Class 11 · Physics · Chapter 11

Thermodynamics

This chapter sets up the language of heat, work and internal energy through the zeroth and first laws, then uses the second law and the Carnot engine to explain why some processes never reverse and why no engine can be 100% efficient.

13 sections13 NCERT questions23 board questions15-question quiz

Before you startWhat the board asks from this chapter

No Class 11 board exam exists, so this chapter carries school-final marks only (commonly 6-10 marks). It is not weighted by CBSE at this stage, but it directly underlies the Class 12 board unit on thermodynamics and kinetic theory, and is a regular, high-yield chapter in JEE and NEET.

Question typeMarksUsually asked
MCQ1Sign of work done, nature of a force, units/dimensions, P-V graph reading.
Assertion–Reason1Reversible vs irreversible, why Carnot efficiency < 1, why ΔU is path-independent.
Very short answer2State the first/second law, define adiabatic/isothermal, define a state variable.
Short answer3Derive work in isothermal/adiabatic process, small numericals with ΔU = Q − W.
Long answer5Carnot cycle efficiency derivation, full labelled P-V diagram, Cp − Cv = R.
Case-based4Gas on a P-V diagram through a cyclic/multi-step process; work, heat, ΔU at each step.

Hot topics, in order

  1. First law of thermodynamics and sign convention
  2. Work done in isothermal and adiabatic processes (with derivation)
  3. Carnot engine and its efficiency (derivation)
  4. Reading work done from a P-V diagram (including cyclic processes)
  5. Second law of thermodynamics — Kelvin-Planck and Clausius statements
  6. Reversible and irreversible processes
  7. Specific heat capacity of a gas at constant volume and constant pressure
  8. Refrigerators and heat pumps, coefficient of performance

The whole chapter, in orderNotes

01 11.1 Introduction

Thermodynamics studies how heat and mechanical work convert into each other, and the direction in which processes are allowed to run. It deals only with macroscopic, bulk variables — pressure, volume, temperature — not with the motion of individual molecules.

A thermodynamic system is the part of the universe under study (e.g., a gas in a cylinder); everything else is the surroundings. A system is open (exchanges matter and energy), closed (only energy), or isolated (neither). The boundary may be adiabatic (blocks heat exchange) or diathermic (allows it).

Exam tip Examiners like one-mark questions that simply ask you to classify a given system as open, closed or isolated — read the scenario carefully for whether matter can cross the boundary.

02 11.2 Thermal Equilibrium

Two systems in contact through a diathermic wall exchange heat until their macroscopic variables stop changing with time — they are then in thermal equilibrium. This is what lets us define temperature: the common value shared once two systems stop exchanging heat.

A system is in full thermodynamic equilibrium only when three equilibria hold together: mechanical (no unbalanced force or pressure difference), chemical (no net reaction or diffusion), and thermal (uniform temperature). Only then do P, V, T have single well-defined values.

Exam tip A question may describe a gas with a piston still moving, or two chambers of a divided box at different pressures — recognise this as NOT in equilibrium, since mechanical equilibrium fails.

03 11.3 Zeroth Law of Thermodynamics

Let systems A and B each be in thermal contact with a third system C (through diathermic walls) but not with each other. Experiment shows A and B are then also in thermal equilibrium with each other. This is the zeroth law: 'Two systems in thermal equilibrium with a third system separately are in thermal equilibrium with each other.'

The zeroth law justifies temperature as a measurable quantity: thermal equilibrium is transitive, so a single number (temperature) can be assigned to every system such that two systems are in equilibrium if and only if they have the same temperature. This is exactly how a thermometer (system C) works — it need never touch A and B together.

Exam tip The zeroth law is asked almost every year as a 1 or 2 mark 'state the law' question. Memorise the exact wording: 'in thermal equilibrium with a third system separately are in thermal equilibrium with each other.'

04 11.4 Heat, Internal Energy and Work

Internal energy (U) is the sum of the kinetic and potential energies of all a system's molecules. For an ideal gas, with no intermolecular forces, U is purely kinetic and depends only on temperature. U is a state variable — it depends only on the current state, not on how the system got there.

U changes in exactly two ways: by heat (energy transferred due to a temperature difference, flowing spontaneously hot→cold) or by work (energy transferred any other way — for a gas, usually by a volume change against external pressure).

Both heat and work are path functions: their value depends on the process, unlike the state function U. This state-vs-path distinction is the most tested idea in the chapter.

dW = P dV (work done by the gas in a small expansion)
Ppressure of the gasdVsmall change in volume
Exam tip Learn this list by heart for a 1-mark 'identify' question: state functions — U, P, V, T, S, enthalpy; path functions — Q, W.

05 11.5 First Law of Thermodynamics

The first law is energy conservation applied to a thermodynamic system. If heat ΔQ is supplied and the system does work ΔW on the surroundings, its internal energy changes by

ΔQ = ΔU + ΔW

Sign convention (NCERT): ΔQ positive when heat is added; ΔW positive when the system does work (expansion), negative when work is done on it (compression); ΔU positive when internal energy rises.

The law is pure energy bookkeeping — it applies to every process, reversible or not, because ΔU depends only on initial and final states.

Special cases: isochoric — ΔW = 0 so ΔQ = ΔU; cyclic — ΔU = 0 so ΔQ = ΔW; adiabatic — ΔQ = 0 so ΔU = −ΔW.

ΔQ = ΔU + ΔW
ΔQheat supplied to the system (J)ΔUchange in internal energy (J)ΔWwork done by the system (J)
Exam tip A 3-mark numerical almost always gives ΔQ and ΔW (or a P-V graph to read ΔW from) and asks for ΔU, or vice-versa. Write the sign convention explicitly before substituting — it is a marking-scheme step.

06 11.6 Specific Heat Capacity

Heat capacity S = ΔQ/ΔT. Specific heat capacity s = S/m; molar specific heat C = S/μ. Water's high specific heat (4186 J kg−1 K−1) is why it is used as a coolant and why coastal climates stay mild.

For a gas, the heat needed for a given ΔT depends on the process, since a gas can also do work by expanding. NCERT defines CV (constant volume — all heat raises U) and CP (constant pressure — some heat also does work). So CP is always greater than CV.

At constant volume, dQ = dU = CVdT. At constant pressure, dQ = dU + PdV = CVdT + RdT (using PV = RT for 1 mole, so PdV = RdT), and dQ = CPdT, giving CP − CV = R (Mayer's relation).

Calorimetry, the measurement of heat exchanged, rests on: heat lost by a hot body = heat gained by a cold body, in an insulated calorimeter with no loss to the surroundings.

S = ΔQ/ΔT
Sheat capacity (J K−1)
s = S/m, C = S/μ
sspecific heat capacity (J kg−1 K−1)Cmolar specific heat capacity (J mol−1 K−1)μnumber of moles
CP − CV = R (Mayer's relation, for 1 mole of an ideal gas)
Runiversal gas constant, 8.314 J mol−1 K−1
Exam tip The derivation of CP − CV = R is a classic 3-mark 'derive' question — always start from dQ = dU + PdV and substitute PV = RT before differentiating.

07 11.7 Thermodynamic State Variables and Equation of State

A system in equilibrium is described by state variables — P, V, T, and amount of matter. These split into extensive variables (mass, volume, internal energy — scale with system size) and intensive variables (pressure, temperature, density — independent of amount).

State variables are linked by an equation of state: for an ideal gas, PV = μRT. So only two of P, V, T are independent for a fixed amount of gas; the third follows. So a fixed quantity of gas is one point on a P-V diagram, and a process is a curve joining two such points.

PV = μRT
μnumber of molesRuniversal gas constantTabsolute temperature (K)
Exam tip A 1-mark question may ask you to classify a quantity as extensive or intensive — mass, volume, internal energy, entropy are extensive; pressure, temperature, density, specific heat are intensive.

08 11.8 Thermodynamic Processes

A process is quasi-static if it proceeds so slowly that the system stays in equilibrium at every instant, so P, V, T are defined throughout. All processes below are assumed quasi-static.

Isothermal: T constant, so PV = constant (Boyle's law) and ΔU = 0 for an ideal gas, giving ΔQ = ΔW. Work in expanding from V1 to V2: W = ∫P dV = μRT ln(V2/V1).

Adiabatic: ΔQ = 0 (insulated, or too fast to exchange heat), so ΔU = −ΔW — expansion cools the gas, compression heats it. Obeys PVγ = constant, γ = CP/CV. Steeper than an isothermal curve through the same point since γ > 1.

Isochoric: V constant, ΔW = 0, ΔQ = ΔU; vertical line on a P-V diagram.

Isobaric: P constant, W = PΔV = μRΔT; horizontal line.

Cyclic: system returns to its initial state, so ΔU = 0 and ΔQ = ΔW — net heat absorbed equals net work done, equal to the area enclosed on the P-V diagram.

Isothermal: W = μRT ln(V2/V1) = 2.303 μRT log10(V2/V1)
Wwork done by the gasV1, V2initial and final volumes
Adiabatic: PVγ = constant, γ = CP/CV
γratio of specific heats
Isobaric: W = PΔV = μRΔT
ΔVchange in volume
Exam tip For a P-V diagram question, remember: area under the curve = work done; for a cyclic process, clockwise loop means net positive work done BY the gas (engine), anticlockwise means net work done ON the gas (refrigerator).

09 11.9 Heat Engines

A heat engine converts heat into work through a cyclic process, using a source (high-temperature reservoir T1, supplies Q1), a working substance taken through a cycle, and a sink (low-temperature reservoir T2, absorbs rejected heat Q2). Since ΔU = 0 over a cycle, the first law gives W = Q1 − Q2.

Efficiency: η = W/Q1 = 1 − Q2/Q1. No engine can reach η = 1 — some heat must always be rejected (Kelvin-Planck statement, section 11.11).

η = W/Q1 = 1 − Q2/Q1
ηefficiency of the engine (no units, expressed as a fraction or %)Q1heat absorbed from the sourceQ2heat rejected to the sink
Exam tip Do not confuse Q1, Q2 here (heat engine, both taken as positive magnitudes by convention) with the ΔQ sign convention of section 11.5 — NCERT deliberately switches notation for this section.

10 11.10 Refrigerators and Heat Pumps

A refrigerator/heat pump is a heat engine run in reverse: external work W extracts heat Q2 from a cold reservoir and rejects a larger Q1 = W + Q2 to a hot reservoir. Its coefficient of performance, α = Q2/W = Q2/(Q1 − Q2), measures heat removed per unit work.

An ideal (Carnot) refrigerator between T1 (hot) and T2 (cold) has α = T2/(T1 − T2). Unlike η, α can exceed 1 — a refrigerator can remove several units of heat per unit of work.

α = Q2/W = Q2/(Q1 − Q2)
αcoefficient of performanceQ2heat extracted from the cold reservoirWwork done on the refrigerator
αCarnot = T2/(T1 − T2)
T1, T2absolute temperatures of hot and cold reservoirs (K)
Exam tip A common numerical gives α and Q2 (or W) and asks for the other, or gives T1, T2 and asks for the maximum possible α — remember α has no upper bound of 1, unlike η.

11 11.11 Second Law of Thermodynamics

The first law tracks energy conservation but does not forbid processes that never actually happen, such as heat flowing spontaneously from cold to hot. The second law supplies this missing direction, with two equivalent NCERT statements:

Kelvin-Planck: 'No process is possible whose sole result is the absorption of heat from a reservoir and the complete conversion of the heat into work.' This is why η = 1 is impossible.

Clausius: 'No process is possible whose sole result is the transfer of heat from a colder object to a hotter object.' A refrigerator moves heat cold→hot, but only with external work, so this is not its sole result — the statement is not violated.

The two statements are logically equivalent: violating one lets you build a device that violates the other.

Exam tip Write both statements exactly as NCERT phrases them — 'sole result' is the key phrase examiners check for in a full-marks answer.

12 11.12 Reversible and Irreversible Processes

A process is reversible if it can be exactly retraced so both system and surroundings return to their original states with no other change anywhere. This requires the process to be quasi-static and free of dissipative effects (friction, viscosity, unrestrained expansion). No real process is perfectly reversible; it is an idealisation setting the theoretical limit on efficiency.

All natural processes are irreversible — heat flow across a finite temperature difference, free expansion into vacuum, and anything with friction cannot be undone without a permanent change elsewhere. This irreversibility is why real engines are always less efficient than the reversible (Carnot) limit.

Exam tip A frequently asked VSA question: 'give two conditions for a process to be reversible' — quasi-static, and free of dissipative forces (friction, viscosity, inelasticity).

13 11.13 Carnot Engine

The Carnot engine is an idealised, reversible engine between two temperatures that gives the maximum possible efficiency for those temperatures, via the four-step Carnot cycle:

  • A→B, isothermal expansion at T1: absorbs Q1, does work W1 = μRT1 ln(V2/V1).
  • B→C, adiabatic expansion: cools T1→T2, doing work at the cost of internal energy.
  • C→D, isothermal compression at T2: rejects Q2; work done on gas W3 = μRT2 ln(V3/V4).
  • D→A, adiabatic compression: heats T2→T1 back to start.

The two adiabatic legs contribute equal and opposite work and cancel, leaving net W = W1 − W3. Using the adiabatic relations for B→C and D→A gives V2/V1 = V3/V4, so the logarithms cancel and Q2/Q1 = T2/T1. Substituting into η = 1 − Q2/Q1:

ηCarnot = 1 − T2/T1

This depends only on the reservoir temperatures, not the working substance, and equals 1 only if T2 = 0 K (unattainable). Carnot's theorem: no engine between the same two temperatures beats a Carnot engine.

ηCarnot = 1 − T2/T1 = 1 − Q2/Q1
T1absolute temperature of the source/hot reservoir (K)T2absolute temperature of the sink/cold reservoir (K)
Exam tip The 5-mark derivation is asked almost every session in some form — practise drawing the P-V diagram (two isothermal curves joined by two steeper adiabatic curves) and labelling A, B, C, D with T1 and T2 before you attempt the algebra.

VocabularyKey terms

Thermodynamic systemThe part of the universe under study; everything else is the surroundings.
Thermal equilibriumThe state two systems reach, through a diathermic wall, when their macroscopic variables stop changing with time.
Zeroth law of thermodynamicsTwo systems separately in thermal equilibrium with a third system are in thermal equilibrium with each other; it establishes temperature as a measurable quantity.
Internal energy (U)The sum of the kinetic and potential energies of all the molecules of a system; a state variable that, for an ideal gas, depends only on temperature.
State variableA quantity (like P, V, T, U) whose value depends only on the current state of the system, not on the path taken to reach it.
Path functionA quantity like heat or work whose value depends on the process, not just the end points.
First law of thermodynamicsStatement of energy conservation for a thermodynamic system: ΔQ = ΔU + ΔW, the heat supplied equals the rise in internal energy plus the work done by the system.
Isothermal processA process carried out at constant temperature, so ΔU = 0 for an ideal gas and PV = constant.
Adiabatic processA process in which no heat enters or leaves the system (ΔQ = 0), so ΔU = −ΔW and PVγ = constant.
Isochoric processConstant-volume process; no work done (ΔW = 0), so all heat changes internal energy.
Isobaric processConstant-pressure process; work done is W = PΔV.
Cyclic processSystem returns to its initial state; ΔU = 0, so net heat absorbed equals net work done.
Heat engineA device that converts heat into work in a cyclic process, using a source, a working substance and a sink.
Efficiency (η)The fraction of heat absorbed by an engine that is converted into useful work, η = 1 − Q2/Q1.
Refrigerator / heat pumpA device that uses external work to move heat from a cold reservoir to a hot one, the reverse of a heat engine.
Coefficient of performance (α)The ratio of heat extracted from the cold reservoir to the work supplied to a refrigerator, α = Q2/W.
Second law of thermodynamicsNo process is possible whose sole result is the complete conversion of heat into work (Kelvin-Planck), or the spontaneous transfer of heat from a colder to a hotter body (Clausius).
Reversible processAn idealised, quasi-static process free of dissipative effects that exactly retraces itself.
Irreversible processA real process (friction, finite temperature gaps, free expansion) that leaves a permanent change behind.
Carnot engineAn idealised, reversible heat engine working in a cycle of two isothermal and two adiabatic steps that has the maximum possible efficiency between two given temperatures.

Keep this on one pageFormula sheet

NameExpressionNotes
Heat capacityS = ΔQ/ΔTamount of heat needed per unit temperature rise, in J K−1
Molar specific heat relation (Mayer's relation)CP − CV = Rfor one mole of an ideal gas
First law of thermodynamicsΔQ = ΔU + ΔWΔQ, ΔW positive when heat is added to / work is done by the system
Equation of state (ideal gas)PV = μRTμ = number of moles, R = 8.314 J mol−1 K−1
Work in an isothermal processW = μRT ln(V2/V1)T constant; equals 2.303 μRT log10(V2/V1)
Adiabatic relationPVγ = constantγ = CP/CV; no heat exchange, ΔQ = 0
Work in an isobaric processW = PΔV = μRΔTP constant
Efficiency of a heat engineη = W/Q1 = 1 − Q2/Q1always less than 1 for a real engine
Coefficient of performanceα = Q2/W = Q2/(Q1 − Q2)for a refrigerator or heat pump; can exceed 1
Carnot efficiencyηCarnot = 1 − T2/T1T1, T2 in kelvin; maximum possible efficiency between these two temperatures
Carnot coefficient of performanceαCarnot = T2/(T1 − T2)maximum possible COP between temperatures T1, T2

Practise with a pencilDiagrams that earn marks

P-V diagram comparing an isothermal and an adiabatic curve

Draw P on the y-axis, V on the x-axis. From a common start point, draw an isothermal curve (label T = constant) and an adiabatic curve (label PVγ = constant) falling more steeply, since γ > 1.

Carnot cycle on a P-V diagram

Draw A, B, C, D clockwise. A→B isothermal expansion at T1 (source, Q1 absorbed); B→C adiabatic expansion (no heat exchange); C→D isothermal compression at T2 (sink, Q2 rejected); D→A adiabatic compression to A. Note enclosed area = net work W = Q1 − Q2.

Schematic of a heat engine (energy-flow diagram)

Draw source T1 (top box) → Q1 → engine (circle) → W (work out, side arrow); engine → Q2 → sink T2 (bottom box). This is the standard sketch before writing η = W/Q1.

Schematic of a refrigerator (energy-flow diagram)

Same three boxes, arrows reversed: Q2 from cold reservoir into the device, W supplied in, Q1 = W + Q2 out to the hot reservoir. Label that Q1 exceeds Q2 by exactly W.

TextbookNCERT questions the board keeps asking

11.1 A geyser heats water flowing at a rate of 3.0 litres per minute from 27 °C to 77 °C. If the geyser operates on a gas burner, what is the rate of consumption of the fuel if its heat of combustion is 4.0 × 104 J g−1?3 m

m = 3.0 kg, ΔT = 50°C, s = 4186 J kg−1 K−1.

Q = msΔT = 3.0 × 4186 × 50 = 6.279 × 105 J per minute.

Fuel/min = Q/(4.0 × 104) ≈ 15.7 g.

11.2 What amount of heat must be supplied to 2.0 × 10−2 kg of nitrogen (at room temperature) to raise its temperature by 45 °C at constant pressure? Molecular mass of N2 = 28; R = 8.3 J mol−1 K−1.3 m

μ = 20/28 = 0.714 mol. N2 diatomic: CP = (7/2)R = 29.05 J mol−1 K−1.

ΔQ = μCPΔT = 0.714 × 29.05 × 45 ≈ 933 J.

11.3 Explain why (a) two bodies at different temperatures T1 and T2 if brought in thermal contact do not necessarily settle to the mean temperature (T1+T2)/2. (b) The coolant in a chemical or a nuclear plant (i.e., the liquid used to prevent the different parts of a plant from getting too hot) should have high specific heat.3 m

(a) The final common temperature depends on the heat capacities (mass × specific heat) of the two bodies, not just their initial temperatures: m1s1(T1 − T) = m2s2(T − T2). T equals the simple mean (T1+T2)/2 only when the two heat capacities are equal.

(b) A coolant with high specific heat absorbs a large quantity of heat for a small rise in its own temperature, so it removes heat efficiently from hot plant parts without itself overheating.

11.4 A cylinder with a movable piston contains 3 moles of hydrogen at standard temperature and pressure. The walls of the cylinder are made of a heat insulator, and the piston is insulated by having a pile of sand on it. By what factor does the pressure of the gas increase if the gas is compressed to half its original volume?3 m

Insulated walls and piston ⟹ adiabatic. P1V1γ = P2V2γ ⟹ P2/P1 = (V1/V2)γ.

H2 diatomic, γ = 1.4; V1/V2 = 2. P2/P1 = 21.4 ≈ 2.64.

11.5 In changing the state of a gas adiabatically from an equilibrium state A to another equilibrium state B, an amount of work equal to 22.3 J is done on the system. If the gas is taken from state A to B via a process in which the net heat absorbed by the system is 9.35 cal, how much is the net work done by the system in the latter case? (Take 1 cal = 4.19 J)5 m

Adiabatic A→B: ΔQ = 0, work done ON system = 22.3 J, so W (by system) = −22.3 J. ΔU = ΔQ − ΔW = 22.3 J (state function, same for any path A→B).

Second path: ΔQ = 9.35 cal = 39.2 J. ΔW = ΔQ − ΔU = 39.2 − 22.3 = 16.9 J.

11.6 Two cylinders A and B of equal capacity are connected to each other via a stopcock. A contains a gas at standard temperature and pressure. B is completely evacuated. The entire system is thermally insulated. The stopcock is suddenly opened. Answer the following: (a) What is the final pressure of the gas in A and B? (b) What is the change in internal energy of the gas? (c) What is the change in the temperature of the gas? (d) Do the intermediate states of the system (before settling to the final equilibrium state) lie on its P-V-T surface?5 m

(a) Free expansion doubles the volume at unchanged T, so by Boyle's law the final pressure is half the initial pressure.

(b) Free expansion: no external pressure opposes it (ΔW = 0), system insulated (ΔQ = 0), so ΔU = 0.

(c) Since ΔU = 0 and U depends only on T for an ideal gas, temperature is unchanged.

(d) No — free expansion is sudden and passes through non-equilibrium states (pressure/density not uniform), so intermediate states cannot be shown on the P-V-T surface; only the initial and final states can.

11.9 A steam engine delivers 5.4 × 108 J of work per minute and services 3.6 × 109 J of heat per minute from its boiler. What is the efficiency of the engine? How much heat is wasted per minute?3 m

W = 5.4 × 108 J/min, Q1 = 3.6 × 109 J/min.

η = W/Q1 = 0.15 = 15%. Heat wasted, Q2 = Q1 − W = 3.06 × 109 J/min.

11.10 An electric heater supplies heat to a system at a rate of 100 W. If the system performs work at a rate of 75 joules per second, at what rate is the internal energy increasing?2 m

ΔU/Δt = ΔQ/Δt − ΔW/Δt = 100 − 75 = 25 W.

Exemplar 11.11 Consider a Carnot's cycle operating between T1 = 500 K and T2 = 300 K producing 1 kJ of mechanical work per cycle. Find the heat transferred to the engine by the reservoirs.3 m

ηCarnot = 1 − 300/500 = 0.4. Q1 = W/η = 1000/0.4 = 2.5 kJ absorbed; Q2 = Q1 − W = 1.5 kJ rejected.

11.12 A person of mass 60 kg wants to lose weight by burning fat by going up and down a 10 m high staircase. Assume he burns twice as much fat while going up than coming down. If 1 kg of fat is burnt on expending 7000 kcal, how many times must he go up and down to reduce his weight by 1 kg? Take g = 9.8 m/s2.5 m

Energy up = mgh = 60 × 9.8 × 10 = 5880 J; down = half = 2940 J. One round trip = 8820 J.

Energy for 1 kg fat = 7000 kcal = 2.93 × 107 J. n = (2.93 × 107)/8820 ≈ 3324 times.

C1 State the zeroth law of thermodynamics and explain its significance.2 m

'Two systems separately in thermal equilibrium with a third system are in thermal equilibrium with each other.' It makes temperature measurable — a thermometer (the third system) can compare two bodies without direct contact, since thermal equilibrium is transitive.

C2 Derive an expression for the work done in an isothermal expansion of an ideal gas.3 m

For μ moles expanding isothermally from V1 to V2, P = μRT/V.

W = ∫V1V2 P dV = μRT ln(V2/V1) = 2.303 μRT log10(V2/V1).

Since T is constant, ΔU = 0, so this is also the heat absorbed.

C3 Derive the expression for the efficiency of a Carnot engine.5 m

See section 11.13 for the full derivation. Net work W = Q1 − Q2; the adiabatic relations reduce the volume ratios so Q2/Q1 = T2/T1; substituting into η = 1 − Q2/Q1 gives η = 1 − T2/T1.

Previous years and frequently askedBoard questions with model answers

MCQ In an isochoric process, the work done by the gas is:
  1. Positive
  2. Negative
  3. Zero
  4. Cannot be determined
Frequently asked1 m
Answer: (c)
In an isochoric (constant-volume) process, ΔV = 0, so W = PΔV = 0.
MCQ A gas is taken through a cyclic process on a P-V diagram. The net heat absorbed by the gas in one complete cycle is 500 J. The net work done by the gas in the cycle is:
  1. 0 J
  2. 500 J
  3. 1000 J
  4. Cannot be determined without more data
Frequently asked1 m
Answer: (b)
For a cyclic process ΔU = 0, so ΔQ = ΔW; hence the net work done equals the net heat absorbed, 500 J.
MCQ Which of the following is a state function?
  1. Heat
  2. Work
  3. Internal energy
  4. Both heat and work
Frequently asked1 m
Answer: (c)
U depends only on state; heat and work are path functions.
MCQ For an ideal gas undergoing an adiabatic process, which relation is correct? (γ = CP/CV)
  1. PV = constant
  2. PVγ = constant
  3. PV2 = constant
  4. P/V = constant
Frequently asked1 m
Answer: (b)
The adiabatic relation for an ideal gas is PVγ = constant.
MCQ A Carnot engine operates between a source at 600 K and a sink at 300 K. Its efficiency is:
  1. 25%
  2. 50%
  3. 75%
  4. 100%
Frequently asked1 m
Answer: (b)
η = 1 − 300/600 = 50%.
MCQ The coefficient of performance of a refrigerator is 5. If the refrigerator releases 6 J of heat to the surroundings per cycle, the work done on the working substance is:
  1. 1 J
  2. 2 J
  3. 5 J
  4. 6 J
Frequently asked1 m
Answer: (a)
Q1 = Q2 + W; α = Q2/W = 5 ⟹ 5W + W = 6 ⟹ W = 1 J.
MCQ On a P-V diagram, an adiabatic curve through a point compared to an isothermal curve through the same point is:
  1. Less steep
  2. Equally steep
  3. More steep
  4. Cannot be compared
Frequently asked1 m
Answer: (c)
The adiabatic curve is steeper than the isothermal curve at the same point because γ > 1.
Assertion–Reason Assertion (A): The internal energy of an ideal gas does not change in an isothermal process. Reason (R): Internal energy of an ideal gas depends only on its temperature.
  1. Both A and R are true and R is the correct explanation of A
  2. Both A and R are true but R is not the correct explanation of A
  3. A is true but R is false
  4. A is false but R is true
Frequently asked1 m
Answer: (a)
Isothermal ⟹ T constant ⟹ ΔU = 0 for an ideal gas; R correctly explains A.
Assertion–Reason Assertion (A): No heat engine can have an efficiency of 100%. Reason (R): Some heat must always be rejected to the sink, as required by the second law of thermodynamics.
  1. Both A and R are true and R is the correct explanation of A
  2. Both A and R are true but R is not the correct explanation of A
  3. A is true but R is false
  4. A is false but R is true
Frequently asked1 m
Answer: (a)
Kelvin-Planck forbids fully converting absorbed heat into work — R correctly explains A.
Assertion–Reason Assertion (A): Free expansion of a gas into vacuum is an irreversible process. Reason (R): During free expansion, the gas does work against the vacuum it expands into.
  1. Both A and R are true and R is the correct explanation of A
  2. Both A and R are true but R is not the correct explanation of A
  3. A is true but R is false
  4. A is false but R is true
Frequently asked1 m
Answer: (c)
A is true (irreversible, non-equilibrium states). R is false: free expansion into vacuum does zero work, no opposing pressure.
Very short answer State the first law of thermodynamics. Write the sign convention used.Frequently asked2 m

The first law states that the heat supplied to a system (ΔQ) equals the sum of the change in its internal energy (ΔU) and the work done by the system (ΔW): ΔQ = ΔU + ΔW.

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

Very short answer Why is CP greater than CV for a gas?Frequently asked2 m

At constant V, all heat raises internal energy. At constant P, the gas also expands and does work, needing extra heat for the same ΔT. So CP > CV; CP − CV = R for one mole.

Very short answer Give two conditions necessary for a thermodynamic process to be reversible.Frequently asked2 m

(i) Quasi-static — infinitely slow, always in equilibrium. (ii) Free of dissipative effects — no friction, viscosity or inelasticity.

Very short answer State the Kelvin-Planck and Clausius statements of the second law of thermodynamics.Frequently asked2 m

Kelvin-Planck: No process's sole result can be absorbing heat from a reservoir and fully converting it to work.

Clausius: No process's sole result can be heat transfer from a colder to a hotter object.

Short answer A gas is compressed isothermally to half its volume. Would the same final pressure result from adiabatic compression to the same volume? Where is more work done on the gas?Frequently asked3 m

No. Isothermal: PV = constant ⟹ P2 = 2P1. Adiabatic: PVγ = constant ⟹ P2′ = 2γP1 > 2P1 since γ > 1.

More work is done on the gas adiabatically: isothermal compression can lose heat to keep T fixed, but adiabatic compression converts all the input work into internal energy (temperature rise), needing more work for the same final volume.

Short answer Derive the relation CP − CV = R for one mole of an ideal gas.Frequently asked3 m

Constant V: dQ = dU = CVdT. Constant P: dQ = dU + PdV = CVdT + PdV. From PV = RT (1 mole), at constant P, PdV = RdT. So dQ = CVdT + RdT = CPdT (definition), giving CP − CV = R.

Short answer A refrigerator is to maintain eatables kept inside it at 9°C. If the room temperature is 36°C, calculate the coefficient of performance of the Carnot refrigerator.Frequently asked3 m

T2 = 9°C = 282 K, T1 = 36°C = 309 K.

αCarnot = T2/(T1 − T2) = 282/27 ≈ 10.4.

Short answer 1 mole of an ideal gas at temperature T1 expands according to the law PV3 = constant. Explain qualitatively why the gas cools during this expansion.Frequently asked3 m

Expansion (V increases) means the gas does positive work. PV3 = constant is steeper than a normal adiabat, so little or no heat is absorbed. With ΔQ ≈ 0 and ΔW > 0, ΔU = ΔQ − ΔW < 0, so temperature falls — the gas cools even more sharply than in ordinary adiabatic expansion.

Long answer Describe the Carnot cycle with the help of a P-V diagram and derive the expression for its efficiency.Frequently asked5 m

Four reversible steps between source T1 and sink T2: (1) isothermal expansion A→B at T1, absorbs Q1; (2) adiabatic expansion B→C, T1→T2; (3) isothermal compression C→D at T2, rejects Q2; (4) adiabatic compression D→A, T2→T1. [P-V diagram: two isothermals joined by two steeper adiabatics, closed loop ABCD.]

W1 = μRT1 ln(V2/V1); W3 (on gas) = μRT2 ln(V3/V4); adiabatic legs cancel in the net.

Adiabatic relations give V2/V1 = V3/V4, so Q2/Q1 = T2/T1.

η = 1 − Q2/Q1 = 1 − T2/T1, the maximum efficiency between T1 and T2.

Long answer (a) State the first law of thermodynamics and explain each term. (b) A gas absorbs 200 J of heat and does 50 J of work on the surroundings. Find the change in internal energy. (c) The same gas, taken along a different path from the same initial to the same final state, absorbs 120 J of heat. Find the work done by the gas along this path.Frequently asked5 m

(a) ΔQ = ΔU + ΔW: ΔQ is heat supplied, ΔU the change in internal energy, ΔW the work done by the system on the surroundings — conservation of energy for the system.

(b) ΔU = ΔQ − ΔW = 200 − 50 = 150 J.

(c) U is a state function, so ΔU = 150 J on any path. ΔW = ΔQ − ΔU = 120 − 150 = −30 J, i.e. 30 J done ON the gas.

Long answer Explain, with a labelled diagram, the working of a heat engine and a refrigerator. Derive the relation between the efficiency of a heat engine and the coefficient of performance of a refrigerator working as its reverse, when both are Carnot devices operating between the same two temperatures.Frequently asked5 m

[Heat engine: source T1 → Q1 → engine → W out, Q2 → sink T2. Refrigerator: arrows reversed, Q2 in from cold body, W in, Q1 = Q2 + W out to hot body.]

Heat engine: absorbs Q1, converts part to W, rejects Q2 = Q1 − W; η = 1 − Q2/Q1.

Refrigerator: uses W to move Q2 from cold to hot, rejecting Q1 = Q2 + W; α = Q2/W.

For Carnot devices, η = 1 − T2/T1 and α = T2/(T1−T2). Since η = (T1−T2)/T1: α = (1 − η)/η — a highly efficient engine corresponds to a poor refrigerator, and vice versa.

Case-based One mole of an ideal gas is taken through cyclic process ABCA: A→B isothermal expansion at T from V0 to 2V0; B→C isochoric, pressure falls to the value at A; C→A isobaric compression back to V0.
(i) What is ΔU for the complete cycle ABCA?
(ii) What is the relation between the net heat absorbed and the net work done over the cycle?
(iii) Is the work done by the gas in step A→B positive or negative? Why?
(iv) In which step, B→C or C→A, is no work done by the gas? Why?Frequently asked4 m

(i) ΔU = 0 for the full cycle — U is a state function and the gas returns to A.

(ii) ΔU = 0 ⟹ ΔQ = ΔW: net heat absorbed = net work done = area enclosed by ABCA.

(iii) Positive — the gas expands (V0→2V0) in A→B, doing work as it pushes outward.

(iv) No work in B→C (isochoric, ΔV = 0, W = PΔV = 0). Work IS done in C→A (isobaric compression 2V0→V0), and it is negative since the gas is compressed.

Case-based 200 g water at 60°C is mixed with 100 g water at 20°C in an insulated calorimeter of negligible heat capacity.
(i) State the principle used to find the final temperature of the mixture.
(ii) Set up the equation for heat lost and heat gained.
(iii) Calculate the final equilibrium temperature. (Specific heat of water = 4186 J kg−1 K−1, or equivalently 1 cal g−1 °C−1)
(iv) Would the final temperature change if the calorimeter itself had a significant heat capacity? Explain briefly.Frequently asked4 m

(i) Calorimetry principle: heat lost by the hotter body = heat gained by the colder body in an insulated system.

(ii) m1s(T1 − T) = m2s(T − T2), m1 = 200 g at 60°C, m2 = 100 g at 20°C, s cancels (same liquid).

(iii) 200(60−T) = 100(T−20) ⟹ 14000 = 300T ⟹ T ≈ 46.7°C.

(iv) Yes — the calorimeter itself would absorb heat too, adding a term mcsc(T − T2) to the heat-gained side, giving a slightly lower final T.

Test yourselfPractice quiz

Marks lost every yearCommon mistakes

  • Confusing the sign convention: NCERT uses ΔQ = ΔU + ΔW with ΔW positive for work done BY the gas — apply it consistently.
  • Treating heat and work as state functions like internal energy — they are path functions, defined only for a process, not a state.
  • Forgetting ΔU = 0 in an isothermal process for an ideal gas even though ΔQ and ΔW are generally non-zero.
  • Mixing up adiabatic (ΔQ = 0) with isothermal (ΔT = 0) — different processes, different P-V curves and work formulas.
  • Using Celsius instead of kelvin in η = 1 − T2/T1 — Carnot formulas need absolute temperature.
  • Assuming a refrigerator's COP is bounded by 1 like efficiency — α often exceeds 1.
  • Stating the second law only as 'entropy increases' without the exact Kelvin-Planck/Clausius wording NCERT expects.
  • Ignoring that the direction of tracing a cycle (clockwise/anticlockwise) decides the sign of net work on a P-V diagram.

How to prepareStrategy for this chapter

Nail definitions exactly as NCERT states them — the zeroth law and the two second-law statements are asked almost verbatim every year. Master the first law numerically: practise problems that give any two of ΔQ, ΔU, ΔW and ask for the third, using process shortcuts (ΔU = 0 isothermal, ΔW = 0 isochoric, ΔQ = 0 adiabatic).

The Carnot cycle derivation is the highest-value 5-mark proof — practise the P-V diagram and the algebra giving Q2/Q1 = T2/T1 by heart, since partial marks are step-wise. Derive CP − CV = R at least twice.

Keep R = 8.314 J mol−1 K−1 and water's specific heat memorised, always convert °C to K before using Carnot or gas-law formulas, and practise reading process type and work done (area under/enclosed) directly off a P-V diagram.

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