Thermodynamics | ISC Class 11 Physics Notes

This note covers thermal equilibrium, the zeroth law, heat, work, internal energy, the first law, heat capacities, thermodynamic processes, pressure-volume graphs, the second law, reversibility, heat engines and the Carnot cycle.

What do thermodynamic states and thermal equilibrium mean?

Thermodynamics studies heat, temperature and the conversion of energy between thermal and other forms. It describes bulk matter through measurable quantities. A system is the matter selected for study; the surroundings are everything outside it with which it may interact.

A macroscopic variable describes a system as a whole. Pressure is normal force per unit area; volume is the space occupied. Pressure, volume, temperature, mass and composition specify the equilibrium state of a gas. Thermodynamics concerns this internal state, rather than the motion of the whole container through space.

How is equilibrium recognised?

In thermodynamic equilibrium, the macroscopic variables describing the system do not change with time. A gas in a closed, rigid, insulated container can have fixed pressure, volume, temperature, mass and composition. Its molecules still move; an unchanging bulk state does not mean motionless molecules.

The nature of the separating wall matters. An adiabatic wall prevents heat transfer. A diathermic wall allows heat transfer. Two systems separated by insulation need not have equal temperatures, even when their individual states remain unchanged.

When systems at different temperatures can exchange heat, energy passes from the hotter system to the colder one. Their states change until thermal equilibrium is reached. At that point, their temperatures are equal and there is no further net heat flow between them.

Definition: Thermal equilibrium is the condition in which systems in thermal contact have equal temperatures and no net heat transfer occurs between them.

Constraints decide which variables change during equilibration. For gases in fixed-volume containers, the volumes cannot change, although pressures can change as temperatures approach equality. Equal temperatures do not require equal pressures, equal volumes or equal masses.

Mechanical equilibrium concerns the balance of forces. Thermal equilibrium concerns temperature equality. Keeping these ideas distinct prevents the mistaken conclusion that a stationary object must have the same temperature as its surroundings.

How does the zeroth law define temperature?

Definition: The zeroth law states that if two systems are separately in thermal equilibrium with a third system, they are also in thermal equilibrium with each other.

Let A, B and C denote three systems. Initially, A and B are separated by an adiabatic wall, while both can exchange heat with C. Allow each to reach thermal equilibrium with C before changing the walls.

Now insulate C from the other systems and replace the wall between A and B with a conducting wall. No further change of their states occurs. A and B were already at the same temperature, even though they had not exchanged heat directly.

The shared property is temperature, denoted by T. It is the thermodynamic quantity equal for systems in thermal equilibrium. Write Tₐ, Tᵦ and T꜀ for the temperatures of A, B and C, respectively.

Tₐ = T꜀ and Tᵦ = T꜀ imply Tₐ = Tᵦ.

The SI unit of temperature is kelvin, K; SI means the International System of Units. In ideal-gas equations and engine-efficiency ratios, T means absolute temperature on the kelvin scale. Celsius temperature values cannot be inserted directly into those ratios.

Two panels show C above A and B. In panel (a), A and B have an insulating partition and conducting contact with C. In panel (b), C is insulated while the partition between A and B conducts heat.

See Fig. 11.2 in your NCERT textbook

The law supplies the basis for comparing temperatures through a third system. Thermal equilibrium expresses equality of temperature, not equality of the total energy stored in the objects. Different quantities of matter can share a temperature without having equal internal energies.

How do heat, internal energy and work differ?

Internal energy, U, is the sum of molecular kinetic and potential energies, measured relative to the frame in which the system's centre of mass is at rest. It excludes the kinetic energy associated with motion of the system as a whole.

Molecular kinetic energy may involve translation, movement from place to place; rotation, turning; and vibration, oscillation within a molecule. Molecular potential energy is energy associated with interactions and configuration. Neglecting intermolecular forces simplifies the internal energy of an ideal gas.

Heat, Q, is energy transferred because of a temperature difference. Work, W, is energy transferred by other means, such as a gas moving a piston. Neither heat nor work is a substance stored inside the system.

Which quantities describe a state?

A state variable depends on the present equilibrium state, not on the history of reaching it. Internal energy is a state variable. The change between fixed initial and final states is therefore fixed, even if different processes connect those states.

The SI unit of heat is the joule, J. The SI unit of work is also the joule. The SI unit of internal energy is the joule. These shared units express the fact that each concerns energy, despite their different physical meanings.

A gas may gain internal energy when it receives heat or when work is done on it. It can lose internal energy by transferring heat outwards or by doing work. The actual change depends on both transfers together.

Note: Say that a system contains internal energy and receives heat. Saying that it contains a quantity of heat confuses stored energy with energy in transit.

How is the first law applied with the correct signs?

The first law of thermodynamics expresses conservation of energy. Define ΔU = U₂ − U₁, where Δ means change, U₁ is initial internal energy and U₂ is final internal energy. With work measured as work done by the system:

Q = ΔU + W, or equivalently ΔU = Q − W.

Heat entering the system is positive. Heat leaving it is negative. Work done by the system is positive; work done on it is negative. A positive internal-energy change indicates an increase in stored internal energy.

How should a calculation be organised?

Identify the system first. Translate descriptions such as “done on” into signed values before substitution. Use one energy unit throughout. Finally, check whether the sign of the answer agrees with the physical direction of energy transfer.

Worked example 1. A heater supplies energy at 100 W while the system does work at 75 J per second. Find the rate at which internal energy increases.

Formula: Let q be heat supplied per second, w work done per second and u internal-energy increase per second. Then u = q − w. A watt, W, is a unit of power: 1 W = 1 J s⁻¹, with s meaning second.

Substitute: u = 100 − 75.

Answer: u = 25 J s⁻¹. The positive result means that energy accumulates inside the system.

Different paths between the same states can have different Q and W. Their difference Q − W remains the same because it equals ΔU. A first-law question can therefore combine information about two different paths without assuming their separate heat and work transfers are equal.

What are principal and molar heat capacities?

Heat capacity, S, is heat required per unit temperature rise for a specified process. Let ΔT mean the final temperature minus the initial temperature. For a small interval, S = Q/ΔT. The SI unit of heat capacity is J K⁻¹.

Specific heat capacity, s, is heat capacity per unit mass: s = Q/(mΔT), where m is mass in kilograms, kg. Here s denotes a material property, while s in a unit such as J s⁻¹ denotes the second.

Molar heat capacity, C, is heat capacity per mole: C = Q/(nΔT), where n is the amount of substance in moles, mol. Its SI unit is J mol⁻¹ K⁻¹; specific heat capacity has unit J kg⁻¹ K⁻¹.

Heat capacities depend on the substance, temperature and conditions of heating. For a gas, the principal heat capacities are those at constant pressure and constant volume. Use Cₚ and Cᵥ for the corresponding molar values. Heating at constant pressure allows expansion work; heating at constant volume does not.

Derivation: How are the two molar heat capacities related?

An ideal gas obeys PV = nRT, where P is pressure, V volume and R the universal gas constant. For a fixed amount, its internal energy depends only on temperature. Consider one mole and a small temperature change.

1. At constant volume, expansion work is zero, so Q = ΔU and ΔU = CᵥΔT.
2. At constant pressure, the first law gives CₚΔT = ΔU + PΔV, where ΔV is the change in volume.
3. The equation PV = RT for one mole gives PΔV = RΔT at constant pressure.
4. Substitute and divide by ΔT to obtain Cₚ = Cᵥ + R.

Cₚ − Cᵥ = R. The relation uses molar heat capacities of an ideal gas. The ratio γ = Cₚ/Cᵥ, read as gamma, is dimensionless because numerator and denominator have the same units.

How are calories converted?

A calorie, cal, corresponds to heating 1 g of water from 14.5 °C to 15.5 °C, where g means gram and °C denotes degrees Celsius. Water's specific heat varies slightly with temperature, making the specified interval important.

Worked example 2. Express the heat needed to raise 1 g of water from 14.5 °C to 15.5 °C in joules. Use 1 cal = 4.186 J exactly.

Formula: This specified heating defines 1 cal.

Answer: Q = 4.186 J. The conversion changes the unit, not the physical amount of energy.

Use 1 cal = 4.186 J as a unit conversion. Do not insert an additional mechanical-equivalent factor into an equation whose heat, work and internal-energy terms already use the same units.

Worked example 3. Heat 2.0 × 10⁻² kg of nitrogen through 45 K at constant pressure. Treat it as an ideal rigid diatomic gas, with Cₚ = 7R/2, molar mass M = 28 g mol⁻¹ and R = 8.3 J mol⁻¹ K⁻¹. A diatomic molecule contains two atoms; “rigid” here means neglecting molecular vibration.

Formula: n = m/M; Q = nCₚΔT. Use matching mass units in m/M.

Substitute: n = 20/28 mol; Q = (20/28) × (7/2) × 8.3 × 45.

Answer: Q = 933.75 J, approximately 9.3 × 10² J. The assumed heat capacity applies to the stated molecular model.

How do pressure-volume graphs represent work?

A pressure-volume graph, or PV graph, plots pressure vertically and volume horizontally. An equilibrium state appears as a point. The SI unit of pressure is the pascal, Pa, and the SI unit of volume is the cubic metre, m³.

A quasi-static process proceeds ideally through equilibrium states, with pressure and temperature differences from the surroundings infinitesimally small. “Infinitesimally” describes differences approaching zero. A sufficiently slow process without large gradients or accelerated piston motion can approximate this idealisation.

How does piston displacement give work?

For a small volume increase dV in a quasi-static gas process, dW = P dV, where dW denotes the small work done. Summing these contributions gives W = ∫ P dV, evaluated from initial volume V₁ to final volume V₂.

The integral sign ∫ denotes continuous summation. The area under the PV curve between the end volumes represents work. Expansion gives positive work. Compression gives negative work because the final volume is smaller than the initial volume.

For pressure remaining constant, W = P(V₂ − V₁). If pressure changes, multiplying an arbitrary pressure by the volume change is insufficient. A straight-line pressure variation allows use of the mean of the two endpoint pressures.

D is at volume 2.0 m³ and pressure 600 N m⁻²; E is at 5.0 m³ and 300 N m⁻²; F is at 2.0 m³ and 300 N m⁻². Arrows run D to E, E to F and vertically F to D. N denotes newton, a force unit; N m⁻² equals Pa.

See Fig. 11.11 in your NCERT textbook

Worked example 4. A gas follows a straight line from D, at 2.0 m³ and 600 Pa, to E, at 5.0 m³ and 300 Pa. It then returns to volume 2.0 m³ at 300 Pa, reaching F. Find work over D to E to F.

Formula: W₁ = [(Pᴅ + Pᴇ)/2](Vᴇ − Vᴅ); W₂ = Pᴇ(Vꜰ − Vᴇ); W = W₁ + W₂. Subscripts D, E and F identify states; W₁ and W₂ are the successive works.

Substitute: W₁ = [(600 + 300)/2] × (5.0 − 2.0) = 1350 J; W₂ = 300 × (2.0 − 5.0) = −900 J.

Answer: W = 450 J. The compression subtracts from the expansion work; it must not be counted as another positive area.

A vertical segment has constant volume and contributes no expansion work. For a complete loop, the enclosed area gives the magnitude of net work. The direction of traversal determines its sign.

What happens in an isothermal ideal-gas process?

An isothermal process maintains constant temperature throughout. For a fixed amount of ideal gas, PV = nRT then implies PV = constant. Pressure varies inversely with volume. This is Boyle's law under its constant-temperature condition.

A gas in a conducting cylinder can expand slowly while remaining in thermal contact with a large reservoir, a body whose temperature is effectively unchanged by the heat exchanged. Heat entering the gas replaces the energy it transfers outwards as work.

Derivation: Work in isothermal expansion

Let V₁ and V₂ be the initial and final volumes, and T the constant absolute temperature. Consider a quasi-static process of a fixed amount n of ideal gas.

1. Begin with the work integral W = ∫ P dV between V₁ and V₂.
2. Use the ideal-gas relation to substitute P = nRT/V.
3. Since n, R and T remain constant, take nRT outside the integral, giving W = nRT ∫ dV/V.
4. The integral gives the natural logarithm, written ln, so evaluating its endpoint difference gives W = nRT ln(V₂/V₁).

W = nRT ln(V₂/V₁). The ratio inside the logarithm is dimensionless, so both volumes must use matching units. The natural logarithm is the logarithm to base e, the mathematical constant.

For a fixed amount of ideal gas, U depends only on temperature. Therefore ΔU = 0 and Q = W in an isothermal process. During expansion V₂ > V₁, so work and heat supplied are positive.

During isothermal compression V₂ < V₁, work done by the gas is negative and heat leaves it. Temperature can stay constant even when heat is transferred. Constant temperature must therefore not be confused with thermal insulation.

Note: The inference “constant temperature means no change in internal energy” is being used for a fixed amount of ideal gas. Do not extend it indiscriminately to every substance or change of state.

How does an adiabatic process change pressure and temperature?

An adiabatic process has no heat transfer: Q = 0. Consequently, the first law becomes ΔU = −W. Work done during expansion comes from internal energy; compression work increases internal energy.

For a quasi-static adiabatic process of a fixed amount of ideal gas with constant heat-capacity ratio γ, PVᵞ = constant. Thus P₁V₁ᵞ = P₂V₂ᵞ, where subscripts 1 and 2 denote initial and final states. This relation is quoted without deriving it.

Derivation: Work in an adiabatic process

Let A denote the constant PVᵞ in this derivation. It is an algebraic constant, not the system A used in the equilibrium discussion.

1. Write P = A/Vᵞ and substitute into W = ∫ P dV from V₁ to V₂.
2. Integrating gives W = A[V₂¹⁻ᵞ − V₁¹⁻ᵞ]/(1 − γ), for γ different from unity.
3. Since AV¹⁻ᵞ = PV at either endpoint, rearrange to W = (P₁V₁ − P₂V₂)/(γ − 1).
4. Use P₁V₁ = nRT₁ and P₂V₂ = nRT₂ to express the result using initial temperature T₁ and final temperature T₂.

W = nR(T₁ − T₂)/(γ − 1). For expansion with positive work, T₂ is lower than T₁. Compression gives negative work by the gas and raises its temperature.

Pressure P is vertical and volume V horizontal. Two falling curves labelled isothermal are joined by two curves labelled adiabatic. The adiabatic portions connect states on the different isotherms.

See Fig. 11.8 in your NCERT textbook

Worked example 5. An insulated cylinder contains 3 mol of hydrogen initially at 1.013 × 10⁵ Pa. Compress the ideal gas quasi-statically to half its original volume. Take γ = 7/5, neglecting molecular vibration. Find its final pressure.

Formula: P₂ = P₁(V₁/V₂)ᵞ.

Substitute: P₂ = 1.013 × 10⁵ × 2¹·⁴.

Answer: P₂ ≈ 2.67 × 10⁵ Pa, or approximately 267000 Pa. The pressure rises by a factor of approximately 2.64; the amount of gas cancels from this pressure ratio.

Worked example 6. Work of 22.3 J is done on a gas during an adiabatic change from state A to state B. Another path between the same states supplies 9.35 cal of heat. Find work done by the gas on that path, using 1 cal = 4.186 J.

Formula: ΔU = −W for the adiabatic path; Q = 9.35 × 4.186 J; W = Q − ΔU for the second path.

Substitute: The first path has W = −22.3 J, so ΔU = 22.3 J. The second has Q = 39.1391 J.

Answer: W = 39.1391 − 22.3 = 16.8391 J, approximately 16.8 J, done by the gas.

How do constant-volume, constant-pressure and cyclic processes compare?

An isochoric process keeps volume constant. With only pressure-volume work considered, W = 0 because the boundary does not move. Thus Q = ΔU. Heating changes the gas's temperature and pressure without producing expansion work.

An isobaric process keeps pressure constant. The work is W = P(V₂ − V₁). For an ideal gas this is also W = nR(T₂ − T₁). Heat supplied during warming is shared between increased internal energy and expansion work.

A cyclic process returns the system to its original state. Internal energy therefore returns to its original value, making the net ΔU zero. Over the full cycle, net heat absorbed equals net work done.

Why does a cycle differ from an isothermal process?

Both can have zero total change of internal energy, but for different reasons. An ideal-gas isothermal process has no temperature change. A cycle may contain temperature changes in its individual stages, provided the final state equals the initial state.

Consequently, zero ΔU alone does not prove that a whole process was isothermal. Likewise, zero net work does not by itself establish that volume remained fixed at every stage. Identify the defining condition before selecting a special-case formula.

For isochoric heating, use Q = nCᵥΔT; for isobaric heating, use Q = nCₚΔT, when the relevant heat capacity can be treated as constant over the interval. The different coefficients record the different energy requirements of the two paths.

What does the second law say about reversibility and engines?

The first law balances energy but does not determine whether every imagined process can occur. The second law of thermodynamics places further restrictions on the direction of processes and on converting heat into useful work.

What are its two statements?

The Kelvin-Planck statement rules out a process whose sole result is taking heat from a reservoir and converting it completely into work. A cyclic heat engine cannot have complete conversion of all its absorbed heat into work as its only effect.

The Clausius statement rules out a process whose sole result is transferring heat from a colder object to a hotter object. A refrigerator can perform that transfer when external work is supplied. The phrase “sole result” is essential to both statements.

A heat engine operates cyclically, taking heat from a hot source, delivering work and rejecting heat to a colder sink. A source supplies heat; a sink receives rejected heat. The energy balance permits rejected heat and useful work together.

When is a process reversible?

A reversible process can be reversed so that both system and surroundings regain their original states with no other change anywhere. A return of the system alone is insufficient. Reversibility is an idealisation requiring quasi-static operation and no dissipative effects.

Dissipative effects, such as friction and viscosity, convert organised mechanical energy into internal energy. They can be minimised but not fully eliminated. Most processes encountered in practice are therefore irreversible.

Free expansion into a vacuum, gas diffusing through a room, and heat spreading from the hotter base of a vessel are irreversible examples. A vacuum is a region without matter. Diffusion is the spontaneous spreading of gas through available space.

An irreversible process cannot restore both system and surroundings exactly by reversal without leaving another change. Slow operation alone does not guarantee reversibility if friction remains. A quasi-static isothermal expansion with a frictionless piston is an ideal reversible example.

How does a Carnot engine achieve its limiting efficiency?

A Carnot engine is a reversible heat engine operating between a hot reservoir at absolute temperature T₁ and a cold reservoir at T₂. Here T₁ > T₂. The ideal source and sink have infinite thermal capacity, so their temperatures remain fixed.

For an ideal-gas realisation, use a cylinder with a frictionless movable piston. Arrange conducting contact with the appropriate reservoir during heat transfer and thermal insulation during temperature-changing stages. The ideal cycle consists of two isothermal and two adiabatic processes.

What happens in its four stages?

1. Isothermal expansion: At T₁ the gas absorbs heat Q₁ from the source and does work while its internal energy remains unchanged.
2. Adiabatic expansion: With the gas insulated, it continues doing work. Its internal energy falls and its temperature reaches T₂.
3. Isothermal compression: At T₂ the surroundings do work on the gas, which rejects heat Q₂ to the sink.
4. Adiabatic compression: Insulation prevents heat exchange while work on the gas raises its temperature to T₁ and restores its original state.

Here Q₁ and Q₂ are positive magnitudes of absorbed and rejected heat. Thus net heat absorbed is Q₁ − Q₂, and the net work output W equals that difference. This notation does not change the earlier signed convention for Q.

The PV plot shows four labelled states joined into a loop. The upper isotherm connects states 1 and 2 at T₁; the lower connects states 3 and 4 at T₂. Adiabatic curves join 2 to 3 and 4 to 1. Arrows show the engine cycle.

See Fig. 11.9 in your NCERT textbook

How is efficiency calculated?

Efficiency, η, is useful work output divided by heat absorbed from the source. For the complete cycle, η = W/Q₁ = 1 − Q₂/Q₁. For a Carnot engine, the temperature form is η = 1 − T₂/T₁.

These expressions give a dimensionless fraction. Multiply by 100 to express efficiency as a percentage. Temperatures in the ratio must be in kelvin. Use the Carnot temperature expression for a reversible engine or to find the upper limit for an engine between those reservoirs.

No engine operating between the same two temperatures can exceed Carnot efficiency. The limiting efficiency is independent of the working substance. Practical engines involving irreversibility have lower efficiencies. Raising the source temperature or lowering the sink temperature increases the ideal limit when the other temperature remains fixed.

Reversing the cycle produces a reversible refrigerator: work is supplied, heat is extracted from the cold reservoir, and heat is delivered to the hot reservoir. This satisfies the Clausius statement because the transfer is accompanied by work input.

Worked example 7. A Carnot engine operates between a source at 500 K and a sink at 300 K, producing 1 kJ of work per cycle. Find its efficiency, heat absorbed and heat rejected. A kilojoule, kJ, equals 1000 J.

Formula: η = 1 − T₂/T₁; Q₁ = W/η; Q₂ = Q₁ − W.

Substitute: η = 1 − 300/500 = 0.40; Q₁ = 1000/0.40.

Answer: Efficiency is 40%; Q₁ = 2500 J is absorbed from the source and Q₂ = 1500 J is rejected to the sink per cycle.

Glossary

• Thermodynamics — The study of heat, temperature and conversions between thermal and other forms of energy.
• Thermal equilibrium — A condition of equal temperature in which contacting systems exchange no net heat.
• Internal energy — Molecular kinetic and potential energy excluding kinetic energy of motion of the whole system.
• Heat — Energy transferred between a system and its surroundings because of a temperature difference.
• State variable — A quantity fixed by the equilibrium state, independently of the path used to reach it.
• Molar heat capacity — Heat required per mole per unit temperature increase under a specified heating condition.
• Quasi-static process — An ideal process proceeding so slowly that successive states remain equilibrium states.
• Isothermal process — A thermodynamic process in which the system's temperature remains constant throughout the change.
• Adiabatic process — A thermodynamic change during which no heat crosses the boundary of the system.
• Isochoric process — A constant-volume change with no expansion work done by or on the gas.
• Isobaric process — A thermodynamic process in which the pressure remains constant throughout the change.
• Reversible process — An ideal process reversible without leaving any change in the system, surroundings or elsewhere.
• Heat engine — A cyclic device absorbing heat, delivering work and rejecting heat to a colder sink.
• Carnot engine — A reversible engine operating between two reservoir temperatures with the highest possible efficiency between them.

Common errors and misconceptions

• Misconception: Heat is stored inside a gas. Correct: Internal energy is stored; heat is energy transferred because of temperature difference.
• Misconception: Work done on a gas is positive in Q = ΔU + W. Correct: W represents work done by the gas, so work done on it is negative.
• Misconception: An isothermal process cannot involve heat transfer. Correct: An ideal gas can absorb heat while doing equal work and retaining its temperature.
• Misconception: Adiabatic means constant temperature. Correct: It means no heat transfer; ideal-gas expansion cools and compression warms the gas under the stated adiabatic conditions.
• Misconception: PVᵞ = constant applies to every insulated expansion. Correct: The stated relation describes a quasi-static ideal-gas adiabatic path with constant γ, not unrestricted free expansion.
• Misconception: Any slow process is reversible. Correct: Dissipative effects must also be absent, and both system and surroundings must be restorable.
• Misconception: A refrigerator violates the second law by moving heat from cold to hot. Correct: It requires work input, so that heat transfer is not its sole result.
• Misconception: Celsius temperatures can be used in Carnot efficiency. Correct: The ratio requires absolute temperatures in kelvin.

Exam-style questions with model answers

Q1. State the zeroth law and name the quantity it establishes as equal for systems in thermal equilibrium. [2 marks]

1. If two systems are separately in thermal equilibrium with a third, they are in thermal equilibrium with each other.
2. The common thermodynamic quantity is temperature. The law establishes the basis for comparing temperatures through another system.

Q2. A heater supplies heat at 100 J s⁻¹ while a system does work at 75 J s⁻¹. Find the rate of increase of its internal energy, stating the sign convention. [3 marks]

1. Take heat entering the system as positive and work done by the system as positive. Thus the given heat-input and work-output rates are both positive quantities.
2. The first law gives internal-energy increase per second as heat supplied per second minus work done per second.
3. The required rate is 100 − 75 = 25 J s⁻¹. The positive answer indicates increasing internal energy.

Q3. A gas changes adiabatically from A to B while 22.3 J of work is done on it. On another path from A to B, it absorbs 9.35 cal. Find the work done by the gas on the second path. Use 1 cal = 4.186 J. [4 marks]

1. On the adiabatic path, heat transfer Q is zero and work done by the gas W is −22.3 J, since the work is done on it.
2. The first law gives the internal-energy change ΔU = Q − W = 22.3 J. This change is the same for both paths because their endpoints coincide.
3. Heat absorbed on the second path is Q = 9.35 × 4.186 = 39.1391 J.
4. Therefore W = Q − ΔU = 16.8391 J, approximately 16.8 J. Its positive sign means work is done by the gas.

Q4. Derive the work done by n moles of ideal gas in a quasi-static isothermal change from volume V₁ to volume V₂ at absolute temperature T. Here R is the universal gas constant. State the internal-energy change and heat transferred. [5 marks]

1. The ideal-gas relation is PV = nRT, where P is gas pressure and V its instantaneous volume. At fixed n and T, this gives P = nRT/V along the path.
2. For a small volume change dV, the work done is dW = P dV. Summation over the path gives W = ∫ P dV, with limits V₁ and V₂.
3. Substitute pressure and take the constants outside: W = nRT ∫ dV/V. Integration gives W = nRT ln(V₂/V₁), where ln means natural logarithm.
4. The internal energy of a fixed amount of ideal gas depends only on temperature. Since temperature remains constant, the internal-energy change ΔU is zero.
5. The first law, Q = ΔU + W, therefore gives Q = W. Expansion absorbs heat and does positive work; compression releases heat and has negative work done by the gas.

Q5. A gas goes in a straight line on a PV graph from D (2.0 m³, 600 Pa) to E (5.0 m³, 300 Pa), then at 300 Pa to F (2.0 m³, 300 Pa). Calculate work over both stages and the total. [3 marks]

1. Along the straight-line expansion, work is mean pressure multiplied by volume increase: [(600 + 300)/2] × (5.0 − 2.0) = 1350 J, positive because volume increases.
2. During constant-pressure compression, work is pressure multiplied by signed volume change: 300 × (2.0 − 5.0) = −900 J. Work is done on the gas.
3. The total work done by the gas is 1350 − 900 = 450 J. Adding the magnitudes instead would give an incorrect answer.

Q6. Describe the four stages of a Carnot engine using an ideal gas, explain its reservoir arrangement, and state its efficiency. The source and sink have absolute temperatures T₁ and T₂ respectively, with T₁ > T₂. [6 marks]

1. The source and sink have ideal infinite thermal capacity and remain at T₁ and T₂. A frictionless piston and switchable conducting contact or insulation allow reversible operation.
2. First, the gas expands isothermally at T₁ while absorbing heat from the source. It performs work with no internal-energy change during this stage.
3. Second, the insulated gas expands adiabatically. It does further work, loses internal energy and cools until its temperature reaches T₂.
4. Third, the gas is compressed isothermally at T₂. Work is done on it while it rejects heat to the cold sink.
5. Fourth, adiabatic compression raises the temperature back to T₁. The original state is restored, completing a cycle with zero net internal-energy change.
6. Efficiency η, defined as net work output divided by heat absorbed, is η = 1 − T₂/T₁. The temperatures must be in kelvin; this is the limiting efficiency between those reservoirs.

Q7. State the Kelvin-Planck and Clausius forms of the second law. Explain why operating a refrigerator does not contradict the latter. [3 marks]

1. The Kelvin-Planck statement forbids a process whose sole result is taking heat from a reservoir and converting all that heat into work.
2. The Clausius statement forbids a process whose sole result is transferring heat from a colder object to a hotter object.
3. A refrigerator requires external work to transfer heat from cold to hot. Since the heat transfer is accompanied by work input, it is not the sole result and does not violate the law.

Q8. A Carnot engine works between 500 K and 300 K and produces 1000 J of work per cycle. Calculate its efficiency and the heat absorbed and rejected per cycle. [3 marks]

1. Efficiency η is the ratio of work output to heat absorbed. For this Carnot engine, η = 1 − 300/500 = 0.40, or 40%.
2. Let Q₁ be heat absorbed from the source. Since work output W = ηQ₁, the heat absorbed is Q₁ = 1000/0.40 = 2500 J.
3. Let Q₂ be the positive magnitude of heat rejected. The cyclic energy balance gives Q₂ = Q₁ − W = 2500 − 1000 = 1500 J.

Key takeaways

• The zeroth law links thermal equilibrium to temperature equality, providing the basis for temperature comparison.
• Heat and work describe energy transfer; internal energy describes energy associated with the system's molecular constituents.
• In Q = ΔU + W, heat entering and work done by the system are positive.
• For an ideal gas, isothermal change gives ΔU = 0, while adiabatic change requires Q = 0.
• Pressure-volume work is the signed area under the process curve, with compression contributing negative work.
• The molar heat capacities of an ideal gas obey Cₚ − Cᵥ = R under their respective heating conditions.
• Reversibility requires restoration of both system and surroundings, with quasi-static operation and no dissipative effects.
• Carnot efficiency gives the upper limit between two reservoirs and must be calculated using absolute temperatures.

Test yourself

Why can equal-temperature bodies have different internal energies?

Temperature equality does not require equal amounts or kinds of matter. Internal energy depends on the system's state and amount.

What is the sign of work done by a gas during compression?

It is negative under the physics convention, because the surroundings do work on the gas.

Why can an ideal gas absorb heat without warming?

During isothermal expansion, absorbed heat equals work done, leaving internal energy and temperature unchanged.

What supplies the work in adiabatic ideal-gas expansion?

The gas's internal energy decreases to supply the work, and its temperature falls.

Why is Cₚ greater than Cᵥ for an ideal gas?

Constant-pressure heating requires expansion work as well as the internal-energy increase associated with a temperature rise.

What does a vertical line on a PV graph imply about expansion work?

Volume is constant along the line, so pressure-volume work is zero despite any pressure change.

Does return to the original state prove reversibility?

No. The surroundings must also return to their original state with no other change anywhere.

Why are the temperature-changing stages of a Carnot cycle adiabatic?

They change the gas temperature reversibly without exchanging heat with additional reservoirs between the source and sink temperatures.