Exercise Solutions Available
Step-by-step answers for all textbook questions in this chapter
Quick Summary
Thermodynamics deals with heat, work, and energy transformations. This chapter is a top scorer for NEET and is revisited in Class 12 Physics and Chemistry. Mastering thermodynamic processes and laws is essential.
Thermodynamic State and Processes
Types of Processes
| Process | Condition | Work done |
|---|---|---|
| Isothermal | T = constant | W = nRT ln(V₂/V₁) |
| Adiabatic | Q = 0 | W = (P₁V₁ − P₂V₂)/(γ−1) |
| Isobaric | P = constant | W = P(V₂ − V₁) |
| Isochoric | V = constant | W = 0 |
Laws of Thermodynamics
Zeroth Law
Systems in thermal equilibrium share the same temperature — basis of thermometry.
First Law
Q = ΔU + W
- Q = heat added to the system
- ΔU = change in internal energy
- W = work done BY the system
For an ideal gas: ΔU = nCᵥΔT (depends only on temperature)
Second Law
- Kelvin-Planck statement: No engine can convert all heat into work.
- Clausius statement: Heat cannot flow spontaneously from cold to hot body.
- Entropy: Entropy of an isolated system never decreases (ΔS ≥ 0).
Third Law
As temperature approaches absolute zero, entropy of a perfect crystal approaches zero.
Carnot Engine
An ideal reversible engine working between temperatures T₁ (hot) and T₂ (cold).
Efficiency: η = 1 − T₂/T₁ = (T₁ − T₂)/T₁
Carnot cycle: 1. Isothermal expansion at T₁ (heat absorbed) 2. Adiabatic expansion (T₁ → T₂) 3. Isothermal compression at T₂ (heat released) 4. Adiabatic compression (T₂ → T₁)
No real engine can exceed Carnot efficiency for the same temperature limits.
Specific Heats of Gases
- Cᵥ: specific heat at constant volume
- Cₚ: specific heat at constant pressure
- γ = Cₚ/Cᵥ (ratio of specific heats)
- Mayer's relation: Cₚ − Cᵥ = R
For monatomic gas: γ = 5/3; Diatomic: γ = 7/5
Important Questions
One Mark
- State the first law of thermodynamics.
- What is an adiabatic process?
- Define efficiency of a heat engine.
- What happens to entropy in an irreversible process?
- State the second law (Kelvin-Planck statement).
Short Answer (3 marks)
- Compare the four thermodynamic processes in terms of Q, W, and ΔU.
- Derive Mayer's relation Cₚ − Cᵥ = R.
- A Carnot engine works between 227°C and 27°C. Find its efficiency.
Long Answer (5 marks)
- Describe the Carnot cycle step by step. Derive the expression for its efficiency. Explain why no real engine can exceed Carnot efficiency.
- State and explain the first and second laws of thermodynamics. Explain entropy and its significance.
Common Mistakes
- Using Celsius instead of Kelvin in efficiency formula — always convert to Kelvin.
- Confusing work done ON the system with work done BY the system — sign convention matters.
- Assuming ΔU = 0 for all isothermal processes (true only for ideal gases).
Board Exam Tips
- Carnot efficiency numericals are the most common calculation in this chapter.
- Learn the P-V diagrams for all four processes — drawing them correctly earns marks.
- NEET: Thermodynamics is a guaranteed source of 2–3 questions annually.