Last Updated: April 2026
Thermodynamics (Chapter 12, Class 11 Physics) contributes 2-3 questions in JEE Main 2027 and 1-2 questions in JEE Advanced. The chapter is mathematical and concept-heavy — but with the right approach, it can be mastered to near-perfection. In JEE Main 2024, thermodynamics questions focused on the efficiency of Carnot engines, entropy changes, and the relationship between PV diagrams and work done.
Thermodynamics — JEE Main Chapter Snapshot
| Sub-Topic | JEE Main Questions | Priority |
|---|---|---|
| First Law of Thermodynamics | 1 | HIGHEST |
| Thermodynamic Processes (PV diagrams) | 1 | HIGHEST |
| Carnot Engine Efficiency | 0-1 | HIGH |
| Second Law + Entropy | 0-1 | HIGH |
| Heat Engines and Refrigerators | 0-1 | MEDIUM |
Laws of Thermodynamics
Zeroth Law
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 defines temperature.
First Law — Conservation of Energy
ΔQ = ΔU + W
- ΔQ = heat added to the system
- ΔU = change in internal energy (depends only on temperature for ideal gas)
- W = work done BY the system (W = ∫P dV)
- For ideal gas: ΔU = nCvΔT (for any process)
Second Law
Heat cannot spontaneously flow from a colder body to a hotter body (Clausius statement). No engine can have 100% efficiency (Kelvin-Planck statement). Entropy of an isolated system never decreases.
Third Law
The entropy of a perfect crystal approaches zero as temperature approaches absolute zero (0 K).
Thermodynamic Processes — Comparison Table
| Process | Constant | Work Done | ΔU | ΔQ | PV Relation |
|---|---|---|---|---|---|
| Isothermal | T | nRT ln(V2/V1) | 0 | = W | PV = constant |
| Adiabatic | Q (no heat) | nCv(T1-T2) = -ΔU | -W | 0 | PV^γ = constant |
| Isochoric | V | 0 | nCvΔT | = ΔU | P/T = constant |
| Isobaric | P | PΔV = nRΔT | nCvΔT | nCpΔT | V/T = constant |
Key Formulas — Quick Reference
- Cp – Cv = R (for ideal gas)
- γ = Cp/Cv (adiabatic index: monatomic = 5/3; diatomic = 7/5)
- Carnot Efficiency = 1 – T2/T1 = (T1-T2)/T1 (T in Kelvin)
- Coefficient of Performance (COP) of refrigerator = T2/(T1-T2)
- ΔS = ΔQ/T (entropy change for reversible isothermal process)
- Work in cyclic process = Area enclosed in PV diagram
Carnot Engine — JEE Focus
The Carnot engine is the most efficient heat engine operating between two temperatures T1 (hot source) and T2 (cold sink):
- Operates in 4 reversible steps: Isothermal expansion → Adiabatic expansion → Isothermal compression → Adiabatic compression
- Efficiency η = 1 – T2/T1
- No real engine can exceed Carnot efficiency
- A Carnot engine working in reverse = Carnot refrigerator
JEE Solved Example:
A Carnot engine operates between 600K and 300K. What is its efficiency?
η = 1 – 300/600 = 1 – 0.5 = 50%
If engine absorbs 1000 J, work done = 1000 × 0.5 = 500 J; heat rejected = 500 J.
Entropy — Conceptual Understanding
- Entropy measures disorder (randomness) in a system
- Entropy increases in irreversible processes
- For isothermal reversible process: ΔS = ΔQ/T
- For cyclic reversible process: ΔS = 0
- Universe’s entropy always increases (second law consequence)
Practice MCQs — JEE Main Thermodynamics
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Frequently Asked Questions
Q1. Most important formula in Thermodynamics for JEE?
First law: ΔQ = ΔU + W; Carnot efficiency: η = 1 – T2/T1; Cp – Cv = R.
Q2. Adiabatic vs Isothermal — key difference?
Adiabatic: ΔQ = 0, steeper PV slope. Isothermal: ΔT = 0, ΔU = 0.
Q3. How to find work from PV diagram?
Work = area under PV curve (for process) or area enclosed (for cyclic).
Q4. JEE practice for Thermodynamics?
Visit JEE Gurukul Mock Tests for chapter-wise thermodynamics practice.
Q5. How is refrigerator different from heat engine?
Heat engine: converts heat → work. Refrigerator: uses work to move heat from cold to hot (Carnot cycle in reverse).