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Basic Concepts Zeroth Law First Law Work Done Processes Cp & Cv Heat Engines Carnot Engine Entropy

Thermodynamics JEE Main & Advanced 2025-26

Master the Laws of Thermodynamics, Heat Engines, Carnot Cycle, and Entropy with complete notes. Includes 200+ solved problems, PV diagrams, and all JEE shortcuts for guaranteed high scores.

📚 9 Complete Sections
✍️ 200+ Solved Examples
🎯 PV Diagram Mastery
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Thermodynamics JEE notes, Formulas, PYQs
Thermodynamics - Complete JEE Physics Notes with PV Diagrams
1

Basic Concepts of Thermodynamics

Thermodynamics is the branch of physics that deals with heat, work, temperature, and energy transformations. It forms the foundation for understanding engines, refrigerators, and natural phenomena. This chapter carries 8-12% weightage in JEE.

1.1 System, Surroundings, and Universe

System

The part of universe under thermodynamic study.

  • Gas in a cylinder
  • Water in a container
  • Ice melting
Surroundings

Everything outside the system that can interact with it.

  • Atmosphere
  • Heat reservoir
  • External pressure
Universe

System + Surroundings = Universe

For isolated universe:

Total energy = constant

Total entropy ≥ 0

1.2 Types of Thermodynamic Systems

System Type Matter Exchange Energy Exchange Example
Open System Yes ✓ Yes ✓ Boiling water in open pan
Closed System No ✗ Yes ✓ Gas in cylinder with piston
Isolated System No ✗ No ✗ Perfect thermos flask

1.3 State Variables and State Functions

State Variables

Macroscopic properties that define the state of a system:

  • Pressure (P): N/m² or Pa
  • Volume (V): m³
  • Temperature (T): Kelvin
  • Internal Energy (U): Joule
  • Entropy (S): J/K
State Functions vs Path Functions

State Functions (Path Independent):

P, V, T, U, S, Enthalpy (H)

Depend only on initial & final states

Path Functions (Path Dependent):

Heat (Q), Work (W)

Depend on the process/path taken

1.4 Equation of State for Ideal Gas

Ideal Gas Equation

\[PV = nRT = NkT\]

Where:

  • P = Pressure (Pa)
  • V = Volume (m³)
  • n = Number of moles
  • R = Gas constant = 8.314 J/mol·K
  • T = Temperature (K)

Alternative forms:

  • N = Number of molecules
  • k = Boltzmann constant
  • k = R/Nₐ = 1.38 × 10⁻²³ J/K
  • Nₐ = 6.022 × 10²³ /mol

1.5 Internal Energy

Internal Energy (U)

Total energy of all molecules in a system (kinetic + potential). For ideal gases, U depends only on temperature.

For Ideal Gas:

\[U = nC_vT = \frac{f}{2}nRT\]

where f = degrees of freedom

Change in Internal Energy:

\[\Delta U = nC_v\Delta T\]

Valid for ALL processes of ideal gas

💡 JEE Key Points

  • Internal energy U is a state function - depends only on T for ideal gas
  • ΔU = 0 for isothermal process of ideal gas
  • Heat and Work are path functions - their values depend on the process
  • Real gases: U depends on both T and V
2

Zeroth Law of Thermodynamics

The Zeroth Law establishes the concept of thermal equilibrium and provides the basis for temperature measurement. Though simple, it's the foundation of thermometry.

Statement of Zeroth Law

"If two systems A and B are separately in thermal equilibrium with a third system C, then A and B are also in thermal equilibrium with each other."

In simple terms:

If T_A = T_C and T_B = T_C, then T_A = T_B

2.1 Thermal Equilibrium

Conditions for Thermal Equilibrium
  • No net heat flow between systems
  • Same temperature throughout
  • Macroscopic properties remain constant with time
  • Thermal contact must exist (or have existed)

💡 Why "Zeroth" Law?

This law was formulated AFTER the First and Second Laws but was considered more fundamental. Since it logically precedes them, it was named the "Zeroth" Law.

JEE Significance: Forms the basis for thermometer function - thermometer (C) is in equilibrium with body (A) and calibration standard (B).

3

First Law of Thermodynamics

The First Law is essentially the Law of Conservation of Energy applied to thermodynamic systems. It quantifies the relationship between heat, work, and internal energy. This is the MOST IMPORTANT concept for JEE - expect 3-4 questions from this alone.

Statement of First Law

"Heat supplied to a system is used to increase its internal energy and to do work by the system."

\[Q = \Delta U + W\]

Differential form:

\[dQ = dU + dW\]

Alternative form:

\[\Delta U = Q - W\]

3.1 Sign Conventions (Critical for JEE)

Quantity Positive (+) Negative (-)
Heat (Q) Heat absorbed by system Heat released by system
Work (W) Work done BY system (expansion) Work done ON system (compression)
ΔU Temperature increases Temperature decreases

⚠️ JEE Trap Alert - Sign Convention

Some textbooks use different sign conventions. Always check:

Physics Convention (NCERT, JEE):

\[Q = \Delta U + W\]

W positive for expansion (work done BY system)

Chemistry Convention:

\[\Delta U = Q + W\]

W positive for compression (work done ON system)

3.2 First Law for Cyclic Process

In a Cyclic Process

System returns to initial state, so internal energy change is zero:

\[\Delta U_{\text{cycle}} = 0 \implies Q_{\text{net}} = W_{\text{net}}\]

Key Points:

  • Net heat absorbed = Net work done by system
  • Work = Area enclosed in PV diagram
  • Clockwise cycle → W > 0 (heat engine)
  • Anticlockwise cycle → W < 0 (refrigerator)

📝 Solved Example 1

Question: A system absorbs 500 J of heat and does 200 J of work. What is the change in internal energy?

Solution:

Given:

  • Q = +500 J (heat absorbed, positive)
  • W = +200 J (work done by system, positive)

Using First Law:

\[Q = \Delta U + W\]
\[\Delta U = Q - W = 500 - 200\]
\[\Delta U = +300 \text{ J}\]

Internal energy increases by 300 J, meaning temperature increases.

4

Work Done in Thermodynamic Processes

Work in thermodynamics is done when the system expands or compresses against external pressure. Understanding work calculations is essential - every PV diagram problem in JEE uses these concepts.

4.1 Basic Work Formula

Work Done by Gas

\[W = \int_{V_1}^{V_2} P \, dV = \text{Area under P-V curve}\]

For Expansion (V₂ > V₁):

W > 0 (work done BY gas)

For Compression (V₂ < V₁):

W < 0 (work done ON gas)

4.2 Work Done in Different Processes

Process Condition Work Done (W)
Isobaric P = constant W = PΔV = P(V₂-V₁) = nRΔT
Isochoric V = constant W = 0
Isothermal T = constant W = nRT ln(V₂/V₁) = nRT ln(P₁/P₂)
Adiabatic Q = 0 W = (P₁V₁ - P₂V₂)/(γ-1) = nCᵥΔT
Polytropic PVⁿ = constant W = (P₁V₁ - P₂V₂)/(n-1)

4.3 PV Diagram Analysis

Volume (V) → Pressure (P) → Isothermal (PV = const) Adiabatic (PVᵞ = const) A (P₁, V₁) Adiabatic curve is steeper than Isothermal

Key: Adiabatic slope = γ × Isothermal slope (at same point)

📝 Solved Example 2

Question: 2 moles of an ideal gas at 300 K expands isothermally from 1 L to 10 L. Find the work done by the gas. (R = 8.314 J/mol·K)

Solution:

Given:

  • n = 2 mol
  • T = 300 K (constant - isothermal)
  • V₁ = 1 L, V₂ = 10 L

For isothermal process:

\[W = nRT \ln\left(\frac{V_2}{V_1}\right)\]
\[W = 2 \times 8.314 \times 300 \times \ln\left(\frac{10}{1}\right)\]
\[W = 2 \times 8.314 \times 300 \times \ln(10)\]
\[W = 2 \times 8.314 \times 300 \times 2.303\]
\[W = 11,488 \text{ J} \approx 11.5 \text{ kJ}\]

💡 Quick Tricks for Work Calculation

1. For any process, work done = Area under PV curve

Positive area (expansion) → W > 0, Negative area (compression) → W < 0

2. For cyclic process:

W_net = Area enclosed by cycle. Clockwise = positive, Anticlockwise = negative

3. Remember ln(10) ≈ 2.303 and ln(2) ≈ 0.693

These appear frequently in isothermal problems

5

Thermodynamic Processes

A thermodynamic process is a transition from one equilibrium state to another. Understanding each type of process and applying the First Law correctly is the key to solving 80% of JEE thermodynamics problems.

5.1 Isobaric Process (P = constant)

Characteristics
  • Pressure remains constant
  • Charles' Law: V ∝ T
  • PV diagram: Horizontal line
  • Example: Heating water in open container
Formulas

Work:

\[W = P\Delta V = nR\Delta T\]

Heat:

\[Q = nC_p\Delta T\]

Internal Energy:

\[\Delta U = nC_v\Delta T\]

5.2 Isochoric Process (V = constant)

Characteristics
  • Volume remains constant
  • Gay-Lussac's Law: P ∝ T
  • PV diagram: Vertical line
  • Example: Heating gas in rigid container
Formulas

Work:

\[W = 0\]

Heat = Internal Energy Change:

\[Q = \Delta U = nC_v\Delta T\]

5.3 Isothermal Process (T = constant)

Characteristics
  • Temperature remains constant
  • Boyle's Law: PV = constant
  • PV diagram: Rectangular hyperbola
  • Requires slow process with thermal contact
Formulas

Internal Energy Change:

\[\Delta U = 0 \text{ (for ideal gas)}\]

Work = Heat:

\[W = Q = nRT\ln\frac{V_2}{V_1}\]

Relation:

\[P_1V_1 = P_2V_2\]

5.4 Adiabatic Process (Q = 0)

Characteristics
  • No heat exchange with surroundings
  • PVᵞ = constant
  • PV diagram: Steeper than isothermal
  • Fast process or insulated container
Formulas

Heat:

\[Q = 0\]

Work:

\[W = -\Delta U = \frac{P_1V_1 - P_2V_2}{\gamma - 1}\]

Relations:

\[PV^\gamma = \text{const}, \quad TV^{\gamma-1} = \text{const}, \quad P^{1-\gamma}T^\gamma = \text{const}\]

5.5 Complete Comparison Table

Process Constant Equation ΔU Q W
Isobaric P V/T = const nCᵥΔT nCₚΔT PΔV = nRΔT
Isochoric V P/T = const nCᵥΔT nCᵥΔT 0
Isothermal T PV = const 0 nRT ln(V₂/V₁) nRT ln(V₂/V₁)
Adiabatic Q = 0 PVᵞ = const nCᵥΔT 0 -nCᵥΔT

📝 Solved Example 3 (JEE Advanced Type)

Question: A monoatomic ideal gas (γ = 5/3) at pressure P and volume V is compressed adiabatically to volume V/8. Find the final pressure and work done.

Solution:

Given:

  • γ = 5/3 (monoatomic)
  • V₁ = V, V₂ = V/8
  • P₁ = P

For adiabatic process: PVᵞ = constant

\[P_1V_1^\gamma = P_2V_2^\gamma\]
\[P \times V^{5/3} = P_2 \times \left(\frac{V}{8}\right)^{5/3}\]
\[P_2 = P \times \left(\frac{V}{V/8}\right)^{5/3} = P \times 8^{5/3}\]
\[P_2 = P \times (2^3)^{5/3} = P \times 2^5 = 32P\]

Work done (compression → negative):

\[W = \frac{P_1V_1 - P_2V_2}{\gamma - 1} = \frac{PV - 32P \times \frac{V}{8}}{\frac{5}{3} - 1}\]
\[W = \frac{PV - 4PV}{\frac{2}{3}} = \frac{-3PV}{\frac{2}{3}} = -\frac{9PV}{2}\]
\[P_2 = 32P, \quad W = -\frac{9PV}{2}\]

Negative work indicates work is done ON the gas (compression).

6

Specific Heat Capacities (Cₚ & Cᵥ)

The two specific heat capacities of gases - at constant pressure (Cₚ) and at constant volume (Cᵥ) - are fundamental to thermodynamics. Their ratio γ = Cₚ/Cᵥ appears in all adiabatic process equations.

6.1 Definitions

Molar Heat Capacity at Constant Volume (Cᵥ)
\[C_v = \frac{1}{n}\left(\frac{dQ}{dT}\right)_V = \frac{1}{n}\frac{dU}{dT}\]

At constant V, W = 0, so Q = ΔU

Unit: J/mol·K

Molar Heat Capacity at Constant Pressure (Cₚ)
\[C_p = \frac{1}{n}\left(\frac{dQ}{dT}\right)_P\]

At constant P, Q = ΔU + PΔV

Always: Cₚ > Cᵥ

6.2 Mayer's Relation

Mayer's Formula (Most Important)

\[C_p - C_v = R\]

where R = 8.314 J/mol·K

Physical Meaning:

The extra heat required at constant pressure (compared to constant volume) is used to do work against external pressure during expansion.

6.3 Values of Cₚ, Cᵥ, and γ for Different Gases

Gas Type DOF (f) Cᵥ Cₚ γ = Cₚ/Cᵥ Examples
Monoatomic 3 (3/2)R (5/2)R 5/3 = 1.67 He, Ne, Ar
Diatomic 5 (5/2)R (7/2)R 7/5 = 1.4 H₂, N₂, O₂, Air
Polyatomic (Linear) 7 (7/2)R (9/2)R 9/7 = 1.29 CO₂, N₂O
Polyatomic (Non-linear) 6 3R 4R 4/3 = 1.33 H₂O, NH₃, CH₄

General Formulas

In terms of f (degrees of freedom):

\[C_v = \frac{f}{2}R\]
\[C_p = \frac{f+2}{2}R\]

Ratio γ:

\[\gamma = \frac{C_p}{C_v} = \frac{f+2}{f} = 1 + \frac{2}{f}\]

As f increases, γ decreases

💡 JEE Memory Tricks

For γ values:

  • Monoatomic: γ = 5/3 ≈ 1.67 (largest)
  • Diatomic: γ = 7/5 = 1.4 (most common in JEE - air)
  • Polyatomic: γ = 4/3 ≈ 1.33 (smallest)

For mixture of gases:

\[\gamma_{mix} = \frac{n_1C_{p1} + n_2C_{p2}}{n_1C_{v1} + n_2C_{v2}}\]
7

Heat Engines & Second Law

Heat engines convert thermal energy into mechanical work. Understanding their efficiency and the Second Law of Thermodynamics is crucial for JEE - expect at least 2 questions on this topic.

7.1 Heat Engine Basics

Working Principle

A heat engine operates in a cyclic process:

  1. Absorbs heat Q₁ from hot source (T₁)
  2. Converts part to work W
  3. Rejects heat Q₂ to cold sink (T₂)
  4. Returns to initial state

Energy Balance:

\[Q_1 = W + Q_2\]

or

\[W = Q_1 - Q_2\]

7.2 Efficiency of Heat Engine

Efficiency Formula

\[\eta = \frac{\text{Work output}}{\text{Heat input}} = \frac{W}{Q_1} = \frac{Q_1 - Q_2}{Q_1} = 1 - \frac{Q_2}{Q_1}\]

Key Points:

  • η is always less than 1 (< 100%)
  • η = 1 only if Q₂ = 0 (impossible)
  • Expressed as percentage: η × 100%

For cyclic process:

  • ΔU = 0 (returns to initial state)
  • W_net = Area of cycle on PV diagram
  • Q_net = W_net

7.3 Second Law of Thermodynamics

Kelvin-Planck Statement

"It is impossible to construct an engine that, operating in a cycle, produces no effect other than the extraction of heat from a single reservoir and the performance of an equivalent amount of work."

In simple terms:

100% efficient heat engine is impossible

Some heat must always be rejected

Clausius Statement

"It is impossible for heat to flow spontaneously from a colder body to a hotter body without external work being done."

In simple terms:

Heat naturally flows from hot to cold

Refrigerator needs external work

⚠️ Both Statements are Equivalent

Kelvin-Planck and Clausius statements are different forms of the same law. If one is violated, the other is also violated. This equivalence is often asked in JEE.

7.4 Refrigerator & Heat Pump

Refrigerator

Transfers heat from cold to hot using external work:

Coefficient of Performance (COP):

\[\beta = \frac{Q_2}{W} = \frac{Q_2}{Q_1 - Q_2}\]

β can be > 1 (unlike efficiency)

Heat Pump

Heats a room by extracting heat from outside:

Coefficient of Performance:

\[\beta_{HP} = \frac{Q_1}{W} = \frac{Q_1}{Q_1 - Q_2}\]

Relation: β_HP = β_refrigerator + 1

8

Carnot Engine & Carnot Cycle

The Carnot engine is an idealized, reversible heat engine with the maximum possible efficiency. Understanding the Carnot cycle is essential - it's the benchmark for all real heat engines and appears in 90% of JEE thermodynamics papers.

8.1 Carnot Cycle - Four Stages

Volume (V) → Pressure (P) → 1→2: Isothermal Expansion 2→3: Adiabatic Expansion 3→4: Isothermal Compression 4→1: Adiabatic Compression 1 2 3 4 T₁ (Hot) T₂ (Cold)

Carnot Cycle: Two isothermal + Two adiabatic processes

Stage Process Description Heat (Q) Work (W)
1 → 2 Isothermal Expansion Gas expands at T₁, absorbs Q₁ Q₁ = nRT₁ ln(V₂/V₁) W₁ = Q₁
2 → 3 Adiabatic Expansion Gas expands, T drops to T₂ Q = 0 W₂ = nCᵥ(T₁-T₂)
3 → 4 Isothermal Compression Gas compressed at T₂, rejects Q₂ Q₂ = nRT₂ ln(V₃/V₄) W₃ = -Q₂
4 → 1 Adiabatic Compression Gas compressed, T rises to T₁ Q = 0 W₄ = -nCᵥ(T₁-T₂)

8.2 Carnot Engine Efficiency

Carnot Efficiency (Maximum Possible)

\[\eta_{\text{Carnot}} = 1 - \frac{T_2}{T_1} = \frac{T_1 - T_2}{T_1}\]

Key Properties:

  • Depends only on temperatures
  • T must be in Kelvin
  • η = 1 only if T₂ = 0 K (impossible)
  • No real engine can exceed this

For Carnot Cycle:

\[\frac{Q_1}{T_1} = \frac{Q_2}{T_2}\]

This is the key relation for Carnot engine problems

📝 Solved Example 4 (JEE Main Type)

Question: A Carnot engine operates between a hot reservoir at 527°C and a cold reservoir at 127°C. Calculate the efficiency and work done per cycle if 1000 J of heat is absorbed from the hot reservoir.

Solution:

Given:

  • T₁ = 527°C = 800 K
  • T₂ = 127°C = 400 K
  • Q₁ = 1000 J

Carnot efficiency:

\[\eta = 1 - \frac{T_2}{T_1} = 1 - \frac{400}{800} = 1 - \frac{1}{2} = \frac{1}{2} = 50\%\]

Work done per cycle:

\[W = \eta \times Q_1 = 0.5 \times 1000 = 500 \text{ J}\]

Heat rejected:

\[Q_2 = Q_1 - W = 1000 - 500 = 500 \text{ J}\]
\[\eta = 50\%, \quad W = 500 \text{ J}, \quad Q_2 = 500 \text{ J}\]

💡 JEE Important Points

  • Carnot engine is reversible - running it backwards gives ideal refrigerator
  • All reversible engines operating between same temperatures have same efficiency
  • To increase efficiency: Increase T₁ or decrease T₂
  • Carnot refrigerator COP: β = T₂/(T₁ - T₂)
  • Real engine efficiency is always less than Carnot efficiency
9

Entropy

Entropy is a measure of disorder or randomness in a system. It provides another way to state the Second Law: natural processes tend to increase the total entropy of the universe.

9.1 Definition of Entropy

Change in Entropy

\[dS = \frac{dQ_{rev}}{T}\]

For a finite process (reversible):

\[\Delta S = \int \frac{dQ_{rev}}{T}\]

Properties:

  • Entropy is a state function
  • SI unit: J/K
  • Extensive property

Important Note:

ΔS depends only on initial and final states, not on path. But to calculate ΔS, we use reversible path.

9.2 Entropy Change in Various Processes

Process Entropy Change (ΔS)
Isothermal (Ideal Gas) \(\Delta S = nR \ln\frac{V_2}{V_1} = nR \ln\frac{P_1}{P_2}\)
Isobaric \(\Delta S = nC_p \ln\frac{T_2}{T_1}\)
Isochoric \(\Delta S = nC_v \ln\frac{T_2}{T_1}\)
Adiabatic Reversible \(\Delta S = 0\) (isentropic)
Phase Change at constant T \(\Delta S = \frac{mL}{T}\) or \(\frac{Q}{T}\)
Heating (constant c) \(\Delta S = mc \ln\frac{T_2}{T_1}\)

9.3 Second Law in Terms of Entropy

Entropy Form of Second Law

For an Isolated System:

\[\Delta S_{universe} \geq 0\]
  • ΔS = 0 for reversible processes
  • ΔS > 0 for irreversible processes

Universe Entropy:

\[\Delta S_{universe} = \Delta S_{system} + \Delta S_{surroundings}\]

Natural processes always increase total entropy

📝 Solved Example 5

Question: Calculate the entropy change when 1 mole of ideal gas expands isothermally to double its volume at 300 K.

Solution:

Given:

  • n = 1 mol
  • V₂ = 2V₁
  • Isothermal process

For isothermal process:

\[\Delta S = nR \ln\frac{V_2}{V_1}\]
\[\Delta S = 1 \times 8.314 \times \ln(2)\]
\[\Delta S = 8.314 \times 0.693\]
\[\Delta S = 5.76 \text{ J/K}\]

Positive entropy change indicates increased disorder (gas occupies more volume).

💡 Physical Interpretation of Entropy

Entropy measures disorder:

  • Gas > Liquid > Solid (entropy)
  • Higher T → Higher entropy
  • Larger volume → Higher entropy
  • Mixing increases entropy

Examples of entropy increase:

  • Ice melting (ΔS > 0)
  • Free expansion of gas (ΔS > 0)
  • Heat flow from hot to cold (ΔS_universe > 0)
  • Mixing of gases (ΔS > 0)

📝 Previous Year Questions Analysis (2015-2024)

JEE Main Pattern

  • ✓ First Law Applications: 30%
  • ✓ Work Done in Processes: 25%
  • ✓ Heat Engine & Carnot: 25%
  • ✓ Cp, Cv & γ: 15%
  • ✓ Entropy: 5%

JEE Advanced Pattern

  • ✓ PV Diagram Analysis: 30%
  • ✓ Cyclic Process Problems: 25%
  • ✓ Combined Kinetic Theory: 20%
  • ✓ Carnot Engine & Entropy: 15%
  • ✓ Multi-concept Problems: 10%

Top 15 Most Repeated Question Types

  1. Work done in cyclic process from PV diagram (area calculation)
  2. First Law application - finding Q, W, or ΔU given other two
  3. Carnot engine efficiency and work output
  4. Adiabatic process - final pressure/temperature calculation
  5. Isothermal expansion work calculation
  6. Ratio of work done in different processes (isothermal vs adiabatic)
  7. Heat capacity Cp, Cv for mixture of gases
  8. Finding γ for unknown gas from process equations
  9. Efficiency comparison of different heat engines
  10. Refrigerator COP calculations
  11. Slope comparison on PV diagram (adiabatic vs isothermal)
  12. Free expansion problems (Q = W = 0, ΔU = 0)
  13. Entropy change in irreversible processes
  14. Work done in polytropic process (PVⁿ = constant)
  15. Heat engine running in reverse (refrigerator/heat pump)

Weightage Analysis

JEE Main: 8-12 marks (2-3 questions)
JEE Advanced: 12-18 marks (3-5 questions)
Difficulty Level: Medium to Hard
Time Required: 20-25 hours study

Year-wise Trend Analysis

Year JEE Main Focus JEE Advanced Focus
2024 Carnot efficiency, Work in isothermal Complex PV cycles, Entropy change
2023 First Law, Adiabatic compression Multi-process cycles, Heat engine chains
2022 Cp/Cv for mixtures, Work done Reversibility, Carnot with real gas
2021 Cyclic process area, Efficiency Combined thermo + kinetic theory
2020 Isothermal vs Adiabatic work ratio Entropy in irreversible process
Download Complete PYQ PDF (2015-2024) with Solutions

🎯 Practice Problem Set

Level 1: Basic (JEE Main Foundation)

  1. A gas absorbs 200 J of heat and does 50 J of work. Find ΔU.
  2. In an isochoric process, 100 J of heat is added to a gas. Find work done and ΔU.
  3. Calculate work done when 2 moles of ideal gas expand isobarically from 10 L to 20 L at 1 atm.
  4. Find work done in isothermal expansion of 1 mol gas from V to 2V at 300 K.
  5. A Carnot engine operates between 500 K and 300 K. Find its efficiency.
  6. For a monoatomic ideal gas, find Cp, Cv, and γ.
  7. In a cyclic process, if Q₁ = 500 J absorbed and Q₂ = 300 J rejected, find work done.
  8. A refrigerator has COP = 4. If work input is 100 J, find heat extracted from cold reservoir.

Level 2: Intermediate (JEE Main Standard)

  1. A diatomic gas (γ = 1.4) at pressure P and volume V is compressed adiabatically to V/4. Find the final pressure.
  2. In a process PV² = constant, find the molar heat capacity in terms of R.
  3. Two moles of helium at 27°C undergo isothermal expansion to triple the volume. Find Q, W, and ΔU.
  4. A Carnot engine absorbs 1000 J from source at 627°C and rejects to sink at 27°C. Find work done and heat rejected.
  5. One mole of an ideal gas goes from state A(P, V) to state B(2P, 2V). Find ΔU, W, Q if the path is: (a) A→C→B (isobaric then isochoric) (b) A→D→B (isochoric then isobaric)
  6. A mixture contains 2 moles of O₂ and 3 moles of Ar. Find γ for the mixture.
  7. An ideal gas expands adiabatically from (P₁, V₁, T₁) to (P₂, V₂, T₂). Show that W = nCv(T₁ - T₂).
  8. The efficiency of a heat engine is 1/6. When temperature of sink is reduced by 62°C, efficiency doubles. Find T₁ and T₂.

Level 3: Advanced (JEE Advanced/Olympiad)

  1. An ideal gas undergoes a cycle consisting of: isothermal expansion A→B, isobaric compression B→C, and isochoric heating C→A. Draw PV diagram and derive expression for efficiency in terms of V_A, V_B, and γ.
  2. A monoatomic gas undergoes process where P = P₀(1 - V²/V₀²). Find work done as gas expands from V = 0 to V = V₀.
  3. Two Carnot engines work in series. First engine takes Q₁ from T₁ and rejects to T₂. Second takes this rejected heat and rejects to T₃. If both have same efficiency, find T₂ in terms of T₁ and T₃.
  4. In a cyclic process shown as right-angled triangle on PV diagram with vertices at (P₀, V₀), (P₀, 2V₀), and (2P₀, V₀), calculate work done and heat absorbed if the process is clockwise.
  5. An ideal gas at temperature T is heated at constant pressure to temperature 2T. Then it's compressed isothermally to original volume. Then cooled at constant volume to original temperature. Draw PV diagram and find net work in terms of nRT.
  6. A diatomic gas undergoes a process P = αV where α is constant. Find the molar specific heat for this process.
  7. Calculate the entropy change of the universe when 100 g of ice at 0°C is mixed with 100 g of water at 100°C in an isolated container. (L_f = 80 cal/g, c_water = 1 cal/g·°C)
  8. An ideal gas engine operates in a cycle consisting of two isobaric and two adiabatic processes. If P_high/P_low = 4 and γ = 1.5, find the efficiency of the engine.

Quick Practice: Formula Verification

Verify these relationships (useful for MCQ elimination):

  1. Show that for isobaric process: Q : ΔU : W = γ : (γ-1) : 1
  2. Prove that work done in adiabatic process = -ΔU
  3. For isothermal expansion to double volume: W = 0.693nRT
  4. Derive: (∂P/∂V)_adiabatic / (∂P/∂V)_isothermal = γ
  5. Show that for Carnot cycle: Q₁/T₁ = Q₂/T₂
Get Complete Solutions PDF with Detailed Working

📋 Quick Formula Reference Card

First Law

Q = ΔU + W

ΔU = nCvΔT (always)

W = ∫PdV

Isobaric (P = const)

W = PΔV = nRΔT

Q = nCpΔT

V/T = constant

Isochoric (V = const)

W = 0

Q = ΔU = nCvΔT

P/T = constant

Isothermal (T = const)

ΔU = 0

Q = W = nRT ln(V₂/V₁)

PV = constant

Adiabatic (Q = 0)

W = -ΔU = nCv(T₁-T₂)

PVᵞ = constant

TVᵞ⁻¹ = constant

Heat Engine

η = W/Q₁ = 1 - Q₂/Q₁

η_Carnot = 1 - T₂/T₁

β_ref = Q₂/W = T₂/(T₁-T₂)

Cp and Cv

Cp - Cv = R

Cv = fR/2

γ = Cp/Cv = 1 + 2/f

γ Values

Monoatomic: 5/3 = 1.67

Diatomic: 7/5 = 1.4

Polyatomic: 4/3 = 1.33

Entropy

ΔS = Q_rev/T

ΔS_universe ≥ 0

ΔS_iso = nR ln(V₂/V₁)

Related Physics Notes

Related Chapter

Kinetic Theory of Gases

Molecular theory, Pressure derivation, Degrees of freedom, Mean free path

Read Notes →
Foundation

Thermal Properties of Matter

Heat transfer, Calorimetry, Thermal expansion, Newton's cooling

Read Notes →
Next Topic

Heat Transfer

Conduction, Convection, Radiation, Stefan-Boltzmann law

Read Notes →
View All Physics Notes →

Thermodynamics - Complete Guide for JEE 2025-26

Why Thermodynamics is Most Important for JEE?

Thermodynamics is one of the highest-scoring chapters in JEE Physics with 8-12% weightage. It's conceptually straightforward once you understand the laws and processes.

  • First Law - Foundation for 60% of problems (Q = ΔU + W)
  • Thermodynamic Processes - Know all 4 types thoroughly
  • PV Diagrams - Work = Area under curve (visual problems)
  • Heat Engines - Efficiency calculations (guaranteed questions)
  • Carnot Engine - Maximum efficiency benchmark
  • Cp, Cv, γ - Essential for adiabatic process problems

In JEE Advanced, expect complex cyclic processes and combined problems with kinetic theory.

Master These Formulas for 100% Accuracy

1. The Golden Equation

Q = ΔU + W (First Law)

ΔU = nCvΔT for ALL processes of ideal gas

2. Process Equations

  • • Isothermal: PV = const, W = nRT ln(V₂/V₁)
  • • Adiabatic: PVᵞ = const, W = (P₁V₁-P₂V₂)/(γ-1)
  • • Isobaric: W = PΔV = nRΔT
  • • Isochoric: W = 0

3. Heat Engine Efficiency

η = 1 - Q₂/Q₁ = 1 - T₂/T₁ (Carnot)

4. Mayer's Relation

Cp - Cv = R

📚 How to Master Thermodynamics for JEE?

For JEE Main Students:

  1. Time Required: 15-18 days (3 hours/day)
  2. Master the First Law and sign conventions first
  3. Practice 100+ problems on work done in different processes
  4. Memorize γ values: 5/3 (mono), 7/5 (dia), 4/3 (poly)
  5. Focus on Carnot engine efficiency problems
  6. Learn to read PV diagrams and calculate area
  7. Solve all NCERT examples and exercises

For JEE Advanced Students:

  1. Time Required: 20-25 days (4 hours/day)
  2. Deep dive into polytropic processes (PVⁿ = const)
  3. Master complex cyclic processes with multiple stages
  4. Practice deriving efficiency for custom cycles
  5. Study entropy and Second Law applications
  6. Solve combined problems (thermo + kinetic theory)
  7. Practice integer-type and matrix-match questions
  8. Solve last 15 years JEE Advanced PYQs

⚠️ Top 10 Mistakes to Avoid in JEE Exam

  • ❌
    Sign convention confusion: W positive for expansion (work BY system), Q positive for heat absorbed
  • ❌
    Using ΔU = nCpΔT: Wrong! ΔU = nCvΔT for ALL processes
  • ❌
    Temperature in Celsius: Always use Kelvin in thermodynamics formulas
  • ❌
    Forgetting Q = 0 in adiabatic: So W = -ΔU (work done at expense of internal energy)
  • ❌
    Confusing η and COP: η < 1 always, but COP can be > 1
  • ❌
    Wrong γ value: Air is diatomic (γ = 1.4), not monoatomic
  • ❌
    Cyclic process ΔU: ΔU = 0 for complete cycle, so Q_net = W_net
  • ❌
    Carnot efficiency confusion: η = 1 - T₂/T₁ (not T₂/T₁)
  • ❌
    Work in cyclic process: W = Area enclosed (clockwise +, anticlockwise -)
  • ❌
    Free expansion: Q = W = 0, ΔU = 0, ΔT = 0 (not adiabatic compression!)

🔗 How Thermodynamics Connects with Other Topics

With Kinetic Theory of Gases:

Internal energy U = (f/2)nRT comes from kinetic theory. Cp, Cv values derived from degrees of freedom. Many JEE Advanced questions combine both chapters.

With Thermal Properties of Matter:

Specific heat, latent heat from thermal properties are used in thermodynamics. Calorimetry problems often use First Law concepts.

With Chemistry (Physical Chemistry):

Enthalpy (H = U + PV), Gibbs free energy, Hess's law - all based on thermodynamics. Sign conventions differ - be careful!

With Mechanical Properties:

Bulk modulus appears in sound velocity formula with γ. Adiabatic bulk modulus B_ad = γP.

⚡ Last Minute Revision Points (Night Before Exam)

Laws & Definitions

  • • Zeroth Law: Thermal equilibrium
  • • First Law: Q = ΔU + W
  • • Second Law: η < 1 always
  • • Internal energy is state function
  • • Q, W are path functions

Key Numbers to Remember

  • • R = 8.314 J/mol·K
  • • γ_mono = 5/3 = 1.67
  • • γ_dia = 7/5 = 1.4
  • • ln(2) = 0.693
  • • ln(10) = 2.303

Quick Checks

  • • Isothermal: ΔU = 0, Q = W
  • • Adiabatic: Q = 0, W = -ΔU
  • • Isochoric: W = 0, Q = ΔU
  • • Isobaric: All three non-zero
  • • Cyclic: ΔU = 0, Q = W

🧠 Complete Chapter Mind Map

THERMODYNAMICS

Zeroth Law

Thermal Equilibrium

Temperature Definition

First Law

Q = ΔU + W

Energy Conservation

Processes

Isothermal, Adiabatic

Isobaric, Isochoric

Second Law

Heat Engines

Entropy, Carnot

Cp - Cv = R

η = 1 - T₂/T₁

ΔS ≥ 0