Thermodynamics is Chapter 11 of CBSE Class 11 Physics — the chapter that explains how heat, work, and energy flow between a system and its surroundings. From a car engine to a refrigerator to the steam that once powered the industrial revolution, every heat machine runs on the laws you will learn here. Master these notes and you can predict whether a process is possible, how efficient an engine can ever be, and why heat refuses to flow uphill on its own.
By the end of these notes you will be able to apply the first law of thermodynamics to any process, distinguish isothermal from adiabatic changes, use the relation Cp − Cv = R, and calculate the efficiency of a Carnot engine. This is a high-weightage chapter carrying roughly 8–10 marks in boards, and a favourite source of questions in JEE and NEET.
Table of Contents
- Key Concepts — Thermal equilibrium, internal energy, the laws of thermodynamics, processes, Carnot engine
- Weightage in Board & Entrance Exams
- Important Definitions
- Solved Examples
- Important Questions for Board Exams
- Quick Revision Points
Key Concepts
1. Thermodynamic System, Surroundings and State Variables
A thermodynamic system is the part of the universe we choose to study — for example, the gas trapped inside a cylinder. Everything outside it that can exchange energy or matter with it is called the surroundings, and the imaginary boundary between them is the wall.
The condition of a system is described by measurable quantities called thermodynamic state variables — pressure (P), volume (V), temperature (T), and internal energy (U). These depend only on the present state, not on how the system reached it.
- Extensive variables depend on the amount of matter: volume, internal energy, mass.
- Intensive variables do not depend on the amount: pressure, temperature, density.
2. Thermal Equilibrium and the Zeroth Law
Two systems are in thermal equilibrium when they are at the same temperature, so there is no net flow of heat between them when they are placed in contact.
The Zeroth Law of Thermodynamics states: if two systems A and B are each in thermal equilibrium with a third system C, then A and B are in thermal equilibrium with each other.
Key idea: The zeroth law tells us that temperature is the property that decides the direction of heat flow — it is what gives temperature its physical meaning.
3. Heat, Internal Energy and Work
Internal energy (U) is the total energy stored inside a system — the sum of the kinetic and potential energies of all its molecules. For an ideal gas it depends only on temperature.
Heat (Q) is energy transferred between a system and its surroundings because of a temperature difference. Work (W) is energy transferred when the system pushes its boundary through a distance.
- Heat and work are path functions — they depend on the route taken between two states.
- Internal energy is a state function — it depends only on the initial and final states.
- SI unit of all three (Q, W, U): the joule (J).
4. Work Done by a Gas
When a gas expands and pushes a piston, it does work on the surroundings. For a small volume change dV at pressure P, the work done is:
W = ∫ P dV
- On a P–V diagram, the work done equals the area under the curve.
- Work done by the gas (expansion) is positive; work done on the gas (compression) is negative.
5. First Law of Thermodynamics
The first law is simply the law of conservation of energy applied to heat. The heat supplied to a system is used partly to increase its internal energy and partly to do external work.
ΔQ = ΔU + ΔW
- ΔQ is positive when heat is added to the system.
- ΔW is positive when work is done by the system.
- ΔU is positive when the internal energy (and temperature) increases.
Sign convention matters: always decide first whether heat is going in and whether the gas is expanding before substituting numbers.
6. Specific Heat Capacity
Specific heat capacity is the heat required to raise the temperature of unit mass of a substance by 1 K. Molar specific heat (C) uses one mole instead of unit mass.
For a gas, the specific heat depends on the conditions, so there are two principal values:
- Cp — molar specific heat at constant pressure.
- Cv — molar specific heat at constant volume.
Since heating a gas at constant pressure also does work pushing back the surroundings, Cp is always greater than Cv.
7. Mayer’s Relation (Cp − Cv = R)
For one mole of an ideal gas, the two molar specific heats are linked by Mayer’s relation:
Cp − Cv = R
where R = 8.314 J mol⁻¹ K⁻¹ is the universal gas constant. The extra R represents the work done by the gas when it expands at constant pressure.
The ratio of specific heats γ = Cp/Cv is an important number: γ = 5/3 for a monatomic gas and 7/5 for a diatomic gas.
8. Thermodynamic Processes
A thermodynamic process is any operation that changes a system from one state to another. Four special processes are central to this chapter.
| Process | Constant Quantity | First-Law Result |
|---|---|---|
| Isothermal | Temperature (ΔU = 0) | ΔQ = ΔW = nRT ln(V₂/V₁) |
| Adiabatic | No heat exchange (ΔQ = 0) | ΔW = −ΔU; PVγ = constant |
| Isobaric | Pressure | ΔW = PΔV = nRΔT |
| Isochoric | Volume (ΔW = 0) | ΔQ = ΔU = nCvΔT |
[DIAGRAM: A P–V diagram showing four curves from the same starting point — a flat horizontal line (isobaric), a vertical line (isochoric), a gentle hyperbola (isothermal), and a steeper hyperbola (adiabatic).]
9. Isothermal and Adiabatic Processes in Detail
In an isothermal process the temperature is held constant, so the gas obeys Boyle’s law PV = constant and all the heat supplied is converted into work.
Wiso = nRT ln(V₂/V₁) = 2.303 nRT log(V₂/V₁)
In an adiabatic process no heat enters or leaves (ΔQ = 0), so any work done comes entirely from the internal energy — the gas cools on expansion and heats on compression.
- PVγ = constant and TVγ−1 = constant
- Wadia = (P₁V₁ − P₂V₂)/(γ − 1) = nR(T₁ − T₂)/(γ − 1)
Important: on a P–V diagram an adiabatic curve is always steeper than an isothermal curve through the same point.
10. Reversible and Irreversible Processes
A reversible process is an idealised process carried out so slowly (quasi-statically) that the system stays in equilibrium throughout and can be exactly retraced, leaving no change in the surroundings.
An irreversible process cannot be retraced — almost every real process (sudden expansion, friction, heat flow across a finite temperature gap) is irreversible.
Key idea: a truly reversible process is a useful limit that no real engine ever reaches, but it sets the ceiling on efficiency.
11. Second Law of Thermodynamics
The first law allows energy conversions but says nothing about their direction. The second law supplies that missing direction.
- Kelvin–Planck statement: no engine can convert heat completely into work in a cyclic process — some heat must always be rejected.
- Clausius statement: heat cannot flow on its own from a colder body to a hotter body.
The two statements are equivalent, and together they forbid a 100% efficient engine and a perfect refrigerator.
12. Carnot Engine and Efficiency
A heat engine takes heat Q₁ from a hot source at T₁, converts part of it into work W, and rejects the rest Q₂ to a cold sink at T₂.
Efficiency η = W/Q₁ = 1 − Q₂/Q₁
The Carnot engine is an ideal reversible engine working in a cycle of two isothermal and two adiabatic processes. Its efficiency is the maximum any engine can have between the same two temperatures:
η = 1 − T₂/T₁ (temperatures in kelvin)
- Carnot efficiency depends only on the two temperatures, not on the working substance.
- Efficiency is 100% only if T₂ = 0 K, which is unattainable.
[DIAGRAM: An energy-flow block diagram — source at T₁ giving Q₁ to the engine, engine delivering work W, and rejecting Q₂ to the sink at T₂.]
13. Refrigerator and Heat Pump
A refrigerator is a heat engine run in reverse: external work W is used to pull heat Q₂ from a cold body and dump Q₁ into a warmer room.
Its performance is measured by the coefficient of performance (COP):
β = Q₂/W = Q₂/(Q₁ − Q₂) = T₂/(T₁ − T₂)
- A higher COP means a more efficient refrigerator.
- Because work is needed, the Clausius statement of the second law is obeyed — heat never flows from cold to hot without an external agent.
Weightage in Board & Entrance Exams
| Exam | Typical Weightage | Most-Tested Areas |
|---|---|---|
| CBSE Board (Class 11) | 8–10 marks | First law, isothermal/adiabatic work, Cp − Cv = R, Carnot efficiency |
| JEE Main / Advanced | 2–4 questions | P–V work, adiabatic relations, efficiency, mixed processes |
| NEET | 2–3 questions | First law sign convention, Carnot engine, refrigerator COP |
[TABLE: Question-type split — VSA (1 mark): laws & definitions; SA (2–3 marks): first-law numericals, Cp − Cv = R, process identification; LA (5 marks): Carnot cycle derivation, efficiency and COP problems.]
Important Definitions
| Term | Definition |
|---|---|
| Internal energy (U) | Total kinetic + potential energy of all molecules; a state function |
| Heat (Q) | Energy transferred due to a temperature difference |
| Work (W) | Energy transferred when a system moves its boundary: W = ∫P dV |
| Zeroth law | Bodies in thermal equilibrium with a third body are in equilibrium with each other |
| First law | ΔQ = ΔU + ΔW — conservation of energy for heat processes |
| Isothermal process | Process at constant temperature; ΔU = 0, so ΔQ = ΔW |
| Adiabatic process | Process with no heat exchange; ΔQ = 0, PVγ = constant |
| Mayer’s relation | Cp − Cv = R for one mole of an ideal gas |
| Second law | Heat cannot fully convert to work in a cycle; no spontaneous cold-to-hot flow |
| Carnot efficiency | Maximum engine efficiency: η = 1 − T₂/T₁ |
| Coefficient of performance | Refrigerator measure: β = T₂/(T₁ − T₂) |
Solved Examples
Example 1
A gas absorbs 200 J of heat and does 80 J of work on its surroundings. Find the change in its internal energy.
Answer: ΔU = ΔQ − ΔW = 200 − 80 = 120 J (internal energy increases).
Example 2
One mole of an ideal gas expands isothermally at 300 K from volume V to 2V. Find the work done. (R = 8.314 J mol⁻¹ K⁻¹)
Answer: W = nRT ln(V₂/V₁) = 1 × 8.314 × 300 × ln 2 = 2494.2 × 0.693 = ≈ 1729 J.
Example 3
A Carnot engine works between a source at 500 K and a sink at 300 K. Find its efficiency.
Answer: η = 1 − T₂/T₁ = 1 − 300/500 = 1 − 0.6 = 0.4 = 40%.
Example 4
In an adiabatic process the internal energy of a gas decreases by 150 J. How much work is done by the gas?
Answer: For an adiabatic process ΔQ = 0, so ΔW = −ΔU = −(−150) = 150 J (work is done by the gas at the cost of its internal energy).
Example 5
The molar specific heat of a gas at constant volume is Cv = 20.8 J mol⁻¹ K⁻¹. Find Cp and γ. (R = 8.314)
Answer: Cp = Cv + R = 20.8 + 8.314 = 29.1 J mol⁻¹ K⁻¹. γ = Cp/Cv = 29.1/20.8 = ≈ 1.4 (a diatomic gas).
Example 6
A refrigerator maintains its interior at 270 K while the room is at 300 K. Find the maximum coefficient of performance.
Answer: β = T₂/(T₁ − T₂) = 270/(300 − 270) = 270/30 = 9.
Important Questions for Board Exams
1-Mark Questions (VSA)
- State the zeroth law of thermodynamics.
- Why is the internal energy of an ideal gas a function of temperature only?
- What is the value of work done in an isochoric process?
- Name the process in which no heat is exchanged with the surroundings.
- Can the efficiency of a heat engine ever be 100%? Justify.
2–3-Mark Questions (SA)
- State the first law of thermodynamics and explain the sign convention used for ΔQ, ΔW and ΔU.
- Distinguish between isothermal and adiabatic processes with one example of each.
- Derive the relation Cp − Cv = R for one mole of an ideal gas.
- Why is Cp greater than Cv for a gas? Explain physically.
5-Mark Questions (LA)
- Describe the four strokes of a Carnot cycle and derive the expression for its efficiency η = 1 − T₂/T₁.
- State both the Kelvin–Planck and Clausius statements of the second law and show that they are equivalent.
- Explain the working of a refrigerator and derive its coefficient of performance in terms of T₁ and T₂.
Quick Revision Points
- State variables: P, V, T, U — depend only on the present state
- Zeroth law defines temperature; bodies at the same temperature are in thermal equilibrium
- First law: ΔQ = ΔU + ΔW (conservation of energy)
- Heat and work are path functions; internal energy is a state function
- Isothermal: ΔU = 0, ΔQ = ΔW = nRT ln(V₂/V₁)
- Adiabatic: ΔQ = 0, PVγ = constant, ΔW = nR(T₁ − T₂)/(γ − 1)
- Isobaric: ΔW = PΔV; Isochoric: ΔW = 0, ΔQ = nCvΔT
- Mayer’s relation: Cp − Cv = R; γ = Cp/Cv (5/3 monatomic, 7/5 diatomic)
- Second law: no 100% engine; heat will not flow cold → hot on its own
- Heat engine efficiency: η = 1 − Q₂/Q₁; Carnot: η = 1 − T₂/T₁
- Refrigerator COP: β = T₂/(T₁ − T₂)
Next Chapter: Chapter 12 — Kinetic Theory
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Related Chapters in Class 11 Physics
- Kinetic Theory Class 11 Notes
- Thermal Properties of Matter Class 11 Notes
- Laws of Motion Class 11 Notes
Practice What You Learned
Take your thermal physics further with our Class 12 Physics notes once you are board-ready.