① First Law of Thermodynamics
The most-tested law in CBSE, NEET & JEE
Statement: The change in internal energy of a system equals the heat added to the system minus the work done by the system.
Common Mistake Alert
This is where most students lose marks. Sign convention is critical. In the IUPAC convention (used in NCERT/JEE): Q is +ve when heat is absorbed, W is +ve when work is done BY the system. Some textbooks use ΔU = Q + W where W is work done ON the system. Never mix conventions.
Work Done by a Gas
For a gas expanding by volume dV against pressure P:
Exam Insight
In PV diagrams, work done by the gas = area enclosed under the curve from V₁ to V₂. If the gas expands (V₂ > V₁), work is positive. If compressed, work is negative. For cyclic processes, net work = area enclosed by the cycle (positive for clockwise, negative for anticlockwise).
🔋 Internal Energy
U depends only on state, not on path
Internal energy U is the sum of all microscopic kinetic and potential energies of molecules in a system. It is a state function — depends only on the current state (T, P, V), not on how the system got there.
For an ideal gas, internal energy depends only on temperature. This is a crucial fact: if T is constant (isothermal), ΔU = 0.
Monoatomic ideal gas: only 3 translational degrees of freedom. No rotational or vibrational modes at ordinary temperatures.
Diatomic gases: 3 translational + 2 rotational degrees of freedom. Vibrational modes are neglected at room temperature (quantum effects).
Polyatomic non-linear: f = 6. Linear polyatomic (CO₂): f = 5. This distinction matters in JEE Advanced.
Degrees of Freedom Table
| Gas Type | Examples | f (DoF) | U = f/2·nRT | Cv = f/2·R | γ = Cp/Cv |
|---|---|---|---|---|---|
| Monoatomic | He, Ar | 3 | 3/2 nRT | 3/2 R | 5/3 ≈ 1.67 |
| Diatomic | H₂, N₂, O₂ | 5 | 5/2 nRT | 5/2 R | 7/5 = 1.4 |
| Polyatomic (linear) | CO₂ | 5 | 5/2 nRT | 5/2 R | 7/5 = 1.4 |
| Polyatomic (non-linear) | H₂O | 6 | 3 nRT | 3R | 4/3 ≈ 1.33 |
📊 Heat Capacities: Cp, Cv, and γ
This is where most derivation questions in JEE come from
Cv — Molar Heat Capacity at Constant Volume
At constant volume: W = 0, so all heat goes into ΔU. Thus Q = ΔU = nCvΔT
Cp — Molar Heat Capacity at Constant Pressure
At constant P, heat goes into both ΔU and work PΔV. So Cp > Cv always.
Mayer's Relation — Derivation
Strategy Tip
If the question gives you γ, you can find f = 2/(γ−1) and then identify the gas type. This reverse-calculation frequently appears in JEE. Also: Cp/Cv = γ does NOT depend on n or T — it's a property of the gas type alone.
② Second Law of Thermodynamics
Why heat engines cannot be 100% efficient
It is impossible to construct a heat engine that, operating in a complete cycle, will produce no other effect than the extraction of heat from a reservoir and the performance of an equal amount of work.
In simple terms: You cannot convert 100% of heat into work in a cycle. There must always be some heat rejected to a cold reservoir.
It is impossible to construct a device that, operating in a cycle, will produce no effect other than the transfer of heat from a colder body to a hotter body without external work.
In simple terms: Heat cannot flow spontaneously from cold to hot. A refrigerator needs external work (electricity).
Both statements are equivalent. Violation of one implies violation of the other. This is a standard proof question in JEE Advanced (it appears as a paragraph or multiple-select problem).
Proof direction: Assume Kelvin-Planck is violated → show Clausius is violated, and vice versa.
Exam Insight — NEET & JEE
NEET asks: "Which of the following violates the second law?" JEE Main asks for efficiency calculations. JEE Advanced asks conceptual equivalence proofs. Know both statements + what each one forbids.
🏎️ Carnot Engine & Efficiency
The most efficient engine theoretically possible
The Carnot engine is an ideal reversible heat engine operating between two temperature reservoirs. No real engine can exceed Carnot efficiency.
Carnot Cycle (4 Steps)
Isothermal Expansion (A→B)
Gas absorbs heat Q₁ from hot reservoir at temperature T_H. ΔU = 0. Work done by gas = Q₁.
Adiabatic Expansion (B→C)
No heat exchange (Q = 0). Gas expands, temperature falls from T_H to T_C. Work done by gas = −ΔU.
Isothermal Compression (C→D)
Gas releases heat Q₂ to cold reservoir at temperature T_C. ΔU = 0. Work done ON gas = Q₂.
Adiabatic Compression (D→A)
No heat exchange. Temperature rises back from T_C to T_H. Cycle is complete.
Common Mistake Alert
Temperature MUST be in Kelvin for the Carnot efficiency formula. Using Celsius will give wrong answers. T_C = 27°C = 300 K (NOT 27). This mistake costs marks in NEET and JEE every year.
Carnot Refrigerator (COP)
♾️ Entropy — The Measure of Disorder
Conceptually important for JEE Advanced
Entropy is a state function — depends only on the current state.
For reversible processes: ΔS_universe = 0
For irreversible processes: ΔS_universe > 0
Entropy never decreases for isolated systems: ΔS ≥ 0
Adiabatic reversible process: ΔS = 0 (isentropic)
In the Carnot cycle: ΔS₁ = +Q₁/T_H (isothermal expansion), ΔS₃ = −Q₂/T_C (isothermal compression). For the cycle to be reversible: ΔS_total = 0 → Q₁/T_H = Q₂/T_C → Q₁/Q₂ = T_H/T_C. This directly gives the Carnot efficiency formula.
Thinking Step
JEE Advanced 2019 asked: "Which process has maximum entropy change?" The answer requires knowing that in irreversible processes, entropy increases more than in reversible ones. Free expansion (into vacuum) is irreversible with Q = 0, W = 0, ΔU = 0, but ΔS > 0 — a classic JEE trap.