Thermodynamics – Class 11 Physics | NCERT Complete Guide
Chapter 11 · Physics

NCERT · Class XI · Physics

Thermodynamics

The science of heat, work, and energy transformations — from steam engines to black holes. Complete resource hub for CBSE Boards, JEE Main, JEE Advanced & NEET.

First LawSecond LawIsothermalAdiabaticCarnot EngineEntropyHeat EnginesRefrigeratorsQuasi-static Process
36
NCERT Questions
120+
MCQs
12
Core Concepts
25+
PYQs Covered

Everything You Need to Master Thermodynamics

🎯 Learning Objectives

  • Understand thermal equilibrium and the Zeroth Law
  • State and apply the First Law of Thermodynamics
  • Distinguish isothermal, adiabatic, isobaric & isochoric processes
  • Analyse heat engine efficiency using Carnot cycle
  • Understand entropy and the Second Law
  • Evaluate refrigerators and heat pumps

📚 Chapter At a Glance

  • Thermal equilibrium & Zeroth Law
  • Internal energy concept
  • First Law: ΔU = Q − W
  • Work done in thermodynamic processes
  • Specific heat capacities: Cₚ, Cᵥ, γ = Cₚ/Cᵥ
  • Carnot theorem & efficiency η = 1 − T₂/T₁
  • Second Law: Kelvin-Planck & Clausius statements
  • Entropy — a measure of disorder

Core Concepts Explained

01 Zeroth Law of Thermodynamics

If two systems are each in thermal equilibrium with a third system, they are in thermal equilibrium with each other. This law establishes the concept of temperature as a measurable property.

A ↔ C and B ↔ C ⟹ A ↔ B
02 First Law of Thermodynamics

Energy is conserved. The change in internal energy of a system equals the heat supplied to the system minus the work done by the system. It rules out the possibility of a perpetual motion machine of the first kind.

ΔU = Q − W
03 Isothermal Process

Temperature remains constant (T = const). For an ideal gas, internal energy does not change (ΔU = 0), so all heat absorbed equals work done by the gas. The curve on a PV diagram is a hyperbola.

W = nRT ln(V₂/V₁)
04 Adiabatic Process

No heat exchange with surroundings (Q = 0). The first law gives ΔU = −W. For rapid processes or thermally insulated systems. The PV curve is steeper than the isothermal curve.

PVᵞ = constant | W = (P₁V₁ − P₂V₂)/(γ−1)
05 Carnot Cycle & Efficiency

The Carnot engine operates between two heat reservoirs in a reversible cycle: isothermal expansion → adiabatic expansion → isothermal compression → adiabatic compression. It is the most efficient engine possible operating between those temperatures.

η = 1 − T_cold / T_hot = 1 − Q_cold / Q_hot
06 Second Law of Thermodynamics

Kelvin-Planck: No process is possible in which the sole result is the absorption of heat from a reservoir and conversion entirely into work. Clausius: Heat cannot spontaneously flow from a colder to a hotter body.

dS ≥ 0 (entropy of an isolated system never decreases)

Key Formulae at a Glance

Quantity Formula Remarks
First LawΔU = Q − WSign convention: Q +ve into system, W +ve by system
Work (isothermal)W = nRT ln(V₂/V₁)Ideal gas, T = constant
Work (isobaric)W = PΔV = P(V₂−V₁)Constant pressure
Work (adiabatic)W = (P₁V₁−P₂V₂)/(γ−1)No heat exchange
Adiabatic relationPVᵞ = const; TVᵞ⁻¹ = constγ = Cₚ/Cᵥ
Carnot efficiencyη = 1 − T₂/T₁T in Kelvin; T₂ < T₁
COP (refrigerator)COP = T₂/(T₁−T₂)Ideal; T₂ = cold reservoir
Entropy changeΔS = Q_rev / TReversible process
Mayer's relationCₚ − Cᵥ = RIdeal gas
Board & Competitive Exam Tips
  • Always state the sign convention clearly in First Law problems
  • Carnot efficiency formula requires absolute (Kelvin) temperatures
  • Adiabatic curve is always steeper than isothermal on a PV diagram
  • Second Law statements (both) are frequently asked in MCQs
  • For NEET: focus on Cₚ, Cᵥ, γ values for monoatomic & diatomic gases
  • PYQ trend: Entropy conceptuals appear almost every JEE Advanced year
  • Draw PV diagrams to visualise work done as enclosed area
  • Reversible vs Irreversible processes — common 2-mark board question

PYQ Highlights

JEE 2023
An ideal gas undergoes a cyclic process. The work done by the gas in the cycle is 500 J. The heat absorbed in the isothermal part is 800 J. Find the heat rejected during the process.
JEE Main · 4 marks
NEET 2022
A Carnot engine operates between 600 K and 300 K. If it absorbs 800 J from the hot reservoir, the work done per cycle is:
NEET · 1 mark MCQ
CBSE 2024
State the Second Law of Thermodynamics (Clausius statement). Why is a perfect refrigerator not possible?
Board Exam · 3 marks
JEE Adv 2021
One mole of an ideal monoatomic gas undergoes an adiabatic expansion. If temperature falls from 300 K to 200 K, find the work done by the gas. (Cᵥ = 12.5 J/mol·K)
JEE Advanced · 3 marks
CBSE 2023
Draw a labelled PV diagram of the Carnot cycle and derive an expression for its thermal efficiency.
Board Exam · 5 marks
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