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Thermodynamics NCERT Solutions, Complete PDF Content and Important Questions
NCERT Solutions for Class 11 Physics Chapter 11 Thermodynamics PDF Download (2026) + Important Questions Welcome to one of the most exciting and universally applicable chapters in your physics syllabus! Whether you are marveling at how a car engine works, wondering how your refrigerator keeps food cold, or looking at the grand scale of stars and the universe, you are looking at Thermodynamics in action. For students preparing for the CBSE Class 11 Exams in 2026 , as well as competitive entrance tests like JEE Main and NEET, this chapter is an absolute must-master. Unlike pure mechanics, Thermodynamics deals with the macroscopic world-heat, temperature, and the conversion of energy from one form to another. It bridges physics and chemistry, giving you an edge in both subjects! In this comprehensive guide, we will break down the entire chapter into simple, student-friendly notes.
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ChapterThermodynamics
SubjectPhysics
Class11
BoardCBSE
DifficultyModerate to High
Exam Weightage7 - 9 Marks
Learning Objectives
- Understand the concepts of thermal equilibrium and the Zeroth Law of Thermodynamics.
- Apply the First Law of Thermodynamics to.
- Calculate heat, internal energy, and work done.
- Differentiate between isothermal, adiabatic, isochoric, and isobaric processes.
- Analyze P-V (Pressure-Volume) indicator diagrams and.
- Calculate work done from the area under the curve.
- State the Second Law of Thermodynamics (Kelvin-Planck and Clausius statements).
- Calculate the efficiency of a Carnot Engine and the coefficient of performance of a refrigerator.
Key Concepts, Definitions and Formulas
- Thermodynamic System: A distinct part of the universe enclosed by a boundary (real or imaginary) across which heat and work can be exchanged
- Zeroth Law of Thermodynamics: If systems A and B are each in thermal equilibrium with a third system C, then A and B are in thermal equilibrium with each other. (This introduces the concept of
- Temperature). Internal Energy (U): The sum of the kinetic and potential energies of the molecules of the system. It is a state variable (depends only on the current state, not the path taken)
- First Law of Thermodynamics: The heat supplied to a system is equal to the sum of the increase in its internal energy and the work done by the system. Delta Q= Delta U+
- Delta W (Sign Convention in Physics: Heat supplied is positive, work done BY the gas is positive)
- Work Done by a Gas: W=P Delta V (
- Or area under the P-V curve). Mayer's Formula: Relationship between specific heat at constant pressure (Cp ) and constant volume (Cv ) for an ideal gas. Cp -Cv =
- Isothermal: Temperature remains constant (Delta T=0)
- Adiabatic: No heat exchange with surroundings (Delta Q=0)
- Equation: PV gamma= constant (where gamma=
- Cp /Cv ). Isochoric: Volume remains constant (Delta V=0,W=0)
- Isobaric: Pressure remains constant (Delta P=0)
- Second Law of Thermodynamics: Heat cannot spontaneously flow from a colder body to a hotter body (Clausius). No engine can be 100% efficient (
- Kelvin-Planck). Carnot Engine Efficiency (): =1- T1 T2 (where T1 is source temperature and T2 is sink temperature in Kelvin). To help you visualize how pressure and volume interact under different conditions, check out this interactive P-V Diagram Simulator! Show me the visualisation
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Full NCERT Solutions
NCERT Q11.1: A geyser heats water flowing at the rate of 3.0 litres per minute from 27 C to 77 C. If the geyser operates on a gas burner, what is the rate of consumption of the fuel if its heat of combustion is 4.0 x 104 J/g?
Answer: Given: Flow rate of water = 3.0 L/min = 3.0 kg/min (since density of water is 1 kg/L). Initial temperature T1 =27 C Final temperature T2 =77 C Rise in temperature Delta T=77-27=50 C Specific heat of water C = 4.2 \times 10^3 \text{ J kg}^{-1} ^\circ\text{C}^{-1} Heat of combustion of fuel H=4.0 x 104 J/g=4.0 x 107 J/kg. Step 1: Calculate total heat required per minute (Delta Q) Delta Q=mC Delta T Delta Q=3.0 x (4.2 x 103) x 50=6.3 x 105 J/min Step 2: Calculate the rate of fuel consumption Let the rate of fuel consumption be mf kg/min. Heat produced by fuel = Delta Q mf x H=6.3 x 105 mf x (4.0 x 107)=6.3 x 105 mf =4.0 x 1076.3 x 105 =1.575 x 10-2 kg/min=15.75 g/min. Result: The rate of combustion of fuel is 15.75 g/min .
NCERT Q11.2: What amount of heat must be supplied to 2.0 x 10-2 kg of nitrogen (at room temperature) to raise its temperature by 45 C at constant pressure? (Molecular mass of N2 = 28; R=8.3 J mol-1K-1).
Answer: Given: Mass of N2 , m=2.0 x 10-2 kg=20 g. Rise in temp, Delta T=45 C=45 K. Molecular mass M=28 g/mol. Step 1: Find the number of moles (n) n=Mm =2820 =0.714 mol. Step 2: Find molar specific heat at constant pressure (Cp ) Nitrogen is a diatomic gas. For a diatomic gas, Cp =27 R. Cp =27 x 8.3=29.05 J mol-1K-1. Step 3: Calculate heat supplied (Delta Q) Delta Q=nCp Delta T Delta Q=0.714 x 29.05 x 45 approx 933.4 J. Result: The heat supplied must be 933.4 J .
NCERT Q11.3: Explain why: (a) Two bodies at different temperatures T1 and T2 if brought in thermal contact do not necessarily settle to the mean temperature (T1 +T2 )/2. (b) The coolant in a chemical or a nuclear plant (i.e., the liquid used to prevent the different parts of a plant from getting too hot) should have high specific heat. (c) Air pressure in a car tyre increases during driving.
Answer: (a) The final equilibrium temperature depends not just on the initial temperatures, but also on the masses and the specific heat capacities of the two bodies. They will settle at (T1 +T2 )/2 ONLY if their masses and specific heat capacities are exactly identical. (b) A coolant needs to absorb a massive amount of heat without its own temperature rising dangerously high (which could cause it to boil or damage pipes). Since Q=ms Delta T Delta T=msQ , a high specific heat (s) ensures that the temperature rise (Delta T) remains small. (c) During driving, friction between the tyres and the road generates heat, increasing the temperature of the air inside the tyre. According to Gay-Lussac's Law (P T at constant volume), an increase in temperature causes a direct increase in air pressure inside the closed tyre.
NCERT Q11.4: A cylinder with a movable piston contains 3 moles of hydrogen at standard temperature and pressure. The walls of the cylinder are made of a heat insulator, and the piston is insulated by having a pile of sand on it. By what factor does the pressure of the gas increase if the gas is compressed to half its original volume?
Answer: Given: Since the cylinder and piston are insulated, no heat is exchanged with the surroundings (Delta Q=0). The process is Adiabatic . For hydrogen (a diatomic gas), the ratio of specific heats gamma=57 =1.4. Initial volume = V1 . Final volume V2 =2V1 . Equation for an Adiabatic Process: P1 V1 gamma = P2 V2 gamma P2 =P1 (V2 V1 ) gamma P2 =P1 (V1 /2V1 )1.4=P1 (2)1.4 P2 =P1 x 2.639 Result: The pressure increases by a factor of approximately 2.64 .
NCERT Q11.5: In changing the state of a gas adiabatically from an equilibrium state A to another equilibrium state B, an amount of work equal to 22.3 J is done on the system. If the gas is taken from state A to B via a process in which the net heat absorbed by the system is 9.35 cal, how much is the net work done by the system in the latter case? (Take 1 cal = 4.19 J)
Answer: Given: First Case (Adiabatic): Heat exchanged Delta Q=0. Work done ON the system Delta W=-22.3 J (Negative by convention). Using First Law of Thermodynamics: Delta Q= Delta U+ Delta W 0=Delta U+(-22.3) Delta U=22.3 J. Since Internal Energy (U) is a state function, the change Delta U from A to B is fixed at 22.3 J regardless of the path taken. Second Case: Heat absorbed Delta Q=9.35 cal=9.35 x 4.19 J=39.18 J. Change in internal energy Delta U=22.3 J. Using First Law: Delta Q= Delta U+ Delta W 39.18=22.3+Delta W Delta W=39.18-22.3=16.88 J. Result: The net work done by the system in the second case is 16.88 J .
NCERT Q11.6: Two cylinders A and B of equal capacity are connected to each other via a stopcock. A contains an ideal gas at standard temperature and pressure. B is completely evacuated. The entire system is thermally insulated. The stopcock is suddenly opened. Answer the following: (a) What is the final pressure of the gas in A and B? (b) What is the change in internal energy of the gas? (c) What is the change in the temperature of the gas? (d) Do the intermediate states of the system (before settling to the final equilibrium state) lie on its P-V-T surface?
Answer: This process is an example of Free Expansion . (a) Since B was evacuated and has equal capacity, the volume of the gas instantly doubles (V2 =2V1 ). During free expansion of an ideal gas, temperature remains constant. By Boyle's Law (P1 V1 =P2 V2 ), the pressure halves. Final Pressure = 0.5 atm. (b) In free expansion, no work is done (Delta W=0 against a vacuum) and no heat is exchanged (Delta Q=0). By the First Law, Delta U=0. The change in internal energy is Zero . (c) Since the internal energy of an ideal gas depends only on temperature, and Delta U=0, the change in temperature is Zero . (d) No. Free expansion is a rapid, irreversible, and non-equilibrium process. Intermediate states are not in equilibrium and therefore cannot be described by thermodynamic variables (P, V, T) and do not lie on the P-V-T surface.
NCERT Q11.7: A steam engine delivers 5.4 x 108 J of work per minute and services 3.6 x 109 J of heat per minute from its boiler. What is the efficiency of the engine? How much heat is wasted per minute?
Answer: Given: Work Output per minute (W) = 5.4 x 108 J Heat Input per minute (Q1 ) = 3.6 x 109 J 1. Efficiency (): = Heat InputWork Output x 100% =3.6 x 1095.4 x 108 x 100%=365.4 x 100%=0.15 x 100%=15%. 2. Heat Wasted (Q2 ): Heat Wasted = Heat Input - Work Output Q2 =Q1 -W Q2 =(3.6 x 109)-(0.54 x 109)=3.06 x 109 J. Result: The efficiency is 15% and the heat wasted per minute is 3.06 x 109 J .
NCERT Q11.8: An electric heater supplies heat to a system at a rate of 100 W. If the system performs work at a rate of 75 Joules per second. At what rate is the internal energy increasing?
Answer: Given: Rate of heat supplied Delta t Delta Q =100 W=100 J/s. Rate of work done by system Delta t Delta W =75 J/s. Using the First Law of Thermodynamics: Delta Q= Delta U+ Delta W Dividing by Delta t: Delta t Delta Q = Delta t Delta U + Delta t Delta W 100=Delta t Delta U +75 Delta t Delta U =100-75=25 J/s=25 W. Result: The internal energy is increasing at a rate of 25 W .
Extra Important Questions for 2026 Exams
Important Q1: The First Law of Thermodynamics is a statement of: A) Conservation of Momentum B) Conservation of Mass C) Conservation of Energy D) Conservation of Charge
Answer: C) Conservation of Energy Explanation: Delta Q= Delta U+ Delta W simply states that energy supplied is converted into internal energy and external work.
Important Q2: During an isothermal expansion of an ideal gas, its internal energy: A) Increases B) Decreases C) Remains constant D) Becomes zero
Answer: C) Remains constant Explanation: For an ideal gas, internal energy depends only on temperature. Since T is constant in an isothermal process, Delta U=0.
Important Q3: The efficiency of a Carnot engine working between 127 C and 27 C is: A) 25% B) 50% C) 75% D) 100%
Answer: A) 25% Explanation: Convert to Kelvin: T1 =400 K, T2 =300 K. =1-( T2 /T1 )=1-(300/400)=1/4=25%.
Important Q4: Define an adiabatic process and state its essential conditions.
Answer: An adiabatic process is one in which no heat enters or leaves the system (Delta Q=0). Essential conditions: The system must be perfectly insulated from its surroundings, and the process must be carried out very rapidly so there is no time for heat exchange.
Important Q5: Why is Cp always greater than Cv for a gas?
Answer: At constant volume (Cv ), all heat supplied increases the internal energy (temperature). At constant pressure (Cp ), the gas expands, so heat supplied is used to increase internal energy AND do external work against the surrounding pressure. Hence, more heat is required.
Important Q6: What is a reversible process? Give one example.
Answer: A reversible process is one that can be retraced in the opposite direction so that the system and surroundings pass through exactly the same intermediate states. Example: Extremely slow (quasi-static) isothermal expansion/compression of an ideal gas.
Important Q7: Can a Carnot engine have 100% efficiency?
Answer: No. For =100% (or 1), the sink temperature T2 must be 0 K (Absolute Zero), which is practically unattainable. Hence, a 100% efficient engine is impossible.
Important Q8: State the First Law of Thermodynamics and apply it to an Isothermal and an Adiabatic process.
Answer: Statement: Heat supplied to a system equals the sum of the increase in internal energy and work done by the system (Delta Q= Delta U+ Delta W). Application to Isothermal: Temperature is constant, so Delta U=0. Therefore, Delta Q= Delta W. All heat supplied is used to do work. Application to Adiabatic: No heat is exchanged, so Delta Q=0. Therefore, 0=Delta U+ Delta W Delta W=- Delta U. If gas expands (work done is positive), internal energy decreases, resulting in cooling.
Important Q9: Derive Mayer's formula (Cp -Cv =R) for an ideal gas.
Answer: ). To make this happen, the compressor motor does electrical work on the system. The Coefficient of Performance (COP), , is defined as the heat extracted divided by the work done.
Important Q11: Does a refrigerator violate the Second Law of Thermodynamics?
Answer: No. The Clausius statement says heat cannot flow from cold to hot spontaneously . A refrigerator requires external work (from the compressor) to make this happen, which perfectly aligns with the Second Law.
Important Q12: If a refrigerator extracts 3000 J of heat from the freezer and the compressor does 1000 J of work, how much heat is rejected to the room?
Answer: Heat rejected (Q1 ) = Heat extracted (Q2 ) + Work done (W) Q1 =3000+1000=4000 J.
Important Q13: Calculate the Coefficient of Performance () for the above refrigerator.
Answer: = WQ2 =10003000 =3. (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): In an adiabatic expansion, a gas cools down. Reason (R): In an adiabatic process, the system does work at the expense of its own internal energy.
Answer: A . Since no external heat is supplied (Delta Q=0), expansion work is done using internal energy (Delta W=- Delta U). As internal energy drops, temperature falls.
Important Q15: Assertion (A): The area under a P-V graph gives the internal energy of the gas. Reason (R): Work done by a gas is the product of pressure and change in volume.
Answer: D . The assertion is false; the area under a P-V graph gives the Work Done , not internal energy. The reason statement is true.
FAQ Section
Is Thermodynamics Chapter 11 important for JEE/NEET 2026?
Absolutely! It is highly scoring, and the concepts overlap heavily with Physical Chemistry, giving you a dual advantage in competitive exams.
Where can I download the Class 11 Physics NCERT PDF for Thermodynamics?
You can download the official, rationalized NCERT Physics textbook directly from the official NCERT website (ncert.nic.in) under the 'Publications' -> 'PDFs' section.
What is the difference between an Isothermal and an Adiabatic process?
In an Isothermal process, temperature remains constant and heat is slowly exchanged with the surroundings. In an Adiabatic process, no heat is exchanged, and the temperature changes as the gas works rapidly.
Can heat transfer from a cold body to a hot body?
Yes, but not spontaneously . According to the Clausius statement of the Second Law, external work (like a compressor in an AC or fridge) is required to pump heat from a cold body to a hot body.
What is the Zeroth law of Thermodynamics?
It states that if Body A is in thermal equilibrium with Body B, and Body B is in thermal equilibrium with Body C, then Body A and C are in thermal equilibrium with each other. It defines the concept of temperature
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