Thermodynamics for JEE Main: Laws & Entropy Guide

Thermodynamics is one of those chapters where a strong grasp of two or three core ideas unlocks every question in the paper. JEE Main tests it in four to five questions each session: work done by gas in various processes, first and second law applications, heat engine efficiency, and the Carnot cycle. The numerical calculations are typically straightforward once you know which formula to apply. This guide builds that recognition skill systematically.

The Zeroth and First Laws

The zeroth law establishes thermal equilibrium as a transitive relation: if A is in equilibrium with B and B with C, then A is in equilibrium with C. This defines temperature as a meaningful quantity. The first law — ΔU = Q − W — is energy conservation for a thermodynamic system. ΔU is the change in internal energy, Q is heat added to the system, and W is work done by the system. Sign conventions matter: work done by the gas is positive when the gas expands.

For an ideal gas, internal energy depends only on temperature: ΔU = nCᵥΔT for any process, not just constant-volume ones. This is a critical insight that resolves many multi-step problems. Work done by the gas is W = ∫P dV, which equals nRΔT for isobaric, zero for isochoric, and nRT ln(V₂/V₁) for isothermal processes. For adiabatic processes (Q = 0), W = −ΔU = −nCᵥΔT. Memorise these four process-specific work expressions and the first law becomes a plug-and-play tool. Check your fluency with a free mock test on thermodynamics.

Standard Processes and the P-V Diagram

The P-V diagram is the visual language of thermodynamics. The area under any curve on a P-V diagram equals the work done by the gas. An isothermal curve is a hyperbola (PV = constant); an adiabatic curve is steeper than the isothermal at every point (PV^γ = constant, γ > 1). These two curves never cross if they share a starting point, which is why adiabatic expansion produces more temperature drop than isothermal expansion at the same volume ratio.

For a cyclic process, ΔU = 0 over the complete cycle, so Q_net = W_net = area enclosed by the loop. Clockwise loops represent engines (net work done by gas); anticlockwise loops represent refrigerators (net work done on gas). The efficiency of any heat engine is η = W/Q_H = 1 − Q_C/Q_H. This connects seamlessly to our kinetic theory guide through the ideal gas assumptions.

The Carnot Cycle and Second Law

The Carnot cycle is the most efficient possible engine operating between two fixed temperatures T_H and T_C. Its four steps are: isothermal expansion at T_H, adiabatic expansion, isothermal compression at T_C, and adiabatic compression. Its efficiency is η_Carnot = 1 − T_C/T_H (temperatures in Kelvin). No real engine exceeds Carnot efficiency — this is the second law. JEE frequently gives T_H and T_C and asks for efficiency, or gives efficiency and one temperature and asks for the other.

The coefficient of performance (COP) for a refrigerator is COP = Q_C/W = T_C/(T_H − T_C). The Carnot COP sets the upper bound for real refrigerators. Entropy ΔS = Q_rev/T for reversible processes and increases (ΔS > 0) for irreversible processes, which is another statement of the second law. JEE rarely tests entropy calculations directly but may ask whether a process is reversible based on entropy change. For the full physics scoring framework, read our Physics 100+ strategy and our 30-day revision plan.