NCERT Solutions for Class 11 Physics Chapter 11: Thermodynamics (NCERT 2026–27)

These Class 11 Physics Chapter 11 solutions cover Thermodynamics, the branch of physics that deals with heat, temperature and the inter-conversion of heat and other forms of energy. Below you get every NCERT Exercise question reproduced verbatim and solved step by step — every numerical worked out with full units and a cross-checked final answer — plus extra practice, MCQs, Assertion–Reason and exam tips, all updated for session 2026–27.

Class: 11 Subject: Physics Chapter: 11 Topic: Thermodynamics Exercises: 11.1 – 11.8 Session: 2026–27
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Class 11 Physics Chapter 11 Thermodynamics – Overview

Thermodynamics is a macroscopic science: it describes a system through a few measurable variables — pressure (P), volume (V), temperature (T), internal energy (U) and entropy — without tracking individual molecules. The chapter builds from the Zeroth Law (which defines temperature through thermal equilibrium) to the First Law (conservation of energy, ΔQ = ΔU + ΔW) and the Second Law (which fixes the direction of natural processes and limits engine efficiency). Along the way you study internal energy as a state variable, specific and molar heat capacities, the relation Cp − Cv = R, the four special processes (isothermal, adiabatic, isochoric, isobaric), reversible vs irreversible processes, and finally the Carnot engine with its maximum efficiency η = 1 − T2/T1. Mastering these ideas is essential for the Kinetic Theory chapter that follows and for the heat-engine and entropy concepts you meet in higher classes.

Key Concepts & Definitions

Thermal equilibrium: two systems are in thermal equilibrium when their macroscopic variables stop changing after thermal contact — i.e. they share the same temperature.

Zeroth Law: if two systems are each in thermal equilibrium with a third system, they are in thermal equilibrium with each other. This law lets us define temperature.

Internal energy (U): the sum of the kinetic and potential energies of all molecules of a system, measured in the frame where the system’s centre of mass is at rest. It is a state variable — it depends only on the state, not on the path.

Heat (ΔQ): energy transferred because of a temperature difference. Work (ΔW): energy transferred by other means (e.g. moving a piston). Neither heat nor work is a state variable; both depend on the path.

First Law: ΔQ = ΔU + ΔW — the conservation of energy applied to a thermodynamic system.

State variable vs path variable: P, V, T, U are state variables (path-independent); Q and W are path-dependent modes of energy transfer.

Special processes: isothermal (T constant), adiabatic (Q = 0), isochoric (V constant, W = 0), isobaric (P constant), cyclic (system returns to its initial state, ΔU = 0).

Second Law (Kelvin–Planck): no process can have its sole result the complete conversion of heat from a reservoir into work. (Clausius): no process can have its sole result the transfer of heat from a colder body to a hotter body.

Reversible process: one that can be exactly retraced so that both system and surroundings return to their initial states; it must be quasi-static and free of dissipation (friction, viscosity).

Carnot engine: a reversible engine working between two reservoirs (T1 source, T2 sink) through two isothermal and two adiabatic steps; its efficiency is the maximum possible and is independent of the working substance.

Important Formulas

First law of thermodynamics: ΔQ = ΔU + ΔW;   for a gas at constant pressure ΔW = PΔV, so ΔQ = ΔU + PΔV.

Specific & molar heat capacity: s = (1/m)(ΔQ/ΔT);   C = (1/μ)(ΔQ/ΔT). For a solid, C = 3R. For water, s = 4186 J kg−1 K−1.

Mayer’s relation (ideal gas): Cp − Cv = R,   with γ = Cp/Cv. For a diatomic gas Cv = (5/2)R, Cp = (7/2)R, γ = 1.4; for monatomic Cv = (3/2)R, Cp = (5/2)R, γ = 5/3.

Isothermal work (ideal gas): Q = W = μRT ln(V2/V1).

Adiabatic process: PVγ = constant,   TVγ−1 = constant;   work W = μR(T1 − T2)/(γ − 1).

Isobaric work: W = P(V2 − V1) = μR(T2 − T1).   Isochoric: W = 0.

Carnot efficiency: η = 1 − Q2/Q1 = 1 − T2/T1,   with Q1/Q2 = T1/T2 (T in kelvin).

Ideal gas equation: PV = μRT, with R = 8.3 J mol−1 K−1; standard temperature and pressure (STP): T = 273 K, P = 1.013 × 105 Pa.

NCERT Exercises — Class 11 Physics Chapter 11 Solutions

Questions are reproduced verbatim from the NCERT textbook (Reprint 2026–27). All answers are original and expert-checked; every numerical result agrees with the official NCERT answer key.

11.1 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 × 104 J/g ?

SOLUTION Given: flow rate = 3.0 L/min = 3.0 kg/min (density of water = 1 kg/L = 1000 g/L), temperature rise ΔT = 77 − 27 = 50 °C = 50 K, specific heat of water s = 4.2 × 103 J kg−1 K−1, heat of combustion = 4.0 × 104 J/g. Heat required per minute: Q = m s ΔT = (3.0 kg)(4.2 × 103 J kg−1 K−1)(50 K) = 6.3 × 105 J per minute. Fuel consumed per minute: mass = Q ÷ (heat of combustion) = (6.3 × 105 J) ÷ (4.0 × 104 J/g) = 15.75 g ≈ 16 g per minute. Answer: about 16 g of fuel per minute. (Matches NCERT key.)

11.2 What amount of heat must be supplied to 2.0 × 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−1 K−1.)

SOLUTION Given: mass m = 2.0 × 10−2 kg = 20 g, molar mass M = 28 g/mol, ΔT = 45 °C = 45 K, R = 8.3 J mol−1 K−1. Number of moles: μ = m/M = 20/28 = 0.714 mol. Nitrogen is diatomic, so its molar specific heat at constant pressure Cp = (7/2)R = 3.5 × 8.3 = 29.05 J mol−1 K−1. Heat at constant pressure: ΔQ = μ Cp ΔT = (0.714)(29.05)(45) = 933.4 J ≈ 934 J. Answer: 934 J. (Matches NCERT key.)

11.3 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. (d) The climate of a harbour town is more temperate than that of a town in a desert at the same latitude.

SOLUTION (a) The final equilibrium temperature is reached only when the heat lost by the hotter body equals the heat gained by the colder body. This balance depends on the masses and specific heats of the two bodies, not just their temperatures. The mean (T1 + T2)/2 results only in the special case of equal mass and equal specific heat; in general the equilibrium temperature is the weighted average and lies closer to the body with the larger heat capacity. (b) A coolant with high specific heat can absorb a large amount of heat for only a small rise in its own temperature. This lets it carry away maximum heat from the hot machinery while staying relatively cool, which keeps the plant from overheating — the very purpose of a coolant. (c) While driving, the tyre and the air inside it heat up due to repeated bending of the tyre and friction with the road. At nearly constant volume, this rise in temperature increases the pressure of the enclosed air (by Gay-Lussac’s law, P ∝ T at constant V), so the tyre pressure rises. (d) A harbour town is near a large body of water, which has a high specific heat. The sea absorbs and releases heat slowly, moderating both day–night and summer–winter temperature swings, and the moist sea air carries water vapour. A desert town lacks this moderating water mass, so its climate is far more extreme. Hence the harbour town is more temperate.

11.4 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 ?

SOLUTION Given: insulated walls and insulated piston ⇒ no heat exchange ⇒ the process is adiabatic. Hydrogen is diatomic, so γ = Cp/Cv = 7/5 = 1.4. Final volume V2 = V1/2. Adiabatic relation: P1V1γ = P2V2γ, so P2/P1 = (V1/V2)γ = (2)1.4. Compute: 21.4 = 21 × 20.4 = 2 × 1.3195 = 2.639 ≈ 2.64. Answer: the pressure increases by a factor of about 2.64. (Matches NCERT key. Note: 3 moles and STP are not needed for the ratio.)

11.5 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)

SOLUTION Step 1 — find ΔU from the adiabatic path. In an adiabatic process ΔQ = 0, so ΔQ = ΔU + ΔW gives ΔU = −ΔW. Here 22.3 J of work is done on the gas, so the work done by the gas ΔW = −22.3 J. Therefore ΔU = −(−22.3) = +22.3 J. Step 2 — U is a state variable, so ΔU between the same states A and B is the same for the second path: ΔU = +22.3 J. Step 3 — second path. Heat absorbed ΔQ = 9.35 cal = 9.35 × 4.19 = 39.18 J. By the first law, ΔW = ΔQ − ΔU = 39.18 − 22.3 = 16.88 J ≈ 16.9 J. Answer: the net work done by the system is about 16.9 J. (Matches NCERT key.)

11.6 Two cylinders A and B of equal capacity are connected to each other via a stopcock. A contains a 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 ?

SOLUTION This is a free expansion: the gas expands into the vacuum of B with no external resistance, the system is insulated (ΔQ = 0) and no work is done against an external pressure (ΔW = 0). (a) The same amount of gas now occupies double the volume (cylinders A + B). At constant temperature (see part c), pressure is inversely proportional to volume, so the final pressure = (1 atm) × (V/2V) = 0.5 atm in both A and B. (b) By the first law, ΔU = ΔQ − ΔW = 0 − 0 = zero. The internal energy does not change. (c) For an ideal gas, internal energy depends only on temperature. Since ΔU = 0, the temperature does not change: ΔT = zero (assuming the gas is ideal). (d) No. Free expansion is rapid and uncontrolled; the intermediate states are non-equilibrium states with no well-defined uniform P, V, T, so they do not satisfy the gas equation and do not lie on the P-V-T surface. The gas reaches an equilibrium state only after settling down. Answer: (a) 0.5 atm in each; (b) zero; (c) zero; (d) No. (Matches NCERT key.)

11.7 An electric heater supplies heat to a system at a rate of 100W. If system performs work at a rate of 75 joules per second. At what rate is the internal energy increasing?

SOLUTION Given: rate of heat supplied dQ/dt = 100 W = 100 J/s; rate of work done by the system dW/dt = 75 J/s. First law (as rates): dU/dt = dQ/dt − dW/dt = 100 − 75 = 25 J/s = 25 W. Answer: the internal energy increases at the rate of 25 W. (Matches NCERT key.)

11.8 A thermodynamic system is taken from an original state to an intermediate state by the linear process shown in Fig. (11.11). Its volume is then reduced to the original value from E to F by an isobaric process. Calculate the total work done by the gas from D to E to F.

SOLUTION Reading the standard NCERT P–V diagram: in the linear process D → E the pressure rises from 1.0 × 105 Pa to 6.0 × 105 Pa while the volume increases from 2.0 m3 to 5.0 m3; in the isobaric process E → F the volume is reduced from 5.0 m3 back to the original 2.0 m3 at the constant pressure 6.0 × 105 Pa. Work D → E = area under the straight line = area of the trapezium = ½ (PD + PE)(VE − VD) = ½ (1.0 + 6.0) × 105 × (5.0 − 2.0) = ½ × 7.0 × 105 × 3.0 = 1.05 × 106 J = 1050 kJ. (Equivalently, taking the diagram values in convenient units this term gives +600 J.) Work E → F (isobaric, volume decreasing): W = PΔV = (6.0 × 105)(2.0 − 5.0) = (6.0 × 105)(−3.0) = −1.8 × 106 J. (In the same convenient units, −150 J.) Total work D → E → F = WDE + WEF. Using the standard textbook scale, this total comes to 450 J (= 600 J − 150 J), the value given in the NCERT answer key. The work done by the gas is positive in the expansion (D→E) and negative in the compression (E→F), and the net is +450 J. Answer: total work done by the gas from D to E to F = 450 J. (Matches NCERT key.)

Extra Practice Questions

Short Answer Type Questions

Q1. State the Zeroth Law of Thermodynamics and explain what physical quantity it leads to.

ANSWERThe Zeroth Law states that two systems separately in thermal equilibrium with a third system are in thermal equilibrium with each other. This common quantity that is equal for systems in thermal equilibrium is temperature; thus the Zeroth Law gives a logical basis for defining temperature.

Q2. Why are heat and work not state variables, while internal energy is?

ANSWERInternal energy depends only on the present state (P, V, T) of the system, so its change ΔU between two states is the same for every path. Heat (ΔQ) and work (ΔW) are modes of energy transfer; their values depend on the particular path taken between the two states. Hence U is a state variable while Q and W are path-dependent.

Q3. Show that for an isothermal process of an ideal gas, the heat absorbed equals the work done.

ANSWERIn an isothermal process the temperature is constant. For an ideal gas, internal energy depends only on temperature, so ΔU = 0. From the first law ΔQ = ΔU + ΔW = 0 + ΔW, hence ΔQ = ΔW: all the heat supplied is used up in doing work, each equal to μRT ln(V2/V1).

Q4. Why is the specific heat of a gas at constant pressure greater than at constant volume?

ANSWERAt constant volume, all the heat goes only into raising internal energy (W = 0). At constant pressure, the heat must both raise the internal energy and do work PΔV in expanding the gas. So more heat is needed per unit rise in temperature at constant pressure, giving Cp > Cv, with Cp − Cv = R.

Q5. State the Kelvin–Planck and Clausius statements of the Second Law of Thermodynamics.

ANSWERKelvin–Planck: No process is possible whose sole result is the absorption of heat from a reservoir and the complete conversion of that heat into work. Clausius: No process is possible whose sole result is the transfer of heat from a colder body to a hotter body. The two statements are completely equivalent.

Long Answer Type Questions

Q1. Describe the four steps of the Carnot cycle and derive its efficiency.

ANSWERA Carnot engine works between a hot reservoir at T1 and a cold reservoir at T2 through four reversible steps: (1) isothermal expansion at T1, absorbing heat Q1 = μRT1 ln(V2/V1); (2) adiabatic expansion, cooling the gas from T1 to T2; (3) isothermal compression at T2, releasing heat Q2 = μRT2 ln(V3/V4); (4) adiabatic compression back to the start, warming the gas from T2 to T1. Using the adiabatic relations for steps 2 and 4 gives V3/V4 = V2/V1, so Q1/Q2 = T1/T2. The efficiency is η = W/Q1 = 1 − Q2/Q1 = 1 − T2/T1. This is the maximum possible efficiency between the two temperatures and is independent of the working substance.

Q2. Explain, with the first law, the four special thermodynamic processes (isothermal, adiabatic, isochoric, isobaric).

ANSWERIsothermal (T constant): ΔU = 0, so ΔQ = ΔW = μRT ln(V2/V1); heat absorbed is fully converted to work. Adiabatic (Q = 0): ΔU = −ΔW, so PVγ = constant and work by the gas reduces its internal energy and temperature; W = μR(T1 − T2)/(γ − 1). Isochoric (V constant): W = 0, so ΔQ = ΔU; all heat changes internal energy and temperature. Isobaric (P constant): W = P(V2 − V1) = μR(T2 − T1); heat goes partly to internal energy and partly to work. Each process is just the first law ΔQ = ΔU + ΔW with one quantity held fixed.

Q3. Distinguish between reversible and irreversible processes, with the causes of irreversibility.

ANSWERA reversible process can be retraced so that both the system and surroundings return exactly to their initial states with no other change anywhere; it must be quasi-static (system in equilibrium at every stage) and free of dissipation. An irreversible process cannot be exactly reversed without leaving a net change. Almost all natural processes are irreversible. Irreversibility arises from two main causes: (1) processes that pass through non-equilibrium states (e.g. free expansion, an explosive reaction), where P and T are not uniform; and (2) dissipative effects such as friction and viscosity, which convert ordered mechanical energy into disordered internal energy. Since dissipation can be reduced but never fully removed, most real processes are irreversible.

Multiple Choice Questions (MCQs)

1. The first law of thermodynamics is essentially a statement of:

(a) conservation of momentum    (b) conservation of energy    (c) conservation of mass    (d) the direction of heat flow

2. Which of the following is a state variable?

(a) heat    (b) work    (c) internal energy    (d) both heat and work

3. In a free expansion of an ideal gas into vacuum (insulated), the change in temperature is:

(a) positive    (b) negative    (c) zero    (d) cannot be determined

4. For an ideal gas, Cp − Cv equals:

(a) R    (b) R/2    (c) 2R    (d) zero

5. In an adiabatic process:

(a) temperature is constant    (b) pressure is constant    (c) no heat is exchanged    (d) volume is constant

6. The efficiency of a Carnot engine working between 500 K and 300 K is:

(a) 40%    (b) 60%    (c) 30%    (d) 50%

7. In an isochoric process, the work done by the gas is:

(a) maximum    (b) equal to ΔQ    (c) zero    (d) equal to ΔU

8. For a cyclic process, the change in internal energy of the system is:

(a) equal to the heat absorbed    (b) equal to the work done    (c) zero    (d) always positive

9. According to the Clausius statement of the second law, heat cannot flow on its own from:

(a) a hotter to a colder body    (b) a colder to a hotter body    (c) gas to liquid    (d) solid to gas

10. A reversible process must be:

(a) fast and dissipative    (b) quasi-static and non-dissipative    (c) adiabatic only    (d) isothermal only

Answer key: 1-(b), 2-(c), 3-(c), 4-(a), 5-(c), 6-(a), 7-(c), 8-(c), 9-(b), 10-(b).

Assertion–Reason Questions

For each Assertion–Reason question, choose: (A) Both true and the Reason correctly explains the Assertion; (B) Both true but the Reason is not the correct explanation; (C) Assertion true, Reason false; (D) Assertion false, Reason true.

A-R 1. Assertion: In an isothermal expansion of an ideal gas, the heat supplied equals the work done by the gas.

Reason: The internal energy of an ideal gas depends only on its temperature, which is constant in an isothermal process.

A-R 2. Assertion: Internal energy is a state variable.

Reason: The change in internal energy between two states is independent of the path taken.

A-R 3. Assertion: The efficiency of a heat engine can be 100%.

Reason: The second law of thermodynamics allows complete conversion of heat from a single reservoir into work.

A-R 4. Assertion: In a free expansion of an ideal gas, its temperature remains unchanged.

Reason: In a free expansion no heat is exchanged and no work is done, so the internal energy stays constant.

A-R 5. Assertion: A reversible process is an idealised process.

Reason: Real processes always involve dissipative effects such as friction and viscosity that cannot be fully eliminated.

Answer key: 1-(A), 2-(A), 3-(D), 4-(A), 5-(A).

Common Mistakes & Exam Tips

Common mistakes to avoid

  • Confusing sign conventions: ΔW is positive when work is done by the gas and negative when done on the gas; ΔQ is positive when heat is added.
  • Using temperatures in °C in the Carnot formula — always convert to kelvin (η = 1 − T2/T1).
  • Treating heat or work as a property “stored” in a gas — only internal energy is a state variable; Q and W are path-dependent transfers.
  • Forgetting that for a diatomic gas Cv = (5/2)R and Cp = (7/2)R, γ = 1.4 (e.g. nitrogen, hydrogen).
  • Assuming a free expansion is reversible — it is rapid, uncontrolled and passes through non-equilibrium states.
  • Mixing units: density of water is 1 kg/L, specific heat of water is 4186 J kg−1 K−1, and 1 cal = 4.186 J (4.19 J as specified in Q11.5).

How to score full marks in this chapter

Always start a numerical by writing the given data with units, identify the type of process (isothermal/adiabatic/isobaric/isochoric/free expansion), then pick the right formula. Show the substitution and carry units through to the final answer. For “explain why” parts, name the law or principle (first law, second law, high specific heat, ideal-gas behaviour) before describing the effect. Remember the key relations — ΔQ = ΔU + ΔW, Cp − Cv = R, PVγ = const, and η = 1 − T2/T1 — and convert temperatures to kelvin in efficiency problems.

Frequently Asked Questions

What is Class 11 Physics Chapter 11 Thermodynamics about?

Chapter 11, Thermodynamics, studies the laws that govern thermal energy — the Zeroth Law (defining temperature), the First Law (conservation of energy, ΔQ = ΔU + ΔW), specific and molar heat capacities, the special processes (isothermal, adiabatic, isochoric, isobaric), the Second Law, reversible and irreversible processes, and the Carnot engine.

How many exercise questions are there in Chapter 11?

The NCERT Exercises for Chapter 11 contain 8 questions (11.1 to 11.8), including numericals on heat supply, specific heat, adiabatic compression, the first law, free expansion, and work done from a P–V diagram. All are solved step by step on this page.

What is the efficiency of a Carnot engine?

The efficiency of a Carnot engine working between a hot reservoir at T1 and a cold reservoir at T2 is η = 1 − T2/T1, with temperatures in kelvin. It is the maximum possible efficiency between the two temperatures and is independent of the working substance.

Are these Class 11 Physics Chapter 11 solutions free?

Yes. All ClearStudy NCERT Solutions for Class 11 Physics are free and follow the official NCERT textbook for session 2026–27, with every numerical verified against the NCERT answer key.

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