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Kinetic Theory NCERT Solutions, Complete PDF Content and Important Questions
NCERT Solutions for Class 11 Physics Chapter 12 PDF Download (2026) + Important Questions: Kinetic Theory Welcome to one of the most interesting chapters in Class 11 Physics! Have you ever wondered why a balloon bursts when left in the hot sun? Or why you can smell a hot delicious meal from the other room but not a cold one? The answers to these everyday phenomena lie hidden in the invisible, fast-moving world of molecules. In the CBSE Class 11 syllabus, Chapter 12 is officially known as the Kinetic Theory of Gases (and yes, it is all about the kinetic energy of these tiny particles!). If you are preparing for your 2026 board exams , or targeting competitive exams like JEE and NEET, mastering this chapter is a must. It beautifully connects mechanics with thermodynamics.
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ChapterKinetic Theory (Focus on Kinetic Energy of Gases)
SubjectPhysics
Class11
BoardCBSE
DifficultyModerate
Exam Weightage~4 to 5 Marks
Learning Objectives
- Understand the behaviour of gases and the assumptions of the Kinetic Theory.
- Derive the expression for the pressure exerted by an ideal gas.
- Relate the macroscopic property of temperature to the microscopic average kinetic energy of molecules.
- Apply the Law of Equipartition of Energy to find the specific heat capacities of monoatomic, diatomic, and polyatomic gases.
- Calculate the Mean Free Path of gas molecules.
Key Concepts, Definitions and Formulas
- Ideal Gas Equation: Relates pressure (P), volume (V), and temperature (T). PV= RT (where is the number of moles and R is the Universal Gas Constant = 8.314
- J/molK) Pressure of an Ideal Gas: P=31 nmv2=31 rho v2 (where n is number density, m is mass of one molecule, rho is density, and v2 is mean square velocity)
- Kinetic Interpretation of Temperature: The average translational kinetic energy of a molecule is directly proportional to its absolute temperature. E=23 kB T (where kB is the Boltzmann constant = 1.38 x 10-23
- J/K) Root Mean Square (RMS) Speed: vrms =m3kB T =
- M3RT Degrees of Freedom (f): The number of independent ways a molecule can possess energy
- Law of Equipartition of Energy: In thermal equilibrium, the total energy of a dynamical system is shared equally among all its active degrees of freedom, and the energy associated with each degree of freedom is 21 kB
- T per molecule. Mean Free Path (l): The average distance a molecule travels between two successive collisions. l=2 n pi d21 To truly understand how temperature affects the kinetic energy and pressure of a gas, let's explore this interactive Kinetic Theory Simulator! Show me the visualisation
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Full NCERT Solutions
NCERT Q12.1: 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 A .
Answer: Given: Diameter of oxygen molecule, d=3 A =3 x 10-10 m. Radius, r=1.5 x 10-10 m. At Standard Temperature and Pressure (STP), the actual volume occupied by 1 mole of any ideal gas (Vactual ) is 22.4 Litres=22.4 x 10-3 m3. Step 1: Calculate the volume of one oxygen molecule V1 =34 pi r3=34 x 3.14 x (1.5 x 10-10)3 V1 =4.187 x 3.375 x 10-30 approx 14.13 x 10-30 m3. Step 2: Calculate the molecular volume of 1 mole of oxygen gas 1 mole contains Avogadro's number of molecules (NA =6.023 x 1023). Molecular volume Vmolecular =NA x V1 Vmolecular =6.023 x 1023 x 14.13 x 10-30 approx 8.51 x 10-6 m3. Step 3: Find the ratio Ratio = Vactual Vmolecular =22.4 x 10-38.51 x 10-6 approx 3.8 x 10-4. Result: The molecular volume is roughly 3.8 x 10-4 times the actual volume occupied by the gas. This shows that most of the gas is just empty space!
NCERT Q12.2: Molar volume is the volume occupied by 1 mole of any (ideal) gas at standard temperature and pressure (STP: 1 atmospheric pressure, 0 C). Show that it is 22.4 litres.
Answer: Given: Pressure P=1 atm=1.013 x 105 N/m2. Temperature T=0 C=273.15 K. Number of moles =1 mole. Universal Gas Constant R=8.314 J mol-1K-1. Step 1: Use the Ideal Gas Equation PV= RT V=P RT Step 2: Substitute the values V=1.013 x 1051 x 8.314 x 273.15 V=1.013 x 1052271.12 V approx 22.4 x 10-3 m3 Step 3: Convert to Litres Since 1 m3=1000 Litres, V=22.4 x 10-3 x 1000=22.4 Litres. Hence proved.
NCERT Q12.3: An oxygen cylinder of volume 30 litres 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-1K-1, molecular mass of O2 =32 u).
Answer: Given: Volume V=30 L=30 x 10-3 m3 (remains constant). Initial absolute pressure P1 =Gauge pressure+Atmospheric pressure=15+1=16 atm. P1 =16 x 1.013 x 105=16.208 x 105 Pa. Initial Temp T1 =27 C=300 K. Final absolute pressure P2 =11+1=12 atm=12 x 1.013 x 105=12.156 x 105 Pa. Final Temp T2 =17 C=290 K. Step 1: Find initial moles (1 ) 1 = RT1 P1 V =8.31 x 30016.208 x 105 x 30 x 10-3 =249348624 approx 19.5 moles. Step 2: Find final moles (2 ) 2 = RT2 P2 V =8.31 x 29012.156 x 105 x 30 x 10-3 =2409.936468 approx 14.7 moles. Step 3: Moles withdrawn and mass taken out Moles taken out Delta=1 -2 =19.5-14.7=4.8 moles. Mass of oxygen withdrawn = Delta x M=4.8 x 32 g=153.6 g. Result: Approximately 0.153 kg of oxygen is withdrawn.
NCERT Q12.4: Calculate the rms speed of oxygen molecules at 27 C.
Answer: Given: Temperature T=27 C=300 K. Molar mass of Oxygen (O2 ), M=32 g/mol=32 x 10-3 kg/mol. Gas constant R=8.31 J mol-1K-1. Calculation: Formula for rms speed is vrms =M3RT . vrms =32 x 10-33 x 8.31 x 300 vrms =32 x 10-37479 vrms =233.7 x 103 =233700 approx 483.4 m/s. Result: The rms speed of oxygen molecules is approximately 483.4 m/s .
NCERT Q12.5: 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 A . Compare the collision time with the time the molecule moves freely between two successive collisions.
Answer: Given: Pressure P=2.0 atm=2.0 x 1.013 x 105 Pa=2.026 x 105 Pa. Temperature T=17 C=290 K. Radius r=1.0 A =10-10 m Diameter d=2 x 10-10 m. Boltzmann constant kB =1.38 x 10-23 J/K. Step 1: Number density (n) From P=nkB T, we get n=kB TP . n=1.38 x 10-23 x 2902.026 x 105 =400.2 x 10-232.026 x 105 approx 5.06 x 1025 m-3. Step 2: Mean free path (l) l=2 pi d2n1 l=1.414 x 3.14 x (2 x 10-10)2 x 5.06 x 10251 l=1.414 x 3.14 x 4 x 10-20 x 5.06 x 10251 l=898.8 x 1051 approx 1.11 x 10-7 m. Step 3: Collision frequency (nu) First, find vrms for N2 (M=28 x 10-3 kg/mol). vrms =M3RT =28 x 10-33 x 8.31 x 290 approx 508.26 m/s. Collision frequency nu= lvrms =1.11 x 10-7508.26 approx 4.58 x 109 Hz.
Extra Important Questions for 2026 Exams
Important Q1: The average kinetic energy of a gas molecule depends only on its: A) Pressure B) Volume C) Absolute Temperature D) Nature of the gas
Answer: C) Absolute Temperature Explanation: Average kinetic energy E=23 kB T. It is strictly dependent on the absolute temperature and nothing else.
Important Q2: The number of degrees of freedom for a rigid diatomic molecule is: A) 3 B) 5 C) 6 D) 7
Answer: B) 5 Explanation: A rigid diatomic gas has 3 translational and 2 rotational degrees of freedom.
Important Q3: If the temperature of a gas is doubled, its RMS speed will become: A) Half B) Double C) Four times D) 2 times
Answer: D) 2 times Explanation: vrms T . If T becomes 2T, vrms becomes 2 x vrms .
Important Q5: Define Mean Free Path. Name two factors on which it depends.
Answer: The average distance travelled by a gas molecule between two successive collisions is called the mean free path. It depends on the size (diameter) of the gas molecules and the number density of the gas.
Important Q6: State the Law of Equipartition of Energy.
Answer: It states that for a dynamic system in thermal equilibrium, the total energy is distributed equally amongst all the degrees of freedom, and the energy associated with each molecule per degree of freedom is 21 kB T.
Important Q7: Why does a gas exert pressure?
Answer: Gas molecules are in constant random motion. They continuously collide with each other and with the walls of the container. These collisions exert a steady force on the walls, and this force per unit area is measured as the pressure of the gas.
Important Q8: Derive the expression for the pressure exerted by an ideal gas on the walls of its container.
Answer: Consider a cubical box of side L containing an ideal gas. Let a molecule of mass m move with velocity vx along the x-axis towards a wall. Initial momentum = mvx . It collides elastically and rebounds with velocity -vx . Final momentum = -mvx . Change in momentum = -mvx -(mvx )=-2mvx . Momentum imparted to the wall = 2mvx . Time taken between of freedom f=5. Total internal energy U=5 x 21 RT=25 RT. Cv =dTdU =25 R. Cp =Cv +R=25 R+R=27 R. gamma= Cv Cp =5/27/2 =57 =1.4.
Important Q10: What is an ideal gas? At what conditions do real gases behave like ideal gases?
Answer: An ideal gas is a theoretical gas composed of randomly moving point particles that do not interact (no intermolecular forces) and perfectly obey the ideal gas law PV=nRT under all conditions of temperature and pressure. Real gases behave like ideal gases at high temperatures and low pressures . At high temperatures, the kinetic energy of molecules is so high that intermolecular forces become negligible. At low pressures, the volume is large, making the actual volume of molecules negligible compared to the total volume. A student is conducting an experiment in a chemistry lab. They have two identical sealed flasks, Flask A and Flask B. Flask A is filled with Helium (a monoatomic gas) and Flask B is filled with Hydrogen (a diatomic gas). Both flasks are kept at exactly 300 K.
Important Q11: Which gas molecules have a higher average translational kinetic energy?
Answer: They both have the SAME average translational kinetic energy. Translational KE = 23 kB T, which depends only on temperature.
Important Q12: Which gas will have a higher total internal energy per mole? Why?
Answer: Hydrogen (Flask B) will have higher total internal energy. Being a diatomic gas, it has 5 degrees of freedom, whereas Helium (monoatomic) only has
Important Q13: If the temperature of Flask A is increased to 1200 K, what will happen to the RMS speed of the Helium molecules?
Answer: The temperature increases from 300 K to 1200 K (4 times). Since vrms T , the RMS speed will become 4 =2 times its original value. (Directions: Select A if both are true and Reason explains Assertion. Select B if both are true but Reason does not explain Assertion. Select C if Assertion is true, Reason is false. Select D if Assertion is false, Reason is true.)
Important Q14: Assertion (A): The pressure of a gas depends on the square of the RMS speed of its molecules. Reason (R): When molecules collide with the container walls, the change in momentum determines the force and thus the pressure.
Answer: A . Both are true, and the derivation of pressure (P=31 rho vrms2 ) is entirely based on the change of momentum of molecules hitting the wall.
Important Q15: Assertion (A): The mean free path of gas molecules decreases with an increase in temperature at constant volume. Reason (R): The mean free path is inversely proportional to the number density of the gas.
Answer: D . The assertion is false. Mean free path l=2 pi d2n1 . At constant volume, the number density (n) remains constant, so l does not change with temperature. The reason statement is true.
FAQ Section
Is Kinetic Theory an important chapter for Class 11 boards?
Yes. Though it is a small chapter, it carries a solid 4-5 marks. Questions are typically very direct, making it a high-scoring chapter if your formulas are clear.
Where can I download the Class 11 Physics NCERT PDF for Chapter 12?
You can download the official and latest NCERT textbook PDF directly from the official NCERT website (ncert.nic.in) under their E-Books section.
What is the difference between an Ideal Gas and a Real Gas?
An ideal gas perfectly follows PV=nRT at all temperatures and pressures, assuming molecules have zero volume and no attractive forces. Real gases have molecular volume and forces, deviating from ideal behavior, especially at low temps and high pressures.
What does the Kinetic Energy of a gas mean?
It refers to the energy gas molecules possess due to their random motion. The average translational kinetic energy of a gas molecule depends purely on the absolute temperature of the gas.
How is Chapter 12 connected to Thermodynamics?
While Thermodynamics (Chapter 11) deals with bulk, macroscopic properties like Heat and Work, Kinetic Theory (Chapter 12) explains those exact same properties at a microscopic level by looking at the bouncing of individual atoms
More Class 11 Physics Chapters
Ch 1: Units and MeasurementsCh 2: Motion in a Straight LineCh 3: Motion in a PlaneCh 4: Laws of MotionCh 5: Work, Energy and PowerCh 6: System of Particles and Rotational MotionCh 7: GravitationCh 8: Mechanical Properties of SolidsCh 9: Mechanical Properties of FluidsCh 10: Thermal Properties of MatterCh 11: ThermodynamicsCh 12: Kinetic TheoryCh 13: OscillationsCh 14: Waves