PhysicsNCERT Class 11
🔥

Thermodynamics Notes

Study Notes

6 Topics32 Formulas42 PYQs6 Videos45 Key Points

Topics

6
1

📖 1. Chapter Overview

Overview

Thermodynamics studies heat, work, temperature and internal energy in macroscopic systems such as gases, engines and refrigerators. It begins with thermal equilibrium and the zeroth law, which gives the scientific basis of temperature measurement. The first law of thermodynamics is energy conservation applied to heat and work: heat supplied to a system changes internal energy and may do work. Different thermodynamic processes such as isothermal, adiabatic, isobaric and isochoric processes are best understood using PV diagrams. The second law explains why natural processes have a direction and why heat engines cannot convert all absorbed heat into work. Entropy measures disorder or energy dispersal. Carnot engine gives the maximum possible efficiency between two temperatures, making this chapter very important for NEET.

Key Points7
  • 1Thermodynamics deals with macroscopic variables like pressure, volume, temperature and internal energy.
  • 2A thermodynamic system may be open, closed or isolated depending on exchange of matter and energy.
  • 3Work done by gas is positive during expansion and negative during compression under the common NEET convention.
  • 4Isothermal ideal gas process has ΔU = 0 because temperature is constant.
  • 5Adiabatic process has ΔQ = 0 because no heat is exchanged.
  • 6Entropy of an isolated system never decreases in natural processes.
  • 7Carnot engine is reversible and has maximum efficiency for given source and sink temperatures.
Memory Tricks2

First Law

Heat given becomes internal energy plus work: Q goes to U and W.

Processes

Iso means same: isothermal same temperature, isobaric same pressure, isochoric same volume.

Examples2

Daily Life Example

A car engine takes heat from burning fuel, converts part of it into mechanical work and rejects remaining heat to surroundings.

NEET Quick Example

If 500 J heat is supplied to a gas and it does 200 J work, then ΔU = 500 - 200 = 300 J.

Reference Tables2
Sign up
Common Mistakes3

Confusing Heat with Internal Energy

Heat is energy in transit due to temperature difference; internal energy belongs to the state of the system.

Using Celsius in Carnot Efficiency

Carnot efficiency requires absolute temperatures in kelvin.

Ignoring Sign Convention

Under NEET convention, work done by gas during expansion is positive and during compression is negative.

Formula Cards4
First Law of Thermodynamics

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

Variables

ΔQ=

Heat supplied to the system

ΔU=

Change in internal energy

ΔW=

Work done by the system

Work Done by Gas

Work done by a gas equals the area under the PV curve.

Variables

W=

Work done by gas

P=

Pressure

dV=

Small change in volume

Diagrams3
Sign up
Quick Revision
2

🌡️ 2. Thermal Equilibrium & Zeroth Law

Overview

Thermal equilibrium is the condition in which two bodies in thermal contact do not exchange net heat. This happens when they have the same temperature. Temperature is therefore the property that decides the direction of heat flow. The zeroth law of thermodynamics states that if two systems are separately in thermal equilibrium with a third system, then they are in thermal equilibrium with each other. This law makes temperature measurement possible because a thermometer acts as the third system. Thermodynamic systems may be open, closed or isolated depending on exchange of matter and energy. A state of a system is described by state variables such as pressure, volume, temperature and internal energy.

Key Points6
  • 1Thermal equilibrium does not mean equal internal energy; it means equal temperature.
  • 2Temperature is a state variable.
  • 3Pressure, volume, temperature and internal energy define thermodynamic state for many systems.
  • 4State variables depend only on current state, not on path.
  • 5A system plus surroundings together form the universe in thermodynamics.
  • 6Zeroth law logically precedes the first and second laws because it defines temperature comparison.
Memory Tricks2

Zeroth Law

If A matches C and B matches C, then A matches B.

System Types

Open exchanges all, closed exchanges energy only, isolated exchanges nothing.

Examples2

Solved Concept Example

If a thermometer is in equilibrium with water and the same thermometer is in equilibrium with oil, both water and oil have the same temperature.

System Example

Gas in a sealed conducting cylinder is a closed system because it can exchange heat but not matter.

Reference Tables2
Sign up
Common Mistakes3

Confusing Thermal Equilibrium with Equal Heat

Thermal equilibrium means equal temperature, not equal heat or equal internal energy.

Calling Heat a State Variable

Heat is a path function, not a property stored in a system.

Ignoring System Boundary

Whether a system is open, closed or isolated depends on what boundary you choose.

Formula Cards3
Thermal Equilibrium Condition

Two bodies A and B are in thermal equilibrium when their temperatures are equal.

Variables

TA=

Temperature of body A

TB=

Temperature of body B

Zeroth Law Statement

If two systems are each in thermal equilibrium with a third, they are in equilibrium with each other.

Variables

TA, TB, TC=

Temperatures of systems A, B and C

Diagrams3
Sign up
Quick Revision
3

🔥 3. Heat, Work & Internal Energy

Overview

Heat, work and internal energy are central to thermodynamics. Heat is energy transferred due to temperature difference. Work is energy transferred when a system expands or compresses against external pressure. Internal energy is the total microscopic kinetic and potential energy of molecules in the system. For an ideal gas, internal energy depends only on temperature. Heat and work are path functions because their values depend on the process path, not only initial and final states. Internal energy is a state function because its change depends only on initial and final states. In the common NEET convention, heat supplied to the system is positive and work done by the system during expansion is positive.

Key Points6
  • 1A system does not contain heat; heat crosses the boundary during a process.
  • 2PV work occurs when gas volume changes.
  • 3At constant volume, work done by gas is zero.
  • 4Area under PV curve gives work done by gas.
  • 5Internal energy of ideal gas depends only on temperature, not pressure or volume separately.
  • 6Same initial and final states can have different heat and work values along different paths.
Memory Tricks2

State vs Path

State is where you are; path is how you got there. U is state, Q and W are path.

Work Sign

Gas expands and gives work out, so W is positive.

Examples2

Numerical Problem

A gas expands at constant pressure 2 × 10^5 Pa from 1 L to 3 L. W = PΔV = 2×10^5×2×10^-3 = 400 J.

Path Function Example

A gas can go from the same initial state to the same final state by different paths, giving different work values, but the same ΔU.

Reference Tables2
Sign up
Common Mistakes3

Calling Heat a State Function

Heat depends on the path of energy transfer, so it is not a state function.

Forgetting Work Is Zero at Constant Volume

If volume does not change, dV = 0 and W = 0.

Assuming Internal Energy Depends on Volume for Ideal Gas

For an ideal gas, internal energy depends only on temperature.

Formula Cards5
Work Done by Gas

Work done by gas during volume change equals area under PV curve.

Variables

W=

Work done by gas

P=

Gas pressure

dV=

Small change in volume

Constant Pressure Work

Work done during an isobaric process.

Variables

P=

Constant pressure

V1=

Initial volume

V2=

Final volume

Internal Energy Change of Ideal Gas

Change in internal energy of ideal gas depends only on temperature change.

Variables

n=

Number of moles

Cv=

Molar heat capacity at constant volume

ΔT=

Temperature change

Diagrams3
Sign up
Quick Revision
4

⚙️ 4. First Law of Thermodynamics

Overview

The first law of thermodynamics is the law of conservation of energy applied to thermodynamic systems. It states that heat supplied to a system is used partly to increase its internal energy and partly to do work by the system. Mathematically, ΔQ = ΔU + ΔW under the common convention where work done by the system is positive. This law shows that heat and work are two ways of transferring energy across a system boundary. It is used in every thermodynamic process: isothermal, adiabatic, isochoric and isobaric. In numerical problems, identifying which quantity is zero or known is the main step. It is one of the most important NEET formulas.

Key Points6
  • 1The first law does not tell the direction of natural processes; that is done by the second law.
  • 2For an ideal gas, ΔU depends only on ΔT.
  • 3For isothermal ideal gas process, Q = W.
  • 4For adiabatic process, ΔU = -W.
  • 5For isochoric process, Q = ΔU.
  • 6For cyclic process, heat absorbed over cycle equals net work done.
Memory Tricks2

First Law Rearrangement

If Q = U + W, then U = Q - W. Subtract work done by gas from heat supplied.

Zero Conditions

Isochoric means W zero, isothermal ideal gas means U zero, adiabatic means Q zero.

Examples2

Solved Example

A gas absorbs 800 J heat and does 300 J work. From ΔQ = ΔU + ΔW, ΔU = 800 - 300 = 500 J.

Previous NEET Style

In a cyclic process, if gas absorbs 1000 J heat in total, net work done by gas is also 1000 J because ΔU = 0.

Reference Tables2
Sign up
Common Mistakes3

Wrong Work Sign

Expansion work by gas is positive; compression work by gas is negative.

Forgetting ΔU = 0 in a Cycle

Internal energy is a state function, so it returns to initial value after a complete cycle.

Applying ΔU = 0 to All Isothermal Processes

ΔU = 0 for isothermal ideal gas, because ideal gas internal energy depends only on temperature.

Formula Cards5
First Law

Heat supplied equals change in internal energy plus work done by the system.

Variables

ΔQ=

Heat supplied to system

ΔU=

Change in internal energy

ΔW=

Work done by system

Internal Energy of Ideal Gas

Internal energy change of an ideal gas depends only on temperature change.

Variables

n=

Number of moles

Cv=

Molar heat capacity at constant volume

ΔT=

Temperature change

Isochoric Application

At constant volume, no boundary work is done.

Variables

ΔV=

Change in volume

W=

Work done

Diagrams3
Sign up
Quick Revision
5

📈 5. Thermodynamic Processes

Overview

A thermodynamic process is a path by which a system changes from one state to another. In an isothermal process, temperature remains constant; for an ideal gas, internal energy remains constant and heat supplied equals work done. In an adiabatic process, no heat is exchanged with surroundings, so work is done at the cost of internal energy. In an isochoric process, volume remains constant and work done is zero. In an isobaric process, pressure remains constant and work is PΔV. PV diagrams visually show these processes, and the area under a PV curve gives work done by gas. NEET frequently asks process identification, work comparison and first-law applications.

Key Points7
  • 1For ideal gas in isothermal process, ΔU = 0.
  • 2For adiabatic process, PV^γ = constant.
  • 3For isochoric process, heat supplied only changes internal energy.
  • 4For isobaric process, heat changes internal energy and also does work.
  • 5Expansion gives positive area and positive work by gas.
  • 6Compression gives negative work by gas.
  • 7Adiabatic expansion cools the gas; adiabatic compression heats it.
Memory Tricks2

Iso Processes

Thermal = temperature, baric = pressure, choric = volume.

Adiabatic

A-diabatic means no heat can pass across the boundary.

Examples2

Isobaric Work Example

A gas expands at pressure 10^5 Pa from 2 L to 5 L. W = PΔV = 10^5 × 3 × 10^-3 = 300 J.

Isochoric Example

If a gas is heated in a rigid container, volume remains constant, so W = 0 and Q = ΔU.

Reference Tables3
Sign up
Common Mistakes3

Confusing Isothermal and Adiabatic

Isothermal keeps temperature constant; adiabatic keeps heat exchange zero.

Forgetting Isochoric Work Is Zero

Work needs volume change, so at constant volume W = 0.

Using PV = constant for Adiabatic

PV = constant is isothermal for ideal gas; adiabatic uses PV^γ = constant.

Formula Cards5
Isothermal Ideal Gas Work

Work done by ideal gas during reversible isothermal process.

Variables

W=

Work done by gas

n=

Number of moles

R=

Universal gas constant

T=

Constant absolute temperature

V1, V2=

Initial and final volumes

Adiabatic Relation

Pressure-volume relation for reversible adiabatic process of an ideal gas.

Variables

P=

Pressure

V=

Volume

γ=

Ratio of heat capacities Cp/Cv

Adiabatic Work

Work done by gas in a reversible adiabatic process.

Variables

P1, V1=

Initial pressure and volume

P2, V2=

Final pressure and volume

γ=

Adiabatic index

Diagrams3
Sign up
Quick Revision
6

♻️ 6. Second Law & Entropy

Overview

The first law tells us energy is conserved, but it does not tell which processes are naturally possible. The second law of thermodynamics gives the direction of natural processes. One statement says heat cannot spontaneously flow from a colder body to a hotter body without external work. Another says no heat engine can convert all absorbed heat into work in a complete cycle. Reversible processes are ideal and can be exactly retraced, while irreversible processes occur naturally with friction, heat flow through finite temperature difference, mixing and free expansion. Entropy is a state function that measures energy dispersal or disorder. For an isolated system, entropy never decreases, which explains why natural processes are irreversible.

Key Points6
  • 1The second law does not violate the first law; it adds direction and limitations.
  • 2A heat engine must reject some heat to a sink.
  • 3Friction and unrestrained expansion make processes irreversible.
  • 4Entropy change of the universe is zero for reversible processes and positive for irreversible processes.
  • 5Heat cannot flow from cold to hot unless external work is supplied, as in a refrigerator.
  • 6Natural processes move toward greater total entropy.
Memory Tricks2

Second Law

Energy is conserved by first law; direction is decided by second law.

Entropy

Entropy tracks spreading: energy and matter naturally spread out.

Examples2

Concept Example

A hot cup of tea cools in a room, but a cold cup never spontaneously becomes hotter than the room without external work.

Entropy Example

When ice melts in a warm room, entropy of ice-water system may increase, and total entropy of universe increases.

Reference Tables3
Sign up
Common Mistakes3

Thinking First Law Allows 100% Engine

First law allows energy conservation, but second law forbids complete conversion of heat into work in a cycle.

Saying Entropy Always Increases for System Alone

Entropy of a system can decrease, but entropy of the universe cannot decrease.

Calling Fast Real Processes Reversible

Reversible processes are ideal, quasi-static and free from dissipative effects.

Formula Cards4
Entropy Change for Reversible Heat Transfer

Entropy change when heat Qrev is reversibly transferred at absolute temperature T.

Variables

ΔS=

Change in entropy

Qrev=

Heat transferred reversibly

T=

Absolute temperature

Engine Efficiency

Efficiency of a heat engine in terms of work output and heat exchanges.

Variables

η=

Efficiency of engine

W=

Work output

Q1=

Heat absorbed from source

Q2=

Heat rejected to sink

Diagrams3
Sign up
Quick Revision
7

🚗 7. Carnot Engine

Overview

A Carnot engine is an ideal reversible heat engine operating between two reservoirs at temperatures T1 and T2. Its cycle consists of two isothermal and two adiabatic processes. During isothermal expansion at the hot reservoir, the gas absorbs heat Q1. During adiabatic expansion, its temperature falls from T1 to T2. During isothermal compression at the cold reservoir, it rejects heat Q2. During adiabatic compression, its temperature rises back to T1. Carnot theorem states that no engine working between the same two temperatures can be more efficient than a Carnot engine. Its efficiency is η = 1 - T2/T1. A refrigerator is a reversed heat engine that uses work to transfer heat from cold to hot.

Key Points7
  • 1Carnot engine gives maximum theoretical efficiency, not practical engine efficiency.
  • 2A Carnot engine can never have 100% efficiency unless sink temperature is 0 K, which is unattainable.
  • 3The working substance returns to its initial state after each cycle, so ΔU = 0 per cycle.
  • 4Net work per cycle equals Q1 - Q2.
  • 5On PV diagram, area enclosed by cycle gives net work done.
  • 6On TS diagram, heat transfer during reversible process equals area under T-S curve.
  • 7Coefficient of performance is used for refrigerators and heat pumps instead of efficiency.
Memory Tricks2

Carnot Cycle Order

Hot iso expansion, adiabatic cooling, cold iso compression, adiabatic heating.

Efficiency

Carnot efficiency depends only on temperatures, not on working gas.

Examples3

Solved Numerical

A Carnot engine works between 600 K and 300 K. Efficiency η = 1 - 300/600 = 0.5 = 50%.

Refrigerator Example

A Carnot refrigerator works between 300 K and 270 K. COP = T2/(T1 - T2) = 270/(30) = 9.

Previous NEET Style

If source temperature is doubled while sink temperature remains same, Carnot efficiency increases because T2/T1 decreases.

Reference Tables3
Sign up
Common Mistakes3

Using Celsius in η = 1 - T2/T1

Always convert source and sink temperatures to kelvin.

Expecting 100% Efficiency

A heat engine must reject some heat to a sink; 100% efficiency is impossible for finite sink temperature.

Confusing Refrigerator COP with Engine Efficiency

Refrigerator performance is Q removed per work input, not work output per heat input.

Formula Cards6
Carnot Efficiency

Maximum possible efficiency of a heat engine between hot reservoir T1 and cold reservoir T2.

Variables

η=

Efficiency of Carnot engine

T1=

Temperature of source in kelvin

T2=

Temperature of sink in kelvin

Heat Engine Efficiency

Efficiency of any heat engine in terms of heat absorbed, heat rejected and work output.

Variables

W=

Work output per cycle

Q1=

Heat absorbed from hot reservoir

Q2=

Heat rejected to cold reservoir

Carnot Heat Ratio

For a reversible Carnot cycle, heat exchanged is proportional to reservoir temperature.

Variables

Q1, Q2=

Heat absorbed and heat rejected

T1, T2=

Absolute temperatures of reservoirs

Diagrams4
Sign up
Quick Revision

Formula Sheet

10
First Law of Thermodynamics

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

Variables

ΔQ=

Heat supplied to the system

ΔU=

Change in internal energy

ΔW=

Work done by the system

Work Done by Gas

Work done by a gas equals the area under the PV curve.

Variables

W=

Work done by gas

P=

Pressure

dV=

Small change in volume

Ideal Gas Internal Energy

For an ideal gas, internal energy change depends only on temperature change.

Variables

n=

Number of moles

Cv=

Molar heat capacity at constant volume

ΔT=

Change in temperature

Carnot Efficiency

Maximum possible efficiency of a heat engine operating between source temperature T1 and sink temperature T2.

Variables

η=

Efficiency

T1=

Absolute temperature of hot reservoir

T2=

Absolute temperature of cold reservoir

Thermal Equilibrium Condition

Two bodies A and B are in thermal equilibrium when their temperatures are equal.

Variables

TA=

Temperature of body A

TB=

Temperature of body B

5 more formulas locked

Sign up free to access all formulas with variables and explanations.

Quick Revision

12 Sign up to access

Unlock 12 Quick Revision Points

Sign up free to access all content, practice PYQs, and get AI explanations.

Learning Videos

6 Sign up to access

Unlock 6 Learning Videos

Sign up free to access all content, practice PYQs, and get AI explanations.

NEET PYQs — Thermodynamics

42 Sign up to access

Showing 3 of 42 questions. Sign up to practice all with answers, explanations, and AI help.

NEET 2026Set 11EasyQ1

At a certain temperature T(K), during a process, 500 J is absorbed by the system and work of 200 J is done by the system. Then change in internal energy of the system is:

NEET 2026Set 11MediumQ2

Consider the following reaction: 2A(g) + B(g) → 2D(g) ΔU° = −10 kJ mol⁻¹ and ΔS° = −44 J K⁻¹ at 298 K Identify the correct option with ΔG° for the reaction and spontaneity of the reaction at 298 K. (Given: R = 8.31 J mol⁻¹ K⁻¹)

NEET 2026Set 11EasyQ3

An electric heater supplies heat to a system at a rate of 100 W. If the system performs work at a rate of 75 J/s, then the rate at which internal energy increases will be:

Free for all NEET aspirants

Unlock the full Thermodynamics experience

All diagrams, videos, quick revision, PYQ practice with AI explanations — plus mock tests, flashcards, and a personalised study plan.