This article is formulated according to the 12th chapter in the NCERT Syllabus of 11th Std Physics OR 1st PUC Physics.


It is the branch of physics that deals with the concepts of heat and temperature and the inter conversion of heat and other forms of energy. In thermodynamics, the main focus is on the macroscopic quantities of the system such as pressure, volume, temperature, internal energy, entropy, enthalpy etc. Thus, thermodynamics provide macroscopic description of the system.

Thermodynamic system

A collection of an extremely large number of atoms or molecules confined within certain boundaries such that it has certain values of pressure, volume and temperature is called thermodynamic system. The system may be in the form of a solid, liquid, gas or a combination of two or more.


Anything outside the thermodynamic system to which energy or matter is exchanged is called its surroundings. A system may be divided into three groups (types): Open system, Closed system, and Isolated system.

Open system

It can exchange both energy and matter with its surroundings.

Closed system

It can exchange only energy with its surroundings.

Isolated system

It will not exchange both energy and matter with its surroundings.

Adiabatic wall

A wall which does not allow any exchange of energy between the systems is known as adiabatic wall.

Diathermic wall

A wall which allows any exchange of energy between the systems is known as adiabatic wall.

Thermodynamic state variables (State variables)

Variables that are required to specify the state of the thermodynamic system are called thermodynamic state variables. Ex: Pressure, Temperature, Volume, Mass, Composition, Internal energy, etc.

Equation of state

The equation which relates the state variables is called an equation of state. The equation of state for an ideal gas is, 𝑃𝑉 = πœ‡π‘…π‘‡.

Types of thermodynamic state variables

State variables are of two types,

(i) Extensive thermodynamic state variable

(ii) Intensive thermodynamic state variable

Extensive thermodynamic state variables

The variables whose value changes for each part of the system are called Extensive thermodynamic state variables. Ex: Internal energy, volume, and mass.

Intensive thermodynamic state variables

The variables whose value remains unchanged for each part of the system are called intensive thermodynamic state variables. Ex: Temperature, pressure and density.

Note: Extensive state variables depend on the size of the system, but intensive state variables do not.

Thermal equilibrium

Two systems in contact are said to be in thermal equilibrium, if both are at the same temperature. In thermal equilibrium, thermodynamic variables such as pressure, volume, temperature, mass and composition will not change with time, for a closed system. That is the system has mechanical, thermal and chemical equilibrium.

Zeroth law of thermodynamics

When two systems 𝐴 and 𝐡 are separately in thermal equilibrium with a third system 𝐢, then the two systems 𝐴 and 𝐡 are also in thermal equilibrium with each other. Zeroth law was formulated by R. H. Flower.

Significance of Zeroth law

The significance of this law is, all the systems in thermal equilibrium with one another must have a common physical quantity that has the same value for both, called temperature.


The energy that is transferred between a system and its surroundings whenever there is a temperature difference between the system and surroundings is called heat. When energy is transferred to the system from its surroundings, then heat is taken as positive. When energy is transferred to the surroundings from the system, then heat is taken as negative.

Work done

Work is said to be done, if a system moves through a certain distance in the direction of the applied force.

Expression for work done by the system

Let a gas taken in the cylinder. Let the cylinder is fitted with a frictionless piston of area of cross-section 𝐴. Let 𝑃 be the pressure of the gas on the cylinder. The force on the piston, 𝐹 = 𝑃𝐴 Let the piston be displaced through a distance 𝑑π‘₯ during the expansion of the gas. Work done by the gas,

π‘‘π‘Š = 𝐹𝑑π‘₯

π‘‘π‘Š = 𝑃𝐴𝑑π‘₯

π‘‘π‘Š = 𝑃𝑑𝑉

Total work done in which the volume changes from 𝑉𝑖 to 𝑉𝑓 is,

π‘Š = βˆ«π‘‘π‘Š = βˆ«π‘ƒπ‘‘π‘‰

𝑾 = 𝑷(𝑽𝒇 βˆ’ π‘½π’Š)

Sign convention

(i) When a system expands against the external pressure, 𝑑𝑉 = (𝑉𝑓 βˆ’ 𝑉𝑖) is positive. Hence work done by the system and is taken as positive.

(ii) When a system is compressed, 𝑑𝑉 is negative. Hence work done on the system and is taken as negative.

Internal energy

The sum of the kinetic and potential energy of the constituent particles of the system is known as internal energy. It is denoted by π‘ˆ.

π‘ˆ = π‘ˆπΎ + π‘ˆP

Note: In the case of an ideal gas, we assume that intermolecular forces are zero. Thus π‘ˆπ‘ƒ = 0. Hence the internal energy is purely kinetic energy and depends only on temperature. In real gases, intermolecular forces are not zero, π‘ˆπ‘ƒ β‰  0.

Sign Convention

(i) Increase in internal energy is taken as positive and (ii) Decrease in internal energy is taken as negative.

Note: Heat and work are not state variables. They are modes of energy transfer to a system resulting in change in its internal energy.

First law of thermodynamics

When some quantity of heat (𝑑𝑄) is supplied to a system, then the quantity of heat absorbed by the system is equal to the sum of the increases in the internal energy of the system (π‘‘π‘ˆ) and the external work done by the system (π‘‘π‘Š) against the expansion. Mathematically,

𝑑𝑄 = π‘‘π‘ˆ + π‘‘π‘Š or 𝒅𝑸 = 𝒅𝑼 + 𝑷𝒅𝑽

Significance of first law

First law of thermodynamics is law of conservation of energy.

Specific heat capacities (Mayer’s relation)

From first law of thermodynamics, 𝑑𝑄 = π‘‘π‘ˆ + 𝑃𝑑𝑉

For one mole of gas, If 𝑑𝑄 is the heat absorbed at constant volume, then

𝑑𝑉 = 0.

That is, 𝑑𝑄 = 𝐢𝑉𝑑𝑇 or 𝐢𝑉 = 𝑑𝑄/𝑑𝑇

𝐢𝑉 = (𝑑𝑄/𝑑𝑇)𝑉 = (π‘‘π‘ˆ/𝑑𝑇)𝑉 or 𝐢𝑉 = π‘‘π‘ˆ/𝑑𝑇 βˆ’ βˆ’ βˆ’ (1)

Where the subscript V is dropped in the last step, since U of an ideal gas depends only on temperature. If 𝑑𝑄 is the heat absorbed at constant pressure, then 𝑑𝑄 = 𝐢𝑃𝑑𝑇 or 𝐢𝑃 = 𝑑𝑄/𝑑𝑇

𝐢𝑃 = (𝑑𝑄/𝑑𝑇)𝑃 = (π‘‘π‘ˆ/𝑑𝑇)𝑃 + (𝑃𝑑𝑉/𝑑𝑇)𝑃 or 𝐢𝑃 = (π‘‘π‘ˆ/𝑑𝑇) + (𝑃𝑑𝑉/𝑑𝑇)𝑃 βˆ’ βˆ’ βˆ’ (2)

The subscript P is dropped in the first term, since U of an ideal gas depends only on temperature. Now, for a mole of an ideal gas, 𝑃𝑉 = 𝑅𝑇 or 𝑃𝑑𝑉 = 𝑅𝑑𝑇

(𝑃𝑑𝑉/𝑑𝑇)𝑃 = 𝑅 βˆ’ βˆ’ βˆ’ (3)

Using the equations (1) and (3) in (2), We have, 𝐢𝑃 = 𝐢𝑉 + 𝑅

π‘ͺ𝑷 βˆ’ π‘ͺ𝑽 = 𝑹

Thermodynamic process

Any process in which the thermodynamic variables of a system change is known as a thermodynamic process.

Quasi-static process

A process in which the system departs only infinitesimally from the equilibrium state is known as a quasi-static process. In this process, the change in pressure or change in volume or change in temperature of the system is very, very small.

Note: Non-equilibrium states of a system are difficult to deal with. It is, therefore, convenient to imagine an ideal process in which at every stage the system is an equilibrium state. Such a process is infinitesimally slow, hence the name quasi-static.

Isothermal process

A process in which the temperature of the system is kept constant throughout is called an isothermal process. In this case 𝑃 and 𝑉 changes, but 𝑇 = π‘π‘œπ‘›π‘ π‘‘π‘Žπ‘›π‘‘. As the temperature is constant, no change in internal energy, π‘‘π‘ˆ = 0. From the first law of thermodynamics,

𝑑𝑄 = π‘‘π‘ˆ + π‘‘π‘Š

𝑑𝑄 = π‘‘π‘Š or 𝑑𝑄 = 𝑃𝑑𝑉

Heat supplied in an isothermal process is used to do work against the surrounding. Ex: Boiling of a liquid, melting of wax or ice, etc.

Expression for Work done during an Isothermal process

Consider an ideal gas that changes its state from 𝑃𝑖, 𝑉𝑖 to 𝑃𝑓, 𝑉𝑓 at a constant temperature. The work done is given by,

(since 𝑃𝑉 = πœ‡π‘…π‘‡)

π‘Š = πœ‡π‘…π‘‡[ln 𝑉]𝑉𝑖𝑉𝑓

𝑾 = 𝝁𝑹𝑻 π₯𝐧[𝑽𝒇/π‘½π’Š]


(i) If 𝑉𝑓 > 𝑉𝑖 then π‘Š = π‘π‘œπ‘ π‘–π‘‘π‘–π‘£π‘’.

(ii) If 𝑉𝑖 > 𝑉𝑓 then π‘Š = π‘›π‘’π‘”π‘Žπ‘‘π‘–π‘£π‘’.


The pressure-volume curve for a fixed temperature is called an isotherm.

Adiabatic process

The process in which heat energy neither enters nor leaves the system is called adiabatic process. In this case, 𝑃, 𝑉 and 𝑇 change, but 𝑑𝑄 = 0. From first law of thermodynamics,

𝑑𝑄 = π‘‘π‘ˆ + π‘‘π‘Š

π‘‘π‘Š = βˆ’π‘‘π‘ˆ

When gas expands adiabatically, π‘Š is positive. Therefore, π‘‘π‘ˆ must be negative. That is internal energy of the system decreases. Ex: Bursting of an automobile tube inflated with air, propagation of sound waves in a gas.

Expression for work done during an adiabatic process

Let a gas in state 𝑃𝑖, 𝑉𝑖, 𝑇𝑖 be adiabatically expand to the state 𝑃𝑓, 𝑉𝑓, 𝑇𝑓

Work done in the process is,

For an adiabatic process, 𝑃𝑉𝛾 = π‘π‘œπ‘›π‘ π‘‘π‘Žπ‘›t

π‘Š = (π‘π‘œπ‘›π‘ π‘‘π‘Žπ‘›π‘‘/(1 – 𝛾))[𝑉𝑓1βˆ’π›Ύ βˆ’ 𝑉𝑖1βˆ’π›Ύ]

π‘Š = (1/(1 – 𝛾))[(π‘π‘œπ‘›π‘ π‘‘π‘Žπ‘›π‘‘ Γ— 𝑉𝑓1βˆ’π›Ύ) – (π‘π‘œπ‘›π‘ π‘‘π‘Žπ‘›π‘‘ Γ— 𝑉𝑖1βˆ’π›Ύ)]

π‘Š = (1/(1 – 𝛾))[𝑃𝑓𝑉𝛾𝑓𝑉𝑓1βˆ’π›Ύ βˆ’ 𝑃𝑖𝑉𝑖𝛾𝑉𝑖1βˆ’π›Ύ]

π‘Š = (1/(1 – 𝛾))[𝑃𝑓𝑉𝑓 βˆ’ 𝑃𝑖𝑉𝑖]

π‘Š = (1/(1 – 𝛾))[πœ‡π‘…π‘‡π‘“ βˆ’ πœ‡π‘…π‘‡π‘–]

π‘Š = (1/(1 – 𝛾))πœ‡π‘…(𝑇𝑓 βˆ’ 𝑇𝑖)

𝑾 = 𝝁𝑹(π‘»π’Š βˆ’ 𝑻𝒇)/(𝜸 – 𝟏)


(i) If 𝑇𝑓 < 𝑇𝑖, π‘Š > 0 (π‘π‘œπ‘ π‘–π‘‘π‘–π‘£π‘’), Temperature decreases when the gas expands.

(ii) If 𝑇𝑓 > 𝑇𝑖, π‘Š < 0 (π‘›π‘’π‘”π‘Žπ‘‘π‘–π‘£π‘’), Temperature increases when the gas compressed.

Isochoric process

A thermodynamic process that takes place at constant volume is called isochoric process. As the volume is kept constant, π‘‘π‘Š = 0 From first law of thermodynamics,

𝑑𝑄 = π‘‘π‘ˆ + π‘‘π‘Š

𝑑𝑄 = π‘‘π‘ˆ

If heat is absorbed by a system at constant volume, its internal energy increases. Ex: Melting of a solid into liquid.

Isobaric process

A thermodynamic process that takes place at constant pressure is called isobaric process. Work done by the gas is,

π‘Š = 𝑃(𝑉𝑓 βˆ’ 𝑉𝑖)

π‘Š = πœ‡π‘…(𝑇𝑓 βˆ’ 𝑇𝑖)

Ex: Heating any liquid at atmospheric pressure, heating a gas at constant pressure.

Cyclic process

It is the process in which the system returns to its initial state after a number of changes. In cyclic process, change in internal energy is zero. From the first law of thermodynamics,

𝑑𝑄 = π‘‘π‘ˆ + π‘‘π‘Š

𝑑𝑄 = π‘‘π‘Š

Net work done during a cyclic process must be equal to the amount of heat energy transferred.

Reversible process

It is a process which can be made to proceed in the opposite direction with same ease so that the system and the surroundings pass through exactly the same intermediate state as in the direct process. Ex: Conversion of ice to water and vice versa, under ideal conditions.

Irreversible process

A process in which the system cannot be retraced to its original state is called an irreversible process. Ex: A body moving on a rough surface from one point to another.

Heat engine

Heat engine is a device which converts heat energy into mechanical energy.

Parts of heat engine


Source is an agency which can provide any amount of heat. It is maintained at constant, high temperature.

Working substance

It is an enclosed thermodynamic system which can do work. A mixture of fuel vapour and air is used as working substance in gasoline or diesel engine.


It can absorb any amount of heat. It is maintained at constant temperature. Temperature of the sink is less than the temperature of the source.

Principle: Working substance absorbs heat from the source and converts a part of it into work and rejects the rest to the sink.

Working: Let 𝑄1 amount of heat is absorbed by the working substance from the source at temperature 𝑇1. Let π‘Š is the total work done by the working substance. 𝑄2 is the amount of heat rejected to the sink at temperature 𝑇2. Net amount of heat absorbed, 𝑑𝑄 = 𝑄1 βˆ’ 𝑄2.

As the working substance returns to the original state, π‘‘π‘ˆ = 0 From the first law of thermodynamics,

𝑑𝑄 = π‘‘π‘ˆ + π‘Š

𝑑𝑄 = π‘Š

π‘Š = 𝑄1 βˆ’ 𝑄2

The efficiency of heat engine

It is defined as the ratio of mechanical work done to the heat absorbed. It is denoted by πœ‚.

πœ‚ = π‘Š/𝑄1 = (𝑄1 βˆ’ 𝑄2)/𝑄1

𝜼 = 𝟏 – (π‘ΈπŸ/π‘ΈπŸ)

From the above equation, it is clear that efficiency of heat engine is less than one or 100%.


The process of removing heat from bodies colder than their surroundings is called refrigeration.


The device which does the process of refrigeration is called refrigerator. In refrigerator heat is absorbed at low temperature and rejected at higher temperature. The working substances are ammonia and Freon.

Heat pump

Refrigerator is a heat engine working backward and it is also called Heat pump. Working steps:

  • Sudden expansion of gas from high to low pressure which cools the working substance.
  • Absorption of heat from the region to be cooled by the working substance.
  • Heating up of the working substance.
  • Release of heat by the working substance to surroundings and returns to the initial state.

Co-efficient of performance

Co-efficient of performance is given by,

𝛼 = π»π‘’π‘Žπ‘‘ 𝑒π‘₯π‘‘π‘Ÿπ‘Žπ‘π‘‘π‘’π‘‘ π‘“π‘Ÿπ‘œπ‘š π‘‘β„Žπ‘’ π‘π‘œπ‘™π‘‘ π‘π‘œπ‘‘π‘¦/π‘Šπ‘œπ‘Ÿπ‘˜ π‘‘π‘œπ‘›π‘’ π‘œπ‘› π‘‘β„Žπ‘’ π‘€π‘œπ‘Ÿπ‘˜π‘–π‘›π‘” π‘ π‘’π‘π‘ π‘‘π‘Žπ‘›π‘π‘’

𝜢 = π‘ΈπŸ/𝑾 = π‘ΈπŸ/(π‘ΈπŸ – π‘ΈπŸ)

𝛼 can be greater than π‘œπ‘›π‘’ and cannot be 𝑖𝑛𝑓𝑖𝑛𝑖𝑑𝑒.

Second law of thermodynamics

This law specifies the condition of transformation of heat into work.

Kelvin-Planck statement

No process is possible whose only result is the absorption of heat from a reservoir and the complete conversion of heat into work.

Clausius statement

No process is possible whose only result is the transfer of heat from a colder object into a hotter object.

Carnot engine

Sadi Carnot introduced the concept of an ideal heat engine called Carnot engine.

Construction: Parts of the Carnot engine are,

Source: It is maintained at a fixed higher temperature 𝑇1.

Sink: It is maintained at fixed low temperature 𝑇2 than the source.

Working substance: A perfect gas acts as working substance. The container is fitted with a piston which can slide without friction, and it is also non-conducting. Container has conducting base and non-conducting side wall.

Insulated stand: It is used to provide complete thermal isolation for working substance that can undergo adiabatic operation.

Carnot cycle:

The working substance of the Carnot engine is taken through a cycle of isothermal and adiabatic process known as Carnot cycle. Carnot cycle for a heat engine with an ideal gas as the working substance is as shown.

Steps of Carnot cycle

Step 1 – Isothermal expansion: The cylinder with gas having pressure 𝑃1, volume 𝑉1 and temperature 𝑇1 is kept on the source at temperature 𝑇1. The gas is allowed to expand isothermally slowly. The temperature tends to decrease, but it is maintained at constant temperature 𝑇1 by absorbing heat from source. Let the pressure and volume change to 𝑃2 and 𝑉2 respectively. For an isothermal process, 𝑑𝑄 = 𝑑W

𝑄1 = π‘Š1 = πœ‡π‘…π‘‡1 ln(𝑉2/𝑉1)

Step2-Adiabatic expansion: The cylinder is placed on the non-conducting stand and the gas is allowed to expand adiabatically until the temperature falls to 𝑇2. Work done during the expansion is, π‘Š2 = (πœ‡π‘…/(𝛾 – 1))(𝑇1 βˆ’ 𝑇2). To make the gas recover its capacity for doing work, it should be brought to the original condition. This is done in next steps.

Step 3 – Isothermal compression: The cylinder is kept on sink at temperature 𝑇2. The gas is compressed isothermally. Let 𝑄2 amount of heat is rejected to the sink. Let the pressure 𝑃3 and volume 𝑉3 change to 𝑃4 and 𝑉4 respectively. For an isothermal process,

𝑑𝑄 = π‘‘π‘Š

𝑄2 = π‘Š3 = βˆ’πœ‡π‘…π‘‡2 ln(𝑉4/𝑉3) βˆ’π‘£π‘’ sign indicates work is done on the system.

𝑄2 = π‘Š3 = πœ‡π‘…π‘‡2 ln(𝑉3/𝑉4)

Step4 – Adiabatic Compression: The cylinder is placed on the non-conducting stand and the gas is compressed adiabatically till the pressure 𝑃4, volume 𝑉4 changes to 𝑃1 and 𝑉1 and temperature 𝑇1. Work done,

π‘Š4 = (πœ‡π‘…/(𝛾 – 1))(𝑇1 βˆ’ 𝑇2)

Total work done during the complete cycle is,

π‘Š = π‘Š1 + π‘Š2 βˆ’ π‘Š3 βˆ’ π‘Š4

Since π‘Š2 = π‘Š4, π‘Š = π‘Š1 βˆ’ π‘Š3 = 𝑄1 βˆ’ 𝑄2.

Note: Work done, π‘Š = π΄π‘Ÿπ‘’π‘Ž π‘’π‘›π‘‘π‘’π‘Ÿ π‘‘β„Žπ‘’ π‘”π‘Ÿπ‘Žπ‘β„Ž

Efficiency of Carnot engine

πœ‚ = (𝑄1 βˆ’ 𝑄2)/𝑄1 = 1 – (𝑄2/𝑄1)

πœ‚ = 1 – [{πœ‡π‘…π‘‡2 ln(𝑉3/𝑉4)}/{πœ‡π‘…π‘‡1 ln(𝑉2/𝑉1)]

πœ‚ = 1 – [{𝑇2 ln(𝑉3/𝑉4)}/{𝑇1 ln(𝑉2/𝑉1)}] βˆ’ βˆ’ βˆ’ (1)

For the step 2, we have adiabatic equation, 𝑇1𝑉2π›Ύβˆ’1 = 𝑇2𝑉3π›Ύβˆ’1

𝑇2/𝑇1 = 𝑉2(π›Ύβˆ’1)/𝑉3(π›Ύβˆ’1) βˆ’ βˆ’ βˆ’ (2)

For Step 4, we have adiabatic equation, 𝑇2𝑉4π›Ύβˆ’1 = 𝑇1𝑉1π›Ύβˆ’1

𝑇2/𝑇1 = 𝑉1π›Ύβˆ’1/𝑉4π›Ύβˆ’1 βˆ’ βˆ’ βˆ’ (3)

From equation (1) and (2), 𝑉2π›Ύβˆ’1/𝑉3π›Ύβˆ’1 = 𝑉1π›Ύβˆ’1/𝑉4π›Ύβˆ’1

(𝑉2/𝑉3)π›Ύβˆ’1 = (𝑉1/𝑉4)π›Ύβˆ’1

𝑉2/𝑉3 = 𝑉1/𝑉4 or 𝑉2/𝑉1 = 𝑉3/𝑉4

ln(𝑉2/𝑉1) = ln(𝑉3/𝑉4)

Equation (1) becomes, 𝜼 = 𝟏 – (π‘»πŸ/π‘»πŸ)


  • Efficiency of Carnot engine depends upon the temperature of the source and sink.
  • Efficiency is independent of the nature of the working substance.
  • Since we cannot have a sink at absolute zero, so a heat engine with 100% efficiency is not possible to realise in practice.

Important Questions for Exams.

One Mark.

1. Which physical quantity is conserved for an iso-thermal process?

2. State the Zeroth law of thermodynamics.

3. What is the significance of zeroth law of thermodynamics?

4. State the first law of thermodynamics.

5. Mention the significance of I law of thermodynamics.

6. What is Quasi-static process?

7. What is equation of state for adiabatic process?

8. Define efficiency of heat engine.

Two Marks.

1. Mention the quantities remaining constant during isobaric and isochoric processes.

2. Mention an example for isothermal and adiabatic process each.

3. What are Diathermic and adiabatic wall?

4. What are state variables? Give two examples.

Three Marks.

1. Explain the parts of Heat engine.

2. What is the principle of heat engine? Sketch the diagrammatic representation of working of heat engine.

3. Draw schematic diagram of the refrigerator. Define its coefficient of performance.

Five Marks.

1. Draw a schematic diagram of pressure versus volume for a Carnot cycle with an ideal gas as working substance. Write an expression for efficiency of a Carnot engine.

2. What is isothermal process? Derive the expression for work done in isothermal process.

3. Explain different stages of Carnot’s cycle with P-V diagram. or Explain Carnot’s Cycle.

4. Discuss the applications of first law of Thermodynamics.

Numerical Problems.

1. A steam engine delivers 5.4 Γ— 108𝐽 of work per minute and services 3.6 Γ— 109𝐽 of heat per minute from its boiler. What is (i) efficiency of the engine. (ii) Work done per cycle. (iii) Heat rejected to sink per cycle. (iv) How much heat is wasted per minute?

2. A Carnot engine has an efficiency of 0.3, when the temperature of the sink is 350K. Find the change in temperature of the source when the efficiency becomes 0.5.

3. A steam engine delivers 7.5Γ—108J of work per minute and services 3.6 Γ— 109J of heat per minute from its boiler. What is the efficiency of the engine? How much heat is wasted per minute? Also find the ratio of temperature of sink to the source.

Leave a Comment

Your email address will not be published.

Scroll to Top