Class 11 Physics · NCERT Chapter 11

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Thermodynamics Class 11 Physics Notes

Complete, exam-ready notes on thermodynamics: thermal equilibrium and the zeroth law, heat, work and internal energy, the first law of thermodynamics, specific heats of a gas, thermodynamic processes (isothermal, adiabatic, isobaric, isochoric and cyclic), the second law of thermodynamics, heat engines and refrigerators — every NCERT topic with MCQs, mark-wise questions and solved numericals for CBSE, JEE and NEET.

Chapter:ThermodynamicsClass:Class 11Subject:PhysicsBoard:CBSEExams:CBSE · JEE · NEET

8 Key Formulas3 Solved Examples40 Practice MCQs

DWritten byDeep Narayan

UpdatedSeptember 2026

Key Concept Summary

State the first law of thermodynamics and give the work done by an ideal gas.

The first law is the principle of conservation of energy for thermodynamic systems: ΔQ = ΔU + ΔW, where ΔQ is the heat supplied to the system, ΔU the change in internal energy and ΔW the work done by the system. For an ideal gas the work done in an isothermal expansion is ΔW = nRT ln(V₂/V₁).

Thermal Equilibrium and the Zeroth Law

Definition

Thermal equilibrium and the zeroth law

Two bodies are in thermal equilibrium when they have the same temperature and no heat flows between them. The zeroth law of thermodynamics states that if two bodies A and B are separately in thermal equilibrium with a third body C, then A and B are in thermal equilibrium with each other. This law is the basis of temperature measurement.

Because of the zeroth law, a thermometer can be used to compare temperatures: a thermometer reads the same value when placed in contact with two bodies in thermal equilibrium. Thermal equilibrium is reached by allowing heat to flow until the temperatures equalise.

The zeroth law justifies the concept of temperature.
A thermometer measures its own temperature, which matches the body's at equilibrium.
Three bodies in mutual contact eventually reach a common temperature.

The 'zeroth' law comes first

It was named the zeroth law because it is even more fundamental than the first law — it must precede it as the logical foundation of temperature.

MCQ Questions

Two bodies are in thermal equilibrium when they have —

The zeroth law of thermodynamics is the basis of —

If A and B are each in thermal equilibrium with C, then A and B are —

Heat flows between two bodies only when —

1 Mark Questions

1
Define thermal equilibrium.
2
State the zeroth law of thermodynamics.
3
Why is the zeroth law the basis of thermometry?

2 Mark Questions

1
State the zeroth law of thermodynamics and explain how it justifies the use of a thermometer.
2
Explain the meaning of thermal equilibrium and give two examples.
3
A and B are each in thermal equilibrium with C. What can you conclude about A and B?

3 Mark Questions

1
State and explain the zeroth law of thermodynamics, and show how it leads to the definition of temperature.
2
Discuss the conditions for thermal equilibrium between two bodies and explain why heat flows only when temperatures differ.

Heat, Work and Internal Energy

Definition

Heat and work

Heat Q is the energy transferred between a system and its surroundings because of a temperature difference. Work W is the energy transferred in organised motion (such as expansion of a gas against an external pressure). Both are path functions — the energy transferred depends on how the change occurs.

Definition

Internal energy

Internal energy U is the total energy of all the molecules of a system — their kinetic and potential energies. For an ideal gas, U depends only on temperature: ΔU = nC_V ΔT. Internal energy is a state function: its change depends only on the initial and final states, not the path.

Work done by a gas and internal energy of an ideal gas

Δ W = P Δ V, Δ U = n C_{V} Δ T

Work done by a gas in expansion: ΔW = PΔV (positive if volume increases).
Heat and work are path functions; internal energy is a state function.
In a volume change the work is the area under the P–V curve.

Path vs state functions

Q and W depend on the path between two states, while U (and its change ΔU) depends only on the states themselves. For a given change the same ΔU can be reached with different combinations of Q and W.

MCQ Questions

Which of the following is a path function? —

For an ideal gas, the internal energy depends only on —

The work done by a gas expanding at constant pressure P through volume ΔV is —

Which of the following is a state function? —

1 Mark Questions

1
Distinguish between heat and work as forms of energy transfer.
2
Define internal energy.
3
Write the expression for the work done by a gas at constant pressure.

2 Mark Questions

1
Distinguish between path functions and state functions, giving examples.
2
A gas expands from 2 m³ to 5 m³ at a constant pressure of 4 × 10⁵ Pa. Find the work done by the gas.
3
Explain why the internal energy of an ideal gas depends only on temperature.

3 Mark Questions

1
Explain the difference between heat, work and internal energy, and classify them as path or state functions with justification.
2
A gas is compressed at constant pressure of 2 × 10⁵ Pa from 4 m³ to 2 m³. Find the work done on the gas and state its sign.

The First Law of Thermodynamics

Definition

First law of thermodynamics

The first law is the conservation of energy for a system: the heat ΔQ supplied to a system partly increases its internal energy ΔU and partly is used by the system to do external work ΔW. Mathematically ΔQ = ΔU + ΔW. For an infinitesimal change, dQ = dU + dW.

First law of thermodynamics

Δ Q = Δ U + Δ W, ideal gas: d U = n C_{V} d T

Sign conventions are crucial: heat added to the system is positive, work done by the system is positive. In an isochoric process (constant volume, ΔW = 0) all the heat goes into internal energy; in an isothermal process (constant temperature, ΔU = 0) all the heat becomes work.

For an isochoric process (volume constant): ΔQ = ΔU.
For an isothermal process (temperature constant): ΔQ = ΔW.
For a cyclic process the net ΔU = 0, so the net heat equals the net work.

The first law is energy conservation

Of the heat supplied, the part not stored as internal energy must come out as work. Perpetual-motion machines of the first kind, which produce work without an energy input, are impossible.

MCQ Questions

The first law of thermodynamics is a statement of —

For a process at constant volume, the heat supplied equals —

In an isothermal process, ΔU for an ideal gas is —

In a cyclic process, the net change in internal energy of the system is —

1 Mark Questions

1
State the first law of thermodynamics.
2
Write the mathematical form of the first law with the sign convention.
3
What is the value of ΔU in an isothermal process for an ideal gas?

2 Mark Questions

1
In a certain process 1000 J of heat is supplied to a gas and it does 600 J of work. Find the change in its internal energy.
2
Explain the first law for an isothermal and an adiabatic process.
3
State the sign conventions for heat and work in the first law.

3 Mark Questions

1
State and explain the first law of thermodynamics, and apply it to isothermal, isobaric and adiabatic processes.
2
A gas absorbs 400 J of heat and does 250 J of work. Find the change in internal energy. If the gas instead expands doing 400 J of work while absorbing 600 J, find the new ΔU.

Thermodynamic Processes

Definition

The four thermodynamic processes

Isothermal: temperature constant (ΔU = 0), e.g. slow expansion against a large heat reservoir. Adiabatic: no heat exchange (ΔQ = 0), e.g. rapid compression or expansion. Isobaric: pressure constant. Isochoric: volume constant (no work). Each is described by the first law and the appropriate condition.

Process equations for an ideal gas

P V = constant (isothermal), P V^{γ} = constant (adiabatic)

For an ideal gas, the isothermal paths follow PV = constant and the adiabatic paths follow PV^γ = constant (γ = C_P/C_V). The work done during an isothermal process is ΔW = nRT ln(V₂/V₁), and during an adiabatic change the temperature and volume relate by TV^(γ−1) = constant.

Isothermal work and adiabatic relation

Δ W_{isothermal} = n R T ln \frac{V _{2}}{V _{1}}, T V^{γ - 1} = constant

An adiabatic curve is steeper than an isothermal curve on a P–V diagram.
An ideal gas in a cyclic process returns to its starting state, so net ΔU = 0.
The work in a cyclic process equals the area enclosed by the loop on the P–V diagram.

Isothermal vs adiabatic

Isothermal processes are slow enough for heat exchange to hold T constant; adiabatic processes are so fast that no heat enters or leaves. On a P–V diagram, adiabatics cut across isothermals with a steeper slope.

MCQ Questions

In an isothermal process, the quantity that remains constant is —

For an adiabatic process of an ideal gas, —

A process with constant volume is called —

The work done by an ideal gas in an isothermal expansion from V₁ to V₂ is —

1 Mark Questions

1
Define an isothermal process.
2
What is an adiabatic process?
3
Write the equation for an ideal gas under isothermal and adiabatic conditions.

2 Mark Questions

1
Distinguish between isothermal and adiabatic processes.
2
One mole of an ideal gas expands isothermally at 300 K from 1 m³ to 3 m³. Find the work done. (R = 8.31 J mol⁻¹ K⁻¹.)
3
Why is an adiabatic curve steeper than an isothermal curve?

3 Mark Questions

1
Explain isothermal, adiabatic, isobaric and isochoric processes, writing the equation for each.
2
A gas is expanded adiabatically from a pressure of 2 × 10⁵ Pa and volume 0.5 m³ so that its pressure becomes 8 × 10⁴ Pa. If γ = 1.4, find the new volume.

Specific Heats of a Gas

Definition

Specific heats of a gas

A gas has two specific heats: C_V at constant volume and C_P at constant pressure. Their difference for an ideal gas is the Mayer relation C_P − C_V = R. Their ratio is γ = C_P/C_V (about 1.4 for a diatomic gas).

Mayer's relation and the ratio of specific heats

C_{P} - C_{V} = R, γ = \frac{C _{P}}{C _{V}}

At constant pressure a gas does work as it expands, so C_P is greater than C_V (more heat is needed for the same temperature rise). Typical values are C_V = 3R/2 for a monatomic gas and C_V = 5R/2 for a diatomic gas (neglecting vibrations).

C_P > C_V for gases because at constant pressure the gas does external work.
γ = 5/3 for monatomic, 7/5 for diatomic gases.
The internal energy change uses C_V: ΔU = nC_VΔT even for a constant-pressure process.

Use C_V for internal energy

Even in a constant-pressure process, the change in internal energy of an ideal gas is ΔU = nC_VΔT — internal energy depends only on temperature, not on how it was changed.

MCQ Questions

Mayer's relation for an ideal gas is —

For a diatomic gas (neglecting vibrations), γ is approximately —

C_P is greater than C_V because —

For a monatomic gas, C_V is approximately —

1 Mark Questions

1
Define C_V and C_P for a gas.
2
State Mayer's relation.
3
Write γ for monatomic and diatomic gases.

2 Mark Questions

1
Explain why C_P is greater than C_V for a gas.
2
For a diatomic gas with C_V = 5R/2, find C_P and γ.
3
Find the change in internal energy of 2 moles of a monatomic gas heated from 300 K to 400 K. (R = 8.31 J mol⁻¹ K⁻¹.)

3 Mark Questions

1
Derive Mayer's relation C_P − C_V = R for an ideal gas.
2
A diatomic gas of 3 moles is heated at constant pressure from 300 K to 400 K. Find the heat supplied given C_P = 7R/2 and R = 8.31 J mol⁻¹ K⁻¹.

Heat Engines, Refrigerators and the Second Law

Definition

Heat engine

A heat engine is a device that converts heat into work. It draws heat Q₁ from a hot reservoir at temperature T₁, converts part of it into work W and rejects the rest Q₂ to a cold reservoir at T₂. Its efficiency is η = W/Q₁ = 1 − Q₂/Q₁. A reversible (Carnot) engine has the maximum possible efficiency η = 1 − T₂/T₁.

Efficiency of a heat engine and the Carnot limit

η = \frac{W}{Q _{1}} = 1 - \frac{Q _{2}}{Q _{1}}, η_{Carnot} = 1 - \frac{T _{2}}{T _{1}}

A refrigerator is a reversed heat engine: it takes heat Q₂ from a cold reservoir and uses external work W to reject heat Q₁ to the hot reservoir. Its coefficient of performance is COP = Q₂/W. The second law of thermodynamics has two equivalent statements: heat cannot flow spontaneously from a cold body to a hot body (Clausius), and it is impossible to construct a heat engine that converts all heat into work without rejecting some heat (Kelvin-Planck).

No heat engine can be 100% efficient; some heat must always be rejected to the cold reservoir.
A refrigerator transfers heat from cold to hot at the cost of external work.
The Carnot efficiency depends only on the two reservoir temperatures.

The second law forbids perfect engines

A perpetual-motion machine of the second kind would extract heat from one reservoir and convert it entirely into work — this is impossible because Q₂ can never be zero.

MCQ Questions

The efficiency of a heat engine is defined as —

A Carnot engine's efficiency depends on —

A refrigerator operates by —

According to the second law of thermodynamics, —

1 Mark Questions

1
Define a heat engine.
2
Write the formula for the efficiency of a Carnot engine.
3
State the second law of thermodynamics (Clausius statement).

2 Mark Questions

1
A heat engine absorbs 1000 J from a hot reservoir and rejects 600 J to the cold reservoir. Find its efficiency.
2
Compute the Carnot efficiency of an engine operating between 400 K and 300 K.
3
Explain why a heat engine cannot be 100% efficient.

3 Mark Questions

1
Describe the working of a heat engine and derive its efficiency, and state the Carnot efficiency limit.
2
A Carnot engine works between temperatures 500 K and 300 K. Find (a) its efficiency and (b) the heat rejected per cycle if it absorbs 4000 J per cycle.
3
Explain how a refrigerator works and define its coefficient of performance.

NCERT Exercise Questions

From the textbook

All NCERT questions for this chapter — NCERT Physics Part II, Chapter 11 (Thermodynamics). Questions keep their NCERT numbering so you can match them to the book.

Exercises (NCERT Q11.1–Q11.8)

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⁻¹?
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.)
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.
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?
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.)
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?
11.7) An electric heater supplies heat to a system at a rate of 100 W. If system performs work at a rate of 75 joules per second. At what rate is the internal energy increasing?
11.8) A thermodynamic system is taken from an original state to an intermediate state by the linear process shown in Fig. (11.13). 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.

Answers at a glance

1
Q11.1 → The geyser consumes fuel at about 15.75 g min⁻¹ (≈ 16 g min⁻¹).
2
Q11.2 → About 9.33 × 10² J of heat must be supplied.
3
Q11.3 → (a) the mean applies only with equal thermal capacities (b) high-specific-heat coolants absorb more heat per degree (c) driving heats the tyre air and so raises pressure (d) the sea's high specific heat moderates a harbour's climate.
4
Q11.4 → The pressure increases by a factor of 2.64 (2^1.4).
5
Q11.5 → The net work done by the system is about 16.9 J.
6
Q11.6 → (a) 0.5 atm in each (b) zero (c) zero (d) no, the intermediate states are non-equilibrium.
7
Q11.7 → Internal energy increases at 25 W.
8
Q11.8 → Total work done by the gas from D to E to F = 450 J.

CBSE Pattern Questions

How CBSE asks this chapter

Physics board paper structure: Section A (MCQ/VSA 1M), Section B (2M), Section C (3M), Section D (5M). Thermodynamics typically carries ~7 marks: 1-mark MCQs on the first law and process types, a 2-mark numerical on work done, a 3-mark derivation of the efficiency of a Carnot engine, and a 5-mark question comparing isothermal and adiabatic processes or proving the second law.

MCQ Questions

The first law of thermodynamics is a statement of the conservation of

In an adiabatic process, the quantity that remains constant is

The work done by an ideal gas in an isothermal expansion is

The efficiency of a Carnot engine operating between temperatures T₁ and T₂ (T₁ > T₂) is

1 Mark Questions

1
State the first law of thermodynamics.
2
Define an isothermal process and an adiabatic process.
3
Write the relation between the principal molar specific heats Cp and Cv of an ideal gas. (Ans: Cp − Cv = R)
4
State the second law of thermodynamics (Kelvin–Planck form).

2 Mark Questions

1
One mole of an ideal gas expands isothermally at 300 K from 2 L to 4 L. Find the work done. (Ans: W = nRT ln(V₂/V₁) ≈ 1729 J)
2
In a cyclic process, heat absorbed is 400 J and work done by the gas is 400 J. Find the change in internal energy. (Ans: ΔU = Q − W = 0)
3
For an ideal gas, Cp = 29 J/mol K and Cv = 21 J/mol K. Find the gas constant R. (Ans: R = 8 J/mol K)

3 Mark Questions

1
Derive the relation between Cp and Cv for an ideal gas. (Ans: Cp − Cv = R)
2
Describe the Carnot cycle with a p–V diagram and derive the expression for its efficiency. (Ans: η = 1 − T₂/T₁)
3
Distinguish between isothermal and adiabatic processes with a p–V illustration and the equations of state of each.

5 Mark Questions

1
Explain the four steps of a Carnot cycle and derive the efficiency of a Carnot engine, showing why no engine can be more efficient than the Carnot engine operating between the same temperatures.
2
Apply the first law of thermodynamics to isochoric, isobaric, isothermal and adiabatic processes, derive the work done in each, and show that an adiabatic process follows pV^γ = constant.

PDF: Assignments & Tests

Printable study material

Free downloadable PDFs for Thermodynamics — two topic-wise assignments covering every concept of the chapter, and two full CBSE-pattern tests (each 90 minutes, 40 marks) mirroring the board paper. No signup required.

Free PDF Downloads & Assignments

No Signup Required

Practice path: complete Assignment 1 first (topic-by-topic), then Assignment 2 (mixed difficulty with full solutions), then attempt the timed tests.

Thermodynamics Assignment 1 — Topic-wise Questions

Covers all six topics: thermal equilibrium and the zeroth law; heat, work and internal energy; the first law of thermodynamics; thermodynamic processes; specific heats of a gas; and heat engines, refrigerators and the second law. Arranged topic by topic.

thermodynamics-class-11-notes-assignment-1.pdf

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Thermodynamics Assignment 2 — Challenge + Integrated Problems

Higher-difficulty set with multi-topic integration problems (first law, isothermal and adiabatic processes, Carnot efficiency, molar specific heats) and full worked answers. Includes an answer key.

thermodynamics-class-11-notes-assignment-2.pdf

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Thermodynamics Test 1 — CBSE Pattern (90 min, 40 marks)

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.

thermodynamics-class-11-notes-test-1-cbse.pdf

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Thermodynamics Test 2 — CBSE Pattern (90 min, 40 marks)

Second full paper with different questions, same 40-mark / 90-minute blueprint. Includes answer key.

thermodynamics-class-11-notes-test-2-cbse.pdf

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Quick Revision

Key formulas at a glance

Memorise these equations — direct application numericals and derivations in CBSE & JEE frequently hinge on these.

First law

Δ Q = Δ U + Δ W

Work by a gas

Δ W = P Δ V

Internal energy

Δ U = n C_{V} Δ T

Isothermal

P V = constant, W = n R T ln (V_{2} / V_{1})

Adiabatic

P V^{γ} = constant, T V^{γ - 1} = constant

Mayer's relation

C_{P} - C_{V} = R, γ = C_{P} / C_{V}

Engine efficiency

η = 1 - \frac{Q _{2}}{Q _{1}}

Carnot efficiency

η = 1 - \frac{T _{2}}{T _{1}}

Exam Strategy

How this chapter is asked

High-yield question patterns observed across CBSE boards, JEE Main & Advanced, and NEET.

First law: ΔQ = ΔU + ΔW (energy conservation).
ΔU = nC_VΔT for an ideal gas always.
Isothermal: ΔU = 0, W = nRT ln(V₂/V₁); adiabatic: ΔQ = 0, PV^γ = const.
Cyclic process: net ΔU = 0, net Q = net W.
C_P − C_V = R; γ = C_P/C_V (5/3 monatomic, 7/5 diatomic).
η = 1 − Q₂/Q₁; Carnot η = 1 − T₂/T₁.
Second law: no perfect engine, no spontaneous cold-to-hot heat flow.

Practice Numericals

Step-by-step worked examples

Standard exam-level problems solved with clear conceptual breakdowns.

Change in internal energy

Problem StatementA gas absorbs 500 J of heat and does 200 J of work. Find the change in internal energy.

Step-by-Step Solution

1First law: ΔU = ΔQ − ΔW = 500 − 200.
2ΔU = 300 J.

Final Answer

300 J

Isothermal work

Problem StatementOne mole of an ideal gas expands isothermally at 300 K from 1 m³ to 2 m³. Find the work done. (R = 8.31 J mol⁻¹ K⁻¹.)

Step-by-Step Solution

1W = nRT ln(V₂/V₁) = 1 × 8.31 × 300 × ln(2).
2W = 2493 × 0.693 ≈ 1728 J.

Final Answer

≈ 1.73 × 10³ J

Carnot efficiency

Problem StatementA Carnot engine operates between 600 K and 300 K. Find its efficiency and the heat rejected if it absorbs 2000 J per cycle.

Step-by-Step Solution

1η = 1 − T₂/T₁ = 1 − 300/600 = 0.5 (50%).
2Q₂ = Q₁(1 − η) = 2000 × 0.5 = 1000 J.

Final Answer

50% efficiency; 1000 J rejected

FAQ

Frequently asked questions

What is the first law of thermodynamics?

It is the statement of energy conservation for a thermodynamic system: the heat ΔQ supplied partly increases the internal energy ΔU and partly is used to do external work ΔW, i.e. ΔQ = ΔU + ΔW.

Why can no heat engine be 100% efficient?

A working engine must reject some heat Q₂ to a cold reservoir to complete its cycle — the second law forbids converting all the heat Q₁ into work. The maximum possible efficiency is the Carnot limit η = 1 − T₂/T₁, which is always less than 1.

What is the difference between isothermal and adiabatic processes?

An isothermal process keeps the temperature constant (ΔU = 0, so all heat becomes work, PV = constant). An adiabatic process exchanges no heat (ΔQ = 0, so work changes internal energy, PV^γ = constant). An adiabatic curve is steeper than an isothermal.

Why is C_P greater than C_V?

At constant pressure the gas expands and does external work, so more heat is needed to produce the same temperature rise. Hence C_P = C_V + R (Mayer's relation), and γ = C_P/C_V > 1.

What does the second law of thermodynamics state?

Equivalent statements: heat cannot flow spontaneously from a cold body to a hot body (Clausius), and it is impossible to build a heat engine that converts all heat into work without rejecting any (Kelvin-Planck). It implies no process is perfectly efficient and entropy of an isolated system never decreases.

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