📌 Snapshot
- Establishes thermodynamics as a macroscopic science of heat–work inter-conversion, dealing with bulk variables P, V, T, U rather than molecular detail.
- Builds the temperature concept through the Zeroth Law and develops the First Law (conservation of energy) ΔQ = ΔU + ΔW, with ΔU as a state variable.
- Defines specific heat capacities for solids and gases, gives Mayer's relation Cp − Cv = R, and analyses isothermal, adiabatic, isobaric, isochoric and cyclic processes.
- States the Second Law (Kelvin–Planck and Clausius forms), classifies processes as reversible/irreversible, and derives the Carnot engine efficiency η = 1 − T₂/T₁ as the upper limit for any heat engine.
- A perennial CUET favourite — pure numerical questions on first-law accounting, isothermal/adiabatic work and Carnot efficiency dominate.
📖 Detailed Notes
2.1 Core concepts
- Thermodynamics is the branch of physics dealing with heat, temperature and inter-conversion of heat and other forms of energy; it is a macroscopic science using few variables like P, V, T, m, composition and U, avoiding the molecular description (NCERT §11.1, p. 227).
- Historically, heat was thought to be a fluid called "caloric" flowing from hotter to colder bodies; Count Rumford's brass-cannon boring experiment (1798) showed heat produced depended on work done and not on the sharpness of the drill, establishing heat as a form of energy (NCERT §11.1, p. 226).
- A system is in thermodynamic equilibrium when its macroscopic variables do not change with time; the nature of the wall — adiabatic (insulating) or diathermic (conducting) — decides whether two systems can reach thermal equilibrium (NCERT §11.2, p. 227–228).
- Zeroth Law of Thermodynamics: two systems in thermal equilibrium with a third system separately are in thermal equilibrium with each other; the common physical quantity having the same value is called temperature T. R.H. Fowler formulated it in 1931, after the First and Second laws were already numbered (NCERT §11.3, p. 228).
- Internal energy U is the sum of kinetic and potential energies of the molecules in the frame in which the centre of mass of the system is at rest; it excludes the bulk kinetic energy of the system and depends only on the state, not on the path (NCERT §11.4, p. 229).
- Heat and work are the two modes of energy transfer that alter U; heat arises from a temperature difference, work is transferred by other means (e.g. moving a piston). Heat and work are NOT state variables — they are "energy in transit" (NCERT §11.4, p. 229–230).
- First Law of Thermodynamics: ΔQ = ΔU + ΔW where ΔQ is heat supplied to the system, ΔW is work done by the system on the surroundings, and ΔU is the change in internal energy. It is the energy-conservation principle applied to thermodynamic systems (NCERT §11.5, p. 230, Eq. 11.1).
- For a gas in a cylinder against constant pressure P, ΔW = P ΔV; hence ΔQ = ΔU + P ΔV (NCERT §11.5, p. 231, Eq. 11.3). Worked illustration: for 1 g of water vaporising at atmospheric pressure ΔQ = 2256 J, ΔW = 169.2 J, so ΔU = 2086.8 J — most heat goes into raising U.
- Specific heat capacity s = (1/m)(ΔQ/ΔT) and molar specific heat C = (1/µ)(ΔQ/ΔT); both depend on the substance and conditions of heating (NCERT §11.6, p. 231, Eqs. 11.5–11.6).
- For an ideal gas at constant volume Cv = (ΔU/ΔT) and at constant pressure Cp = (ΔU/ΔT) + R, leading to Mayer's relation Cp − Cv = R (NCERT §11.6, p. 232, Eq. 11.8). For a mole of solid using equipartition, U = 3RT giving C = 3R, which agrees with experiment at ordinary temperatures (carbon is an exception).
- 1 calorie is defined as the heat needed to raise 1 g of water from 14.5 °C to 15.5 °C; in SI, the specific heat capacity of water is 4186 J kg⁻¹ K⁻¹, and 1 cal = 4.186 J (NCERT §11.6, p. 232).
- State variables fall into two kinds — extensive (U, V, M scale with size) and intensive (P, T, ρ remain unchanged on subdivision). The equation of state connects state variables; for an ideal gas PV = µRT (NCERT §11.7, p. 232–233).
- A quasi-static process is infinitely slow so the system remains in equilibrium with its surroundings at every stage; the pressure and temperature differ from those of the surroundings only infinitesimally (NCERT §11.8.1, p. 233–234).
- Isothermal process: T fixed, PV = constant (Boyle's law); work done W = µRT ln(V₂/V₁); since ΔU = 0 for an ideal gas, Q = W (NCERT §11.8.2, p. 234, Eq. 11.12). In isothermal expansion the gas absorbs heat and does work; in compression the reverse.
- Adiabatic process: Q = 0, PV^γ = constant where γ = Cp/Cv; work done W = (P₁V₁ − P₂V₂)/(γ − 1) = µR(T₁ − T₂)/(γ − 1); positive work by the gas lowers its temperature (NCERT §11.8.3, p. 234–235, Eqs. 11.13–11.16).
- Isochoric process: V constant ⇒ W = 0, so all heat absorbed goes to ΔU (and hence temperature change) governed by Cv (NCERT §11.8.4, p. 235).
- Isobaric process: P fixed ⇒ W = P(V₂ − V₁) = µR(T₂ − T₁); heat absorbed goes partly into ΔU and partly into work, governed by Cp (NCERT §11.8.5, p. 235, Eq. 11.17).
- Cyclic process: the system returns to its initial state; ΔU = 0, so total Q = total W (NCERT §11.8.6, p. 235).
- Second Law of Thermodynamics — Kelvin-Planck statement: no process is possible whose sole result is the absorption of heat from a reservoir and the complete conversion of that heat into work. Clausius statement: no process is possible whose sole result is the transfer of heat from a colder to a hotter body. The two statements are completely equivalent (NCERT §11.9, p. 236).
- A reversible process can be turned back so that both the system and surroundings return to their original states with no other change; it must be quasi-static AND free of dissipative effects like friction and viscosity. Spontaneous natural processes are irreversible (NCERT §11.10, p. 236–237).
- Carnot engine: a reversible heat engine operating between a hot reservoir at T₁ and a cold reservoir at T₂; its cycle has four steps — isothermal expansion at T₁, adiabatic expansion to T₂, isothermal compression at T₂, adiabatic compression back to T₁ (NCERT §11.11, p. 237–238).
- Carnot efficiency: η = 1 − Q₂/Q₁ = 1 − T₂/T₁; Carnot's theorem proves (a) no engine working between two given temperatures can be more efficient than the Carnot engine, and (b) the Carnot efficiency is independent of the working substance (NCERT §11.11, p. 238–239, Eq. 11.27). The universal relation Q₁/Q₂ = T₁/T₂ defines a thermodynamic temperature scale (Eq. 11.28).
2.2 Definitions to memorise
| Term | Definition | Page |
|---|---|---|
| Thermodynamics | Branch of physics dealing with heat, temperature and inter-conversion of heat with other forms of energy at the macroscopic level. | 227 |
| Thermal equilibrium | State in which two systems separated by a diathermic wall show no further spontaneous change in their macroscopic variables. | 228 |
| Adiabatic wall | An insulating wall that does not allow flow of heat between the systems it separates. | 228 |
| Diathermic wall | A conducting wall that allows heat to flow from one system to another. | 228 |
| Zeroth Law | Two systems in thermal equilibrium with a third system separately are in thermal equilibrium with each other. | 228 |
| Temperature | The thermodynamic variable whose value is equal for two systems in thermal equilibrium. | 228 |
| Internal energy U | Sum of kinetic and potential energies of the molecular constituents in the frame in which the centre of mass of the system is at rest. | 229 |
| Heat | Energy transferred between system and surroundings due to a temperature difference; energy in transit, not a state variable. | 229 |
| Work (thermodynamic) | Energy transfer brought about by means that do not involve a temperature difference (e.g. moving a piston). | 229 |
| First Law of Thermodynamics | ΔQ = ΔU + ΔW — conservation of energy applied to a thermodynamic system. | 230 |
| Specific heat capacity s | s = (1/m)(ΔQ/ΔT); heat needed per unit mass per unit temperature rise. Unit: J kg⁻¹ K⁻¹. | 231 |
| Molar specific heat C | C = (1/µ)(ΔQ/ΔT); unit J mol⁻¹ K⁻¹. | 231 |
| Cp, Cv | Molar specific heats at constant pressure and constant volume; satisfy Cp − Cv = R for an ideal gas. | 232 |
| γ (gamma) | Ratio of specific heats, γ = Cp/Cv; appears in adiabatic relations. | 234 |
| Calorie | Heat required to raise 1 g of water from 14.5 °C to 15.5 °C; 1 cal = 4.186 J. | 232 |
| Equation of state | Relation connecting the state variables of a system, e.g. PV = µRT for an ideal gas. | 233 |
| Extensive variable | A variable whose value scales with the size of the system (e.g. U, V, M). | 233 |
| Intensive variable | A variable that is independent of the size of the system (e.g. P, T, ρ). | 233 |
| Quasi-static process | An infinitely slow process in which the system remains in thermal and mechanical equilibrium with its surroundings at every stage. | 234 |
| Isothermal process | A process at constant temperature; for an ideal gas PV = constant. | 234 |
| Adiabatic process | A process in which no heat is exchanged with the surroundings (Q = 0); PV^γ = constant. | 234 |
| Isochoric process | A process at constant volume; W = 0. | 235 |
| Isobaric process | A process at constant pressure; W = PΔV. | 235 |
| Cyclic process | A process in which the system returns to its initial state, so ΔU = 0. | 235 |
| Kelvin-Planck statement | No process is possible whose sole result is the absorption of heat from a reservoir and the complete conversion of the heat into work. | 236 |
| Clausius statement | No process is possible whose sole result is the transfer of heat from a colder object to a hotter object. | 236 |
| Reversible process | A process that can be turned back so that both the system and surroundings return to their original states with no other change in the universe. | 237 |
| Irreversible process | A spontaneous natural process that cannot be reversed without producing some change in the surroundings. | 236 |
| Carnot engine | A reversible heat engine working between two reservoirs at T₁ and T₂ via two isothermal and two adiabatic processes. | 237 |
| Carnot efficiency | η = 1 − T₂/T₁; the maximum possible efficiency of a heat engine between two given temperatures. | 238 |
2.3 Diagrams / processes to remember
- Fig. 11.1, p. 227: Two gases A and B separated first by an adiabatic wall (no thermal equilibrium attained) and then by a diathermic wall (equilibrium attained with new P, V values).
- Fig. 11.2, p. 228: Schematic illustrating the Zeroth Law — A and B each in contact with C via conducting walls, then connected directly; they remain in equilibrium.
- Fig. 11.3, p. 229: Internal energy U of a gas as sum of kinetic + potential energies of molecules (translational, rotational, vibrational), and exclusion of bulk motion.
- Fig. 11.4, p. 229: Heat (from temperature difference) and work (via a piston with weight) as two distinct modes of energy transfer.
- Fig. 11.5, p. 232: Variation of specific heat capacity of water with temperature in the range 0–100 °C, motivating the precise calorie definition.
- Fig. 11.6, p. 233: Non-equilibrium states — free expansion of a gas and an explosive chemical reaction — that cannot be described by state variables.
- Fig. 11.7, p. 234: Quasi-static process — surrounding reservoir and external pressure differ from system values only infinitesimally.
- Fig. 11.8, p. 235: P–V curves comparing isothermal and adiabatic processes; the adiabatic curve is steeper because γ > 1.
- Fig. 11.9, p. 237: Carnot cycle on a P–V diagram showing the four steps — isothermal expansion 1→2, adiabatic expansion 2→3, isothermal compression 3→4, adiabatic compression 4→1.
- Fig. 11.10, p. 239: Irreversible engine I coupled to a reversible refrigerator R — used to prove Carnot's theorem by contradicting the Kelvin-Planck statement.
2.4 Common confusions / NTA trap points
- Heat vs internal energy: a body "contains" internal energy, not heat. Heat is energy in transit during a process, so statements like "a gas in a given state has a certain amount of heat" are meaningless (NCERT §11.4, p. 230).
- Sign convention in First Law: in NCERT, ΔW is the work done BY the system on the surroundings, so ΔQ = ΔU + ΔW; isothermal expansion gives W > 0 and isothermal compression gives W < 0 (NCERT §11.5/§11.8, p. 230, 234). Mixing this up with the "work done on the system" convention is a classic trap.
- Adiabatic ≠ isothermal: in adiabatic Q = 0 (the system is insulated) and PV^γ = const; in isothermal T = const and PV = const. Don't conflate the two equations or the corresponding work formulas.
- For an isothermal process of an ideal gas ΔU = 0 (because U depends only on T), so Q = W; but ΔU = 0 does NOT imply Q = 0 (NCERT §11.8.2, p. 234).
- Carnot efficiency uses absolute (Kelvin) temperatures. Plugging in Celsius values is a guaranteed wrong answer (NCERT §11.11, Eq. 11.27, p. 238).
- The Carnot efficiency is independent of the working substance — NTA likes to put "depends on the gas used" as a distractor (NCERT §11.11, p. 239).
- Zeroth Law was formulated AFTER the First and Second Laws (in 1931, by R.H. Fowler) — but logically it precedes them (NCERT §11.3, p. 228).
🎯 Practice MCQs
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Q1. The Zeroth Law of Thermodynamics leads to the concept of which of the following physical quantities?
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Answer: C
The Zeroth Law states that two systems in thermal equilibrium with a third are in thermal equilibrium with each other; the common variable having the same value is defined as temperature. Internal energy is a separate concept arising in §11.4.
Q2. According to the First Law of Thermodynamics in the form used in the NCERT, the equation ΔQ = ΔU + ΔW has ΔW meaning
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Answer: B
The NCERT defines ΔW explicitly as work done by the system on the surroundings, with ΔQ as heat supplied to the system. The "work done on the system" convention belongs to an alternative form not used in this chapter.
Q3. An electric heater supplies heat to a system at the rate of 100 W. If the system performs work at the rate of 75 J s⁻¹, the rate at which the internal energy of the system increases is
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Answer: A
From the First Law, dU/dt = dQ/dt − dW/dt = 100 − 75 = 25 W. The distractor 175 W comes from incorrectly adding the two rates.
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Q4. For 1 g of water converted from liquid to vapour at atmospheric pressure (Vₗ = 1 cm³, Vg = 1671 cm³) the measured latent heat is 2256 J. The change in internal energy is approximately
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Answer: B
ΔW comes out to 169.2 J. By the First Law, ΔU = ΔQ − ΔW = 2256 − 169.2 = 2086.8 J. Option C is the work done, option A is the latent heat itself — both classic traps.
Q5. Mayer's relation for an ideal gas connecting the molar specific heat capacities at constant pressure and constant volume is
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Answer: B
Starting from ΔQ = ΔU + PΔV, for one mole of an ideal gas Cp = Cv + R, i.e. Cp − Cv = R. The ratio γ = Cp/Cv is a separate quantity.
Q6. The molar specific heat capacity of a solid predicted by the law of equipartition of energy (Dulong-Petit type result quoted in this chapter) is
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Answer: C
For a mole of solid, total energy U = 3RT (3 oscillators in three dimensions, each contributing kBT); since ΔV is negligible for a solid, ΔQ ≈ ΔU, giving C = 3R. The values (3/2)R and (5/2)R refer to monatomic and diatomic ideal gases discussed in the next chapter.
Q7. One mole of an ideal gas expands isothermally at temperature T from volume V₁ to volume V₂. The work done by the gas is
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Answer: B
Integrating ΔW = P ΔV with P = RT/V for one mole gives W = RT ln(V₂/V₁). The "zero work" option fits an isochoric (not isothermal) process; the linear form ignores the changing pressure.
Q8. Two moles of an ideal gas expand isothermally at 300 K from a volume of 2 L to 6 L. Taking R = 8.3 J mol⁻¹ K⁻¹ and ln 3 ≈ 1.099, the heat absorbed by the gas is closest to
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Answer: B
Q = W = 2 × 8.3 × 300 × ln 3 ≈ 2 × 8.3 × 300 × 1.099 ≈ 5473 J ≈ 5.47 kJ. Option C is the one-mole answer; option D doubles it incorrectly.
Q9. For a quasi-static adiabatic process of an ideal gas, the correct relation between pressure and volume is (γ = Cp/Cv)
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Answer: C
For an adiabatic process Q = 0 and one quotes the standard result PV^γ = constant where γ = Cp/Cv. The isothermal relation PV = constant is a distractor.
Q10. An ideal monatomic gas (γ = 5/3) at 27 °C is compressed adiabatically to one-eighth of its initial volume. Using TV^(γ − 1) = constant, the final temperature is
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Answer: C
T₁ V₁^(γ−1) = T₂ V₂^(γ−1). With γ − 1 = 2/3 and V₂/V₁ = 1/8, T₂/T₁ = (V₁/V₂)^(2/3) = 8^(2/3) = 4. So T₂ = 4 × 300 K = 1200 K.
Q11. A Carnot engine operates between a source at 500 K and a sink at 300 K. Its efficiency is
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Answer: B
η = 1 − T₂/T₁ = 1 − 300/500 = 0.40 = 40 %. Temperatures must be in kelvin; using Celsius values is the standard NTA trap.
Q12. Which of the following statements is the Clausius statement of the Second Law of Thermodynamics?
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Answer: B
Option B is the Clausius statement (about refrigerators). Option A is the Kelvin-Planck statement (about heat engines). Option C is the First Law, and option D is a property of ideal-gas internal energy.
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