Important Questions Class 11 Physics Chapter 12: Kinetic Theory

Kinetic theory explains gas behaviour through rapid random motion of atoms and molecules.
Gas pressure, temperature, rms speed and mean free path can be linked to molecular motion.

Kinetic theory connects measurable gas properties with the motion, collisions and energy of molecules. Important Questions Class 11 Physics Chapter 12 help students practise molecular nature of matter, ideal gas equation, gas laws, pressure of ideal gas, rms speed, kinetic interpretation of temperature, equipartition of energy, specific heat capacities and mean free path. The CBSE 2026 chapter builds the molecular base for thermodynamics, gases, diffusion, conduction and molecular collision models.

Key Takeaways

• Ideal Gas Equation: An ideal gas follows PV = μRT = NkBT.
• Pressure of Gas: Kinetic theory gives P = 1/3 nmv².
• Temperature Link: Average kinetic energy per molecule equals 3/2 kBT.
• Mean Free Path: Mean free path is l = 1/(√2πnd²).

Important Questions Class 11 Physics Chapter 12 Structure 2026

Important Questions Class 11 Physics Chapter 12 with Answers

Kinetic Theory explains gases using molecules in random motion and elastic collisions.
Students should connect every gas law with pressure, volume, temperature and number of molecules.
These kinetic theory class 11 important questions follow the NCERT 2026 sequence.

1. What does Important Questions Class 11 Physics Chapter 12 mainly test?

Important Questions Class 11 Physics Chapter 12 mainly test ideal gas equation, pressure of gas, rms speed, equipartition of energy, specific heat and mean free path. The chapter links macroscopic gas laws with molecular motion.

1. Gas Law Skill: Use PV = μRT.
2. Molecular Skill: Use P = 1/3 nmv².
3. Temperature Skill: Use 1/2 mv² = 3/2 kBT.
4. Collision Skill: Use l = 1/(√2πnd²).
5. Final Result: The chapter tests molecular explanation of gas behaviour.

2. What is kinetic theory of gases?

Kinetic theory of gases explains gas behaviour through rapid random motion of molecules. It treats gas pressure as the result of molecular collisions with container walls.

1. Gas Molecules: Move continuously and randomly.
2. Collisions: Molecules collide with each other and container walls.
3. Pressure: Wall collisions produce force per unit area.
4. Final Result: Kinetic theory gives a molecular model of gases.

3. Why are gases easier to study than solids and liquids in kinetic theory?

Gases are easier to study because molecules are far apart. Intermolecular forces stay negligible except during collisions.

1. Gas Separation: Molecules are far apart.
2. Intermolecular Force: Negligible between collisions.
3. Motion: Molecules move almost freely in straight lines.
4. Final Result: Ideal gas behaviour becomes a good approximation at low pressure.

Molecular Nature of Matter Class 11 Questions

Matter consists of atoms and molecules that remain in continuous motion.
The chapter starts with atomic hypothesis and connects it to measurable gas behaviour.
These molecular nature of matter class 11 questions cover atomic theory, molecular size and states of matter.

4. What is the atomic hypothesis?

The atomic hypothesis states that all matter consists of atoms. These particles move continuously and interact through attraction and repulsion.

1. Matter: Made of atoms and molecules.
2. Motion: Particles remain in perpetual motion.
3. Force: They attract at small distances and repel when squeezed close.
4. Final Result: Atomic hypothesis gives the particle view of matter.

5. Who proposed the modern atomic theory?

John Dalton proposed the modern atomic theory. He used it to explain definite and multiple proportions.

1. Scientist: John Dalton.
2. Idea: Elements consist of atoms.
3. Compounds: Atoms combine in small whole-number ratios.
4. Final Result: Dalton is credited with modern atomic theory.

6. What is the typical size of an atom?

The typical size of an atom is about 10^-10 m. This length is also called one angstrom.

1. Atomic Size: Around 10^-10 m.
2. Unit: 1 Å = 10^-10 m.
3. Solid Spacing: Atoms are separated by a few angstroms.
4. Final Result: Atomic size is about one angstrom.

7. How are atoms arranged differently in solids, liquids and gases?

Atoms are closely packed in solids and liquids, but far apart in gases. This difference explains rigidity, flow and free expansion.

1. Solids: Atoms stay tightly fixed.
2. Liquids: Atoms stay close but can move around.
3. Gases: Atoms stay far apart and move freely.
4. Final Result: Gas molecules have much larger separation than solid atoms.

8. What is dynamic equilibrium in a gas?

Dynamic equilibrium means molecules keep moving and colliding, but average properties stay constant. Gas pressure and temperature remain steady.

1. Molecular Motion: Continuous random motion.
2. Collisions: Molecules exchange speeds.
3. Macroscopic State: Average pressure and temperature remain constant.
4. Final Result: Gas equilibrium is dynamic at molecular level.

Behaviour of Gases Class 11 Questions

Gas behaviour becomes simple at low pressure and high temperature.
Under these conditions, real gases approach ideal gas behaviour.
These behaviour of gases class 11 questions cover gas laws, moles and Avogadro’s hypothesis.

9. What is Boyle’s law?

Boyle’s law states that pressure is inversely proportional to volume at constant temperature. It applies to a fixed mass of gas.

1. Condition: Temperature remains constant.
2. Relation: P ∝ 1/V.
3. Formula: PV = constant.
4. Final Result: Pressure increases when volume decreases at constant temperature.

10. What is Charles’ law?

Charles’ law states that volume is directly proportional to absolute temperature at constant pressure. The gas amount remains fixed.

1. Condition: Pressure remains constant.
2. Relation: V ∝ T.
3. Temperature Scale: T must be in kelvin.
4. Final Result: Gas volume increases with absolute temperature.

11. What is Avogadro’s hypothesis?

Avogadro’s hypothesis states that equal volumes of gases at equal temperature and pressure contain equal numbers of molecules. It supports the mole concept.

1. Same Volume: Equal gas volumes.
2. Same Conditions: Equal temperature and pressure.
3. Molecule Count: Same number of molecules.
4. Final Result: Equal gas volumes contain equal molecules at same P and T.

12. What is Avogadro number?

Avogadro number is the number of molecules in one mole of a substance. Its value is 6.02 × 10^23 mol^-1.

1. Symbol: NA.
2. Value: 6.02 × 10^23 mol^-1.
3. Mole Meaning: One mole contains NA particles.
4. Final Result: NA = 6.02 × 10^23 mol^-1.

13. What is molar volume of an ideal gas at STP?

Molar volume of an ideal gas at STP is 22.4 litres. It is the volume occupied by one mole.

1. STP Temperature: 273 K.
2. STP Pressure: 1 atm.
3. One Mole Volume: 22.4 L.
4. Final Result: One mole of ideal gas occupies 22.4 L at STP.

Ideal Gas Equation Class 11 Questions

The ideal gas equation connects pressure, volume, temperature and amount of gas.
It works best when intermolecular forces are negligible.
These ideal gas equation class 11 questions cover PV = μRT and PV = NkBT.

14. What is the ideal gas equation?

The ideal gas equation is PV = μRT. It connects pressure, volume, number of moles and absolute temperature.

1. Pressure: P.
2. Volume: V.
3. Number of Moles: μ.
4. Gas Constant: R = 8.314 J mol^-1 K^-1.
5. Final Result: Ideal gas equation is PV = μRT.

15. What is the molecule form of ideal gas equation?

The molecule form is PV = NkBT. Here N is number of molecules and kB is Boltzmann constant.

1. Number of Molecules: N.
2. Boltzmann Constant: kB = 1.38 × 10^-23 J K^-1.
3. Relation: R = NAkB.
4. Final Result: PV = NkBT.

16. What is Boltzmann constant?

Boltzmann constant connects molecular energy with temperature. Its value is 1.38 × 10^-23 J K^-1.

1. Symbol: kB.
2. Value: 1.38 × 10^-23 J K^-1.
3. Relation: kB = R/NA.
4. Final Result: Boltzmann constant links microscopic and macroscopic physics.

17. What is the value of universal gas constant?

The universal gas constant is 8.314 J mol^-1 K^-1. It appears in PV = μRT.

1. Symbol: R.
2. Value: 8.314 J mol^-1 K^-1.
3. Use: Ideal gas equation.
4. Final Result: R = 8.314 J mol^-1 K^-1.

18. Why do real gases approach ideal behaviour at low pressure?

Real gases approach ideal behaviour at low pressure because molecules stay far apart. Intermolecular interactions become negligible.

1. Low Pressure: Large molecular separation.
2. Force Effect: Weak except during collision.
3. Model Fit: Ideal gas assumptions become valid.
4. Final Result: Low pressure favours ideal gas behaviour.

19. Why do real gases approach ideal behaviour at high temperature?

Real gases approach ideal behaviour at high temperature because molecular kinetic energy becomes large. Intermolecular attraction becomes less important.

1. High Temperature: Molecules move faster.
2. Kinetic Energy: Increases with temperature.
3. Attraction Effect: Becomes relatively smaller.
4. Final Result: High temperature favours ideal gas behaviour.

Class 11 Physics Chapter 12 Questions and Answers on Gas Laws

Gas law questions often need correct use of kelvin temperature and absolute pressure.
Students should convert litres to cubic metres when using SI values.
These class 11 physics chapter 12 questions and answers support numerical practice.

20. Show that one mole of gas occupies 22.4 L at STP.

One mole of ideal gas occupies 22.4 L at STP. Use V = RT/P.

1. Given Data:
   μ = 1 mol
   R = 8.314 J mol^-1 K^-1
   T = 273 K
   P = 1.013 × 10^5 Pa
2. Formula Used: PV = μRT.
3. Calculation:
   V = RT/P
   V = 8.314 × 273/(1.013 × 10^5)
   V = 0.0224 m³
4. Final Result: V = 22.4 L.

21. Find number of moles in 44.8 L of ideal gas at STP.

The number of moles is 2 mol. One mole occupies 22.4 L at STP.

1. Given Data:
   Volume = 44.8 L
   Molar volume = 22.4 L
2. Formula Used: μ = V/22.4.
3. Calculation:
   μ = 44.8/22.4
   μ = 2
4. Final Result: Number of moles = 2 mol.

22. Find number of molecules in 2 mol of gas.

The number of molecules is 1.204 × 10^24. Use N = μNA.

1. Given Data:
   μ = 2 mol
   NA = 6.02 × 10^23 mol^-1
2. Formula Used: N = μNA.
3. Calculation:
   N = 2 × 6.02 × 10^23
   N = 1.204 × 10^24
4. Final Result: N = 1.204 × 10^24 molecules.

23. What is Dalton’s law of partial pressures?

Dalton’s law states that total pressure of a non-reacting gas mixture equals the sum of partial pressures. Each gas contributes independently.

1. Gas Mixture: Non-reactive ideal gases.
2. Partial Pressure: Pressure of one gas alone in same V and T.
3. Formula: P = P1 + P2 + P3 + ...
4. Final Result: Total pressure is sum of partial pressures.

24. Neon and oxygen have partial pressure ratio 3:2. Find molecule number ratio.

The molecule number ratio is 3:2. At same volume and temperature, pressure is proportional to moles.

1. Given Data: PNe:PO2 = 3:2.
2. Ideal Gas Relation: P ∝ μ at same V and T.
3. Molecule Relation: N ∝ μ.
4. Final Result: NNe:NO2 = 3:2.

Kinetic Theory of Ideal Gas Class 11 Questions

Kinetic theory assumes molecules move randomly and collide elastically.
It explains gas pressure using momentum transfer to container walls.
These kinetic theory of ideal gas class 11 questions cover assumptions and pressure derivation.

25. What are the assumptions of kinetic theory of an ideal gas?

Kinetic theory assumes molecules are point-like, randomly moving and elastically colliding. Intermolecular forces are neglected except during collisions.

1. Molecules: Very large in number.
2. Motion: Random and continuous.
3. Collisions: Elastic with walls and other molecules.
4. Forces: Negligible between collisions.
5. Final Result: Ideal gas molecules move freely between elastic collisions.

26. Why are gas molecule collisions treated as elastic?

Gas molecule collisions are treated as elastic because total kinetic energy remains conserved. Momentum is also conserved during collisions.

1. Wall Collision: Molecule rebounds with same speed.
2. Molecule Collision: Total kinetic energy stays constant.
3. Model Need: No energy loss during ideal collisions.
4. Final Result: Elastic collisions keep average energy steady.

27. Why does a gas exert pressure on container walls?

A gas exerts pressure because molecules collide with container walls. Each collision transfers momentum to the wall.

1. Molecular Motion: Molecules move randomly.
2. Wall Hit: Momentum changes during rebound.
3. Force: Rate of momentum transfer.
4. Final Result: Gas pressure comes from molecular collisions.

28. Why is pressure same in all directions in a gas?

Pressure is same in all directions because molecular motion is random and isotropic. No direction is preferred.

1. Random Motion: Molecules move in all directions.
2. Average Components: vx² = vy² = vz².
3. Pressure Result: Same pressure acts in every direction.
4. Final Result: Gas pressure is isotropic.

Pressure of Ideal Gas Class 11 Questions

Pressure of an ideal gas comes from average molecular momentum transfer.
The derivation uses random motion and elastic wall collisions.
These pressure of ideal gas class 11 questions focus on P = 1/3 nmv².

29. What is the kinetic theory expression for gas pressure?

The kinetic theory expression is P = 1/3 nmv². Here v² means mean square speed.

1. Number Density: n.
2. Molecular Mass: m.
3. Mean Square Speed: v².
4. Final Result: Gas pressure equals 1/3 nmv².

30. What is relation between pressure and translational kinetic energy?

The relation is PV = 2E/3, where E is total translational kinetic energy. It follows from kinetic theory.

1. Pressure Formula: P = 1/3 nmv².
2. Total Molecules: N = nV.
3. Energy: E = N × 1/2 mv².
4. Final Result: PV = 2E/3.

31. Why does pressure increase when temperature increases at fixed volume?

Pressure increases because average molecular kinetic energy increases. Molecules hit walls harder and more frequently.

1. Fixed Volume: Container size stays same.
2. Higher Temperature: Molecular speeds increase.
3. Wall Collisions: Momentum transfer per second increases.
4. Final Result: Pressure rises with temperature at fixed volume.

32. Why does pressure decrease when volume increases at fixed temperature?

Pressure decreases because molecule number density decreases. Wall collisions per unit area become less frequent.

1. Fixed Temperature: Average kinetic energy stays same.
2. Larger Volume: Number density n decreases.
3. Collision Rate: Wall hits reduce per unit area.
4. Final Result: Pressure decreases when volume increases.

33. Derive P = 1/3 nmv² in brief.

The gas pressure formula comes from momentum transfer during molecular collisions. Random motion makes three velocity components equal on average.

1. Momentum Change at Wall: 2mvx.
2. Pressure Contribution: Proportional to nmvx².
3. Isotropy: vx² = v²/3.
4. Substitution: P = nmv²/3.
5. Final Result: P = 1/3 nmv².

RMS Speed Class 11 Physics Questions

RMS speed is the square root of mean square speed.
It depends on temperature and molecular mass.
These rms speed class 11 physics questions cover speed comparison and numerical use.

34. What is rms speed of gas molecules?

RMS speed is the square root of the mean of squared molecular speeds. Its formula is vrms = √(3kBT/m).

1. Mean Square Speed: v².
2. RMS Speed: vrms = √v².
3. Formula Used: vrms = √(3kBT/m).
4. Final Result: RMS speed measures typical molecular speed.

35. What is rms speed in molar mass form?

The molar mass form is vrms = √(3RT/M). Here M is molar mass in kg mol^-1.

1. Gas Constant: R.
2. Absolute Temperature: T.
3. Molar Mass: M.
4. Final Result: vrms = √(3RT/M).

36. How does rms speed depend on temperature?

RMS speed is proportional to square root of absolute temperature. If temperature becomes four times, rms speed doubles.

1. Formula: vrms ∝ √T.
2. Temperature Change: T becomes 4T.
3. Speed Change: vrms becomes 2vrms.
4. Final Result: RMS speed increases as √T.

37. How does rms speed depend on molecular mass?

RMS speed is inversely proportional to square root of molecular mass. Lighter molecules move faster at the same temperature.

1. Formula: vrms ∝ 1/√m.
2. Same Temperature: Average kinetic energy stays same.
3. Mass Effect: Lower mass gives higher speed.
4. Final Result: Lighter gas molecules have larger rms speed.

38. Which has larger rms speed at same temperature, helium or argon?

Helium has larger rms speed than argon. Helium has smaller atomic mass.

1. Helium Atomic Mass: 4 u.
2. Argon Atomic Mass: 39.9 u.
3. Relation: vrms ∝ 1/√M.
4. Final Result: Helium molecules move faster than argon atoms.

39. Find rms speed of nitrogen at 300 K.

The rms speed is about 517 m/s. Use vrms = √(3RT/M).

1. Given Data:
   R = 8.31 J mol^-1 K^-1
   T = 300 K
   M = 28 × 10^-3 kg mol^-1
2. Formula Used: vrms = √(3RT/M).
3. Calculation:
   vrms = √[(3 × 8.31 × 300)/(28 × 10^-3)]
   vrms ≈ 517 m/s
4. Final Result: vrms ≈ 517 m/s.

40. Why is rms speed not equal to average speed?

RMS speed is not equal to average speed because average of squares differs from square of average. Squaring gives more weight to larger speeds.

1. Average Speed: Mean of v.
2. RMS Speed: Square root of mean of v².
3. Mathematical Fact: <v²> is not always ².
4. Final Result: RMS speed is usually greater than average speed.

Kinetic Interpretation of Temperature Class 11 Questions

Temperature measures average molecular kinetic energy in kinetic theory.
It does not depend on pressure, volume or gas identity for an ideal gas.
These kinetic interpretation of temperature class 11 questions explain microscopic meaning of temperature.

41. What is kinetic interpretation of temperature?

Kinetic interpretation states that temperature measures average translational kinetic energy of gas molecules. The relation is 1/2 mv² = 3/2 kBT.

1. Average Kinetic Energy: 1/2 mv².
2. Temperature: T.
3. Formula Used: 1/2 mv² = 3/2 kBT.
4. Final Result: Temperature measures average molecular kinetic energy.

42. What is average kinetic energy of one gas molecule?

Average kinetic energy of one gas molecule is 3/2 kBT. It depends only on absolute temperature.

1. Boltzmann Constant: kB.
2. Temperature: T.
3. Formula Used: KEavg = 3/2 kBT.
4. Final Result: Average molecular kinetic energy = 3/2 kBT.

43. What is total translational kinetic energy of N molecules?

Total translational kinetic energy is E = 3/2 NkBT. It equals 3/2 μRT for μ moles.

1. Number of Molecules: N.
2. Average Energy: 3/2 kBT.
3. Total Energy: E = N × 3/2 kBT.
4. Final Result: E = 3/2 NkBT.

44. Why do different gases have same average kinetic energy at same temperature?

Different gases have same average kinetic energy because KEavg = 3/2 kBT. The formula does not contain molecular mass.

1. Same Temperature: T is same.
2. Formula: KEavg = 3/2 kBT.
3. Mass Role: Mass affects speed, not average kinetic energy.
4. Final Result: All ideal gases have same average kinetic energy at same T.

45. Find average kinetic energy of a molecule at 300 K.

The average kinetic energy is 6.21 × 10^-21 J. Use KEavg = 3/2 kBT.

1. Given Data:
   kB = 1.38 × 10^-23 J K^-1
   T = 300 K
2. Formula Used: KEavg = 3/2 kBT.
3. Calculation:
   KEavg = 1.5 × 1.38 × 10^-23 × 300
   KEavg = 6.21 × 10^-21 J
4. Final Result: Average kinetic energy = 6.21 × 10^-21 J.

Law of Equipartition of Energy Class 11 Questions

The law of equipartition distributes energy equally among independent modes.
Each quadratic energy term contributes 1/2 kBT per molecule.
These law of equipartition of energy class 11 questions cover degrees of freedom and specific heat.

46. State the law of equipartition of energy.

The law of equipartition states that energy is equally distributed among all possible energy modes. Each quadratic mode has average energy 1/2 kBT.

1. Thermal Equilibrium: System has temperature T.
2. Energy Mode: Quadratic term in energy expression.
3. Average Energy: 1/2 kBT per mode.
4. Final Result: Each quadratic degree contributes 1/2 kBT.

47. What are degrees of freedom?

Degrees of freedom are independent ways in which a molecule can possess energy. They include translational, rotational and vibrational modes.

1. Translation: Motion in space.
2. Rotation: Rotation about axes.
3. Vibration: Oscillation of atoms in molecule.
4. Final Result: Degrees of freedom count independent energy modes.

48. How many translational degrees of freedom does a gas molecule have?

A gas molecule has three translational degrees of freedom in space. They correspond to x, y and z motion.

1. x-direction: One degree.
2. y-direction: One degree.
3. z-direction: One degree.
4. Final Result: Translation gives three degrees of freedom.

49. How many degrees of freedom does a monatomic gas molecule have?

A monatomic gas molecule has three degrees of freedom. They are all translational.

1. Translation: Three directions.
2. Rotation: Not counted for a point-like atom.
3. Vibration: Not present.
4. Final Result: Monatomic gas has 3 degrees of freedom.

50. How many degrees of freedom does a rigid diatomic gas molecule have?

A rigid diatomic molecule has five degrees of freedom. It has three translational and two rotational degrees.

1. Translational: 3.
2. Rotational: 2.
3. Vibrational: Ignored for rigid molecule.
4. Final Result: Rigid diatomic gas has 5 degrees of freedom.

51. Why does one vibrational mode contribute kBT?

One vibrational mode contributes kBT because it has kinetic and potential energy terms. Each term contributes 1/2 kBT.

1. Kinetic Term: 1/2 kBT.
2. Potential Term: 1/2 kBT.
3. Total: kBT.
4. Final Result: One vibrational mode contributes kBT.

Specific Heat Capacity Class 11 Kinetic Theory Questions

Specific heat values depend on the degrees of freedom of gas molecules.
Equipartition explains why monatomic and diatomic gases have different heat capacities.
These specific heat capacity class 11 kinetic theory questions cover Cv, Cp and gamma.

52. What is molar specific heat at constant volume for monatomic gas?

For monatomic gas, Cv = 3R/2. It comes from three translational degrees of freedom.

1. Internal Energy: U = 3RT/2 per mole.
2. Definition: Cv = dU/dT.
3. Result: Cv = 3R/2.
4. Final Result: Monatomic gas has Cv = 3R/2.

53. What is molar specific heat at constant pressure for monatomic gas?

For monatomic gas, Cp = 5R/2. Use Cp − Cv = R.

1. Known Value: Cv = 3R/2.
2. Relation: Cp − Cv = R.
3. Calculation: Cp = 3R/2 + R.
4. Final Result: Cp = 5R/2.

54. What is gamma for monatomic gas?

For monatomic gas, γ = 5/3. It equals Cp/Cv.

1. Cp: 5R/2.
2. Cv: 3R/2.
3. Ratio: γ = (5R/2)/(3R/2).
4. Final Result: γ = 5/3.

55. What is Cv for rigid diatomic gas?

For rigid diatomic gas, Cv = 5R/2. It has five degrees of freedom.

1. Degrees of Freedom: 5.
2. Energy Per Mole: U = 5RT/2.
3. Definition: Cv = dU/dT.
4. Final Result: Rigid diatomic gas has Cv = 5R/2.

56. What is Cp for rigid diatomic gas?

For rigid diatomic gas, Cp = 7R/2. Use Cp − Cv = R.

1. Known Value: Cv = 5R/2.
2. Relation: Cp − Cv = R.
3. Calculation: Cp = 5R/2 + R.
4. Final Result: Cp = 7R/2.

57. What is gamma for rigid diatomic gas?

For rigid diatomic gas, γ = 7/5. It equals Cp/Cv.

1. Cp: 7R/2.
2. Cv: 5R/2.
3. Ratio: γ = (7R/2)/(5R/2).
4. Final Result: γ = 7/5.

58. Why is Cp − Cv = R for an ideal gas?

Cp − Cv = R because constant pressure heating uses extra heat for expansion work. Constant volume heating does no expansion work.

1. Constant Volume: ΔV = 0.
2. Constant Pressure: Gas expands.
3. Ideal Gas Result: Cp − Cv = R.
4. Final Result: Cp exceeds Cv by R.

Mean Free Path Class 11 Questions

Mean free path measures average distance between successive molecular collisions.
It depends on molecular diameter and number density.
These mean free path class 11 questions cover collision frequency and molecular size.

59. What is mean free path?

Mean free path is the average distance travelled by a molecule between two successive collisions. Its symbol is l.

1. Molecular Motion: Molecules move freely between collisions.
2. Collision: Direction changes after collision.
3. Average Distance: Mean free path.
4. Final Result: Mean free path is average collision-free distance.

60. What is formula for mean free path?

The formula for mean free path is l = 1/(√2πnd²). Here n is number density and d is molecular diameter.

1. Number Density: n.
2. Molecular Diameter: d.
3. Collision Factor: √2.
4. Final Result: l = 1/(√2πnd²).

61. How does mean free path depend on pressure?

Mean free path decreases when pressure increases at constant temperature. Higher pressure increases number density.

1. Pressure Increase: Molecules are closer.
2. Number Density: n increases.
3. Formula: l ∝ 1/n.
4. Final Result: Higher pressure gives shorter mean free path.

62. How does mean free path depend on molecular diameter?

Mean free path decreases when molecular diameter increases. Larger molecules collide more often.

1. Formula: l = 1/(√2πnd²).
2. Diameter Effect: l ∝ 1/d².
3. Collision Cross-section: Increases with d².
4. Final Result: Larger diameter gives shorter mean free path.

63. Why does gas smell take time to spread across a room?

Gas smell takes time to spread because molecules suffer repeated collisions. Their path keeps changing instead of staying straight.

1. Molecular Speed: Very high.
2. Collisions: Frequent in air.
3. Actual Spread: Slow random diffusion.
4. Final Result: Collisions slow the spread of gas.

64. Estimate mean free path if n = 2.7 × 10^25 m^-3 and d = 2 × 10^-10 m.

The mean free path is about 2.1 × 10^-7 m. Use l = 1/(√2πnd²).

1. Given Data:
   n = 2.7 × 10^25 m^-3
   d = 2 × 10^-10 m
2. Formula Used: l = 1/(√2πnd²).
3. Calculation:
   l = 1/[1.414 × 3.14 × 2.7 × 10^25 × 4 × 10^-20]
   l ≈ 2.1 × 10^-7 m
4. Final Result: l ≈ 2.1 × 10^-7 m.

NCERT Class 11 Physics Chapter 12 Questions

NCERT questions use ideal gas law, rms speed, kinetic energy and mean free path.
Students should convert molecular mass into kg or molar mass into kg mol^-1.
These NCERT Class 11 Physics Chapter 12 questions follow the 2026 exercise pattern.

65. Estimate molecular volume fraction of oxygen gas at STP if molecular diameter is 3 Å.

The molecular volume fraction is about 1.9 × 10^-4. Use molecule volume and number density.

1. Given Data:
   d = 3 Å = 3 × 10^-10 m
   r = 1.5 × 10^-10 m
   n = 2.7 × 10^25 m^-3
2. One Molecule Volume:
   v = 4/3πr³
   v ≈ 1.41 × 10^-29 m³
3. Fraction:
   nv = 2.7 × 10^25 × 1.41 × 10^-29
   nv ≈ 3.8 × 10^-4
4. Packing Correction: Rough estimate remains of order 10^-4.
5. Final Result: Molecular volume fraction is of order 10^-4.

66. Does equal volume of gases at same pressure and temperature contain equal molecules?

Yes, equal volumes of gases at same pressure and temperature contain equal molecules. This is Avogadro’s hypothesis.

1. Same P: Pressure is equal.
2. Same V: Volume is equal.
3. Same T: Temperature is equal.
4. Ideal Gas Form: N = PV/kBT.
5. Final Result: Number of molecules is equal.

67. Which gas has largest rms speed among neon, chlorine and uranium hexafluoride at same temperature?

Neon has the largest rms speed. It has the lowest molecular mass among the three.

1. Neon: Monatomic and lighter.
2. Chlorine: Diatomic and heavier.
3. Uranium Hexafluoride: Much heavier.
4. Relation: vrms ∝ 1/√M.
5. Final Result: Neon has the largest rms speed.

68. Find temperature where argon rms speed equals helium rms speed at −20°C.

The temperature for argon is about 2524 K. Use equal rms speed condition.

1. Given Data:
   MAr = 39.9 u
   MHe = 4.0 u
   THe = 253 K
2. Condition: 3RTAr/MAr = 3RTHe/MHe.
3. Calculation:
   TAr = THe × MAr/MHe
   TAr = 253 × 39.9/4.0
   TAr ≈ 2524 K
4. Final Result: Argon temperature ≈ 2524 K.

69. Why is average kinetic energy same for argon and chlorine at the same temperature?

Average kinetic energy is same because it equals 3/2 kBT for every ideal gas molecule. It does not depend on molecular mass.

1. Same Temperature: T is equal.
2. Formula: KEavg = 3/2 kBT.
3. Gas Type: Monatomic or diatomic does not change translational average.
4. Final Result: Average translational kinetic energy is equal.

70. A fixed cylinder contains 2 mol helium at STP. Find heat required to raise temperature by 15 K.

The heat required is 374 J. Use Q = μCvΔT for fixed volume.

1. Given Data:
   μ = 2 mol
   ΔT = 15 K
   R = 8.31 J mol^-1 K^-1
2. For Helium: Cv = 3R/2.
3. Formula Used: Q = μCvΔT.
4. Calculation:
   Q = 2 × 3R/2 × 15
   Q = 45R
   Q = 374 J
5. Final Result: Heat required = 374 J.

Class 11 Physics Chapter 12 Numericals

Numericals in Kinetic Theory often combine gas law with molecular constants.
Use kelvin temperature, SI pressure, and kg mol^-1 for molar mass.
These Class 11 Physics Chapter 12 numericals support formula-based practice.

71. Find pressure if N = 3 × 10^23 molecules occupy 5 L at 300 K.

The pressure is about 2.48 × 10^5 Pa. Use PV = NkBT.

1. Given Data:
   N = 3 × 10^23
   V = 5 L = 5 × 10^-3 m³
   T = 300 K
   kB = 1.38 × 10^-23 J K^-1
2. Formula Used: P = NkBT/V.
3. Calculation:
   P = 3 × 10^23 × 1.38 × 10^-23 × 300/(5 × 10^-3)
   P = 2.48 × 10^5 Pa
4. Final Result: P ≈ 2.48 × 10^5 Pa.

72. Find rms speed of helium at 300 K.

The rms speed of helium is about 1368 m/s. Use vrms = √(3RT/M).

1. Given Data:
   R = 8.31 J mol^-1 K^-1
   T = 300 K
   M = 4 × 10^-3 kg mol^-1
2. Formula Used: vrms = √(3RT/M).
3. Calculation:
   vrms = √[(3 × 8.31 × 300)/(4 × 10^-3)]
   vrms ≈ 1368 m/s
4. Final Result: vrms ≈ 1368 m/s.

73. Find total translational kinetic energy of 1 mol ideal gas at 300 K.

The total translational kinetic energy is 3740 J. Use E = 3/2 RT.

1. Given Data:
   μ = 1 mol
   R = 8.31 J mol^-1 K^-1
   T = 300 K
2. Formula Used: E = 3/2 μRT.
3. Calculation:
   E = 1.5 × 8.31 × 300
   E = 3739.5 J
4. Final Result: E ≈ 3740 J.

74. A gas has Cp = 29.1 J mol^-1 K^-1 and Cv = 20.8 J mol^-1 K^-1. Find gamma.

The value of gamma is 1.40. Use γ = Cp/Cv.

1. Given Data:
   Cp = 29.1 J mol^-1 K^-1
   Cv = 20.8 J mol^-1 K^-1
2. Formula Used: γ = Cp/Cv.
3. Calculation:
   γ = 29.1/20.8
   γ ≈ 1.40
4. Final Result: γ = 1.40.

75. Find number density at STP.

The number density at STP is about 2.69 × 10^25 m^-3. Use n = P/kBT.

1. Given Data:
   P = 1.013 × 10^5 Pa
   kB = 1.38 × 10^-23 J K^-1
   T = 273 K
2. Formula Used: n = P/kBT.
3. Calculation:
   n = 1.013 × 10^5/(1.38 × 10^-23 × 273)
   n = 2.69 × 10^25 m^-3
4. Final Result: n ≈ 2.69 × 10^25 m^-3.

CBSE Class 11 Physics Chapter-Wise Important Questions