Last Updated: April 2026
Thermodynamics JEE Main 2026 is a chapter with consistent 2–3 question presence every session. The chapter (NCERT Class 11 Physics, Chapter 12) covers the laws of thermodynamics, thermodynamic processes, the Carnot cycle, and entropy — all of which connect directly to Chemistry’s chemical thermodynamics (Hess’s Law, Gibbs energy). Mastering thermodynamics gives you returns in both Physics and Chemistry. This guide covers all concepts, key formulae, derivations, and 50 MCQs.
Thermodynamic Systems and State Variables
| Concept | Definition | JEE Relevance |
|---|---|---|
| System | Part of universe under study | Identify open/closed/isolated correctly |
| Surroundings | Everything else | Heat flow direction depends on system definition |
| Open system | Exchanges both matter and energy | e.g., open flask |
| Closed system | Exchanges only energy | e.g., sealed piston |
| Isolated system | No exchange of matter or energy | e.g., thermos flask |
| State variables | P, V, T, n, U, S — describe current state | Path functions vs state functions distinction |
Zeroth Law of Thermodynamics
If body A is in thermal equilibrium with body C, and body B is in thermal equilibrium with body C, then A and B are in thermal equilibrium with each other. This forms the basis for temperature measurement.
First Law of Thermodynamics
Statement: Energy cannot be created or destroyed — only converted from one form to another.
Mathematical form: ΔU = Q − W
- ΔU = change in internal energy of system
- Q = heat absorbed by system (positive when absorbed, negative when released)
- W = work done BY the system (positive when done by system)
Key Thermodynamic Processes
| Process | Condition | Work Done | ΔU | Q |
|---|---|---|---|---|
| Isothermal | T = constant | W = nRT ln(V₂/V₁) | ΔU = 0 (ideal gas) | Q = W |
| Adiabatic | Q = 0 (no heat exchange) | W = −ΔU = nCᵥ(T₁−T₂) | ΔU = −W | Q = 0 |
| Isobaric | P = constant | W = PΔV = nRΔT | ΔU = nCᵥΔT | Q = nCₚΔT |
| Isochoric | V = constant | W = 0 | ΔU = Q = nCᵥΔT | Q = nCᵥΔT |
| Cyclic | Returns to initial state | W = Area enclosed on P-V diagram | ΔU = 0 | Q = W |
Molar Heat Capacities for Ideal Gas
- Cᵥ = heat capacity at constant volume = (f/2)R where f = degrees of freedom
- Cₚ = heat capacity at constant pressure = Cᵥ + R (Mayer’s relation)
- γ (gamma) = Cₚ/Cᵥ = (f+2)/f
| Gas Type | Degrees of Freedom (f) | Cᵥ | Cₚ | γ |
|---|---|---|---|---|
| Monatomic (He, Ar) | 3 | (3/2)R | (5/2)R | 5/3 ≈ 1.67 |
| Diatomic (N₂, O₂) | 5 | (5/2)R | (7/2)R | 7/5 = 1.4 |
| Polyatomic | 6 | 3R | 4R | 4/3 ≈ 1.33 |
Second Law of Thermodynamics
Kelvin-Planck Statement: No heat engine can convert all heat absorbed from a source into work — some heat must always be rejected to a sink (cold reservoir).
Clausius Statement: Heat cannot spontaneously flow from a cold body to a hot body without external work.
Entropy (S): ΔS = Q_reversible / T. For irreversible processes: ΔS > 0 (entropy of universe always increases).
Carnot Engine: Maximum Efficiency
The Carnot cycle is the most efficient cycle possible between two temperatures. It consists of two isothermal processes and two adiabatic processes.
| Parameter | Formula |
|---|---|
| Carnot Efficiency (η) | η = 1 − (T₂/T₁) = 1 − (T_cold/T_hot) |
| Work done by Carnot engine | W = Q₁ − Q₂ = Q₁(1 − T₂/T₁) |
| COP of Carnot refrigerator | COP = T₂/(T₁ − T₂) = Q₂/W |
Key points on Carnot efficiency:
- η depends only on temperatures of source and sink — not on working substance
- η = 100% only if T₂ = 0 K (absolute zero — impossible in practice)
- No engine operating between T₁ and T₂ can be more efficient than Carnot engine
- Efficiency increases when T₁ is increased OR T₂ is decreased
Adiabatic Process — Important Relations
- PVᵞ = constant
- TV^(γ−1) = constant
- T^γ P^(1−γ) = constant
- Work in adiabatic: W = (P₁V₁ − P₂V₂)/(γ−1) = nCᵥ(T₁−T₂)
JEE Main 2026 Thermodynamics: High-Yield Topics
| Topic | Frequency in JEE Main | Key Formula/Concept |
|---|---|---|
| Carnot efficiency calculation | Very High (almost every session) | η = 1 − T₂/T₁ |
| First law: ΔU, Q, W calculation | High | ΔU = Q − W; sign convention |
| Specific heat capacities of gases | High | γ values for mono/diatomic; Mayer’s relation |
| Adiabatic process equations | Medium-High | PVγ = const; TV(γ-1) = const |
| P-V diagram interpretation | Medium-High | Area under curve = work; slope of adiabat vs isotherm |
| Entropy statements | Medium | ΔS > 0 for irreversible; ΔS = Q_rev/T |
Frequently Asked Questions: Thermodynamics JEE
How many questions come from Thermodynamics in JEE Main?
JEE Main typically has 2–3 questions from Thermodynamics in Physics per session. Combined with Chemical Thermodynamics from Chemistry (2–3 questions), the total yield from thermodynamics topics can be 4–6 questions per session — making it one of the most important chapters for JEE Main strategy.
What is the difference between isothermal and adiabatic processes on a P-V diagram?
On a P-V diagram, both isothermal and adiabatic curves are downward sloping. The adiabatic curve is steeper (more slope) than the isothermal curve at any given point. This is because the adiabatic slope = −γ(P/V) while isothermal slope = −(P/V). Since γ > 1, adiabatic is always steeper.
What is the sign convention for work in JEE thermodynamics?
JEE follows the Physics convention: W is positive when work is done BY the system (expansion). Chemistry follows the opposite: W = −PΔV with expansion = negative work. In JEE Physics problems, use ΔU = Q − W. Always check which convention the question implies.
JEE Thermodynamics Practice Quiz
[cg_quiz data=”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″]
Master Thermodynamics with 500+ JEE-level MCQs in JEE Gurukul’s Physics module.