Thermodynamics ISC Class 11: Notes, Processes and Heat Engines

What Thermodynamics is about

Weightage: Thermodynamics belongs to unit 8 of the ISC Class XI Physics syllabus. CISCE weights units 7-9 together (Properties of Bulk Matter; Heat and Thermodynamics; Behaviour of Perfect Gases and Kinetic Theory of Gases) at 20 of the 70 marks in Paper I (Theory). Class XI is examined by the school, so your school sets the paper within this weightage.Source: CISCE ISC Physics syllabus, examination year 2028, Class XI unit weightage

Key concepts

Thermal equilibrium and the zeroth law

Two systems in thermal contact exchange heat until their macroscopic properties stop changing: they are then in thermal equilibrium. 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. The property that is equal for all systems in mutual thermal equilibrium is called temperature. The law justifies the use of a thermometer: the thermometer (C) is brought into equilibrium with each body in turn, and equal readings mean the bodies would not exchange heat.

Worked example

A thermometer reads 37 °C when held in a patient's mouth and also 37 °C when dipped in a bowl of water. What does the zeroth law say about the patient and the water?

Try it first, then show the solutionSolution

1. The thermometer (C) is in equilibrium with the patient (A) and with the water (B).
2. By the zeroth law, A and B are in thermal equilibrium: if the patient's hand were placed in the water, there would be no net heat flow.

Answer: They are at the same temperature; no net heat would flow between them

Heat, work and internal energy

Internal energy U of a system is the total energy of its molecules: random kinetic energy plus the potential energy of intermolecular forces. For an ideal gas there are no intermolecular forces, so U depends only on temperature. U is a state function: its change between two states is the same whatever path is followed. Heat Q is energy transferred because of a temperature difference; work W is energy transferred by other means (a moving piston). Neither is a property of the system: both depend on the path. A system does not 'contain' heat or work; it contains internal energy. Joule showed that the same rise in temperature can be produced by heat or by work: 1 cal = 4.186 J exactly (the mechanical equivalent of heat).

Worked example

A 10 kg weight falls through 2 m twenty times, turning paddles in 1 kg of water in an insulated container. Find the work done and the rise in temperature. (g = 9.8 m s⁻²; c = 4186 J kg⁻¹ K⁻¹)

Try it first, then show the solutionSolution

1. W = 20 × mgh = 20 × 10 × 9.8 × 2 = 3920 J
2. All of it raises U: ΔT = 3920/(1 × 4186) = 0.94 K

Answer: 3920 J; about 0.94 K (no heat was supplied, yet the water warms)

Work done by a gas and the pV diagram

When a gas at pressure p pushes a piston of area A through dx, it does work dW = pA dx = p dV. Over a change of volume, $W={\int }_{V_1}^{V_2}pdV$, the area under the curve on a pV diagram. Expansion: W positive (done by the gas); compression: W negative (done on the gas). At constant pressure, W = p(V₂ − V₁). For a cyclic process, the net work is the area enclosed by the loop: positive if traced clockwise (an engine), negative if anticlockwise. Because the area depends on the path, so does W.

Worked example

A gas at a constant pressure of $1\times 10^5$ Pa expands from 2 litres to 5 litres. Find the work done by the gas.

Try it first, then show the solutionSolution

1. ΔV = 3 L = 3 × 10⁻³ m³
2. W = pΔV = 10⁵ × 3 × 10⁻³ = 300 J

Answer: 300 J (done by the gas)

First law of thermodynamics

First law: the heat supplied to a system equals the increase in its internal energy plus the work done by the system: $Q=\Delta U+W$ (in differential form, dQ = dU + p dV). It is the law of conservation of energy including heat. Sign convention (as in the syllabus): Q > 0 when heat is given to the system; W > 0 when work is done by the system; ΔU > 0 when internal energy rises. Special cases: adiabatic, Q = 0, so ΔU = −W; isothermal (ideal gas), ΔU = 0, so Q = W; isochoric, W = 0, so Q = ΔU; cyclic, ΔU = 0, so Q = W (net). The first law says energy cannot be created, but does not say which way a process will go: that is the second law's job.

Worked example

A gas absorbs 500 J of heat and does 200 J of work. Find the change in its internal energy. What would ΔU be if 200 J of work were done on the gas while it lost 50 J of heat?

Try it first, then show the solutionSolution

1. ΔU = Q − W = 500 − 200 = 300 J
2. Second case: Q = −50 J, W = −200 J: ΔU = −50 − (−200) = +150 J

Answer: +300 J; +150 J

Isochoric, isobaric and isothermal processes

Isochoric (constant volume): W = 0, so Q = ΔU = nC_vΔT; the pV graph is a vertical line. Isobaric (constant pressure): W = pΔV = nRΔT, Q = nC_pΔT; a horizontal line. Isothermal (constant temperature): for an ideal gas, U is constant, so ΔU = 0 and Q = W; Boyle's law pV = constant holds, and the graph is a rectangular hyperbola. It must happen slowly in a container with conducting walls, in contact with a large reservoir. Work in an isothermal process: $W={\int }_{V_1}^{V_2}\frac{nRT}{V}dV=nRT\ln \frac{V_2}{V_1}=2.303nRT{\log }_{10}\frac{V_2}{V_1}=nRT\ln \frac{p_1}{p_2}$.

Worked example

Two moles of an ideal gas at 300 K expand isothermally to twice their volume. Find the work done and the heat absorbed. (R = 8.31 J mol⁻¹ K⁻¹)

Try it first, then show the solutionSolution

1. W = nRT ln 2 = 2 × 8.31 × 300 × 0.693
2. W = 3456 J; ΔU = 0, so Q = W = 3456 J

Answer: About 3.46 kJ of work, with the same heat absorbed

Adiabatic processes

In an adiabatic process no heat enters or leaves (Q = 0): it happens quickly, or in a thermally insulated container. Then ΔU = −W: a gas that expands adiabatically does work at the cost of its internal energy and cools; one compressed adiabatically heats up (a bicycle pump gets warm). For an ideal gas, $p{V}^{\gamma }$ = constant, $T{V}^{\gamma -1}$ = constant and $\frac{T^{\gamma }}{p^{\gamma -1}}$ = constant, where γ = C_p/C_v (derivation not required). On a pV diagram an adiabatic curve is steeper than an isothermal through the same point. Work done: $W=\frac{p_1{V}_1-p_2{V}_2}{\gamma -1}=\frac{nR(T_1-T_2)}{\gamma -1}$.

Worked example

A monatomic ideal gas (γ = 5/3) at 300 K is compressed adiabatically to one-eighth of its volume. Find its final temperature.

Try it first, then show the solutionSolution

1. $T_2=T_1{\left(\frac{V_1}{V_2}\right)}^{\gamma -1}=300\times 8^{2/3}$
2. $8^{2/3}=4$, so T₂ = 1200 K

Answer: 1200 K (927 °C): the principle of ignition in a diesel engine

Principal and molar heat capacities; Mayer's relation

A gas has two principal specific heats: at constant volume ($c_v$) and at constant pressure ($c_p$). Per mole, these are the molar heat capacities $C_v$ and $C_p$ (J mol⁻¹ K⁻¹). Mayer's relation: at constant volume all the heat goes into U: $n{C}_v\Delta T=\Delta U$. At constant pressure the same ΔU is needed (U depends only on T) plus work pΔV = nRΔT: $n{C}_p\Delta T=n{C}_v\Delta T+nR\Delta T$, so $C_p-C_v=R$. So $C_p>C_v$ always. The ratio $\gamma =\frac{C_p}{C_v}$ is 5/3 for a monatomic gas ($C_v=\frac{3}{2}R$), 7/5 for a diatomic gas at room temperature ($C_v=\frac{5}{2}R$). In terms of specific heats, $c_p-c_v=\frac{R}{M}$, with M the molar mass.

Worked example

Find the heat needed to raise the temperature of 2 mol of oxygen (diatomic) by 10 K (a) at constant pressure, (b) at constant volume. Explain the difference. (R = 8.31 J mol⁻¹ K⁻¹)

Try it first, then show the solutionSolution

1. (a) $Q_p=n{C}_p\Delta T=2\times \frac{7}{2}\times 8.31\times 10=581.7J$
2. (b) $Q_v=n{C}_v\Delta T=2\times \frac{5}{2}\times 8.31\times 10=415.5J$
3. Difference = 166.2 J = nRΔT, the work done in expanding at constant pressure.

Answer: (a) 582 J (b) 416 J

Second law; reversible and irreversible processes

Kelvin-Planck statement: it is impossible to construct an engine which, working in a cycle, absorbs heat from a single source and converts it entirely into work. (Some heat must always be rejected to a colder sink.) Clausius statement: it is impossible for heat to flow by itself from a colder body to a hotter body; a refrigerator needs external work. The two statements are equivalent. A reversible process can be retraced exactly in the opposite direction, with the system and surroundings returning to their original states: it must be infinitely slow (quasi-static), with no friction, viscosity or heat flow across a finite temperature difference. All real processes (free expansion of a gas, heat conduction, friction, mixing) are irreversible; reversible processes are ideals.

Worked example

State whether each process is reversible or irreversible: (a) a gas rushing into a vacuum, (b) a very slow isothermal expansion against a frictionless piston, (c) a block sliding to rest on a rough floor.

Try it first, then show the solutionSolution

1. (a) Irreversible: the gas never collects itself back into half the container.
2. (b) Reversible (ideally): it passes through equilibrium states and can be retraced.
3. (c) Irreversible: friction turns work into heat, which cannot spontaneously push the block back.

Answer: (a) irreversible (b) reversible (c) irreversible

Heat engines and the Carnot cycle

A heat engine works in a cycle: it takes heat $Q_1$ from a hot source at $T_1$, converts part into work W and rejects $Q_2$ to a cold sink at $T_2$; $W=Q_1-Q_2$. Efficiency $\eta =\frac{W}{Q_1}=1-\frac{Q_2}{Q_1}$. The Carnot cycle is an ideal reversible cycle for an ideal gas: (1) isothermal expansion at $T_1$, absorbing $Q_1$; (2) adiabatic expansion, cooling from $T_1$ to $T_2$; (3) isothermal compression at $T_2$, rejecting $Q_2$; (4) adiabatic compression back to the start. On a pV diagram it is bounded by two isothermals and two steeper adiabatics; the enclosed area is the net work. For a Carnot engine $\frac{Q_2}{Q_1}=\frac{T_2}{T_1}$, so $\eta =1-\frac{T_2}{T_1}$ (no derivation needed). Carnot's theorem: no engine working between the same two temperatures is more efficient than a reversible one. η = 1 would need $T_2=0$ K, which is unattainable.

Worked example

A Carnot engine works between 527 °C and 27 °C and takes 1000 J of heat per cycle from the source. Find its efficiency, the work done and the heat rejected per cycle.

Try it first, then show the solutionSolution

1. T₁ = 800 K, T₂ = 300 K: η = 1 − 300/800 = 0.625
2. W = 0.625 × 1000 = 625 J
3. Q₂ = 1000 − 625 = 375 J

Answer: 62.5%; 625 J; 375 J

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.

Topics in this chapter

• First Law and Thermodynamic Processes
• Heat Engines, Refrigerators and the Carnot Cycle

Thermodynamics formula sheet

First law

$Q=\Delta U+W,W=\int pdV$

Q > 0 in; W > 0 by the system.

Mechanical equivalent of heat

$1cal=4.186J$

Exact.

Isobaric work

$W=p(V_2-V_1)=nR\Delta T$

Q = nC_pΔT.

Isothermal process

$pV=\text{constant},W=nRT\ln \frac{V_2}{V_1}=2.303nRT{\log }_{10}\frac{V_2}{V_1}$

ΔU = 0, Q = W.

Adiabatic process

$p{V}^{\gamma }=\text{constant},T{V}^{\gamma -1}=\text{constant}$

Q = 0; derivation not required.

Adiabatic work

$W=\frac{p_1{V}_1-p_2{V}_2}{\gamma -1}=\frac{nR(T_1-T_2)}{\gamma -1}$

ΔU = −W.

Mayer's relation

$C_p-C_v=R,\gamma =\frac{C_p}{C_v}$

Monatomic 5/3; diatomic 7/5.

Engine efficiency

$\eta =\frac{W}{Q_1}=1-\frac{Q_2}{Q_1}$

W = Q₁ − Q₂.

Carnot efficiency

$\eta =1-\frac{T_2}{T_1}$

Kelvin temperatures; no derivation.

Thermodynamics: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

How the school exam asks Thermodynamics

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

Tricks to remember Thermodynamics

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

• Four processes, four zeros

  Isochoric W = 0; isothermal ΔU = 0; adiabatic Q = 0; cyclic ΔU = 0.

  Use it when: Applying the first law quickly.

• Adiabatic is steeper

  pV^γ falls faster than pV because γ > 1.

  Use it when: Comparing curves on pV diagrams.

• Mayer's R

  Cp needs an extra R to pay for pushing the piston.

  Use it when: Remembering Cp − Cv = R.

• Carnot order

  'Iso-Adia-Iso-Adia': expand hot, expand and cool, compress cold, compress and warm.

  Use it when: Describing the Carnot cycle.

• Clockwise works

  Clockwise loop on pV: net work done by the gas (engine).

  Use it when: Sign of net work in cycles.

Thermodynamics: exam-level solved questions

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

Case-study practice (Section D style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 5-mark question split into 1 + 1 + 2 + 1 marks.

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.

Where students lose marks in Thermodynamics

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

• ✗Mistake: Mixing up sign conventions

  Students take W as positive when work is done on the gas and then add instead of subtract.

  Wrong working

  200 J of work done on a gas that absorbs 50 J: ΔU = 50 − 200 = −150 J.

  Right working

  W = −200 J (done on the gas): ΔU = Q − W = 50 + 200 = +250 J.

  ✓Do this: State the convention (Q in +, W by the gas +) at the top of every answer.

  Usually costs: 1-2 marks.

• ✗Mistake: Using Celsius in efficiency and gas formulae

  Students write η = 1 − 27/527.

  Wrong working

  η = 1 − 27/527 = 0.95.

  Right working

  η = 1 − 300/800 = 0.625.

  ✓Do this: Convert every temperature to kelvin first.

  Usually costs: 1-2 marks.

• ✗Mistake: Saying no heat is exchanged in an isothermal process

  Constant temperature is confused with zero heat flow.

  Wrong working

  Isothermal: T is constant, so Q = 0.

  Right working

  Isothermal: ΔU = 0, so Q = W; heat flows in during expansion. It is the adiabatic process that has Q = 0.

  ✓Do this: Isothermal = constant T; adiabatic = no heat.

  Usually costs: 1 mark.

• ✗Mistake: Writing a vague second law

  Students write 'heat flows from hot to cold' without the essential condition.

  Wrong working

  Clausius: heat always flows from hot to cold.

  Right working

  Clausius: heat cannot flow by itself from a colder body to a hotter one without external work.

  ✓Do this: Include 'by itself' (Clausius) and 'working in a cycle … entirely into work' (Kelvin-Planck).

  Usually costs: 1 mark.

• ✗Mistake: Using pV = constant for an adiabatic change

  Students apply Boyle's law to a sudden compression or expansion.

  Wrong working

  Sudden halving of volume: p doubles.

  Right working

  pV^γ = constant: p increases by 2^γ (2.64 times for γ = 1.4).

  ✓Do this: Sudden or insulated means adiabatic: use the power γ.

  Usually costs: 1-2 marks.

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.

1. Q1 · 1 mark · MCQExam level

   In an adiabatic process:
   (a) ΔU = 0
   (b) Q = 0
   (c) W = 0
   (d) p is constant

   Hint

   No heat exchange.

   Answer

   (b)

2. Q2 · 1 mark · MCQExam level

   1 calorie is equal to:
   (a) 4.186 J
   (b) 0.239 J
   (c) 41.86 J
   (d) 1 J

   Hint

   Mechanical equivalent of heat.

   Answer

   (a)

3. Q3 · 1 mark · Very short answerExam level

   State the zeroth law of thermodynamics.

   Hint

   Two systems and a third.

   Answer

   If two systems are each in thermal equilibrium with a third system, they are in thermal equilibrium with each other.

4. Q4 · 1 mark · Very short answerExam level

   State the Kelvin-Planck form of the second law.

   Hint

   Single source.

   Answer

   No engine working in a cycle can absorb heat from a single source and convert it entirely into work.

5. Q5 · 2 marks · Short answerExam level

   Distinguish between isothermal and adiabatic processes (two points).

   Hint

   What is constant? How fast?

   Answer

   Isothermal: T constant, heat exchanged, slow, conducting walls, pV = const. Adiabatic: no heat exchanged, T changes, fast or insulated, pV^γ = const.

6. Q6 · 2 marks · Short answerHarder exam level

   One mole of a monatomic ideal gas is heated at constant volume from 300 K to 400 K. Find the heat supplied. (R = 8.31 J mol⁻¹ K⁻¹)

   Hint

   C_v = 3R/2.

   Answer

   Q = nC_vΔT = 1.5 × 8.31 × 100 = 1246.5 J (all of it raises U).

7. Q7 · 2 marks · Short answerHarder exam level

   A gas absorbs 1000 J of heat at a constant pressure of 1.5 × 10⁵ Pa and expands by 2 L. Find the work done and ΔU.

   Hint

   W = pΔV.

   Answer

   W = 1.5 × 10⁵ × 2 × 10⁻³ = 300 J; ΔU = 1000 − 300 = 700 J.

8. Q8 · 2 marks · Short answerHarder exam level

   In a diesel engine, air (γ = 1.4) at 1.0 × 10⁵ Pa and 300 K is compressed adiabatically to one-tenth of its volume. Find its final pressure and temperature.

   Hint

   pV^γ = const and TV^(γ−1) = const.

   Answer

   p₂ = 10⁵ × 10^1.4 = 2.5 × 10⁶ Pa; T₂ = 300 × 10^0.4 = 754 K (about 480 °C, hot enough to ignite the fuel).

9. Q9 · 2 marks · Short answerHarder exam level

   A Carnot engine with its sink at 27 °C has an efficiency of 25%. Find the source temperature.

   Hint

   η = 1 − T₂/T₁.

   Answer

   T₁ = 300/0.75 = 400 K = 127 °C.

10. Q10 · 3 marks · Short answerHarder exam level

    Two moles of an ideal gas at 300 K are compressed isothermally from 20 L to 10 L. Find the work done on the gas and the heat released. (R = 8.31 J mol⁻¹ K⁻¹)

    Hint

    W = nRT ln(V₂/V₁) is negative.

    Answer

    W = 2 × 8.31 × 300 × ln(1/2) = −3456 J: 3456 J of work is done on the gas, and since ΔU = 0, 3456 J of heat is released.

11. Q11 · 3 marks · Short answerHarder exam level

    Find the heat needed to raise the temperature of 3 mol of helium (monatomic) by 25 K at constant pressure, and split it into the rise in internal energy and the work done. (R = 8.31 J mol⁻¹ K⁻¹)

    Hint

    C_v = 3R/2 and C_p = 5R/2.

    Answer

    Q = nC_pΔT = 3 × 2.5 × 8.31 × 25 = 1558 J; ΔU = nC_vΔT = 3 × 1.5 × 8.31 × 25 = 935 J; W = nRΔT = 3 × 8.31 × 25 = 623 J (check: 935 + 623 = 1558).

12. Q12 · 5 marks · Long answerHarder exam level

    (a) State the Kelvin-Planck and Clausius statements of the second law. [2]
    (b) What is a reversible process? Give two conditions for it and one example of an irreversible process. [2]
    (c) Why is the efficiency of a heat engine always less than 1? [1]

    Hint

    Single source; by itself.

    Answer

    (a) Kelvin-Planck: no cyclic engine can take heat from a single source and convert it entirely into work. Clausius: heat cannot flow by itself from a colder body to a hotter one. (b) A process that can be retraced exactly so that the system and surroundings return to their original states; it must be quasi-static and free of friction and of heat flow across finite temperature differences. Free expansion of a gas is irreversible. (c) Some heat must be rejected to the sink; η = 1 would need a sink at 0 K.

13. Q13 · 3 marks · Short answerHOTS / competency

    A diatomic ideal gas (γ = 1.4) at 2.0 × 10⁵ Pa occupies 3 L. It expands adiabatically to 6 L. Find its final pressure and the work it does, and explain where that energy comes from.

    Hint

    p₁V₁^γ = p₂V₂^γ; W = (p₁V₁ − p₂V₂)/(γ − 1).

    Answer

    p₂ = 2.0 × 10⁵ × (1/2)^1.4 = 7.58 × 10⁴ Pa. W = (2.0 × 10⁵ × 3 × 10⁻³ − 7.58 × 10⁴ × 6 × 10⁻³)/0.4 = (600 − 455)/0.4 = 363 J. No heat enters, so the work comes from the gas's internal energy (ΔU = −363 J) and the gas cools.

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. 1.State the first law of thermodynamics.Show answer

   Q = ΔU + W: heat supplied = rise in internal energy + work done by the system.

2. 2.Work done by a gas on a pV diagram?Show answer

   The area under the curve.

3. 3.Equation of an adiabatic process?Show answer

   pV^γ = constant.

4. 4.Mayer's relation?Show answer

   C_p − C_v = R.

5. 5.γ for a monatomic gas?Show answer

   5/3.

6. 6.State the Clausius statement.Show answer

   Heat cannot flow by itself from a colder body to a hotter body.

7. 7.Efficiency of a Carnot engine?Show answer

   η = 1 − T₂/T₁.

8. 8.Is internal energy a state function?Show answer

   Yes; heat and work are not.

Quick revision checklist

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

• ☐Can you state the zeroth law and explain why it defines temperature?
• ☐Can you apply Q = ΔU + W with correct signs?
• ☐Can you find work from a pV diagram, including cycles?
• ☐Can you derive the work in an isothermal expansion?
• ☐Can you use pV^γ and TV^(γ−1) for adiabatic changes?
• ☐Can you derive Cp − Cv = R?
• ☐Can you state the Kelvin-Planck and Clausius statements?
• ☐Can you draw the Carnot cycle and find its efficiency?

Thermodynamics: frequently asked questions