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CalibratePhysics

Chapter 4

Thermodynamics

Maharashtra HSC · 11 subtopics · textbook pages 75–108

Formula SheetEvery formula this chapter can be examined on, in one table.

Temperature and the Zeroth Law

FormulaSymbols, units and conditions
T (in K)=t (in ∘C)+273.15≈t+273T\,(\text{in K}) = t\,(\text{in }^\circ\text{C}) + 273.15 \approx t + 273
  • Every thermodynamic formula below uses absolute temperature in kelvin. Converting ∘^\circC to K is the single commonest place marks are dropped in this chapter.

Internal Energy

FormulaSymbols, units and conditions
U=32kT  (monatomic gas, per atom),U=52kT  (diatomic gas, per molecule)U = \dfrac{3}{2}kT \ \ \text{(monatomic gas, per atom)}, \qquad U = \dfrac{5}{2}kT \ \ \text{(diatomic gas, per molecule)}—
ΔU=n CV ΔT\Delta U = n\,C_V\,\Delta T
  • UU is the energy of the random, disordered molecular motion; kk is Boltzmann's constant, TT the absolute temperature, nn the number of moles, CVC_V the molar specific heat at constant volume. ΔU=nCVΔT\Delta U = nC_V\Delta T holds for an ideal gas in every process, not just at constant volume.

Work Done by a System

FormulaSymbols, units and conditions
dW=p dVdW = p\,dV—
W=∫ViVfp dVW = \int_{V_i}^{V_f} p\,dV—
W=p (Vf−Vi)=p ΔV(constant pressure only)W = p\,(V_f - V_i) = p\,\Delta V \qquad \text{(constant pressure only)}
  • pp is the pressure the system exerts on the piston (Pa), dVdV the infinitesimal volume change (m3^3), WW in joules. WW is positive for expansion, negative for compression, zero when VV is constant. On a pp–VV diagram, WW is the area under the curve.
W=J×HW = J \times H
  • Mechanical equivalent of heat: WW is the mechanical work done, HH the heat produced, JJ the mechanical equivalent of heat (Joule's constant).

Heat Supplied to a Substance (from Std XI, still needed here)

FormulaSymbols, units and conditions
Q=m c ΔT(temperature change),Q=m L(change of phase)Q = m\,c\,\Delta T \qquad \text{(temperature change)}, \qquad Q = m\,L \qquad \text{(change of phase)}
  • mm is the mass (kg), cc the specific heat capacity (J kg−1^{-1} K−1^{-1}), LL the latent heat (J kg−1^{-1}). Needed for the standard boiling-water problem, where Q=mLQ = mL, W=p ΔVW = p\,\Delta V and ΔU=Q−W\Delta U = Q - W.

First Law of Thermodynamics

FormulaSymbols, units and conditions
ΔU=Q−WequivalentlyQ=ΔU+W\Delta U = Q - W \qquad \text{equivalently} \qquad Q = \Delta U + W
  • QQ is heat supplied to the system, WW the work done by the system, ΔU\Delta U the change in internal energy. All three in the same energy unit (J or kJ).
  • QuantityPositive whenNegative when
  • QQheat is absorbed by the systemheat is released by the system
  • WWwork is done by the system (expansion)work is done on the system (compression)
  • ΔU\Delta Uinternal energy (temperature) risesinternal energy (temperature) falls

Equation of State

FormulaSymbols, units and conditions
pV=nRTpV = nRT
  • pp in Pa, VV in m3^3, TT in K, nn in mol, R=8.314R = 8.314 J mol−1^{-1} K−1^{-1}. For fixed nn only two of pp, VV, TT are independent.

Isothermal Process

FormulaSymbols, units and conditions
pV=constant,piVi=pfVfpV = \text{constant}, \qquad p_iV_i = p_fV_f—
ΔT=0,ΔU=0,Q=W\Delta T = 0, \qquad \Delta U = 0, \qquad Q = W—
W=nRTln⁡ ⁣(VfVi)=2.303 nRTlog⁡10 ⁣(VfVi)=nRTln⁡ ⁣(pipf)W = nRT\ln\!\left(\dfrac{V_f}{V_i}\right) = 2.303\,nRT\log_{10}\!\left(\dfrac{V_f}{V_i}\right) = nRT\ln\!\left(\dfrac{p_i}{p_f}\right)
  • Constant-temperature change of an ideal gas, slow enough to stay in thermal equilibrium with the surroundings. All the heat absorbed is spent as work.

Isobaric Process

FormulaSymbols, units and conditions
Δp=0,ViTi=VfTf\Delta p = 0, \qquad \dfrac{V_i}{T_i} = \dfrac{V_f}{T_f}—
W=p (Vf−Vi)=nR (Tf−Ti)W = p\,(V_f - V_i) = nR\,(T_f - T_i)—
ΔU=n CV (Tf−Ti),Q=n Cp (Tf−Ti)\Delta U = n\,C_V\,(T_f - T_i), \qquad Q = n\,C_p\,(T_f - T_i)—
Cp=CV+RC_p = C_V + R
  • Constant-pressure change. CpC_p and CVC_V are molar specific heats (J mol−1^{-1} K−1^{-1}). Useful shortcut: if the volume of nn mol doubles isobarically, W=pVi=nRTiW = pV_i = nRT_i.

Isochoric Process

FormulaSymbols, units and conditions
ΔV=0,piTi=pfTf\Delta V = 0, \qquad \dfrac{p_i}{T_i} = \dfrac{p_f}{T_f}—
W=0,Q=ΔU=n CV ΔTW = 0, \qquad Q = \Delta U = n\,C_V\,\Delta T
  • Constant-volume change. No work is done, so all the heat supplied goes into internal energy.

Adiabatic Process

FormulaSymbols, units and conditions
Q=0,ΔU=−WQ = 0, \qquad \Delta U = -W—
pVγ=constant,TVγ−1=constant,p1−γTγ=constantpV^{\gamma} = \text{constant}, \qquad TV^{\gamma-1} = \text{constant}, \qquad p^{1-\gamma}T^{\gamma} = \text{constant}—
γ=CpCV\gamma = \dfrac{C_p}{C_V}—
W=piVi−pfVfγ−1=nR (Ti−Tf)γ−1W = \dfrac{p_iV_i - p_fV_f}{\gamma - 1} = \dfrac{nR\,(T_i - T_f)}{\gamma - 1}
  • No heat is exchanged: the system is perfectly insulated, or the change is sudden. γ\gamma is the adiabatic ratio.
pf=pi(ViVf)γ,Tf=Ti(ViVf)γ−1,(p2p1)γ−1=(T2T1)γp_f = p_i\left(\dfrac{V_i}{V_f}\right)^{\gamma}, \qquad T_f = T_i\left(\dfrac{V_i}{V_f}\right)^{\gamma-1}, \qquad \left(\dfrac{p_2}{p_1}\right)^{\gamma-1} = \left(\dfrac{T_2}{T_1}\right)^{\gamma}
  • Work these from p11−γT1γ=p21−γT2γp_1^{1-\gamma}T_1^{\gamma} = p_2^{1-\gamma}T_2^{\gamma} when the question gives you a temperature ratio and asks for a pressure ratio.
  • Gasγ\gamma
  • Monatomic5/3≈1.675/3 \approx 1.67
  • Diatomic7/5=1.47/5 = 1.4
  • Triatomic4/3≈1.334/3 \approx 1.33

Cyclic Process

FormulaSymbols, units and conditions
ΔU=0,Q=W=area enclosed by the loop\Delta U = 0, \qquad Q = W = \text{area enclosed by the loop}
  • Clockwise loop: WW positive (work done by the system, engine). Anticlockwise loop: WW negative (work done on the system, refrigerator). For NN cycles, total work =N×= N \times (area of one loop).

Free Expansion

FormulaSymbols, units and conditions
Q=0,W=0,ΔU=0,ΔT=0Q = 0, \qquad W = 0, \qquad \Delta U = 0, \qquad \Delta T = 0
  • Sudden uncontrolled expansion of a gas into vacuum inside an insulated container. Cannot be drawn on a pp–VV diagram; only the initial and final points can be marked.

Heat Engine

FormulaSymbols, units and conditions
Q=QH+QC=∣QH∣−∣QC∣Q = Q_H + Q_C = |Q_H| - |Q_C|—
W=QH+QC=∣QH∣−∣QC∣W = Q_H + Q_C = |Q_H| - |Q_C|—
η=WQH=1+QCQH=1−∣QC∣QH\eta = \dfrac{W}{Q_H} = 1 + \dfrac{Q_C}{Q_H} = 1 - \dfrac{|Q_C|}{Q_H}
  • QHQ_H is heat absorbed at the source (positive), QCQ_C heat rejected at the sink (negative in sign convention). η\eta is a pure number, no unit; multiply by 100 for a percentage. η\eta is always less than 1.

Refrigerator, Air Conditioner and Heat Pump

FormulaSymbols, units and conditions
∣QH∣=QC+∣W∣|Q_H| = Q_C + |W|—
K=QCW=QCQH−QCK = \dfrac{Q_C}{W} = \dfrac{Q_C}{Q_H - Q_C}
  • KK is the coefficient of performance (also called quality factor or QQ-value); QCQ_C is heat extracted from the cold reservoir, WW the work done by the compressor, all as magnitudes. KK is dimensionless and can be greater than 1. Typically K≈5K \approx 5 for a household refrigerator.
H=QCt,K=QCW=H tP t=HPH = \dfrac{Q_C}{t}, \qquad K = \dfrac{Q_C}{W} = \dfrac{H\,t}{P\,t} = \dfrac{H}{P}
  • For an air conditioner: HH is the rate of heat removed (heat current, W), PP the electrical power required (W), tt the time. Typical KK for room air conditioners is 2.5 to 3.0.

Carnot Cycle and Carnot Engine

FormulaSymbols, units and conditions
QCQH=TCTH\dfrac{Q_C}{Q_H} = \dfrac{T_C}{T_H}—
η=1−QCQH=1−TCTH=TH−TCTH\eta = 1 - \dfrac{Q_C}{Q_H} = 1 - \dfrac{T_C}{T_H} = \dfrac{T_H - T_C}{T_H}—
KCarnot=QCQH−QC=TCTH−TCK_{\text{Carnot}} = \dfrac{Q_C}{Q_H - Q_C} = \dfrac{T_C}{T_H - T_C}
  • THT_H is the source (hot reservoir) temperature and TCT_C the sink (cold reservoir) temperature, both in kelvin. This is the maximum efficiency any engine working between those two temperatures can have, and KCarnotK_{\text{Carnot}} the largest coefficient of performance any refrigerator between them can have.
QC=QH(TCTH),W=QH−QCQ_C = Q_H\left(\dfrac{T_C}{T_H}\right), \qquad W = Q_H - Q_C
  • Handy working form when a numerical gives QHQ_H, THT_H and TCT_C and asks for heat rejected, work done and efficiency.

Constants and Standard Values

  • QuantityValue
  • Universal gas constant RR8.3148.314 J mol−1^{-1} K−1^{-1}
  • Boltzmann constant kk1.38×10−231.38 \times 10^{-23} J K−1^{-1}
  • 1 atmosphere1.013×1051.013 \times 10^{5} Pa (textbook often uses 1.01×1051.01 \times 10^{5} Pa)
  • 1 litre10−310^{-3} m3^3
  • Latent heat of vaporisation of water22562256 kJ kg−1^{-1}
  • 0 ∘0\,^\circC273273 K
Short NotesThe chapter subtopic by subtopic, in the textbook's own order.11 subtopics
  1. 1Introduction
  2. 2Zeroth Law of Thermodynamics1 board question
  3. 3Heat, Internal Energy and Work2 board questions
  4. 4First Law of Thermodynamics4 board questions
  5. 5Thermodynamic State Variables2 board questions
  6. 6Thermodynamic Process7 board questions
  7. 7Heat Engines
  8. 8Refrigerators and Heat Pumps
  9. 9Second Law of Thermodynamics1 board question
  10. 10Carnot Cycle and Carnot Engine
  11. 11Sterling Cycle
The whole chapter in brief

What This Chapter Is About

  • Thermodynamics is the branch of physics dealing with heat, temperature and the inter-conversion of heat and other forms of energy.
  • Heat is a form of energy, not a fluid. Benjamin Thomson's cannon-boring observation (1798) — heat produced depended on the work done in turning the drill and on how long it bored, not on the sharpness of the drill — established this.
  • Everything in this chapter follows from two ideas: energy is conserved (first law), and heat flow has a preferred direction (second law).

Thermal Equilibrium and the Zeroth Law

  • Two systems in thermal contact are in thermal equilibrium if they do not transfer heat between each other.
  • Zeroth law: if two systems are each in thermal equilibrium with a third system, they are also in thermal equilibrium with each other.
  • Consequence: systems in thermal equilibrium are at the same temperature. This is what lets a thermometer work, and it makes temperature a measurable quantity.
  • It is called the zeroth law because it was formulated after the first and second laws, but is logically prior to them.

Internal Energy

  • Internal energy UU is the energy associated with the random, disordered motion of the molecules of a system. It is a microscopic quantity, quite separate from the ordered macroscopic kinetic energy of a moving body.
  • Monatomic ideal gas: only translational kinetic energy, 32kT\dfrac{3}{2}kT per atom. Diatomic gas: 52kT\dfrac{5}{2}kT per molecule. Polyatomic gases add rotational and vibrational kinetic energy; liquids and solids also need the potential energy of intermolecular attraction.
  • For an ideal gas UU depends only on temperature. So ΔT=0\Delta T = 0 forces ΔU=0\Delta U = 0.

Systems and Surroundings

  • A thermodynamic system is a collection of objects forming a unit that can exchange energy with its surroundings. Anything not part of the system is its surroundings or environment.
SystemExchanges energy?Exchanges matter?Example
OpenYesYesWater boiling in an open kettle
ClosedYesNoWater boiling in a closed boiler
IsolatedNoNoThermos flask

Heat, and the Sign Convention

  • Heat QQ is the energy transferred between a system and its environment because of a temperature difference between the two.
  • Energy transferred to the system is positive (system gains); energy transferred from the system is negative (system loses). If TS=TET_S = T_E there is no transfer and Q=0Q = 0.
  • Internal energy of a system can be changed in exactly two ways: by heating it, or by doing work on it.

First Law of Thermodynamics

  • Statement: the change in the internal energy of a system is the difference between the heat supplied to the system and the work done by the system on its surroundings.

ΔU=Q−WorQ=ΔU+W\Delta U = Q - W \qquad \text{or} \qquad Q = \Delta U + W

  • Equivalent wording: when heat QQ is added to a system, its internal energy increases by ΔU\Delta U and the remainder is lost as work WW done on the surroundings.
  • It is a generalisation of the law of conservation of energy, extended from isolated systems to systems that can exchange energy.
  • QQ and WW can each be positive, negative or zero, so ΔU\Delta U can be positive (QQ exceeds WW), negative (WW exceeds QQ) or zero (Q=WQ = W).
  • Limitations: it does not say which direction a process will go, and it does not forbid 100 per cent conversion of heat into work. Both gaps are closed by the second law.

State Variables and Equilibrium

  • A property (state variable) is any measurable characteristic of a system in equilibrium: pp, VV, TT, density, mass.
Intensive variablesExtensive variables
Do not depend on size of the systemDepend on size of the system
Pressure pp, temperature TT, density ρ\rhoMass MM, volume VV, internal energy UU
Unchanged if the system is halvedHalved if the system is halved
  • Thermodynamic equilibrium requires all three simultaneously: mechanical equilibrium (no unbalanced forces, pressure uniform and steady), chemical equilibrium (no reactions or diffusion, composition uniform and steady), thermal equilibrium (temperature uniform and steady).
  • A system not in equilibrium — a punctured ball, an exploding fuel-air mixture — cannot be described in terms of state variables at all.
  • The mathematical relation between state variables is the equation of state; for an ideal gas, pV=nRTpV = nRT.
  • The graph of the equation of state is the pp–VV diagram or indicator diagram. A constant-temperature pp–VV curve is an isotherm.

Work and Heat Depend on the Path

  • Three different paths from state A (pi,Vi)(p_i, V_i) to state B (pf,Vf)(p_f, V_f) — a smooth curve, isobaric-then-isochoric, isochoric-then-isobaric — enclose three different areas, so give three different amounts of work. The initial and final states are the same in all three.
  • Conclusion: work done by a system depends not only on the initial and final states but on the path taken.
  • Same for heat: slow isothermal expansion absorbs a finite QQ; free expansion of the same gas between the same two states has Q=0Q = 0. So heat transferred also depends on the path.
  • QQ and WW are path quantities; UU is a state quantity.

Quasi-static, Reversible and Irreversible

  • A quasi-static process changes the state variables infinitesimally slowly, so the system stays in thermodynamic equilibrium throughout.
ReversibleIrreversible
Can be retraced in the reverse directionCannot be retraced in the reverse direction
Same path forwards and backwards on the pp–VV diagramDifferent paths forwards and backwards; shows hysteresis
No loss of energy; system returns exactly to its initial statePermanent energy loss to friction or other dissipative forces
Very slow, quasi-static, always in equilibriumOften sudden; passes through non-equilibrium states
Ideal, does not occur in practiceMelting ice reversed by freezing is treated as reversible; two bodies reaching thermal equilibrium, bursting a balloon, rubbing palms, burning a candle are irreversible
  • Two causes of irreversibility: (i) processes such as free expansion or explosive reaction take the system through non-equilibrium states; (ii) friction, viscosity and other dissipative forces are always present and cannot be fully eliminated.
  • All spontaneous natural processes are irreversible; an irreversible process defines the preferred direction of change and is unidirectional.

The Four Standard Processes

ProcessHeld constantEquation of stateWWQQΔU\Delta U
IsothermalTTpV=pV = constnRTln⁡(Vf/Vi)nRT\ln(V_f/V_i)Q=WQ = W00
IsobaricppV/T=V/T = constp ΔVp\,\Delta VnCpΔTnC_p\Delta TnCVΔTnC_V\Delta T
IsochoricVVp/T=p/T = const00nCVΔTnC_V\Delta TQ=ΔUQ = \Delta U
Adiabaticnone (Q=0Q=0)pVγ=pV^{\gamma} = const(piVi−pfVf)/(γ−1)(p_iV_i - p_fV_f)/(\gamma-1)00−W-W
  • Isothermal: needs good thermal contact with the environment and an extremely slow change. Any change of phase (melting ice, boiling water) is isothermal. Neither QQ nor WW is zero.
  • Isobaric: boiling water at atmospheric pressure. None of ΔU\Delta U, QQ, WW is zero. The isobars on a weather map are the same idea — lines of equal pressure.
  • Isochoric: heating a gas in a rigid container, or diffusion in a closed chamber. All the heat supplied raises the internal energy.
  • Adiabatic: puncturing a tyre or balloon; the compression stroke of a diesel engine. Adiabatic expansion cools the gas, adiabatic compression heats it. Usually a sudden change, so there is no time to exchange heat.
  • The adiabatic pp–VV curve is steeper than the isothermal curve through the same point, because γ>1\gamma > 1.

Cyclic Process and Free Expansion

  • Cyclic process: the system returns to its initial state, so ΔU=0\Delta U = 0 and Q=WQ = W. The work done in one cycle is the area of the closed loop; positive for a clockwise loop, negative for an anticlockwise loop. Every heat engine works on a cyclic process.
  • Free expansion: adiabatic, no heat exchanged and no work done, so Q=W=ΔU=0Q = W = \Delta U = 0 and the temperature of an ideal gas is unchanged. It is uncontrolled and instantaneous, the system is never in equilibrium, and it cannot be plotted as a path on a pp–VV diagram.

Heat Engines

  • A heat engine is a device that transforms heat partly into work or mechanical energy, working through a cyclic process.
  • Three elements: a working substance (the system — ideal gas, steam, or a fuel-air mixture), a hot reservoir or source at THT_H and a cold reservoir or sink at TCT_C, and a cylinder with a movable frictionless massless piston.
  • Three basic steps every cycle: absorb heat QHQ_H from the hot reservoir; convert part of it into work WW; reject the remaining heat QCQ_C to the cold reservoir.
  • External combustion engine: working substance heated externally (steam engine). Internal combustion engine: working substance heated internally (petrol or diesel engine).

W=QH+QC=∣QH∣−∣QC∣,η=WQH=1−∣QC∣QHW = Q_H + Q_C = |Q_H| - |Q_C|, \qquad \eta = \dfrac{W}{Q_H} = 1 - \dfrac{|Q_C|}{Q_H}

  • Efficiency is the fraction of the absorbed heat converted into work. It has no unit, so WW, QHQ_H and QCQ_C must all be in the same energy unit.
  • Not all heat absorbed can be converted into work: QCQ_C is never zero, so η\eta is never 1.

Refrigerators, Air Conditioners and Heat Pumps

  • These are heat engines run backwards: they convert mechanical work into heat transport, forcing heat from a cold region to a hot region.
  • Refrigeration is artificial cooling — cooling a space or keeping it below the ambient temperature.
  • Mechanism: a refrigerant (for example isobutane) is made to evaporate in the cooling coil inside (absorbing its heat of vaporisation from the cold region) and then compressed and condensed outside (rejecting that heat to the atmosphere). That is why the back of a working refrigerator feels warm.
  • Four steps of one cycle: (1) the fluid passes through a nozzle into a low-pressure region and turns into cool gas — essentially adiabatic expansion; (2) the cool gas absorbs heat from the fridge contents — isobaric expansion; (3) the compressor compresses the gas adiabatically, heating it and turning it back to liquid; (4) the hot liquid passes through the outside coils and rejects heat to the atmosphere — isobaric compression.

∣QH∣=QC+∣W∣,K=QCW=QCQH−QC|Q_H| = Q_C + |W|, \qquad K = \dfrac{Q_C}{W} = \dfrac{Q_C}{Q_H - Q_C}

  • KK, the coefficient of performance, is dimensionless. Larger KK is better: it means more heat extracted for less work. Typically about 5 for a household fridge, 2.5 to 3.0 for a room air conditioner.
  • Air conditioner: same working as a refrigerator, differing only in the volume of the chamber cooled. Evaporator coils inside the room, condenser outside, a fan circulating the cooled air. Performance is often measured as K=H/PK = H/P, heat current over power.
  • Heat pump: a refrigerator operating inside out — evaporator coils outside absorbing heat from cold outdoor air, condenser coils inside releasing that heat, so the building is warmed.
  • Capacity of a refrigerator is quoted in litres (the internal volume); capacity of an air conditioner in tonnes, historically the mass of ice that would melt in 24 hours to give the same cooling.

Second Law of Thermodynamics

  • Kelvin-Planck (engine) statement: it is impossible to extract an amount of heat QHQ_H from a hot reservoir and use it all to do work WW. Some amount of heat QCQ_C must be exhausted to a cold reservoir. This prohibits the possibility of a perfect heat engine.
  • Clausius (refrigerator) statement: it is not possible for heat to flow from a colder body to a warmer body without any work having been done to accomplish this flow. This rules out a perfect refrigerator.
  • The second law puts constraints on the direction of heat transfer and on the efficiency a heat engine can achieve. Consequences: η\eta can never be 1, and KK can never be infinite.
  • Note the "net" in the Clausius statement: energy can pass from a cold object to a hot one by radiation or energetic particles, but in any spontaneous process the net transfer is hot to cold.

Carnot Cycle and Carnot Engine

  • Sadi Carnot (1824) proposed a hypothetical ideal engine of maximum efficiency, in which every step is reversible.
  • Reversible heat exchange must be isothermal (working substance at THT_H while absorbing, at TCT_C while rejecting); reversible work must be adiabatic. Hence two isothermals and two adiabatics.
  • The four strokes: AB isothermal expansion at THT_H, BC adiabatic expansion, CD isothermal compression at TCT_C, DA adiabatic compression.

η=1−QCQH=1−TCTH,KCarnot=TCTH−TC\eta = 1 - \dfrac{Q_C}{Q_H} = 1 - \dfrac{T_C}{T_H}, \qquad K_{\text{Carnot}} = \dfrac{T_C}{T_H - T_C}

  • Design rule: for maximum efficiency, THT_H as high as possible and TCT_C as low as possible.
  • Carnot's theorem: the Carnot engine is the most efficient heat engine, and all Carnot engines working between the same two temperatures have the same efficiency, whatever their working substance. Likewise a Carnot refrigerator has the greatest coefficient of performance of all refrigerators between the same two temperatures.
  • η=100\eta = 100 per cent would need TC=0T_C = 0 K, which is unattainable. So even the ideal engine cannot be 100 per cent efficient.
  • KCarnotK_{\text{Carnot}} depends only on the temperatures: a small temperature difference gives a very large KK (much heat pumped for little work), a large temperature difference a very small KK.

Sterling Cycle

  • A closed, fully reversible cycle used by the Sterling engine; working substances are air, helium, hydrogen or nitrogen.
  • Two isothermal processes, AB at THT_H and CD at TCT_C, joined by two constant-parameter processes BC and DA.
  • AB isothermal expansion: heat QHQ_H supplied at THT_H, useful work done by the gas. BC: part of the absorbed heat QQ is released to the refrigerator and stored for reuse, gas cools to TCT_C. CD isothermal compression: heat QCQ_C rejected to the coolant (sink) at TCT_C. DA: the compressed gas absorbs the stored heat QQ and its temperature rises to THT_H; the cycle then repeats from A.
  • Heating the gas makes it produce useful work (engine); doing work on the gas runs it as a refrigerator. The reversed Sterling cycle is used extensively in cryogenics to reach very low temperatures and to liquefy gases.

Traps That Cost Marks

  • Converting ∘^\circC to K. Every efficiency and every gas-law step needs kelvin. 727 ∘727\,^\circC is 1000 K, not 727 K.
  • Sign of WW in the first law. WW is the work done by the system. Work done on the system is negative WW, which makes ΔU=Q−W\Delta U = Q - W larger. When a numerical says "work done on the gas" and "heat released", both are negative and the safest route is the board's own form, ΔU=∣Q∣−∣W∣\Delta U = |Q| - |W|.
  • W=p ΔVW = p\,\Delta V is valid only at constant pressure. For an isothermal change you must use nRTln⁡(Vf/Vi)nRT\ln(V_f/V_i).
  • ln⁡\ln versus log⁡\log. The isothermal formula uses natural log; if you use log⁡10\log_{10}, multiply by 2.303.
  • ΔU=nCVΔT\Delta U = nC_V\Delta T holds in every process, including isobaric and adiabatic — not only at constant volume.
  • Free expansion is adiabatic but is not the adiabatic process: pVγ=pV^{\gamma} = constant does not apply to it, because it is not quasi-static.
  • Efficiency is a ratio and has no unit. Do not write it in joules, and do not forget to multiply by 100 when a percentage is asked.
  • Do not use η=1−TC/TH\eta = 1 - T_C/T_H for a real engine; that is the Carnot (maximum) efficiency only.
  • Coefficient of performance is not efficiency and is routinely greater than 1. That is not a violation of anything.
  • In a cyclic process ΔU=0\Delta U = 0, so Q=WQ = W — but QQ and WW are individually non-zero.
  • Litres to m3^3: 1 L =10−3= 10^{-3} m3^3. Forgetting this is worth an entire numerical.
Solved Numerical ExamplesWorked numericals to attempt first, each with a hint before the full solution.Coming soon

These are being added chapter by chapter. They are solved numericals from the prescribed book and the reference guides — not previous board questions, which live under Practice — so that you can work through the method before meeting it in a real paper.

Important DerivationsEvery derive / obtain / prove question in this chapter. 1 have come up in two or more papers.1
  1. 1
    Work done during an isothermal process Repeats
    Derive an expression for the work done during an isothermal process.

    Asked in: February 2023 (Q.29(i), 4 marks); February 2024 (Q.19, 3 marks); June 2025 (Q.27, 4 marks)

    Show the derivation
    Step 1
    Consider the isothermal expansion of an ideal gas.
    Step 2
    Let its initial volume be ViV_i and the final volume be VfV_f.
    Step 3
    The work done in an infinitesimally small isothermal expansion is given by, dW=pdVdW=pdV
    Step 4
    The total work done in bringing out the expansion from the initial volume ViV_i to the final volume VfV_f is given by,
    W=∫ViVfpdV....(1)W=\int_{V_i}^{V_f}pdV\qquad....(1)
    Step 5
    But, for an ideal gas, pV = nRT. Using this in the equation (1) we get,
    W=nRT∫ViVfdVVW=nRT\int_{V_i}^{V_f}\dfrac{dV}{V}
    ∴W=nRTln⁡VfVi....(2)\therefore W=nRT\ln\dfrac{V_f}{V_i}\qquad....(2)

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