to 11 atm and its temperature drops to 17 °C. Estimate the mass of oxygen taken out of the cylinder (R = 8.31 J mol⁻¹ K⁻¹, molecular mass of O₂ = 32 u).
- 12.5) An air bubble of volume 1.0 cm³ rises from the bottom of a lake 40 m deep at a temperature of 12 °C. To what volume does it grow when it reaches the surface, which is at a temperature of 35 °C?
- 12.6) Estimate the total number of air molecules (inclusive of oxygen, nitrogen, water vapour and other constituents) in a room of capacity 25.0 m³ at a temperature of 27 °C and 1 atm pressure.
- 12.7) Estimate the average thermal energy of a helium atom at (i) room temperature (27 °C), (ii) the temperature on the surface of the Sun (6000 K), (iii) the temperature of 10 million kelvin (the typical core temperature in the case of a star).
- 12.8) Three vessels of equal capacity have gases at the same temperature and pressure. The first vessel contains neon (monatomic), the second contains chlorine (diatomic), and the third contains uranium hexafluoride (polyatomic). Do the vessels contain equal number of respective molecules? Is the root mean square speed of molecules the same in the three cases? If not, in which case is v_rms the largest?
- 12.9) At what temperature is the root mean square speed of an atom in an argon gas cylinder equal to the rms speed of a helium gas atom at −20 °C? (atomic mass of Ar = 39.9 u, of He = 4.0 u).
- 12.10) Estimate the mean free path and collision frequency of a nitrogen molecule in a cylinder containing nitrogen at 2.0 atm and temperature 17 °C. Take the radius of a nitrogen molecule to be roughly 1.0 Å. Compare the collision time with the time the molecule moves freely between two successive collisions (Molecular mass of N₂ = 28.0 u).
Answers at a glance
- 1Q12.1 → The fraction is about 3.8 × 10⁻⁴.
- 2Q12.2 → V = RT/P = 0.0224 m³ = 22.4 litres at STP.
- 3Q12.3 → (a) ideal-gas behaviour (b) T₁ > T₂ (c) PV/T = 0.26 J K⁻¹ (d) no; 6.3 × 10⁻⁵ kg of hydrogen gives the same value.
- 4Q12.4 → About 0.14 kg of oxygen is taken out of the cylinder.
- 5Q12.5 → The bubble grows to about 5.3 cm³.
- 6Q12.6 → There are about 6.11 × 10²⁶ air molecules in the room.
- 7Q12.7 → (i) 6.21 × 10⁻²¹ J (ii) 1.24 × 10⁻¹⁹ J (iii) 2.07 × 10⁻¹⁶ J.
- 8Q12.8 → Yes, equal numbers of molecules; v_rms differs, and it is largest for neon, the lightest.
- 9Q12.9 → Approximately 2.52 × 10³ K (2523 K).
- 10Q12.10 → Mean free path = 1.11 × 10⁻⁷ m; collision frequency = 4.58 × 10⁹ s⁻¹; free time ≈ 500 × collision time.
CBSE Pattern Questions
How CBSE asks this chapter
MCQ Questions
1 Mark Questions
- 1Define the mean free path of a gas molecule.
- 2State the ideal gas equation in terms of Boltzmann constant k. (Ans: PV = NkT)
- 3What is the relation between rms speed and the pressure/density of a gas? (Ans: vrms = √(3P/ρ))
- 4State the law of equipartition of energy.
2 Mark Questions
- 1Find the rms speed of oxygen molecules (M = 32 g/mol) at 300 K. (Ans: vrms = √(3RT/M) ≈ 484 m/s)
- 2A cylinder contains 0.2 mol of an ideal gas at 27°C occupying 5 L. Find its pressure. (Ans: P = nRT/V ≈ 9.97 × 10⁴ Pa)
- 3At what temperature are the rms speed of nitrogen molecules double its value at 300 K? (Ans: T′ = 4T = 1200 K)
3 Mark Questions
- 1Using kinetic theory, derive the expression P = ⅓ρc² for the pressure exerted by an ideal gas, defining the symbols.
- 2Explain the mean free path of a gas molecule and write the factors on which it depends. (Ans: λ = 1/(√2 n πd²))
- 3Distinguish between rms, mean and most probable speeds of gas molecules and write their relations with temperature.
5 Mark Questions
- 1Derive the expression for the pressure of an ideal gas using kinetic theory and hence obtain PV = ⅓mnc², and the relation between average kinetic energy of a molecule and the temperature. (Ans: KE = (3/2)kT)
- 2Explain the degrees of freedom of gaseous molecules, apply the law of equipartition of energy to monatomic, diatomic and polyatomic gases, and obtain the ratio γ = Cp/Cv for a diatomic gas. (Ans: γ = 7/5)
PDF: Assignments & Tests
Printable study material
Free PDF Downloads & Assignments
Practice path: complete Assignment 1 first (topic-by-topic), then Assignment 2 (mixed difficulty with full solutions), then attempt the timed tests.
Covers all six topics: ideal gas and its assumptions; pressure exerted by an ideal gas; kinetic interpretation of temperature; rms speed; degrees of freedom and specific heats; and the Maxwell speed distribution. Arranged topic by topic.
Higher-difficulty set with multi-topic integration problems (molecular kinetic energy, rms speeds, equipartition) and full worked answers. Includes an answer key.
Full board-pattern test: Section A (MCQ/VSA 1M × 10), Section B (2M × 4), Section C (3M × 4), Section D (5M × 2) = 40 marks. Time: 90 minutes. Includes answer key.
Second full paper with different questions, same 40-mark / 90-minute blueprint. Includes answer key.
Quick Revision
Key formulas at a glance
Memorise these equations — direct application numericals and derivations in CBSE & JEE frequently hinge on these.
Ideal gas equation
Pressure of a gas
Kinetic interpretation
RMS speed
Equipartition
Most probable speed
Exam Strategy
How this chapter is asked
High-yield question patterns observed across CBSE boards, JEE Main & Advanced, and NEET.
- PV = nRT; k_B = R/N_A ≈ 1.38 × 10⁻²³ J K⁻¹.
- P = (1/3)ρv²_avg: pressure from molecular collisions.
- Average translational KE per molecule = (3/2)k_B T.
- v_rms = √(3RT/M); v_rms ∝ √T, ∝ 1/√M.
- v_mp : v_mean : v_rms ≈ 1 : 1.13 : 1.22.
- Equipartition: ½k_B T per degree of freedom.
- Monatomic C_V = 3R/2, γ = 5/3; diatomic C_V = 5R/2, γ = 7/5.
Practice Numericals
Step-by-step worked examples
Standard exam-level problems solved with clear conceptual breakdowns.
RMS speed of oxygen
Step-by-Step Solution
- 1v_rms = √(3RT/M) = √(3 × 8.31 × 300 / 0.032).
- 2v_rms = √(7479 / 0.032) = √233718 ≈ 483 m/s.
Final Answer
≈ 483 m/s
Average kinetic energy at a temperature
Step-by-Step Solution
- 1KE = (3/2)k_B T = 1.5 × 1.38 × 10⁻²³ × 300.
- 2KE = 6.21 × 10⁻²¹ J.
Final Answer
6.21 × 10⁻²¹ J
Pressure from molecular speed
Step-by-Step Solution
- 1P = (1/3)ρv²_rms, so v²_rms = 3P/ρ.
- 2v²_rms = 3 × 3 × 10⁵ / 2 = 4.5 × 10⁵.
- 3v_rms = √(4.5 × 10⁵) ≈ 671 m/s.
Final Answer
≈ 671 m/s
FAQ
Frequently asked questions
What does the kinetic theory assume about gas molecules?
It assumes a large number of identical molecules in random motion, negligible molecular size compared with intermolecular distances, perfectly elastic collisions of negligible duration, and no intermolecular forces except during collisions.
What is the kinetic interpretation of temperature?
The average translational kinetic energy per molecule equals (3/2)k_B T, so temperature is a direct measure of the average kinetic energy of the molecules. At a fixed temperature, all gases have the same average translational KE per molecule.
Why do lighter gases move faster than heavier ones at the same temperature?
Since average kinetic energy (3/2)k_B T is the same for all gases at a given temperature, lighter molecules must move faster to have the same energy: v_rms = √(3RT/M) is inversely proportional to √M.
What is the law of equipartition of energy?
In thermal equilibrium, energy is shared equally among all degrees of freedom, each contributing (1/2)k_B T per molecule. This gives a monatomic gas C_V = 3R/2 and a diatomic gas C_V = 5R/2 at ordinary temperatures.
Why do molecules in a gas have many different speeds?
Collisions continually redistribute energy among the molecules, so their speeds are distributed rather than uniform. The Maxwell distribution describes this, with a most probable, a mean and an RMS speed, all of which shift higher as temperature rises.
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