NEET Physics · NCERT Class 11

Thermodynamics notes

Thermal Equilibrium, Temperature & Zeroth Law, Thermodynamic Systems, State Variables & Equation of State, Heat, Work & Internal Energy, First Law of Thermodynamics, Thermodynamic Processes: Isothermal, Adiabatic, Isobaric, Isochoric & Cyclic, Heat Capacity, Specific Heat & Mayer Relation, Second Law, Entropy & Irreversibility, Heat Engines, Refrigerators & Heat Pumps, Carnot Engine & Carnot Cycle, Calorimetry, Latent Heat & Thermal Exchange and NEET Problem Strategy: Signs, Graphs & Concept Traps — explained topic by topic with key points, formulas, memory tricks and common mistakes.

  • 11 topics
  • 83 min read
  • 35 formulas
  • 49 PYQs
1

Thermal Equilibrium, Temperature & Zeroth Law

6 min

Thermal equilibrium is the condition in which two bodies in contact stop exchanging net heat because they have the same temperature. Temperature is the physical quantity that tells the direction of heat flow: heat flows naturally from higher temperature to lower temperature. The Zeroth Law states that if two systems are separately in thermal equilibrium with a third system, they are in thermal equilibrium with each other. This law makes temperature measurement possible because a thermometer can act as the third system. For NEET, remember that thermal equilibrium does not mean equal heat content or equal internal energy; it means equal temperature and no net heat flow.

Key points5

  • Temperature is a state variable, while heat is energy in transit.
  • A thermometer works by reaching thermal equilibrium with the body whose temperature is measured.
  • Two objects can have the same temperature but different masses, materials and internal energies.
  • The Zeroth Law is called zeroth because it is more fundamental than the first and second laws.
  • Thermal equilibrium is a macroscopic condition; microscopic molecular motion still continues.

Memory tricks2

Zeroth Law Shortcut: Think: Same as C, same as each other. Thermometer C certifies equality of temperature.

Temperature vs Heat: Temperature tells direction; heat is the energy in motion.

Examples2

Thermometer Example

A clinical thermometer reads body temperature only after it reaches thermal equilibrium with the body.

Metal Spoon in Tea

A spoon in hot tea becomes warm because heat flows from tea to spoon until thermal equilibrium is reached.

Common mistakes2

Confusing Equal Temperature with Equal Heat

A bucket and a cup of water at 30°C have the same temperature, but the bucket contains more internal energy.

Using Celsius in Thermodynamic Formulae

Gas-law and Carnot calculations require kelvin, not Celsius.

Reference tables2

Quick revision6

Formulas2

Celsius to Kelvin Conversion
T(K)=t(°C)+273.15T(K) = t(°C) + 273.15

Used to convert Celsius temperature into absolute temperature scale used in thermodynamics.

T
absolute temperature in kelvin
t
temperature in degree Celsius
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2

Thermodynamic Systems, State Variables & Equation of State

7 min

A thermodynamic system is the part of the universe chosen for study, while everything outside it is the surroundings. Systems may be open, closed or isolated depending on exchange of mass and energy. The macroscopic condition of a system is described by state variables such as pressure, volume, temperature and internal energy. A state function depends only on the state, not on the path followed. For an ideal gas, pressure, volume and temperature are connected by the equation of state PV = nRT. NEET questions often test whether a quantity is a state variable or a path variable, and whether the system is open, closed or isolated.

Key points5

  • A thermodynamic state is fixed when enough independent state variables are known.
  • For a fixed amount of ideal gas, any two of P, V and T determine the third.
  • Equilibrium states are represented as points on a P-V diagram.
  • Quasi-static processes pass through a series of equilibrium states.
  • Internal energy of an ideal gas depends only on temperature.

Memory tricks2

Open-Closed-Isolated: Open opens everything; closed closes mass only; isolated isolates both mass and energy.

State vs Path: State is address; path is route. U has an address, Q and W depend on the route.

Examples2

Pressure Cooker

A pressure cooker is nearly a closed system during cooking because mass exchange is small until steam escapes.

Human Body

The human body is an open thermodynamic system because it exchanges food, gases, heat and work with surroundings.

Common mistakes2

Calling Heat a State Variable

A system does not contain heat; heat is energy transferred because of temperature difference.

Ignoring Kelvin

Always use absolute temperature in PV = nRT.

Reference tables2

Quick revision7

Formulas3

Ideal Gas Equation
PV=nRTPV = nRT

Equation of state for an ideal gas connecting pressure, volume, amount and temperature.

P
pressure of gas
V
volume of gas
n
number of moles
R
universal gas constant
T
absolute temperature
Molecular Form of Ideal Gas Equation
PV= NkBTPV =\text{ NkBT}

Useful when the number of molecules is given instead of moles.

N
number of molecules
kB
Boltzmann constant
T
absolute temperature
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3

Heat, Work & Internal Energy

8 min

Heat, work and internal energy are central quantities in thermodynamics. Internal energy is the total microscopic energy of molecules due to random motion and interactions. Heat is energy transferred between system and surroundings because of temperature difference. Work is energy transferred when a macroscopic force acts through displacement, such as gas pushing a piston. In thermodynamics, work done by a gas during expansion is positive in the NCERT convention commonly used for physics: W = ∫P dV. Heat and work are path dependent, but change in internal energy is path independent. For an ideal gas, internal energy depends only on temperature.

Key points5

  • Internal energy includes kinetic and potential energies of microscopic particles.
  • For an ideal gas, intermolecular potential energy is neglected.
  • Area under a P-V curve gives work done by the gas.
  • Work is zero in an isochoric process because volume does not change.
  • Heat supplied can increase internal energy, produce work, or both.

Memory tricks2

U is Inside: Internal energy is inside the gas; heat and work are ways energy crosses the boundary.

Expansion Positive: When volume expands, gas pushes the world: work by gas is positive.

Examples2

Bicycle Pump

Air becomes warm during compression because work is done on the gas, increasing its internal energy.

Steam Engine

Steam expands and pushes a piston, converting heat into mechanical work.

Common mistakes2

Saying Heat is Stored

A body stores internal energy, not heat. Heat exists only during transfer.

Missing Area Meaning

On P-V graphs, the area under the curve, not the slope, gives work.

Reference tables2

Quick revision6

Formulas3

Work Done by Gas
W=∫PdVW = ∫ P dV

The area under the curve in a P-V diagram gives work done by gas.

W
work done by gas
P
pressure of gas
dV
small change in volume
Constant Pressure Work
W=P(V2−V1)W = P(V2 - V1)

Work done by gas when pressure remains constant.

P
constant pressure
V1
initial volume
V2
final volume
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4

First Law of Thermodynamics

7 min

The First Law of Thermodynamics is the law of conservation of energy applied to thermal systems. In the NCERT physics sign convention, heat supplied to a system is used partly to increase internal energy and partly to do external work: Q = ΔU + W. If heat is supplied but no work is done, internal energy increases. If a gas expands without heat input, its internal energy decreases. The law does not tell whether a process is naturally possible; that is the role of the second law. NEET problems usually involve identifying signs of Q, W and ΔU for different processes and applying the equation carefully.

Key points5

  • The first law connects heat, work and internal energy quantitatively.
  • It applies to all thermodynamic processes.
  • Internal energy change is independent of path, while Q and W depend on path.
  • For ideal gas, ΔU depends only on temperature change.
  • In a cycle, the net heat supplied equals net work done.

Memory tricks2

First Law Sentence: Heat given = energy kept + work spent.

Cycle Shortcut: In a complete cycle, U comes home, so ΔU = 0.

Examples2

Heating Gas in a Cylinder

If gas absorbs 500 J heat and does 200 J work, its internal energy increases by 300 J.

Rapid Expansion

If gas expands adiabatically and does work, its temperature falls because internal energy decreases.

Common mistakes2

Wrong Sign of Work

Check whether the question uses work done by gas or work done on gas. NCERT physics commonly uses Q = ΔU + W, where W is work by gas.

Applying ΔU = nCvΔT to Any Substance

This simple relation is mainly used for ideal gases in this chapter.

Reference tables2

Quick revision7

Formulas3

First Law of Thermodynamics
Q=ΔU+WQ = ΔU + W

Heat supplied to a system equals increase in internal energy plus work done by the system.

Q
heat supplied to system
ΔU
change in internal energy
W
work done by system
Cyclic Process
ΔUcycle =0 and Qnet = WnetΔ\text{Ucycle }= 0\text{ and Qnet }=\text{ Wnet}

Since internal energy is a state function, it returns to the initial value after a complete cycle.

ΔUcycle
net change in internal energy over one cycle
Qnet
net heat absorbed
Wnet
net work done
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5

Thermodynamic Processes: Isothermal, Adiabatic, Isobaric, Isochoric & Cyclic

10 min

A thermodynamic process is a change of a system from one equilibrium state to another. The most important NEET processes are isothermal, adiabatic, isobaric, isochoric and cyclic. In an isothermal process, temperature remains constant and for an ideal gas ΔU = 0. In an adiabatic process, no heat is exchanged and temperature usually changes. In an isobaric process, pressure is constant; in an isochoric process, volume is constant and work is zero. A cyclic process returns to its initial state, so net change in internal energy is zero. P-V graphs are essential because the area under the curve gives work.

Key points5

  • Quasi-static processes are slow enough for the system to remain nearly in equilibrium.
  • Isothermal expansion needs heat input to keep temperature constant.
  • Adiabatic expansion cools the gas; adiabatic compression heats it.
  • No work is done in constant volume heating or cooling.
  • Clockwise P-V cycle gives positive net work by gas.

Memory tricks3

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

Adiabatic A: Adiabatic has A for Absent heat: Q = 0.

Vertical Volume: On P-V graph, vertical line means volume is fixed, so isochoric work is zero.

Examples2

Slow Expansion in Thermal Contact

A gas expanding slowly in contact with a heat reservoir is nearly isothermal.

Sound Wave Compression

Compression and rarefaction in sound waves are approximately adiabatic because they occur rapidly.

Common mistakes3

Mixing Isothermal and Adiabatic

Isothermal has constant temperature and heat exchange; adiabatic has no heat exchange and temperature generally changes.

Forgetting Work in Isobaric Process

At constant pressure, volume can change, so work is usually not zero.

Wrong Cycle Direction

Clockwise cycle gives positive work by gas; anticlockwise gives negative work by gas.

Reference tables2

Quick revision7

Formulas4

Isothermal Ideal Gas
PV= constantPV =\text{ constant}

For a fixed amount of ideal gas at constant temperature.

P
pressure
V
volume
Isothermal Work
W= nRT ln(V2/V1)W =\text{ nRT ln}(V2/V1)

Work done by an ideal gas during reversible isothermal expansion or compression.

n
number of moles
R
universal gas constant
T
absolute temperature
V1, V2
initial and final volumes
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6

Heat Capacity, Specific Heat & Mayer Relation

7 min

Heat capacity measures how much heat is required to raise the temperature of a body. Specific heat capacity is heat required per unit mass per unit temperature rise, while molar heat capacity is heat required per mole per unit temperature rise. For gases, heat required depends strongly on the process. At constant volume, supplied heat only increases internal energy, so Cv is used. At constant pressure, heat also does work of expansion, so Cp is greater than Cv. For an ideal gas, Mayer’s relation is Cp - Cv = R. The ratio γ = Cp/Cv is crucial in adiabatic processes and heat engine questions.

Key points5

  • Cp is greater than Cv for gases because expansion work is done at constant pressure.
  • Specific heat depends on material and sometimes temperature.
  • Molar heat capacity is preferred for gases in thermodynamics.
  • For monoatomic ideal gas, Cv = 3R/2 and Cp = 5R/2.
  • For diatomic ideal gas at ordinary temperature, Cv = 5R/2 and Cp = 7R/2.

Memory tricks2

Cp Greater: P for Push: at constant Pressure the gas Pushes piston, so Cp is larger.

Mayer Memory: Cp is one R more than Cv for an ideal gas: Cp = Cv + R.

Examples2

Water as Coolant

Water is used in cooling systems because its high specific heat allows it to absorb much heat with small temperature rise.

Gas Heating in Balloon

A heated balloon expands, so some heat increases internal energy and some does work against the atmosphere.

Common mistakes2

Using Same Heat Capacity for All Processes

For gases, heat capacity depends on the process; use Cv for constant volume and Cp for constant pressure.

Confusing Heat Capacity and Specific Heat

Heat capacity is for the entire body; specific heat is per kg.

Reference tables2

Quick revision7

Formulas4

Heat Capacity
S=Q/ΔTS = Q/ΔT

Heat required to raise the temperature of a body by one kelvin.

S
heat capacity
Q
heat supplied
ΔT
temperature change
Specific Heat Capacity
Q=mcΔTQ = mcΔT

Heat required depends on mass, specific heat and temperature rise.

m
mass
c
specific heat capacity
ΔT
temperature change
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7

Second Law, Entropy & Irreversibility

8 min

The First Law permits many energy-conserving processes, but the Second Law decides which processes occur naturally. Heat naturally flows from hot to cold, not cold to hot without external work. No heat engine can convert all absorbed heat into work in a cyclic process. Entropy is a measure of energy dispersal or disorder, and the entropy of an isolated system never decreases. Reversible processes are ideal and can be exactly retraced; real processes are irreversible due to friction, turbulence, finite temperature differences and dissipation. For NEET, focus on Kelvin-Planck and Clausius statements, entropy change, and why 100% efficient heat engines are impossible.

Key points5

  • Kelvin-Planck statement: complete conversion of heat from a single reservoir into work is impossible in a cycle.
  • Clausius statement: heat cannot flow from cold to hot without external work.
  • Entropy is a state function.
  • For a reversible process, dS = δQrev/T.
  • Irreversibility is associated with entropy production.

Memory tricks2

Second Law Direction: Second law is the traffic rule of heat: hot to cold is allowed naturally; cold to hot needs work.

Entropy: Entropy likes spreading. Energy and matter tend to disperse spontaneously.

Examples2

Perfume Spreading

Perfume molecules spread through a room spontaneously; they do not collect back into the bottle by themselves.

Ice Melting in Warm Water

Heat flows from warm water to ice, not the reverse, unless work is supplied through a device.

Common mistakes2

Thinking First Law Forbids All Impossible Processes

The first law only checks energy conservation; the second law checks natural possibility.

Saying Entropy Always Increases for a System

Entropy of a system can decrease, but total entropy of system plus surroundings for an isolated universe cannot decrease.

Reference tables2

Quick revision7

Formulas3

Entropy Change for Reversible Heat Transfer
ΔS= Qrev/TΔS =\text{ Qrev}/T

Entropy change when heat is transferred reversibly at constant temperature.

ΔS
change in entropy
Qrev
reversible heat transfer
T
absolute temperature
Infinitesimal Entropy Change
dS=δQrev/TdS = δ\text{Qrev}/T

General definition of entropy change for a reversible infinitesimal process.

dS
small entropy change
δQrev
small reversible heat transfer
T
absolute temperature
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8

Heat Engines, Refrigerators & Heat Pumps

8 min

A heat engine is a cyclic device that absorbs heat from a hot reservoir, converts part of it into work, and rejects the remaining heat to a cold reservoir. Its efficiency is the fraction of absorbed heat converted into work. A refrigerator works in the reverse direction: it removes heat from a cold body and rejects it to a hotter surroundings by using external work. A heat pump is similar but its useful output is heat delivered to the hot region. NEET questions commonly ask for efficiency, coefficient of performance, energy balance, and why an engine cannot have 100% efficiency due to the second law.

Key points5

  • A heat engine must operate between at least two reservoirs.
  • Work output is less than heat absorbed from the source.
  • A refrigerator does not destroy heat; it transfers heat from cold to hot using work.
  • Coefficient of performance can be greater than 1.
  • Efficiency of heat engines is always less than 1.

Memory tricks2

Engine vs Fridge: Engine gives work; fridge takes work.

COP Can Cross One: COP is not efficiency. It compares moved heat with work input, so it can be greater than 1.

Examples2

Car Engine

Fuel combustion provides heat; part becomes mechanical work and the rest is rejected through exhaust and radiator.

Home Refrigerator

A refrigerator removes heat from inside the cabinet and releases more heat into the room because electrical work is also added.

Common mistakes2

Assuming COP is Percentage

COP is a ratio and may exceed 1; do not convert it like efficiency unless asked.

Forgetting Rejected Heat

A heat engine cannot convert all QH into work; some heat QC must be rejected.

Reference tables2

Quick revision7

Formulas4

Heat Engine Work
W=QH−QCW = QH - QC

Net work delivered by a heat engine in one cycle.

W
work output
QH
heat absorbed from hot reservoir
QC
heat rejected to cold reservoir
Heat Engine Efficiency
η=W/QH=1−QC/QHη = W/QH = 1 - QC/QH

Fraction of heat absorbed that is converted into useful work.

η
efficiency
W
work output
QH
heat absorbed
QC
heat rejected
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9

Carnot Engine & Carnot Cycle

9 min

The Carnot engine is an ideal reversible heat engine operating between two reservoirs at temperatures TH and TC. Its cycle has four reversible steps: isothermal expansion at TH, adiabatic expansion, isothermal compression at TC and adiabatic compression. It has the maximum possible efficiency for any engine working between the same two temperatures. Carnot efficiency depends only on reservoir temperatures and is given by η = 1 - TC/TH, where temperatures must be in kelvin. No real engine can exceed Carnot efficiency because real engines have irreversibilities such as friction and finite temperature heat transfer. NEET often asks cycle sequence, efficiency and comparison of engines.

Key points5

  • During isothermal expansion, engine absorbs heat QH from hot reservoir.
  • During adiabatic expansion, temperature falls from TH to TC.
  • During isothermal compression, engine rejects heat QC to cold reservoir.
  • During adiabatic compression, temperature rises from TC back to TH.
  • Carnot theorem forms a foundation for thermodynamic temperature scale.

Memory tricks2

Carnot Sequence: I A I A: Isothermal expansion, Adiabatic expansion, Isothermal compression, Adiabatic compression.

Efficiency Temperature Rule: Hot high, cold low makes efficiency grow: η = 1 - cold/hot.

Examples2

Efficiency Calculation

An engine between 600 K and 300 K has maximum efficiency 1 - 300/600 = 0.5 or 50%.

Improving Engine Efficiency

Raising the source temperature or lowering the sink temperature increases the theoretical maximum efficiency.

Common mistakes2

Using Celsius in Carnot Formula

Carnot efficiency must use kelvin temperatures. Using Celsius gives physically wrong results.

Thinking Carnot Engine is Real

Carnot engine is an ideal reversible model; real engines are always less efficient.

Reference tables2

Quick revision7

Formulas3

Carnot Efficiency
ηCarnot =1−TC/THη\text{Carnot }= 1 - TC/TH

Maximum efficiency of a reversible engine operating between hot and cold reservoirs.

ηCarnot
Carnot engine efficiency
TC
temperature of cold reservoir in kelvin
TH
temperature of hot reservoir in kelvin
Carnot Heat Ratio
QC/QH=TC/THQC/QH = TC/TH

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

QC
heat rejected to cold reservoir
QH
heat absorbed from hot reservoir
TC
cold reservoir temperature
TH
hot reservoir temperature
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10

Calorimetry, Latent Heat & Thermal Exchange

7 min

Calorimetry deals with measurement of heat exchanged between bodies. When hot and cold bodies are mixed in an insulated container, heat lost by the hot body equals heat gained by the cold body, assuming no loss to surroundings. Temperature change without phase change is calculated using Q = mcΔT. During a phase change, temperature remains constant while heat changes the internal molecular arrangement; this heat is called latent heat and is calculated using Q = mL. Although calorimetry is often introduced in thermal properties, it supports thermodynamics by clarifying heat transfer, thermal equilibrium and energy conservation. NEET questions often combine calorimetry with phase change.

Key points5

  • A calorimeter is treated as insulated to avoid heat exchange with surroundings.
  • Specific heat determines how easily temperature changes.
  • Latent heat of fusion is for solid-liquid transition.
  • Latent heat of vaporisation is for liquid-gas transition.
  • Steam at 100°C contains more heat than water at 100°C due to latent heat.

Memory tricks2

Latent Means Hidden: Latent heat is hidden heat: it changes state without showing temperature rise.

Calorimetry Balance: In an insulated mixture: hot loses exactly what cold gains.

Examples2

Steam Burns

Steam at 100°C causes more severe burns than water at 100°C because steam releases latent heat on condensation.

Ice in Water

Ice cools a drink effectively because it absorbs heat both to melt and then to warm the melted water.

Common mistakes2

Adding Temperature Change During Phase Change

During melting or boiling at fixed pressure, temperature remains constant until the phase change is complete.

Ignoring Calorimeter Heat Capacity

If the calorimeter’s water equivalent or heat capacity is given, include its heat gain or loss.

Reference tables2

Quick revision7

Formulas3

Sensible Heat
Q=mcΔTQ = mcΔT

Heat required for temperature change without phase change.

Q
heat exchanged
m
mass
c
specific heat capacity
ΔT
change in temperature
Latent Heat
Q=mLQ = mL

Heat required for phase change at constant temperature.

m
mass undergoing phase change
L
specific latent heat
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11

NEET Problem Strategy: Signs, Graphs & Concept Traps

6 min

Thermodynamics NEET problems become easy when you first identify the process, sign convention and known state variables. Always draw or inspect the P-V graph: area under the curve gives work, and enclosed area gives net work in a cycle. Next use the first law Q = ΔU + W and the ideal-gas fact that ΔU depends only on temperature. For engines and refrigerators, apply energy balance before efficiency or COP. For Carnot questions, convert all temperatures to kelvin. Conceptual traps usually involve confusing heat with internal energy, using Celsius, assuming all processes are reversible, or forgetting that second law limits efficiency.

Key points6

  • If temperature is unchanged for an ideal gas, internal energy change is zero.
  • If volume is unchanged, work is zero.
  • If heat exchange is zero, process is adiabatic.
  • If a graph is cyclic, net work is the enclosed area.
  • Efficiency cannot exceed Carnot efficiency for same reservoirs.
  • COP and efficiency are different quantities.

Memory tricks2

T-V-Q Cycle: T constant means ΔU zero; V constant means W zero; Q zero means adiabatic.

Graph Rule: Under curve is work; inside cycle is net work.

Examples2

One-Line First Law Example

If Q = 100 J and gas does W = 40 J, then ΔU = 60 J.

Cycle Example

If a cyclic engine absorbs 800 J and rejects 500 J, work output is 300 J and efficiency is 37.5%.

Common mistakes3

Not Reading Sign Convention

Some sources use work done on gas. NEET physics NCERT style generally uses work done by gas in Q = ΔU + W.

Skipping Units

Use SI units: pressure in pascal, volume in cubic metre, temperature in kelvin and energy in joule.

Assuming ΔU Depends on Path

Internal energy is a state function; for ideal gas it depends only on temperature.

Reference tables2

Quick revision7

Formulas3

Universal Thermodynamics Solver
Q=ΔU+WQ = ΔU + W

Use this after identifying W and ΔU from process conditions.

Q
heat supplied
ΔU
internal energy change
W
work done by gas
P-V Work Shortcut
W= area under P−V curveW =\text{ area under }P-V\text{ curve}

Graphical method for work done by gas.

W
work done by gas
P-V curve
pressure-volume path of the process
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NEET PYQs from Thermodynamics

The latest 3 of 49 previous-year questions from this chapter.

  1. 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:

    • A400 J
    • B300 J
    • C700 J
    • D500 J
  2. NEET 2026Set 11EasyQ2

    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:

    • A75 W
    • B100 W
    • C125 W
    • D25 W
  3. NEET 2026Set 11MediumQ3

    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)

    • A−1.635 kJ mol⁻¹, spontaneous
    • B−0.63568 kJ mol⁻¹, spontaneous
    • C+0.63568 kJ mol⁻¹, non-spontaneous
    • D+1.635 kJ mol⁻¹, non-spontaneous

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