Chapter 11 of 14

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Madhya Pradesh (MPBSE)Class 11 Physics★ Unit VIII · part of the 20-mark block covering Units VII to IX (Mechanical Properties of Solids, Mechanical Properties of Fluids, Thermal Properties of Matter, Thermodynamics, Kinetic Theory). CBSE does not publish a separate per-chapter figure.

Thermodynamics

The First Law is just bookkeeping — energy in equals energy stored plus energy out. The Second Law is the interesting one: it explains why a hot cup of tea never spontaneously gets hotter by cooling the room, why no engine ever reaches 100% efficiency, and why the Carnot engine's efficiency limit isn't a claim anyone has to take on faith — it can be proved. CBSE Class 11 Physics Chapter 11.

⏱ 21 min readHard difficulty✓ Free · NCERT-aligned

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Study Guide

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Key points, glossary and the questions most likely to show up.

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By the end of this chapter you'll be able to…

1Classify systems as open, closed, or isolated, and state variables as intensive or extensive
2State the zeroth law and explain why it is what makes temperature measurement meaningful
3Distinguish heat, work, and internal energy, and explain why only internal energy is a state variable
4Apply the first law delta-Q = delta-U + delta-W with a fixed sign convention to isothermal, adiabatic, isochoric, isobaric, and cyclic processes
5Derive Cp - Cv = R for an ideal gas from the first law
6State the Kelvin-Planck and Clausius statements of the second law and explain their equivalence
7Explain what makes a process reversible, and why almost no real process is
8Derive the Carnot efficiency eta = 1 - T2/T1 and reproduce the proof that no engine can exceed it

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Why this chapter matters

Every other physical science eventually runs into these two laws. The first law is energy conservation with heat added to the accounting; the second law is the reason a hot cup of tea never spontaneously gets hotter, and the reason no engine ever built has reached 100% efficiency. The Carnot engine turns that limit into a number you can calculate, and the proof behind it is one of the cleanest arguments in the whole syllabus.

Before you start — revise these

A 5-minute refresher here will save you 30 minutes of confusion below.

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Class 11 Physics: Thermal Properties of Matter

Heat, temperature, specific heat capacity, and calorimetry — this chapter builds the laws that govern them

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Class 11 Physics: Mechanical Properties of Fluids

Pressure and the PV relationships already used informally with gases, now formalised as an equation of state

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Class 11 Physics: Work, Energy and Power

Work as force through a displacement — this chapter reuses it as P delta-V for a gas pushing a piston

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Class 11 Physics: Units and Measurements

Dimensional consistency — checking whether both sides of a thermodynamic equation are extensive or intensive uses the same discipline

Thermodynamics

1. What this chapter covers

Textbook section	Topic
11.1	Introduction
11.2	Thermal equilibrium
11.3	Zeroth law of thermodynamics
11.4	Heat, internal energy and work
11.5	First law of thermodynamics
11.6	Specific heat capacity
11.7	Thermodynamic state variables and equation of state
11.8	Thermodynamic processes
11.9	Second law of thermodynamics
11.10	Reversible and irreversible processes
11.11	Carnot engine

Thermodynamics does not care what a system is made of. Whether it is a gas in a cylinder, a chemical reaction, or a star, the same handful of laws governs how heat and work move in and out of it. That is also why this chapter is short on things to memorise and long on things to reason through — every quantity here traces back to just two laws.

2. Systems, state variables, and what "equilibrium" actually means

System type	Exchanges with surroundings	Example
Open	Matter and energy	Tea in an open cup
Closed	Energy only, no matter	Gas in a cylinder with a movable piston
Isolated	Neither	Gas in a rigid, insulated container

A gas in a closed cylinder is described by a handful of state variables — pressure, volume, temperature, mass, composition. Not all state variables behave the same way when you split the system in two. Take a gas in equilibrium and imagine dividing it into two equal halves:

Intensive variables — pressure, temperature, density — stay the same in each half.
Extensive variables — volume, mass, internal energy — halve along with the system.

This is a genuinely useful check, not a classification exercise: in any correct thermodynamic equation, both sides must carry the same extensive/intensive character. In $Δ Q = Δ U + P Δ V$ , every term is extensive — $P$ is intensive, but $P Δ V$ is the product of an intensive and an extensive quantity, so it comes out extensive too, matching $Δ Q$ and $Δ U$ .

A gas obeying $P V = μ R T$ is said to satisfy an equation of state — a relation connecting the state variables so that, for a fixed amount of gas, only two of them are independent.

State variables describe equilibrium states only. A gas expanding freely into a vacuum, or an explosive combustion, passes through states where pressure is not even uniform through the container — those in-between states have no well-defined $(P, V, T)$ and cannot be plotted on a state diagram at all.

3. The zeroth law: how temperature gets a rigorous meaning

Before the zeroth law, "temperature" was just a feeling of hot and cold. The law that finally pins it down mathematically is almost embarrassingly simple, which is exactly why it was named after the first and second laws were already numbered — it was formulated by R.H. Fowler in 1931, decades later, once physicists realised the first and second laws silently assumed it.

Picture systems A and B, each separated from a third system C by a wall that conducts heat, while A and B are kept apart by an insulating wall. Left long enough, A and C reach thermal equilibrium, and separately, B and C reach thermal equilibrium. Now replace the A–B insulating wall with a conducting one. Experimentally, nothing changes — A and B are already in equilibrium with each other.

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

The payoff is that this guarantees a shared quantity exists — call it temperature $T$ — such that $T_{A} = T_{C}$ and $T_{B} = T_{C}$ imply $T_{A} = T_{B}$ . Without the zeroth law, there would be no guarantee that "same temperature" is even a consistent, transitive property to assign to bodies at all.

4. Heat, work, and internal energy — three ideas the chapter insists you keep separate

Internal energy $U$ is the sum of the kinetic and potential energies of a system's molecules, measured in the frame where the system's centre of mass is at rest — so a fast-moving bullet is not "hotter" just because it is fast; its internal energy is about the disordered motion of its molecules, not the bullet's overall flight. Crucially, $U$ is a state variable: its value depends only on the system's current state, never on how it got there.

Heat and work are not state variables. They are the two ways a system's internal energy can change:

Heat — energy transfer caused by a temperature difference between system and surroundings.
Work — energy transfer by any other means (moving a piston, stirring, compressing).

This distinction sounds pedantic until you try to use it: "this gas contains 500 J of heat" is meaningless — heat is not a thing a system has, only a thing that flows. "This gas absorbed 500 J of heat" and "this gas has 500 J more internal energy than before" are both perfectly meaningful, and they are not the same statement.

5. The First Law: energy bookkeeping with a sign convention to fix first

$Δ Q = Δ U + Δ W$

$Δ Q$ — heat supplied to the system by the surroundings (positive when added)
$Δ W$ — work done by the system on the surroundings (positive when the system expands and pushes outward)
$Δ U$ — change in internal energy

Rearranged, this is the more familiar $Δ U = Δ Q - Δ W$ . Both forms say exactly the same thing; the only way to get it wrong is to mix sign conventions mid-problem — for instance treating "work done on the system" as positive $Δ W$ while still using this formula, which silently flips every answer's sign. Fix the convention before touching a number.

Because $U$ is a state variable and $Q$ , $W$ are not, $Δ U$ depends only on the initial and final states — but $Q$ and $W$ individually can differ across two different paths connecting the same two states, even though $Δ Q - Δ W$ stays fixed. This single fact is the whole reason isothermal and adiabatic processes between the same two temperatures give different work done, while $Δ U$ tracks only $T$ .

Worked example (from the textbook). One gram of water turns to steam at atmospheric pressure. Its latent heat is 2256 J/g, so $Δ Q = 2256$ J. At atmospheric pressure, the volume goes from $1 cm^{3}$ (liquid) to $1671 cm^{3}$ (vapour):

$Δ W = P (V_{g} - V_{l}) = 1.013 \times 1 0^{5} \times 1671 \times 1 0^{- 6} \approx 169.2 J$ $Δ U = Δ Q - Δ W = 2256 - 169.2 \approx 2086.8 J$

Most of the heat supplied does not go into pushing back the atmosphere — it goes into pulling the water molecules apart against their own attraction, which is exactly what "internal energy" was defined to capture.

6. Specific heat capacity — and where $C_{p} - C_{v} = R$ actually comes from

For a substance of mass $m$ absorbing heat $Δ Q$ for a temperature rise $Δ T$ , specific heat capacity is $s = \frac{1}{m} \frac{Δ Q}{Δ T}$ . Per mole instead of per unit mass, it is the molar specific heat capacity $C = \frac{1}{μ} \frac{Δ Q}{Δ T}$ .

For solids, the law of equipartition of energy — which is properly the subject of the next chapter, Kinetic Theory, but the textbook borrows the result here — predicts each of $N$ atoms vibrating in three dimensions carries average energy $3 k_{B} T$ , giving total energy $U = 3 R T$ per mole and therefore $C = 3 R$ (the Dulong-Petit law).

It matches experiment well at ordinary temperatures for most solids (carbon is a known exception), and is known to break down at low temperature.

For gases, specific heat depends on how the heat is supplied — at constant volume or constant pressure — because at constant pressure some of the heat also does work pushing the gas's own boundary outward. Starting from $Δ Q = Δ U + P Δ V$ for one mole:

At constant volume, $Δ V = 0$ , so $C_{v} = (\frac{Δ U}{Δ T})_{v}$ — heat only builds internal energy.
At constant pressure, $C_{p} = (\frac{Δ U}{Δ T})_{p} + P (\frac{Δ V}{Δ T})_{p}$ . Since $U$ for an ideal gas depends only on $T$ , the first term is the same $C_{v}$ either way. Using $P V = R T$ for one mole, $P (\frac{Δ V}{Δ T})_{p} = R$ .

$C_{p} - C_{v} = R$

This is Mayer's relation, and it only holds for an ideal gas — the derivation leans on $U$ depending on temperature alone, which is an ideal-gas property.

7. Thermodynamic processes: the quasi-static idealisation, and five named cases

If you suddenly drop the external pressure on a gas, the piston accelerates outward and the gas passes through states where pressure is not even uniform inside the container — not equilibrium states, and therefore not describable by a single $(P, V, T)$ .

To keep the mathematics tractable, thermodynamics idealises processes as quasi-static: infinitely slow, so the system stays in equilibrium with its surroundings — differing in pressure and temperature only infinitesimally — at every instant. A quasi-static process is, strictly, a hypothetical construct; real "slow enough" processes are just good approximations to it.

Process	Held fixed	First law becomes	Work done by gas
Isothermal	$T$	$Δ U = 0, Δ Q = Δ W$	$W = μ R T ln (V_{2} / V_{1})$
Adiabatic	$Q = 0$	$Δ U = - Δ W$	$W = \frac{μ R ( T _{1} - T _{2} )}{γ - 1}$
Isochoric	$V$	$Δ U = Δ Q$	$0$
Isobaric	$P$	$Δ Q = Δ U + P Δ V$	$W = P (V_{2} - V_{1})$
Cyclic	returns to start	$Δ U = 0, Δ Q = Δ W$	net area enclosed on $P$ - $V$ diagram

Isothermal work, derived directly: for an ideal gas at fixed $T$ , $P = μ R T / V$ , so

$W = \int_{V_{1}}^{V_{2}} P d V = μ R T \int_{V_{1}}^{V_{2}} \frac{d V}{V} = μ R T ln (\frac{V _{2}}{V _{1}})$

Because $U$ depends only on $T$ for an ideal gas, $Δ U = 0$ here, so all of this work is paid for entirely by heat absorbed from the surroundings: $Q = W$ .

Adiabatic, by contrast, has $Q = 0$ by definition, so any work the gas does comes directly out of its own internal energy — the gas cools as it expands adiabatically, and heats as it is compressed adiabatically. For an ideal gas undergoing an adiabatic change,

$P V^{γ} = constant, γ = \frac{C _{p}}{C _{v}}$

On a $P$ – $V$ diagram, the adiabatic curve through any point is steeper than the isothermal curve through the same point — an isothermal expansion loses pressure only because volume rises; an adiabatic expansion loses pressure for that and because temperature is falling too, so pressure drops faster for the same $Δ V$ .

8. The Second Law: why the First Law alone allows nonsense

The First Law permits a book on a table to spontaneously cool the table and hop into the air, converting the table's lost internal energy exactly into the book's gained potential energy — total energy is conserved throughout. It never happens. Something beyond energy conservation is needed to rule it out, and that something is the Second Law.

Kelvin-Planck statement: No process is possible whose sole result is the absorption of heat from a reservoir and its complete conversion into work.

Clausius statement: No process is possible whose sole result is the transfer of heat from a colder body to a hotter one.

These read as two different claims but are provably equivalent — each can be shown to imply the other. In practice, they say the same thing two ways: a perfect heat engine (100% efficient) is impossible, and a perfect refrigerator (moving heat uphill for free) is impossible.

9. Reversible and irreversible processes

A process is reversible if it can be run backward so that both system and surroundings return exactly to their original states, with no trace left anywhere in the universe. Nearly nothing in nature qualifies. Two separate reasons break reversibility:

Non-equilibrium states — free expansion, explosive combustion, sudden mixing — the system passes through states with no well-defined pressure or temperature, and there is no way to walk that path backward in the same steps.
Dissipation — friction, viscosity, electrical resistance — always converts ordered mechanical or electrical energy into disordered heat, and that conversion cannot be un-done without external help.

A vessel's base cools by conducting heat to its cooler sides, never the reverse. A stirred liquid's kinetic energy becomes heat in the liquid; the liquid never spontaneously un-stirs itself and cools the stirring rod. Reversibility requires both quasi-static behaviour and zero dissipation — a slow isothermal expansion in a frictionless cylinder is about as close as physics gets.

10. The Carnot engine — and why nothing can beat it

A heat engine absorbs heat $Q_{1}$ from a hot reservoir, delivers work $W$ , and rejects the rest, $Q_{2}$ , to a cold reservoir; efficiency is $η = W / Q_{1} = 1 - Q_{2} / Q_{1}$ . Sadi Carnot asked, in 1824, what the best possible efficiency between two fixed temperatures could be — and worked out the answer before heat was even properly understood as energy.

An ideal reversible engine operating between just two temperatures must use only isothermal steps (to absorb and reject heat without a finite temperature gap) and adiabatic steps (to change temperature without exchanging heat at all) — any other process would need a whole ladder of intermediate reservoirs. That gives the four-step Carnot cycle:

Isothermal expansion at $T_{1}$ : absorbs $Q_{1}$ , does work $μ R T_{1} ln (V_{2} / V_{1})$ .
Adiabatic expansion: $T_{1} \to T_{2}$ , no heat exchanged.
Isothermal compression at $T_{2}$ : rejects $Q_{2}$ , work done on gas $μ R T_{2} ln (V_{3} / V_{4})$ .
Adiabatic compression: $T_{2} \to T_{1}$ , back to the start.

Working through the algebra (the two adiabatic legs force $V_{2} / V_{1} = V_{3} / V_{4}$ ), the volume ratios cancel and the efficiency collapses to a strikingly clean result:

$η_{Carnot} = 1 - \frac{T _{2}}{T _{1}}$

depending on nothing but the two reservoir temperatures — not on the working substance, not on the gas used to derive it.

Why can no engine beat this — and this is provable, not asserted. Suppose an irreversible engine I could exceed a Carnot engine's efficiency, $η_{I} > η_{R}$ , both running between the same two reservoirs. Couple them: let I absorb $Q_{1}$ from the hot reservoir and deliver work $W^{'}$ , and run R backward, as a refrigerator, driven by a smaller work input $W$ , to pump exactly $Q_{1}$ back into the hot reservoir.

Since $W^{'} > W$ (that is what $η_{I} > η_{R}$ means for the same $Q_{1}$ ), the combined I+R system takes heat $W^{'} - W$ from the cold reservoir alone and converts it entirely into work, with the hot reservoir completely unchanged. That is exactly what the Kelvin-Planck statement forbids.

So the assumption fails: no engine, reversible or not, can exceed the Carnot efficiency between the same two temperatures. The same coupling argument shows a reversible engine's efficiency cannot depend on what it is made of — one Carnot engine cannot beat another built from a different substance either, which is why using an ideal gas to derive $η = 1 - T_{2} / T_{1}$ is enough to fix the answer for every Carnot engine.

One more consequence worth carrying forward: since $η = 1 - Q_{2} / Q_{1} = 1 - T_{2} / T_{1}$ for any Carnot engine regardless of substance, $Q_{1} / Q_{2} = T_{1} / T_{2}$ is a relation that involves no material property at all — it can be used to define a temperature scale from heat ratios alone, independent of what thermometer or gas is used to measure it.

Summary

The zeroth law makes "temperature" a rigorous, transitive property: two systems each in equilibrium with a third are in equilibrium with each other.
Internal energy $U$ is a state variable; heat and work are not — they are the two paths by which $U$ changes, and their individual values depend on the path taken even when $Δ U$ does not.
The first law, $Δ Q = Δ U + Δ W$ , is energy conservation with heat included. Fix the sign convention before solving anything.
$C_{p} - C_{v} = R$ holds only for an ideal gas, and follows directly from $Δ Q = Δ U + P Δ V$ applied at constant $V$ and constant $P$ .
Isothermal, adiabatic, isochoric, isobaric, and cyclic processes are five named cases of the same first law, distinguished by what is held fixed.
The second law forbids what the first law alone allows: a heat engine can never be 100% efficient (Kelvin-Planck), and heat never flows spontaneously from cold to hot (Clausius) — the two statements are equivalent.
Reversibility demands both a quasi-static process and zero dissipation; almost nothing in nature meets both conditions.
The Carnot engine's efficiency, $η = 1 - T_{2} / T_{1}$ , is not just the best known efficiency — Carnot's theorem proves it is the best possible one, using nothing but the Kelvin-Planck statement and a thought experiment coupling two engines.

Key formulas & results

Everything you need to memorise, in one card. Screenshot this for revision.

First law of thermodynamics

delta-Q = delta-U + delta-W

Q supplied to system positive; W done BY system positive. Fix this convention before solving anything.

Isothermal work

W = mu R T ln(V2/V1)

T constant, so delta-U = 0 for an ideal gas; Q = W entirely.

Adiabatic relation

P V^gamma = constant, where gamma = Cp/Cv

Q = 0; work is paid for entirely out of internal energy, so temperature changes.

Adiabatic work

W = mu R (T1 - T2) / (gamma - 1)

Positive when the gas expands and cools (T2 < T1).

Isobaric work

W = P (V2 - V1) = mu R (T2 - T1)

Pressure fixed; heat supplied splits between raising U and doing this work.

Mayer's relation

Cp - Cv = R

Ideal gas only — the derivation assumes U depends on T alone.

Adiabatic index

gamma = Cp / Cv

Monatomic 5/3, diatomic (no vibration) 7/5.

Dulong-Petit law (solids)

C = 3R

From the law of equipartition, borrowed forward from Chapter 12; holds at ordinary temperature, breaks down at low T.

Specific heat capacity

s = (1/m) (delta-Q / delta-T)

Per unit mass; unit J per kg per K.

Molar specific heat capacity

C = (1/mu) (delta-Q / delta-T)

Per mole; depends on the process (constant P or constant V) for a gas.

Ideal gas equation of state

P V = mu R T

Fixes only two of P, V, T as independent for a given amount of gas.

Carnot efficiency

eta = 1 - T2/T1 = 1 - Q2/Q1

Temperatures in kelvin. Depends only on the reservoir temperatures, never on the working substance.

Carnot heat-temperature ratio

Q1 / Q2 = T1 / T2

Follows from the Carnot efficiency formula; used to define a thermodynamic temperature scale independent of any thermometer.

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Common mistakes & fixes

These are the exact errors that cost students marks in board exams. Read them once, save yourself the trouble.

WATCH OUT

✗ Mixing up the first-law sign convention mid-problem

✓ This chapter uses delta-Q = delta-U + delta-W with W done BY the system. If a source uses delta-U = Q + W instead (W done ON the system), every sign flips. Pick one convention and hold it for the whole problem.

WATCH OUT

✗ Treating isothermal and adiabatic as interchangeable 'no temperature change' processes

✓ Isothermal: T fixed, so delta-U = 0 and Q = W. Adiabatic: Q = 0, so delta-U = -W and temperature DOES change — often the opposite of what students expect.

WATCH OUT

✗ Applying Cp - Cv = R to a real or non-ideal gas

✓ The derivation assumes internal energy depends only on temperature, which is only true for an ideal gas. Real gases deviate, especially near condensation.

WATCH OUT

✗ Believing an engine could reach 100% efficiency with a good enough design

✓ The second law (Kelvin-Planck) forbids it outright, regardless of engineering — some heat must always be rejected to the cold reservoir.

WATCH OUT

✗ Saying a gas 'contains' a certain amount of heat

✓ Heat is energy in transit, not a property a system has. Only internal energy is a state variable; 'heat supplied' or 'work done' are the meaningful phrases.

WATCH OUT

✗ Forgetting to convert temperature to kelvin before using the Carnot formula

✓ eta = 1 - T2/T1 requires both temperatures in kelvin. Using Celsius directly gives a wrong ratio, not just a wrong-looking number.

WATCH OUT

✗ Using the Carnot efficiency as the actual efficiency of a real engine

✓ Carnot efficiency is the maximum POSSIBLE efficiency between two temperatures. Real engines involve irreversibility and always fall short of it.

WATCH OUT

✗ Confusing isochoric (no work, V fixed) with isobaric (work done, P fixed)

✓ In isochoric processes W = 0 and all heat goes to delta-U. In isobaric processes W = P delta-V, and heat splits between delta-U and W.

WATCH OUT

✗ Assuming Q or W individually depend only on the initial and final states

✓ Only delta-U is path-independent. Q and W separately depend on the path taken between two states, even though delta-Q - delta-W is always the same.

WATCH OUT

✗ Explaining free expansion's delta-U = 0 as 'Q equals W'

✓ In free expansion into a vacuum, both Q = 0 (insulated) and W = 0 (no external pressure to push against) individually — it is not an isothermal-style Q = W situation.

WATCH OUT

✗ Treating the Kelvin-Planck and Clausius statements as two independent laws

✓ They are provably equivalent statements of the same second law — violating one can be shown to violate the other.

WATCH OUT

✗ Assuming a quasi-static process is something that can actually be achieved

✓ It is an idealisation — infinitely slow, always in equilibrium with the surroundings. Real 'slow' processes only approximate it; that approximation is what makes the work formulas usable.

NCERT exercises (with solutions)

Every NCERT exercise from this chapter — what it covers and how many questions to expect.

Ex s

Exercises

Calorimetry-style heat calculations, adiabatic compression, thermal-contact and coolant reasoning, free expansion through a stopcock, rate of change of internal energy, and work done along a linear-then-isobaric P-V path (Q11.1-11.8)

Questions

Practice problems

Work through this chapter's problems as a readiness check — reveal each solution, mark yourself honestly, and get your gap report at the end.

Readiness check

Are you exam-ready for Thermodynamics?

10 problems from this chapter. Try each one, reveal the worked solution, mark yourself honestly — get your gap report at the end.

10 questions~7 min worth ~20 marks in Madhya Pradesh (MPBSE) exams

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

•Systems: open, closed, isolated. State variables: intensive (unchanged on splitting the system) or extensive (halved on splitting).
•Zeroth law: two systems each in equilibrium with a third are in equilibrium with each other — this is what makes temperature well-defined.
•Internal energy U is a state variable; heat and work are not — they are the two modes by which U changes.
•First law: delta-Q = delta-U + delta-W, with W done BY the system taken positive.
•Isothermal: delta-U = 0, Q = W = mu R T ln(V2/V1). Adiabatic: Q = 0, delta-U = -W, P V^gamma = constant.
•Mayer's relation Cp - Cv = R holds for ideal gases only; Dulong-Petit C = 3R for solids is borrowed from the equipartition law in Chapter 12.
•Second law: no 100% efficient engine (Kelvin-Planck); heat never flows spontaneously cold to hot (Clausius) — the two statements are equivalent.
•A process is reversible only if it is quasi-static AND free of dissipation; almost nothing in nature qualifies.
•Carnot efficiency eta = 1 - T2/T1 is the maximum possible between two temperatures, provable from the Kelvin-Planck statement alone, and independent of the working substance.

Madhya Pradesh (MPBSE) marks blueprint

Where the marks come from in this chapter — so you can plan your prep.

Typical chapter weightage: 20 marks across Units VII to IX (Chapters 8-12); no official per-chapter split is published

Question type	Marks each	Typical count	What it tests
First law and process calculations	3-5	1-2	delta-U = Q - W applied to isothermal, adiabatic, isobaric, isochoric, and cyclic processes
Specific heat and Mayer's relation	2-3	1	Cp, Cv, gamma, and the Cp - Cv = R derivation
Second law and Carnot engine	3-5	1	Kelvin-Planck / Clausius statements, reversibility, Carnot efficiency and its derivation

Prep strategy

Fix the first-law sign convention before touching any numbers
Memorise the process table — what is held fixed, and the resulting work formula — for all five named processes
Be able to derive Cp - Cv = R, not just quote it
Practise the Carnot's-theorem proof sketch; it is a standard 5-mark 'prove that no engine can exceed Carnot efficiency' question
Always convert temperatures to kelvin before using any Carnot formula

Where this shows up in the real world

This chapter isn't just an exam topic — it lives in the world around you.

Car engines and power plants

Every combustion engine and steam turbine is a heat engine operating between a hot combustion/boiler temperature and the cooler ambient exhaust, bounded by the same Carnot ceiling derived in this chapter.

Refrigerators, air conditioners, and heat pumps

These are Carnot cycles run in reverse — work is supplied to move heat from a cold space to a hot one, and the coefficient of performance follows directly from the same T1, T2 relations as engine efficiency.

Power plant siting near rivers and coastlines

Steam turbines reject the unavoidable second-law waste heat to cooling water, which is why large thermal power stations are built near a river, lake, or sea rather than an arbitrary location.

Why insulated engines are not automatically more efficient

Reducing heat loss helps, but the second law still caps efficiency at 1 - T2/T1 regardless of how well an engine is insulated — the limit comes from the temperatures involved, not from preventable leakage alone.

Exam strategy

Battle-tested tips from teachers and toppers for this chapter.

State the first-law sign convention explicitly at the start of any numerical answer

Identify the process type first (isothermal, adiabatic, isochoric, isobaric, cyclic) before picking a work formula

Convert every temperature to kelvin before using eta = 1 - T2/T1 or Q1/Q2 = T1/T2

For P-V diagram cycle questions, compute work leg by leg — isochoric legs always contribute zero regardless of pressure change

For a 'prove no engine exceeds Carnot efficiency' question, reproduce the coupled-engine argument using the Kelvin-Planck statement rather than just quoting the formula

Going beyond the textbook

For olympiad aspirants and curious learners — topics that build on this chapter.

STRETCH

Derive the Carnot cycle result Q1/T1 = Q2/T2 fully from the isothermal-work formula and the adiabatic relation, rather than quoting it.

STRETCH

Analyse an Otto or Diesel cycle (used in real petrol and diesel engines) and compare its theoretical efficiency to the Carnot limit between the same two extreme temperatures.

STRETCH

Investigate entropy as delta-S = delta-Q_rev / T, and show that the Carnot cycle has zero net entropy change while any irreversible cycle has positive net entropy production.

STRETCH

Work out the coefficient of performance of a Carnot heat pump used for heating, and explain why it is always at least 1 greater than the corresponding refrigerator's COP between the same two temperatures.

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JEE Main & Advanced practice

Competitive-level problems on this chapter, above the board pattern. Try each one on paper before opening the solution.

JEE MainCyclic process on a P-V diagramGraph-based numeric

One mole of an ideal gas is taken around a rectangular cycle ABCDA on a P-V diagram: A $(P_{0}, V_{0})$ $\to$ B $(2 P_{0}, V_{0})$ (isochoric) $\to$ C $(2 P_{0}, 2 V_{0})$ (isobaric) $\to$ D $(P_{0}, 2 V_{0})$ (isochoric) $\to$ A (isobaric). With $P_{0} = 1 \times 1 0^{5} Pa$ and $V_{0} = 1 \times 1 0^{- 3} m^{3}$ , find the net work done by the gas in one complete cycle.

Stuck? Show the approach

Work is done only on the two isobaric legs — the isochoric legs (constant $V$ ) contribute zero work regardless of the pressure change. Compute B $\to$ C and D $\to$ A separately and add, keeping track of sign (expansion positive, compression negative).

Show the full solution

A $\to$ B (isochoric): $Δ V = 0 \Rightarrow W = 0$ .

B $\to$ C (isobaric expansion at $P = 2 P_{0}$ ): $W_{B C} = 2 P_{0} (2 V_{0} - V_{0}) = 2 P_{0} V_{0}$ .

C $\to$ D (isochoric): $Δ V = 0 \Rightarrow W = 0$ .

D $\to$ A (isobaric compression at $P = P_{0}$ ): $W_{D A} = P_{0} (V_{0} - 2 V_{0}) = - P_{0} V_{0}$ .

Net work $W = W_{B C} + W_{D A} = 2 P_{0} V_{0} - P_{0} V_{0} = P_{0} V_{0}$ .

With the given numbers: $W = (1 \times 1 0^{5}) (1 \times 1 0^{- 3}) = 100 J$ .

Answer: W = 100 J, done by the gas (the cycle runs clockwise on the P-V diagram, so the enclosed area equals the net work output, as for any heat engine).

The trap

The area-equals-work shortcut only gives the net work for the full closed loop. Do not apply it leg-by-leg — each isochoric leg genuinely contributes zero work even though pressure is changing on it, precisely because $Δ V = 0$ there.

JEE MainCarnot engineNumeric, unit conversion

A Carnot engine works between $2 7^{\circ} C$ and $22 7^{\circ} C$ , absorbing 600 cal of heat from the source per cycle. Calculate the work done per cycle and the heat rejected to the sink (take 1 cal = 4.186 J).

Stuck? Show the approach

Convert both temperatures to kelvin first — this is the step most often skipped. Then apply $η = 1 - T_{2} / T_{1}$ , and split $Q_{1}$ into work and rejected heat using that efficiency.

Show the full solution

$T_{1} = 227 + 273 = 500 K$ , $T_{2} = 27 + 273 = 300 K$ .

$η = 1 - \frac{T _{2}}{T _{1}} = 1 - \frac{300}{500} = 0.4$

$W = η Q_{1} = 0.4 \times 600 = 240 cal = 240 \times 4.186 \approx 1005 J$

$Q_{2} = Q_{1} - W = 600 - 240 = 360 cal \approx 1507 J$

Answer: Work done per cycle = 240 cal (about 1005 J). Heat rejected to the sink = 360 cal (about 1507 J).

The trap

Using 27 and 227 directly as if they were kelvin gives a completely wrong efficiency ratio (300/500 vs. an unphysical 27/227) — this single missed conversion is the most common reason Carnot-efficiency answers come out wrong.

JEE MainSpecific heat of a gas mixtureFormula application

1 mole of a monatomic ideal gas ( $C_{v} = 3 R /2$ ) is mixed with 1 mole of a diatomic ideal gas ( $C_{v} = 5 R /2$ , vibration inactive). Find the molar specific heat at constant volume of the mixture.

Stuck? Show the approach

For a mixture, the total internal energy adds, so the effective $C_{v}$ is the mole-weighted average of the individual $C_{v}$ values, not a simple arithmetic average of ratios.

Show the full solution

$C_{v, mix} = \frac{n _{1} C _{v 1} + n _{2} C _{v 2}}{n _{1} + n _{2}} = \frac{1 \times \frac{3 R}{2} + 1 \times \frac{5 R}{2}}{2} = \frac{4 R}{2} = 2 R$

Answer: Cv of the mixture = 2R.

The trap

This is a mole-weighted average of Cv, not of gamma. Averaging the two gamma values (5/3 and 7/5) directly gives a wrong, meaningless number — gamma must be recomputed from the mixture's own Cv and Cp after combining them separately.

JEE AdvancedProcess with P proportional to VMulti-part derivation

An ideal monatomic gas undergoes a process in which pressure is directly proportional to volume, $P = α V$ , taking it from state $(P_{0}, V_{0})$ to $(2 P_{0}, 2 V_{0})$ . Find, in terms of $P_{0}$ and $V_{0}$ : (a) the work done by the gas, (b) the change in internal energy, (c) the heat absorbed.

Stuck? Show the approach

Work is the area under a straight line through the origin on the P-V diagram — a trapezoid, computable directly without needing $α$ explicitly once $P$ and $V$ at both ends are known. For $Δ U$ , avoid solving for $T_{1}$ and $T_{2}$ separately: use $n R Δ T = Δ (P V)$ directly, since $P V = n R T$ at every state.

Show the full solution

(a) Work done by the gas is the trapezoidal area under $P = α V$ from $V_{0}$ to $2 V_{0}$ :

$W = \frac{1}{2} (P_{0} + 2 P_{0}) (2 V_{0} - V_{0}) = \frac{3}{2} P_{0} V_{0}$

(b) Since $P V = n R T$ , $n R T_{1} = P_{0} V_{0}$ and $n R T_{2} = (2 P_{0}) (2 V_{0}) = 4 P_{0} V_{0}$ , so $n R Δ T = 3 P_{0} V_{0}$ .

For a monatomic gas, $Δ U = n C_{v} Δ T = n (\frac{3 R}{2}) Δ T = \frac{3}{2} (n R Δ T) = \frac{3}{2} (3 P_{0} V_{0}) = \frac{9}{2} P_{0} V_{0}$

Answer: Work done by the gas = 1.5 P0V0. Change in internal energy = 4.5 P0V0. Heat absorbed = 6 P0V0.

The trap

This process is not isothermal, isobaric, or isochoric, so it is tempting to assume delta-U is zero or to try and force-fit one of the named-process formulas. Since PV quadruples here, T does too — delta-U must be computed from n R delta-T = delta(PV), not assumed away.

JEE AdvancedCarnot refrigerator and coefficient of performanceNumeric, reversed-cycle application

A Carnot refrigerator maintains a freezer at $- 1 3^{\circ} C$ while rejecting heat to a room at $2 7^{\circ} C$ . If it extracts 200 J of heat from the freezer per cycle, find the work that must be supplied per cycle, and the coefficient of performance.

Stuck? Show the approach

A Carnot refrigerator is a reversed Carnot engine, so the same $Q_{1} / T_{1} = Q_{2} / T_{2}$ relation holds, with $Q_{2}$ now the heat extracted from the cold space and $Q_{1}$ the heat dumped into the hot space. Coefficient of performance is defined as heat extracted divided by work supplied — not heat rejected divided by work.

Show the full solution

$T_{2} = - 13 + 273 = 260 K$ (freezer), $T_{1} = 27 + 273 = 300 K$ (room).

$\frac{Q _{1}}{T _{1}} = \frac{Q _{2}}{T _{2}} \Rightarrow Q_{1} = Q_{2} (\frac{T _{1}}{T _{2}}) = 200 \times \frac{300}{260} \approx 230.8 J$

$W = Q_{1} - Q_{2} = 230.8 - 200 = 30.8 J$

$COP = \frac{Q _{2}}{W} = \frac{T _{2}}{T _{1} - T _{2}} = \frac{260}{300 - 260} = 6.5$

Answer: Work required per cycle is about 30.8 J. Coefficient of performance = 6.5.

The trap

COP is Q2 (heat extracted from the cold space) divided by W, not Q1 divided by W. Swapping in the heat rejected to the hot reservoir instead of the heat extracted from the cold one is the single most common error in refrigerator and heat-pump problems.

Where else this chapter is tested

CBSE board isn't the only one — other exams test this chapter too.

CBSE Class 11 Physics examHigh

JEE Main and Advanced (Thermodynamics)High

NEET PhysicsMedium

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Questions students ask

The real ones — pulled from the Q&A community and tutor sessions.

Internal energy is a property a system HAS at any given moment — it depends only on the system's current state, not on how it got there. Heat is not something a system has at all; it is energy in transit, a name for one particular way internal energy can change. Saying 'this gas has 500 J of heat' is meaningless in the same way 'this gas has 500 J of work' would be; saying 'this gas absorbed 500 J of heat' is a perfectly meaningful statement about a process, not a state.

It follows from a proof, not from trying every possible engine. Assume an engine could beat the Carnot efficiency between the same two reservoirs. Couple it to a reversed Carnot engine (acting as a refrigerator) so that together they return exactly as much heat to the hot reservoir as the first engine took out. The combined system would then extract heat purely from the cold reservoir and convert it entirely into work with no other change anywhere — which is exactly what the Kelvin-Planck statement forbids. So no such engine can exist.

The derivation relies on internal energy depending only on temperature — true for an ideal gas because it has no intermolecular potential energy, but false for a real gas, where U also depends on volume through the (weak) forces between molecules. Real gases have Cp - Cv close to R only when they are dilute enough to behave nearly ideally.

Isothermal keeps temperature fixed by allowing heat to flow freely with the surroundings, so for an ideal gas internal energy does not change and heat supplied equals work done: Q = W. Adiabatic allows no heat exchange at all (Q = 0), so any work the gas does comes directly out of its own internal energy, and temperature changes. On a P-V diagram, the adiabatic curve through a point is always steeper than the isothermal curve through the same point.

They are equivalent statements of one law, even though they read differently. It can be shown that a device violating the Clausius statement (a 'free' refrigerator moving heat uphill) could be used to build a device that violates the Kelvin-Planck statement (a 100% efficient engine), and vice versa. Textbooks state both because each is the natural language for a different kind of device — an engine versus a refrigerator.

A quasi-static process changes so slowly that the system is essentially in equilibrium with its surroundings at every instant — pressure and temperature differ from the surroundings only infinitesimally throughout. Real processes are never truly quasi-static, but the idealisation is necessary because non-equilibrium states have no well-defined pressure or temperature at all, so there would be no way to write a work formula like W = integral of P dV without it — that integral only makes sense if P is well-defined at every intermediate volume.

Because real engines are always well below their Carnot limit, and closing that gap is where the practical gains are. Friction, turbulence, finite-rate heat transfer, and other irreversibilities all cost efficiency independently of the temperature limit. Carnot efficiency sets the ceiling; engineering determines how close a real design gets to it.

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