JEE Main Physics Thermodynamics 2027: Laws, Processes & Carnot Engine
First law, isothermal and adiabatic processes, work, and Carnot efficiency, taught with a full worked example and the sign-convention trap.
Edurack
September 28, 2026
Thermodynamics has a reputation for being 'conceptual and confusing'. In reality it is four processes, one conservation law and one efficiency formula. Get the sign convention right and you will stop losing marks to minus signs.
Most thermodynamics errors are sign errors, not physics errors.
Chapter at a Glance
| Snapshot | Detail |
|---|---|
| NTA unit | Unit 8 of 20 — Thermodynamics |
| Priority (trend-based) | Moderate |
| Typical question style | Process-based numericals and conceptual P-V diagram MCQs |
| Best first step | Get the sign convention right, then learn each process's work |
Priority reflects past-paper trends, not an official NTA weightage.
What the NTA Syllabus Covers
- Thermal equilibrium, concept of temperature, zeroth law
- Heat, work and internal energy, first law of thermodynamics
- Isothermal and adiabatic processes
- Second law, reversible and irreversible processes
Master These Topics
1. The First Law and Its Sign Convention
ΔU = Q − W, where Q is heat added to the gas and W is work done by the gas. Internal energy of an ideal gas depends only on temperature: ΔU = nCvΔT.
The convention used in JEE Main: heat absorbed is positive, and work done by the gas is positive. Compression means negative work by the gas.
2. The Four Processes
| Process | Condition | Work done by gas |
|---|---|---|
| Isochoric | V constant | 0 |
| Isobaric | P constant | PΔV |
| Isothermal | T constant | nRT ln(V₂/V₁) |
| Adiabatic | Q = 0 | (P₁V₁ − P₂V₂)/(γ − 1) |
For an adiabatic process, PV^γ = constant, where γ = Cp/Cv. On a P-V diagram the adiabatic curve is steeper than the isotherm.
Worked example (isobaric heating): One mole of a monatomic ideal gas is heated at constant pressure through 100 K (R = 8.31 J/mol·K).
- Q = nCpΔT = (5/2) × 8.31 × 100 ≈ 2078 J
- W = nRΔT = 831 J
- ΔU = nCvΔT = (3/2) × 8.31 × 100 ≈ 1247 J
Check: Q − W = 2078 − 831 = 1247 J, matching ΔU. Also Cp − Cv = R (Mayer's relation).
Trap: In free expansion into a vacuum, an ideal gas does no work and does not exchange heat, so ΔU = 0 and the temperature stays the same.
3. Second Law and Carnot Engine
The second law states that heat does not flow spontaneously from cold to hot, and no engine can convert heat entirely into work. Real processes are irreversible. A Carnot engine working between temperatures T₁ (hot) and T₂ (cold), in kelvin, has maximum efficiency:
η = 1 − T₂/T₁
Worked example: Between 500 K and 300 K, η = 1 − 300/500 = 40%. For a refrigerator, the coefficient of performance is T₂ / (T₁ − T₂).
For any cyclic process, net work done equals the area enclosed by the loop on the P-V diagram.
Common Traps to Avoid
- Using Celsius temperatures in the Carnot efficiency formula. Always convert to kelvin.
- Losing a sign when the gas is compressed (work by gas is negative).
- Assuming ΔU = 0 in any process that returns to the same pressure. It needs the same temperature.
- Thinking an adiabatic process has no temperature change. It has no heat exchange, but temperature does change.
60-Second Revision Sheet
ΔU = Q − W;Cp − Cv = R;γ = Cp/Cv- Isothermal
W = nRT ln(V₂/V₁); adiabaticPV^γ = const - Carnot
η = 1 − T₂/T₁(kelvin) - Cycle work = area enclosed on P-V diagram
Your Study Plan
- Day 1: first law and sign convention, five problems on each process.
- Day 2: P-V diagram reading and cyclic processes.
- Day 3: adiabatic relations, γ, and comparisons with isothermal curves.
- Day 4: Carnot engine, refrigerator, timed mixed set.
Practice Thermodynamics Questions Free → (opens in a new tab)
Continue Your Physics Journey
- Previous chapter: Properties of Solids & Liquids
- Next chapter: Kinetic Theory of Gases
- All 20 JEE Main Physics chapters
- Complete JEE Main Syllabus 2027 guide
Frequently Asked Questions
Which formula should I remember first in Thermodynamics?
The first law, ΔU = Q − W, together with the work formulas for each of the four processes.
Why is the adiabatic curve steeper than the isothermal on a P-V graph?
Because in an adiabatic process pressure falls faster with volume, PV^γ = constant with γ greater than 1, so its slope is γ times the isothermal slope at the same point.