PhysicsCBSE Class 11 · Chapter 11

Thermodynamics Class 11: Notes, Laws, Processes and Solved Numericals

Updated 14 October 2026 · Fact-checked

Thermodynamics is Chapter 11 of CBSE Class 11 Physics. It covers thermal equilibrium and the zeroth law, heat, work and internal energy, the first law, specific heats of gases, state variables and the equation of state, isothermal, adiabatic, isochoric, isobaric and cyclic processes, reversible and irreversible processes, and the second law.

What Thermodynamics is about

Thermodynamics studies heat, work and energy for a whole system, such as a gas in a cylinder, without worrying about individual molecules. A system is described by a few state variables: pressure, volume, temperature and internal energy.

The zeroth law says that two systems each in thermal equilibrium with a third are in equilibrium with each other, which is what makes temperature (and thermometers) meaningful. The first law, $Δ Q = Δ U + Δ W$ , is energy conservation: heat given to a system either raises its internal energy or lets it do work. Applying it to an ideal gas gives $C_{p} - C_{v} = R$ .

The heart of the chapter is the set of thermodynamic processes: isothermal (constant T, $P V$ constant, ΔU = 0), adiabatic (no heat exchange, $P V^{γ}$ constant), isochoric (constant volume, no work), isobaric (constant pressure) and cyclic (ΔU = 0, work = area of the loop on a P-V diagram). Finally, the second law explains why some processes go only one way: heat does not flow by itself from cold to hot, and heat cannot be turned completely into work with no other effect. Heat engines and refrigerators have been removed for 2026-27, and the Carnot engine, though printed, is not named in the CBSE content. The next chapter, Kinetic Theory, explains these results from the motion of molecules.

Weightage: This chapter is part of Unit VIII, Thermodynamics. The syllabus groups Unit VIII with Unit VII, Properties of Bulk Matter and Unit IX, Behaviour of Perfect Gases and Kinetic Theory of Gases: together they carry 20 of the 70 theory marks in the school annual exam.Source: CBSE Physics (042) syllabus 2026-27, Class XI course structure, Unit VIII Thermodynamics

Key concepts

Thermal equilibrium and the zeroth law

NCERT section 11.2 and 11.3

A system is the part of the universe we study (a gas in a cylinder); everything else is the surroundings. A system is in equilibrium when its macroscopic variables (P, V, T, mass, composition) do not change with time.

Two systems in contact through a diathermic (heat-conducting) wall exchange heat until they reach thermal equilibrium; through an adiabatic (insulating) wall they cannot.

Zeroth law: 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.

The quantity that is equal for all systems in thermal equilibrium is the temperature. The zeroth law is why a thermometer (system C) can compare the temperatures of two bodies without touching them together.

Worked example

A thermometer reads the same when placed in contact with a block of copper and then with a beaker of water. What can you conclude, and which law justifies it? [1 mark]

Try it first, then show the solutionSolution

The thermometer is in equilibrium with each, so by the zeroth law the copper and water are in thermal equilibrium with each other.

Answer: They are at the same temperature (zeroth law); no heat will flow if they touch.

Heat, internal energy and work

NCERT section 11.4

Internal energy U is the sum of the kinetic and potential energies of all the molecules of a system (in its own centre-of-mass frame). It is a state variable: its value depends only on the state (P, V, T), not on how the system got there. For an ideal gas, U depends on temperature alone.

Heat and work are not things a system has; they are two modes of energy transfer:

Heat: energy transferred because of a temperature difference.
Work: energy transferred by other means, such as a piston moving against the gas.

Work done by a gas when its volume changes by a small dV at pressure P: $d W = P d V$ , so $W = \int_{V_{1}}^{V_{2}} P d V$ , the area under the P-V curve. Expansion: W positive (done by the gas). Compression: W negative.

Worked example

A gas at a constant pressure of $2 \times 1 0^{5}$ Pa expands from 1.0 L to 3.0 L. How much work does it do? [1 mark]

Try it first, then show the solutionSolution

$W = P Δ V = 2 \times 1 0^{5} \times (3.0 - 1.0) \times 1 0^{- 3}$
W = 400 J (done by the gas)

Answer: 400 J

The first law of thermodynamics

NCERT section 11.5

Energy is conserved. Heat ΔQ supplied to a system goes partly into raising its internal energy and partly into work done by it:
$Δ Q = Δ U + Δ W$

Sign convention (NCERT): ΔQ is positive when heat is given to the system; ΔW is positive when work is done by the system; ΔU is positive when internal energy increases.

Isolated system: ΔQ = 0 and ΔW = 0, so ΔU = 0.
Since U is a state variable, ΔU between two states is the same by any path, while ΔQ and ΔW separately depend on the path.

Example: boiling water. Heat supplied goes mostly into the internal energy (separating molecules) and partly into work pushing back the atmosphere as steam forms.

Worked example

A gas absorbs 100 J of heat and does 40 J of work on its surroundings. Find the change in its internal energy. [1 mark]

Try it first, then show the solutionSolution

$Δ U = Δ Q - Δ W = 100 - 40$
ΔU = +60 J (internal energy rises)

Answer: +60 J

Specific heat capacities of gases: Cp − Cv = R

NCERT section 11.6

For gases the heat needed depends on the conditions:

Molar specific heat at constant volume $C_{v}$ : no work is done, so all the heat raises U: $Δ Q = μ C_{v} Δ T = Δ U$ .
Molar specific heat at constant pressure $C_{p}$ : the gas also expands and does work.

Mayer's relation: for one mole of an ideal gas at constant pressure, $Δ Q = Δ U + P Δ V$ . Since U depends only on T, $Δ U = C_{v} Δ T$ , and $P V = R T$ gives $P Δ V = R Δ T$ . So $C_{p} Δ T = C_{v} Δ T + R Δ T$ :
$C_{p} - C_{v} = R$

The ratio $γ = \frac{C _{p}}{C _{v}}$ is about $\frac{5}{3}$ for monatomic gases and $\frac{7}{5}$ for diatomic gases at room temperature (Kinetic Theory explains these values).

Worked example

For a monatomic ideal gas $C_{v} = \frac{3}{2} R$ . Find $C_{p}$ and γ. [1 mark]

Try it first, then show the solutionSolution

$C_{p} = C_{v} + R = \frac{5}{2} R$
$γ = \frac{C _{p}}{C _{v}} = \frac{5}{3} \approx 1.67$

Answer: $C_{p} = \frac{5}{2} R$ ; γ = 5/3

State variables and the equation of state

NCERT section 11.7

The equilibrium state of a system is fixed by a few state variables, such as P, V, T, mass and internal energy. They are linked by an equation of state; for an ideal gas it is $P V = μ R T$ , so only two of P, V, T are independent for a given amount.

Extensive variables depend on the size of the system: they double when you put two identical systems together (V, U, mass).
Intensive variables do not (P, T, density).

State variables describe equilibrium states only. During a rapid, turbulent change the gas has no single well-defined pressure or temperature.

Worked example

Classify as extensive or intensive: internal energy, pressure, volume, temperature, density. [1 mark]

Try it first, then show the solutionSolution

Doubling the system doubles U and V, but leaves P, T and density unchanged.

Answer: Extensive: internal energy, volume. Intensive: pressure, temperature, density.

Quasi-static and isothermal processes

NCERT section 11.8.1 and 11.8.2

A quasi-static process is carried out so slowly that the system stays in equilibrium (with well-defined P and T) at every stage. It is an idealisation; real slow processes come close.

Isothermal process: temperature stays constant (gas in a cylinder with a conducting base on a large heat reservoir, changed slowly). For an ideal gas:

$P V$ = constant (Boyle's law); the P-V curve is a hyperbola.
U depends only on T, so ΔU = 0 and ΔQ = ΔW: all heat absorbed is turned into work.
Work done: $W = \int P d V = \int \frac{μ R T}{V} d V = μ R T ln \frac{V _{2}}{V _{1}}$ .

Expansion ( $V_{2} > V_{1}$ ): gas absorbs heat and does work. Compression: work is done on it and it gives out heat.

Worked example

Two moles of an ideal gas at 400 K expand isothermally to three times their volume. Find the work done and the heat absorbed. (R = 8.314 J mol $^{- 1}$ K $^{- 1}$ , ln 3 = 1.099) [2 marks]

Try it first, then show the solutionSolution

$W = μ R T ln \frac{V _{2}}{V _{1}} = 2 \times 8.314 \times 400 \times 1.099 \approx 7.31 \times 1 0^{3} J$
ΔU = 0, so heat absorbed Q = W ≈ 7.31 kJ

Answer: About 7.31 kJ each

Adiabatic, isochoric, isobaric and cyclic processes

NCERT section 11.8.3 to 11.8.6

Adiabatic process: no heat enters or leaves (insulated system, or a very fast change such as a bursting tyre). ΔQ = 0, so $Δ W = - Δ U$ : work done by the gas comes from its internal energy, so it cools on expansion and warms on compression. For an ideal gas:
$P V^{γ} = constant$ , $T V^{γ - 1} = constant$
$W = \frac{μ R ( T _{1} - T _{2} )}{γ - 1}$
On a P-V diagram the adiabatic curve is steeper than the isothermal through the same point.

Isochoric process (constant volume): W = 0, so all heat changes U: $Δ Q = Δ U = μ C_{v} Δ T$ .

Isobaric process (constant pressure): $W = P (V_{2} - V_{1}) = μ R (T_{2} - T_{1})$ and $Δ Q = μ C_{p} Δ T$ .

Cyclic process: the system returns to its initial state, so ΔU = 0 and ΔQ = ΔW. The net work equals the area enclosed by the loop on the P-V diagram (positive if traversed clockwise).

Worked example

Air (γ = 1.4) at 27 °C is compressed adiabatically to one-eighth of its volume. Find its final temperature. [2 marks]

Try it first, then show the solutionSolution

$T_{2} = T_{1} (\frac{V _{1}}{V _{2}})^{γ - 1} = 300 \times 8^{0.4}$
$8^{0.4} \approx 2.30$ , so $T_{2} \approx 689 K$ (about 416 °C)

Answer: About 689 K

Second law; reversible and irreversible processes

NCERT section 11.9 and 11.10

The first law allows many processes that never happen: a cold cup of tea never gets hotter by taking heat from the cooler room. The second law rules them out.

Kelvin-Planck statement: you cannot have a process that does nothing except take heat from one reservoir and turn all of it into work; something else must also change.
Clausius statement: heat cannot pass from a colder body to a hotter one all by itself; some other effect (such as outside work) must go with it.

(In an isothermal expansion all the absorbed heat does become work, but the gas is left with a larger volume, so that is not the 'sole result'.)

A reversible process can be retraced exactly, with both system and surroundings returned to their original states. It needs to be quasi-static and free of dissipation (friction, viscosity, heat flow across a finite temperature difference). Irreversible processes include free expansion of a gas into vacuum, heat flowing from hot to cold, mixing of gases and any process with friction. All real processes are irreversible; reversible ones are useful ideals.

Worked example

Explain why the free expansion of a gas into an evacuated chamber is irreversible. [2 marks]

Try it first, then show the solutionSolution

The gas rushes into the vacuum rapidly and turbulently; it does not pass through equilibrium states.
The gas never collects back into the first chamber by itself; to restore it, work must be done on it from outside, changing the surroundings.

Answer: It cannot be retraced without a change in the surroundings, so it is irreversible.

Physics in real life: Thermodynamics

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Adiabatic compression

Suppose a science teacher demonstrates a 'fire syringe': air at 27 °C in a sealed tube is pushed by a piston to one third of its volume so quickly that no heat escapes.

Estimate the temperature of the compressed air. (γ = 1.4)

Try it first, then show the solutionSolution

$T_{2} = T_{1} (\frac{V _{1}}{V _{2}})^{γ - 1} = 300 \times 3^{0.4}$
$3^{0.4} \approx 1.55$ , so $T_{2} \approx 466 K$ (about 193 °C)

Answer: About 466 K, hot enough to scorch a wisp of cotton placed inside

Memory hook: Squeeze fast, heat up; let out fast, cool down.

Adiabatic expansion

Imagine pressing the nozzle of a deodorant can for several seconds: the can feels cold in your hand.

Suppose the gas expands quickly as it leaves. Use the first law to explain the cooling.

Try it first, then show the solutionSolution

The expansion is too fast for much heat to flow in: ΔQ ≈ 0
The gas does work pushing the atmosphere back, so ΔW > 0
ΔU = −ΔW < 0: internal energy and hence temperature fall

Answer: The gas does work at the expense of its internal energy, so it cools.

Memory hook: No heat in, work out: the energy comes from inside.

Work converted to internal energy

Suppose you rub your hands together, pushing with a friction force of 10 N through 0.2 m on each of 20 strokes.

How much work turns into internal energy, and by roughly how much would it warm 0.1 kg of skin (c ≈ 3500 J kg $^{- 1}$ K $^{- 1}$ ) if none escaped?

Try it first, then show the solutionSolution

Work $= 10 \times 0.2 \times 20 = 40 J$
$Δ T = \frac{40}{0.1 \times 3500} \approx 0.11 K$

Answer: 40 J; about 0.1 K warmer

Memory hook: Work can raise internal energy without any heat being supplied.

Topics in this chapter

Thermodynamics formula sheet

First law: $Δ Q = Δ U + Δ W$
Work done by a gas: $W = \int_{V_{1}}^{V_{2}} P d V$
Mayer's relation and γ: $C_{p} - C_{v} = R, γ = \frac{C _{p}}{C _{v}}$
Isothermal process (ideal gas): $P V = const, Δ U = 0, W = μ R T ln \frac{V _{2}}{V _{1}}$
Adiabatic process (ideal gas): $P V^{γ} = const, T V^{γ - 1} = const, W = \frac{μ R ( T _{1} - T _{2} )}{γ - 1}$
Isochoric process: $W = 0, Δ Q = Δ U = μ C_{v} Δ T$
Isobaric process: $W = P Δ V = μ R Δ T, Δ Q = μ C_{p} Δ T$
Cyclic process: $Δ U = 0, Δ Q = Δ W = area of the loop$

How the school exam asks Thermodynamics

The question patterns to expect, and the method that works for each.

Question type	How it is asked	Method
Laws and definitions	State the zeroth, first and second laws; explain internal energy as a state variable; heat vs work.	Exact statements with the word 'sole' in the second law; give one example each.
First-law numericals	Find ΔU, Q or W in a process; heater supplying heat while the system does work.	Use ΔQ = ΔU + ΔW with NCERT signs; rates work the same way (per second).
Processes	Work, heat and ΔU in isothermal, adiabatic, isochoric and isobaric changes; final temperature or pressure in adiabatic changes.	Pick the process formula; remember ΔU = $μ C_{v} Δ T$ for any process of an ideal gas.
P-V diagrams	Work from the area; compare isothermal and adiabatic curves; work in a cycle.	Area under curve = work; loop area = net work; adiabatic is steeper.
Derivations	Derive $C_{p} - C_{v} = R$ ; derive work in an isothermal process.	First law with $P V = μ R T$ ; integrate $μ R T / V$ .
Second law and reversibility	Kelvin-Planck and Clausius statements; examples of irreversible processes; conditions for reversibility.	Quasi-static and no dissipation; list friction, free expansion, heat flow across finite ΔT.

Tricks to remember Thermodynamics

Short memory aids. Say each one aloud once, then use it on a question the same day.

Q = U + W (heat in = rise + work out)
Money in = savings + spending. Heat in is split between internal energy and work done by the gas.
Use it when: First-law sign questions.
Iso-names
Iso-THERMal: T constant. Iso-CHORic: volume (chamber) constant. Iso-BARic: pressure (bar) constant. A-diabatic: no heat passes.
Use it when: Naming processes.
Zero for each process
Isothermal: ΔU = 0. Adiabatic: Q = 0. Isochoric: W = 0. Cyclic: ΔU = 0.
Use it when: Applying the first law quickly.
Adiabat is steeper
On a P-V graph through the same point, the adiabatic curve falls faster than the isothermal (γ > 1).
Use it when: Graph questions.
Sole result
The second law forbids only processes whose sole result is perfect heat-to-work conversion or cold-to-hot heat flow.
Use it when: Second-law reasoning.

Thermodynamics: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Example 1: Isothermal change of an ideal gas · 1 mark · MCQExam level

In an isothermal process of an ideal gas:
(a) ΔQ = 0
(b) ΔW = 0
(c) ΔU = 0
(d) ΔP = 0

Try it first, then show the solutionSolution

U of an ideal gas depends only on T, which is constant.

Answer: (c) ΔU = 0

Example 2: Heating by compression · 1 mark · Assertion-reasonExam level

Assertion (A): When a gas is compressed adiabatically, its temperature rises.
Reason (R): In an adiabatic process no heat is exchanged, so the work done on the gas goes into its internal energy.
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.

Try it first, then show the solutionSolution

ΔQ = 0 and ΔW < 0 (work done on the gas), so ΔU = −ΔW > 0 and T rises.

Answer: (a)

Example 3: Work in a rectangular cycle · 2 marks · Very short answerHarder exam level

A gas is taken round a rectangular cycle on a P-V diagram between pressures $1 \times 1 0^{5}$ Pa and $3 \times 1 0^{5}$ Pa and volumes 1 L and 3 L, clockwise. Find the net work done and the net heat absorbed. [2 marks]

Try it first, then show the solutionSolution

Net work = area of the rectangle $= (3 - 1) \times 1 0^{5} \times (3 - 1) \times 1 0^{- 3} = 400 J$
Cyclic: ΔU = 0, so net heat absorbed = 400 J

Answer: 400 J each

Example 4: Heating at constant pressure · 3 marks · Short answerHarder exam level

Two moles of a diatomic ideal gas ( $C_{v} = \frac{5}{2} R$ ) are heated at constant pressure from 300 K to 350 K. Find the heat supplied, the work done by the gas and the change in internal energy. (R = 8.314 J mol $^{- 1}$ K $^{- 1}$ ) [3 marks]

Try it first, then show the solutionSolution

$Q = μ C_{p} Δ T = 2 \times \frac{7}{2} \times 8.314 \times 50 \approx 2910 J$
$W = μ R Δ T = 2 \times 8.314 \times 50 \approx 831 J$
$Δ U = μ C_{v} Δ T = 2 \times \frac{5}{2} \times 8.314 \times 50 \approx 2079 J$ (check: 2910 − 831 = 2079 ✓)

Answer: Q ≈ 2910 J; W ≈ 831 J; ΔU ≈ 2079 J

Example 5: Isothermal compression · 2 marks · Very short answerHarder exam level

Two moles of an ideal gas at 300 K are compressed isothermally from 10 L to 5 L. Find the work done by the gas and the heat exchanged. (R = 8.314 J mol $^{- 1}$ K $^{- 1}$ , ln 2 = 0.693) [2 marks]

Try it first, then show the solutionSolution

$W = μ R T ln \frac{V _{2}}{V _{1}} = 2 \times 8.314 \times 300 \times ln 0.5 \approx - 3458 J$
ΔU = 0, so Q = W = −3458 J: the gas gives out about 3.46 kJ of heat

Answer: W ≈ −3.46 kJ; 3.46 kJ of heat released

Example 6: Isothermal versus adiabatic · 5 marks · Long answerHarder exam level

(a) Distinguish between isothermal and adiabatic processes (three points). (b) Sketch both curves on a P-V diagram starting from the same state, and say which is steeper. (c) One mole of a monatomic gas (γ = 5/3) at 400 K expands to eight times its volume, first isothermally and then (starting again) adiabatically. Find the work done in each case. (ln 8 = 2.079, R = 8.314 J mol $^{- 1}$ K $^{- 1}$ ) [5 marks]

Try it first, then show the solutionSolution

(a) Isothermal: T constant, heat exchanged, PV = constant, needs slow change in a conducting vessel. Adiabatic: Q = 0, T changes, $P V^{γ}$ = constant, needs an insulated vessel or a fast change
(b) Both are falling curves; the adiabatic one is steeper (pressure falls faster because the gas also cools)
(c) Isothermal: $W = R T ln 8 = 8.314 \times 400 \times 2.079 \approx 6914 J$
Adiabatic: $T_{2} = T_{1} (\frac{V _{1}}{V _{2}})^{γ - 1} = 400 \times 8^{- 2/3} = \frac{400}{4} = 100 K$
$W = \frac{R ( T _{1} - T _{2} )}{γ - 1} = \frac{8.314 \times 300}{2/3} \approx 3741 J$

Answer: Isothermal ≈ 6.91 kJ; adiabatic ≈ 3.74 kJ

Example 7: Free expansion · 3 marks · Short answerHOTS / competency

An insulated container is divided into two equal halves. One half holds an ideal gas at 300 K and the other is a vacuum. The partition is removed. Find the final temperature and explain using the first law. Is the process reversible? [3 marks]

Try it first, then show the solutionSolution

Insulated: ΔQ = 0. The gas expands into vacuum, pushing nothing: ΔW = 0
So ΔU = 0; for an ideal gas U depends only on T, so T stays 300 K (pressure halves)
It is irreversible: the gas never returns to one half on its own, and the expansion is not quasi-static

Answer: 300 K; irreversible

Example 8: All heat into work? · 2 marks · Very short answerHOTS / competency

In the isothermal expansion of an ideal gas, all the heat absorbed is converted into work. Does this contradict the Kelvin-Planck statement of the second law? Explain. [2 marks]

Try it first, then show the solutionSolution

Kelvin-Planck forbids complete conversion of heat into work as the sole result.
Here the gas also ends up in a different state (larger volume, lower pressure); to use it again it must be compressed, so the conversion is not the only result. No contradiction.

Answer: No: the conversion is not the sole result, since the gas's state has changed.

Case-study practice (Section E style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 4-mark question split into 1 + 1 + 1 + 1 marks.

A gas in a cylinder

Suppose 1 mole of an ideal monatomic gas ( $C_{v} = \frac{3}{2} R$ , R = 8.3 J mol $^{- 1}$ K $^{- 1}$ ) in a cylinder with a piston is heated at constant volume from 300 K to 400 K, and then allowed to expand at constant pressure until its temperature reaches 500 K.

(i) · 1 mark
How much heat is absorbed in the constant-volume step?
Answer
Answer: 1245 J
$μ C_{v} Δ T = 1.5 \times 8.3 \times 100$
(ii) · 1 mark
How much work is done in the constant-volume step?
Answer
Answer: Zero
No change in volume.
(iii) · 1 mark
How much work does the gas do in the constant-pressure step?
Answer
Answer: 830 J
$μ R Δ T = 8.3 \times 100$
(iv) · 1 mark
How much heat is absorbed in the constant-pressure step?
Answer
Answer: 2075 J
$μ C_{p} Δ T = 2.5 \times 8.3 \times 100$

Pumping a tyre

Suppose a cyclist pumps air quickly into a tyre. In each stroke the air in the pump (γ = 1.4, initially at 27 °C and 1 atm) is compressed to half its volume before the valve opens.

(i) · 1 mark
What kind of process is a quick compression?
Answer
Answer: Adiabatic
Too fast for heat to escape.
(ii) · 1 mark
Find the temperature after compression.
Answer
Answer: About 396 K
$300 \times 2^{0.4}$
(iii) · 1 mark
Find the pressure after compression.
Answer
Answer: About 2.64 atm
$2^{1.4} \approx 2.64$
(iv) · 1 mark
Why does the pump barrel get warm?
Answer
Answer: Work done on the air raises its internal energy
ΔQ = 0, so ΔU = −ΔW > 0; the hot air then warms the barrel.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

A system is given 150 J of heat per second by a heater, and does work at 90 J per second. Find the rate of change of its internal energy. What happens to its temperature if it is an ideal gas? [2 marks] (2 marks)

Step	Marks
$\frac{Δ U}{Δ t} = \frac{Δ Q}{Δ t} - \frac{Δ W}{Δ t} = 150 - 90 = 60 J s^{- 1}$	1
U increases, and for an ideal gas U depends only on T, so its temperature rises	1

Derive Mayer's relation $C_{p} - C_{v} = R$ for an ideal gas. [3 marks] (3 marks)

Step	Marks
First law for one mole: $Δ Q = Δ U + P Δ V$ ; at constant volume $Δ Q = Δ U = C_{v} Δ T$	1
At constant pressure: $C_{p} Δ T = Δ U + P Δ V$ , and ΔU is the same ( $C_{v} Δ T$ ) because U depends only on T	1
$P V = R T \Rightarrow P Δ V = R Δ T$ ; so $C_{p} Δ T = C_{v} Δ T + R Δ T$ , i.e. $C_{p} - C_{v} = R$	1

(a) Derive an expression for the work done by μ moles of an ideal gas in an isothermal expansion from $V_{1}$ to $V_{2}$ . (b) One mole of a diatomic ideal gas (γ = 1.4) at 300 K doubles its volume (i) isothermally, (ii) adiabatically. Find the work done in each case. (c) Which is larger, and why? (R = 8.314 J mol $^{- 1}$ K $^{- 1}$ , ln 2 = 0.693, $2^{- 0.4} = 0.758$ ) [5 marks] (5 marks)

Step	Marks
(a) $W = \int_{V_{1}}^{V_{2}} P d V = \int \frac{μ R T}{V} d V = μ R T ln \frac{V _{2}}{V _{1}}$	1
(b)(i) $W = 8.314 \times 300 \times 0.693 \approx 1729 J$	1
(ii) $T_{2} = 300 \times 2^{- 0.4} \approx 227.4 K$	1
$W = \frac{R ( T _{1} - T _{2} )}{γ - 1} = \frac{8.314 \times 72.6}{0.4} \approx 1510 J$	1
(c) Isothermal work is larger: in the adiabatic case the gas cools and its pressure falls faster, so the area under its P-V curve is smaller	1

Where students lose marks in Thermodynamics

The slips examiners see most often, and what to write instead.

Mistake: Mixing sign conventions
Some books use W done on the gas.
Wrong working
Gas compressed with 50 J of work, no heat: ΔU = Q + W = 0 + (−50) = −50 J
Right working
With NCERT signs, W done by gas = −50 J, so ΔU = Q − W = 0 − (−50) = +50 J
Do this: Use ΔQ = ΔU + ΔW with W done by the system, and say so.
Usually costs: 1-2 marks.
Mistake: Saying heat and work are state variables
Thinking a system 'contains heat'.
Do this: Only U (with P, V, T) is a state variable; Q and W depend on the path.
Usually costs: 1 mark.
Mistake: Using $P V$ = constant for an adiabatic change
Mixing up Boyle's law and the adiabatic law.
Do this: Adiabatic: $P V^{γ}$ = constant and $T V^{γ - 1}$ = constant.
Usually costs: Whole numerical.
Mistake: Using Celsius in $T V^{γ - 1}$ or PV = μRT
Habit.
Do this: Always convert to kelvin.
Usually costs: Whole numerical.
Mistake: Writing heat-engine efficiency or refrigerator answers
Old syllabus.
Do this: Heat engines and refrigerators were removed for 2026-27, and the Carnot engine is not named in the CBSE content.
Usually costs: Time.

NCERT exercise guide: Thermodynamics

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Exercises 11.1-11.2 (heat and specific heat)

What it tests: Fuel consumption of a geyser and the heat needed to warm a gas at constant pressure.

How to approach it: Q = mcΔT or $μ C_{p} Δ T$ with $C_{p} = C_{v} + R$ .

Same kind as Exercise 11.2

How much heat is needed to raise the temperature of 0.028 kg of nitrogen (one mole) by 50 K at constant pressure? Take $C_{v} = \frac{5}{2} R$ , R = 8.3 J mol $^{- 1}$ K $^{- 1}$ .

Show the solutionSolution

$C_{p} = \frac{7}{2} R = 29.05 J m o l^{- 1} K^{- 1}$
$Q = μ C_{p} Δ T = 1 \times 29.05 \times 50 \approx 1453 J$

Answer: About 1.45 kJ

Exercises 11.3-11.6 (processes and reasoning)

What it tests: Why coolants have high specific heat, why air pressure in tyres changes, adiabatic compression of hydrogen, adiabatic work independent of path, and free expansion between connected cylinders.

How to approach it: First law with the right zero (Q = 0 adiabatic, ΔU = 0 isothermal); $P V^{γ}$ constant.

Same kind as Exercise 11.4

A monatomic ideal gas is compressed quickly (adiabatically) to half its volume. By what factor does its pressure rise?

Show the solutionSolution

$P_{2} = P_{1} (\frac{V _{1}}{V _{2}})^{γ} = P_{1} \times 2^{5/3}$
$2^{5/3} \approx 3.17$

Answer: About 3.2 times

Exercises 11.7-11.8 (first law and P-V paths)

What it tests: A heater supplying heat while the system does work, and the work done along a path on a P-V diagram.

How to approach it: Rates: dU/dt = dQ/dt − dW/dt; work = area under the path.

Same kind as Exercise 11.8

A gas goes from state A ( $2 \times 1 0^{5}$ Pa, 2 L) at constant volume to B ( $4 \times 1 0^{5}$ Pa, 2 L), then at constant pressure to C ( $4 \times 1 0^{5}$ Pa, 5 L). Find the total work done by the gas.

Show the solutionSolution

A → B: constant volume, W = 0
B → C: $W = 4 \times 1 0^{5} \times 3 \times 1 0^{- 3} = 1200 J$

Answer: 1200 J

Practice questions on Thermodynamics

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQExam level
The concept of temperature follows from the:
(a) zeroth law
(b) first law
(c) second law
(d) Boyle's law
Hint
Thermal equilibrium.
Answer
(a) zeroth law
Q2 · 1 mark · MCQExam level
In which process is no work done by the gas?
(a) isothermal
(b) isobaric
(c) isochoric
(d) adiabatic
Hint
W = PΔV.
Answer
(c) isochoric
Q3 · 1 mark · Assertion-reasonExam level
Assertion (A): For an ideal gas, $C_{p}$ is greater than $C_{v}$ .
Reason (R): At constant pressure part of the heat supplied is used by the gas to do work in expanding.
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.
Hint
Mayer's relation.
Answer
(a)
Q4 · 2 marks · Very short answerExam level
State the first law of thermodynamics and the sign convention for each term.
Hint
Energy conservation.
Answer
ΔQ = ΔU + ΔW. ΔQ is positive when heat is supplied to the system; ΔU is positive when its internal energy increases; ΔW is positive when the system does work on its surroundings.
Q5 · 1 mark · Very short answerHarder exam level
How much heat is needed to raise the temperature of 1 mole of a monatomic ideal gas by 10 K at constant volume? (R = 8.314 J mol $^{- 1}$ K $^{- 1}$ )
Hint
$C_{v} = \frac{3}{2} R$ .
Answer
$Q = 1 \times 1.5 \times 8.314 \times 10 \approx 125$ J.
Q6 · 2 marks · Very short answerHarder exam level
A gas expands from 2 L to 6 L against a constant pressure of $1.5 \times 1 0^{5}$ Pa while absorbing 800 J of heat. Find the change in its internal energy.
Hint
W = PΔV.
Answer
$W = 1.5 \times 1 0^{5} \times 4 \times 1 0^{- 3} = 600$ J; ΔU = 800 − 600 = 200 J.
Q7 · 2 marks · Very short answerHarder exam level
Show that for an ideal gas undergoing an adiabatic process, $T V^{γ - 1}$ is constant, starting from $P V^{γ}$ = constant.
Hint
Substitute P = μRT/V.
Answer
$P V^{γ} = \frac{μ R T}{V} V^{γ} = μ R T V^{γ - 1}$ . Since μR is constant, $T V^{γ - 1}$ is constant.
Q8 · 2 marks · Very short answerHarder exam level
A gas is taken along a triangular cycle on a P-V diagram with corners at ( $1 \times 1 0^{5}$ Pa, 1 L), ( $1 \times 1 0^{5}$ Pa, 4 L) and ( $3 \times 1 0^{5}$ Pa, 1 L). Find the magnitude of the net work done per cycle.
Hint
Area of a triangle.
Answer
Area $= \frac{1}{2} \times (3 \times 1 0^{- 3}) \times (2 \times 1 0^{5}) = 300$ J.
Q9 · 3 marks · Short answerHarder exam level
List three conditions that make a process irreversible, with an example of each.
Hint
Dissipation and non-equilibrium.
Answer
(1) Friction or viscosity, e.g. a block sliding to rest. (2) Heat flow across a finite temperature difference, e.g. a hot cup cooling in a room. (3) Rapid, non-quasi-static change such as free expansion of a gas into vacuum or the bursting of a balloon.
Q10 · 3 marks · Short answerHOTS / competency
Starting from the same state, an ideal gas is expanded to the same final volume (a) isothermally, (b) adiabatically. Which process ends at the higher pressure, and which does more work? Explain with a P-V sketch.
Hint
Adiabat is steeper.
Answer
The isothermal curve lies above the adiabatic one after the common starting point, because the gas cools in the adiabatic expansion. So the isothermal process ends at a higher pressure and has more area under its curve: it does more work.
Q11 · 2 marks · Very short answerHOTS / competency
A gas is compressed adiabatically and then allowed to cool back to its original temperature at constant volume. Explain, using the first law, whether its internal energy at the end is greater than, less than or equal to its initial value.
Hint
U depends only on T for an ideal gas.
Answer
Equal (for an ideal gas). The adiabatic compression raises U by the work done on it; cooling at constant volume removes exactly that much as heat, returning T, and hence U, to the initial value.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.State the zeroth law of thermodynamics.Show answer
If two systems are each in thermal equilibrium with a third, they are in thermal equilibrium with each other.
2.State the first law of thermodynamics.Show answer
ΔQ = ΔU + ΔW: heat supplied equals the rise in internal energy plus the work done by the system.
3.What is Mayer's relation?Show answer
$C_{p} - C_{v} = R$ .
4.What is ΔU in an isothermal process of an ideal gas?Show answer
Zero.
5.Write the equation of an adiabatic process.Show answer
$P V^{γ}$ = constant.
6.What does the area enclosed by a cycle on a P-V diagram represent?Show answer
The net work done in the cycle.
7.State the Clausius statement.Show answer
Heat cannot flow from a colder body to a hotter one with no other effect taking place.
8.Give two conditions for a process to be reversible.Show answer
It must be quasi-static, and free of dissipative effects such as friction.

Best order to study it

Learn system, surroundings, equilibrium and the zeroth law.
Learn internal energy, heat and work, and work as the area under a P-V curve.
Learn the first law with NCERT sign conventions.
Derive $C_{p} - C_{v} = R$ and learn γ.
Learn state variables, the equation of state, extensive and intensive variables.
Study isothermal, adiabatic, isochoric, isobaric and cyclic processes with their formulas.
Finish with the second law and reversible and irreversible processes.

Quick revision checklist

The night before the exam: can you do each of these without looking?

Can you explain how the zeroth law defines temperature?
Can you explain why U is a state variable but Q is not?
Can you apply the first law with correct signs?
Can you derive Mayer's relation?
Can you derive the work done in an isothermal expansion?
Can you find the final temperature after an adiabatic change?
Can you find W, Q and ΔU in an isobaric process?
Can you find the work in a cycle from a P-V diagram?
Can you state the Kelvin-Planck and Clausius statements?
Can you give three examples of irreversible processes?

Thermodynamics: frequently asked questions

Are heat engines and the Carnot engine in the 2026-27 syllabus?

Heat engines and refrigerators (old 12.9 and 12.10) were removed. The NCERT book still prints the Carnot engine (11.11), but the CBSE content for this unit does not name it, so it is not expected in the school exam.

Which sign convention should I use?

NCERT writes ΔQ = ΔU + ΔW, with ΔQ positive for heat given to the system and ΔW positive for work done by the system. Use this convention and state it in your answer.

Why is the adiabatic curve steeper than the isothermal one?

In an adiabatic expansion the gas also cools, so its pressure falls faster than in an isothermal expansion. Mathematically, the slope of PV^\gamma = constant is γ times the slope of PV = constant at the same point.

Does an isothermal expansion violate the second law, since all heat becomes work?

No. The conversion is not the sole result: the gas ends up at a larger volume and lower pressure. To repeat the process, it must be compressed again, which needs work and gives out heat.

How does this chapter link to Kinetic Theory?

Kinetic Theory explains internal energy as molecular kinetic energy, derives C_v = \frac{3}{2}R for monatomic gases, and gives γ = 5/3 and 7/5, values used in adiabatic problems here.

Based on: CBSE Physics (042) syllabus 2026-27 (Classes XI-XII); NCERT Physics Part II, Textbook for Class XI, Chapter 11 (Reprint 2026-27)