CBSE Class 11 Physics Chapter 12 Notes: Thermodynamics

Chapter Introduction

Thermodynamics is the study of heat, work, and the laws governing energy transformations. While Chapter 11 dealt with how heat flows, Chapter 12 asks: what are the fundamental limits on converting heat into useful work? The first law (energy conservation) and the second law (direction of natural processes, limit on efficiency) are among the most powerful principles in all of physics. The Carnot engine sets the theoretical benchmark for efficiency, and the distinction between reversible and irreversible processes underpins why the universe has an arrow of time.

Key Definitions

Detailed Concepts

Thermal Equilibrium and the Zeroth Law

Zeroth law of thermodynamics: 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.

This law provides the basis for temperature measurement — it justifies the use of a thermometer (system C) to compare temperatures of different bodies.

First Law of Thermodynamics

ΔQ = ΔU + ΔW

• ΔQ = heat supplied to the system
• ΔU = change in internal energy of the system
• ΔW = work done by the system

Sign convention (NCERT):

• Heat absorbed by system: ΔQ > 0; heat released: ΔQ < 0
• Work done by system: ΔW > 0; work done on system: ΔW < 0
• Increase in internal energy: ΔU > 0; decrease: ΔU < 0

Work done in a thermodynamic process: For a small change in volume,

ΔW = P ΔV (at constant pressure)

For a general process: W = ∫ P dV (area under the P-V curve)

Thermodynamic Processes

Isothermal vs Adiabatic — key differences:

Heat Engines

A heat engine operates in a cycle: 1. Absorbs heat Q₁ from a hot reservoir at T₁ 2. Converts part of it into work W 3. Rejects heat Q₂ to a cold reservoir at T₂

Efficiency: η = W/Q₁ = (Q₁ – Q₂)/Q₁ = 1 – Q₂/Q₁

No real engine can have 100% efficiency — some heat must always be rejected.

Refrigerators and Heat Pumps

A refrigerator works in reverse of a heat engine: 1. Takes heat Q₂ from the cold reservoir (inside fridge) at T₂ 2. External work W is done on the system 3. Heat Q₁ = Q₂ + W is rejected to the hot reservoir (room) at T₁

Coefficient of performance (COP): β = Q₂/W = Q₂/(Q₁ – Q₂)

Note: COP can be greater than 1 (unlike efficiency, which is always < 1).

Second Law of Thermodynamics

Kelvin-Planck statement: It is impossible to devise a process whose only result is the absorption of heat from a single reservoir and its complete conversion into work. (No 100% efficient engine.)

Clausius statement: It is impossible to devise a process whose only result is the transfer of heat from a colder body to a hotter body without external work. (No self-acting refrigerator.)

Both statements are equivalent — violating one implies violating the other.

Reversible and Irreversible Processes

Reversible process: Can be reversed such that both system and surroundings return to their original states. Conditions:

• Infinitely slow (quasi-static)
• No friction, no dissipation
• No heat flow across finite temperature difference

Irreversible process: Cannot be perfectly reversed. Most natural processes are irreversible. Examples:

• Free expansion of gas into vacuum
• Heat flow from hot to cold
• Mixing of two gases
• Friction, viscosity, electrical resistance
• Inelastic deformation

Carnot Engine

The Carnot engine is an ideal reversible engine operating between two temperatures T₁ (hot) and T₂ (cold). It uses four reversible processes:

Carnot efficiency: η = 1 – T₂/T₁ (in Kelvin)

Carnot theorem: No engine operating between two temperatures can be more efficient than a Carnot engine operating between the same two temperatures.

Carnot COP for refrigerator: β = T₂/(T₁ – T₂)

Important Derivations

1. Work Done in Isothermal Process

For an ideal gas: PV = nRT → P = nRT/V

W = ∫ P dV = nRT ∫ dV/V (from V₁ to V₂) = nRT ln(V₂/V₁) = nRT ln(P₁/P₂)

2. Work Done in Adiabatic Process

Since Q = 0, by first law: ΔU = –W → W = –ΔU = –nC_v(T₂ – T₁) = nC_v(T₁ – T₂)

Using C_v = R/(γ – 1): W = nR(T₁ – T₂)/(γ – 1)

Also: W = (P₁V₁ – P₂V₂)/(γ – 1)

3. Carnot Efficiency

For the Carnot cycle:

• Heat absorbed: Q₁ = nRT₁ ln(V₂/V₁) (isothermal expansion)
• Heat rejected: Q₂ = nRT₂ ln(V₃/V₄) (isothermal compression)

Using the adiabatic relations, one can show: V₂/V₁ = V₃/V₄

Common Mistakes

1. Wrong sign convention: Always follow the NCERT convention (Q absorbed positive, W done by system positive). Mixing conventions leads to wrong answers. 2. Confusing isothermal and adiabatic curves on P-V diagram: The adiabatic curve is steeper (slope = –γP/V vs –P/V for isothermal). In an adiabatic expansion, pressure drops faster. 3. Using Celsius in Carnot efficiency: η = 1 – T₂/T₁ requires Kelvin temperatures. Using Celsius gives wrong results. 4. Assuming ΔU = 0 for all cyclic processes: While ΔU = 0 for the full cycle, it is non-zero for individual steps. ΔU = 0 only for the complete cycle (return to initial state). 5. Confusing efficiency and COP: Efficiency η = W/Q₁ is always < 1 for a heat engine. COP β = Q₂/W can be > 1 for a refrigerator. 6. Forgetting that Q = 0 in adiabatic process: In adiabatic processes, the first law gives ΔU = –W. All work comes from (or goes into) internal energy. No heat is exchanged.

Numerical Practice

Example 1: Isothermal Work

An ideal gas at 300 K expands isothermally from 2 L to 10 L. Find the work done if n = 2 mol.

W = nRT ln(V₂/V₁) = 2 × 8.314 × 300 × ln(10/2) = 4988.4 × ln(5) = 4988.4 × 1.609 ≈ 8026 J

Example 2: Carnot Efficiency

A Carnot engine operates between 500 K and 300 K. Find its efficiency and the heat rejected if it absorbs 1000 J per cycle.

η = 1 – T₂/T₁ = 1 – 300/500 = 0.4 = 40%

W = ηQ₁ = 0.4 × 1000 = 400 J

Q₂ = Q₁ – W = 1000 – 400 = 600 J

Example 3: Adiabatic Compression

A monoatomic ideal gas (γ = 5/3) at 300 K is adiabatically compressed to 1/4 of its original volume. Find the final temperature.

TV^(γ-1) = constant → T₁V₁^(γ-1) = T₂V₂^(γ-1)

T₂ = T₁(V₁/V₂)^(γ-1) = 300 × 4^(5/3 – 1) = 300 × 4^(2/3)

4^(2/3) = (2²)^(2/3) = 2^(4/3) ≈ 2.52

T₂ ≈ 300 × 2.52 = 756 K

Exam Tips

1. First law: Always state ΔQ = ΔU + ΔW with sign convention before solving numericals. 2. P-V diagram questions: Practise sketching isothermal, adiabatic, isobaric, and isochoric curves. Label all intersections. 3. Carnot cycle: Know the four stages and be able to draw the P-V diagram with all four curves labelled. This is a 5-mark question area. 4. Efficiency and COP: Know both formulas and their meanings. Efficiency < 1 always; COP can be > 1. 5. Second law statements: Memorise both Kelvin-Planck and Clausius statements and be able to explain their equivalence. 6. Reversible vs irreversible: Be ready to give 2–3 examples of each and explain why most natural processes are irreversible. 7. Numericals: Use Kelvin temperatures always. Check dimensional consistency.

NCERT Exercise Focus Areas

Conceptual Questions for Practice

These questions test deeper understanding beyond numerical problem-solving. Try answering them in 3–4 sentences each before looking at your notes.

1. Can a Carnot engine operate if the source and sink are at the same temperature? Explain using the Carnot efficiency formula.

2. Why is it impossible to reach absolute zero (0 K) in practice, even though thermodynamics does not explicitly forbid it?

3. A gas undergoes free expansion into a vacuum. Is this process reversible or irreversible? Justify your answer.

4. The COP of a refrigerator is 5. Does this mean the refrigerator is more "efficient" than a heat engine with efficiency 40%? Why or why not?

5. Two identical gases at the same temperature are compressed — one isothermally and one adiabatically — to half their original volumes. Which gas ends up at a higher final pressure, and why?

Quick Revision Notes

Second law: Kelvin-Planck (no 100% engine) and Clausius (no self-acting fridge) — equivalent statements.

Carnot theorem: No engine beats Carnot efficiency between the same temperatures.

Key takeaway: Thermodynamics places fundamental limits — energy is conserved (first law), but not all of it is usable (second law).