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NCERT Class 11 Chemistry Chapter 6 Notes Thermodynamics - Download PDF

NCERT Class 11 Chemistry Chapter 6 Notes Thermodynamics - Download PDF

Edited By Sumit Saini | Updated on Apr 09, 2024 02:53 PM IST

NCERT Class 11 Chemistry chapter 6 notes basically compiles up the different types of forces that exist in between gases and also gives the theoretical approach to find the variation at certain levels. The chapter Thermodynamics Class 11 notes is helpful in conceptualizing the laws that are applied to real gas as well as to ideal gas laws. This NCERT chapter 6 Class 11 includes the topics gives brief visualization and theoretical relation between the system and its surroundings. With detailed explanations and structured content, chemistry class 11 chapter 6 notes pdf provide students with the necessary resources to understand complex concepts effectively.

Introductory Part may include: The NCERT Class 11 chapter 6 notes gives us a brief intro of the system and surroundings, state functions, applications related to it. Enthalpy, heat capacities, extensive and intensive properties are also included to understand the concept more precisely. Hess’s law, spontaneous and non-spontaneous processes, Gibbs energy are the highlighting key feature of Thermodynamics 11 notes. CBSE Class 11 Chemistry chapter 6 notes cover the introductory part of thermodynamics with basic gas laws and derivations to calculate the change that occurred in the system and surroundings CBSE Class 11 Chemistry chapter 6 notes can be easily downloaded at no cost and are accessible offline in the future use as well.

Also, students can refer,

NCERT Class 11 Chapter 6 Notes

Thermodynamics Class 11 Notes- Topic 1:

System and Surroundings:

A system can be defined as where the observations were made, it is the part of the universe.

Surrounding can be defined as the part other than the system is, surrounding.

The system and surroundings can be differentiated by a wall called a boundary.


Types of the system:

The system can be further divided into three categories such as-

  1. Open system- Boundary is imaginary in this system and exchange of mass or matter and energy both can be possible between system and surroundings. Example: Open beaker.

  2. Closed system-No exchange of matter occurs but the exchange of energy can be possible in between the system and surroundings. Example: closed container of steel or copper.

  3. Isolated system- No exchange of matter or energy can be possible. Example: Closed container like a thermos flask.


Thermodynamics Class 11 Notes- Topic 2:

The State of The System

In general concern when we have to make any calculations we require measurable quantities or properties.

Those quantities are temperature, pressure, the volume combines to form the composition of the system. But in order to deal with these quantities, we must have average conclusions of such measurements so we introduce a term called state functions or state variables. These quantities like pressure(p), volume(V), temperature(T), and amount(n).such variables like p, V, T, n are stated as state variables or state functions.

Internal energy: The combined energy or sum of energies such as mechanical, electrical or chemical are so-called internal energy of the system. This can be denoted in thermodynamics by the letter ‘U’. This can be affected by a change in any of the following:

  • Heat goes into and out of the system

  • Work done; it can be done on the system or by the system.

  • Matter; Exchange of matter( entering or leaving)

General case: This can b shown using an equation.


Here the change in internal energy is the sum of heat passing in and out and the work done. It is independent of the change and also depends on the initial and final state of the system. The equation that we get for change in internal energy is the mathematical form of the First law of thermodynamics.

Adiabatic system: It is the system in which the system and surroundings are separated by a wall, where no transfer or exchange of matter or energy occurs.

Thermodynamics Class 11 Notes- Topic 3:


Work: Here we take an example of a cylinder or piston, which is fitted perfectly and is filled with a frictionless container with one mole of gas.

Let us first discuss the variables to calculate the work done.

The pressure of the gas=p

The volume of the gas=V

The external pressure of the gas=pext

Area of cross section=A

The piston is moved to a certain distance, taken as length=l

Change in volume∆V=A×l=(Vf-Vl)

So here, the pressure inside the piston becomes


Case1: Expansion process

In the case of this process, work is done by the system, by expanding the volume of the system. When the work is done by the system then the negative sign convention will be used in it. so the value of work done will be


Now consider the two cases of expansion of ideal gas.


  1. Work is to be done in isothermal and reversible condition, where the expansion of ideal gas occur.

Reversible work is taken from initial volume to final volume.


The value of (dP)dV is so small, so we can write it as;


The cylinder is filled with an ideal gas, so consider the ideal gas equation.



At constant temperature conditions;




B. Isothermal condition, where expansion of ideal gas occur.

  • For irreversible change:


  • For reversible change


  • For adiabatic change



Enthalpy can be denoted by the letter ‘H’.

Enthalpy is defined as the sum of pressure volume and internal energy.

Change in enthalpy can be defined as the heat is evolved or absorbed at constant pressure conditions in a given system.

For the Exothermic process; Heat is given out of the system, whereas for the Endothermic process heat is given to the system through surroundings.

Relationship between Change in enthalpy and change in internal energy of the system



Thermodynamics Class 11 Notes- Topic 4:

  1. Extensive property:

Extensive property can be defined as that property in which the value depends on the quantity or the size that is present in the system.

Examples of extensive properties are mass, enthalpy, volume, etc.

  1. Intensive property:

Intensive property can be defined as that property in which the value does not depend on the quantity or the size that is present in the system.

Examples of intensive properties are temperature, density, etc.

  1. Heat capacity:

Heat capacity may be defined increase in temperature is directly proportional to heat transferred.

  1. Relationship between the value of CpandCv.

Constant volume value of heat capacity; Cv.

Constant pressure value of heat capacity; Cp.


Thermodynamics Class 11th Notes- Topic 5:

Measurement of the value of change in internal energy and change in enthalpy with Bomb calorimeter.


Enthalpy changes occur during the process of phase transformation.

  • Enthalpy of Fusion: Enthalpy of fusion can be defined as the conversion of one mole of solid is accompanied to change it in liquid, condition of fusion is governed at its melting point

  • Enthalpy of Vaporisation: Enthalpy of vaporization can be defined as the conversion of one mole of solid is accompanied to change it in liquid, the condition of fusion is governed at its boiling point.

  • Enthalpy of Sublimation: Enthalpy of sublimation is defined as the change in when one mole of solid is converted to gaseous state at its melting point. But temperature is below at its melting range.

  • Enthalpy of Combustion: Enthalpy of fusion is defined as the change in when one mole of substance burn in excess amount of air.

  • Standard enthalpy of formation: It is defined as in standard condition of temperature at 273K and pressure at 1 atm formation of one mole of substance occur with the help of their constituent elements.

  • Thermochemical equations: It is defined as in chemical reaction when the products and reactants are present with their physical value and value of∆rH present such reaction equation is known as Thermochemical equation.

  • Hess’s Law

It is defined as Hess’s law for constant heat summation.

In general, to solve the questions we can write the equation as follows:

1648799321687 1648799321996


  • On combining the two reactions we get the final result as shown above.

  • The constant heat summation law suggest that whether the heat is absorbed or evolved in one or many steps are involved, the total amount of heat is same.

Thermodynamics Class 11 Notes- Topic 5:

Born-Haber Cycle:


  • Spontaneity:

Spontaneous process: It is the process that has been taken by itself or tendency to occur on its own.

The process is not instantaneous in nature. Speed can vary from slow to fast.

Examples of such process: Common salt dissolves in water on its own.

  • Entropy:

Entropy is defined as the degree of disorders and randomness in a system.

The entropy of the gaseous system is found to be more than in solid state.

Change in entropy∆S=qrevT=∆HT;

Change is entropy can be denoted as ∆S, and is zero for equilibrium condition.



Thermodynamics Class 11 Notes- Topic 6:

Gibb’s Energy and Spontaneity:

It is defined as mathematically ∆G=∆H-T∆S

The above reaction is at constant temperature and pressure.

  1. Value of ∆G is negative or less than zero, the process would be spontaneous.

  2. Value of ∆G is positive or more than zero, the process would be non-spontaneous.

Condition of equilibrium, when all reactants and products are in standard condition, value of ∆G will be:




Significance of NCERT Class 11 Chemistry Chapter 6 Notes

Thermodynamics Class 11 notes will be helpful in revising all concepts more precisely and in a concise manner. With the ease of concern, it can be revised anywhere and anytime with no internet use offline mode as well. These ch 6 chemistry class 11 notes offer a comprehensive overview of important topics covered in the curriculum. The chapter is divided into parts so as to approach the topics directly jump on it or in a stepwise manner which so ever is convenient for the students. Detailed understanding of the basic concepts of thermodynamics can be obtained from cbse class 11 chemistry ch 6 notes.

Also, this NCERT Class 11 Chemistry chapter 6 notes pdf is useful to cover the major as well as the minor topics of the Class 11 CBSE Chemistry Syllabus. CBSE Class 11 Chemistry chapter 6 notes helps to understands the basic laws governed in system and surroundings. Class 11 Thermodynamic notes contain the laws, different types of process like enthalpy of fusion, vaporisation, sublimation, combustion and also the concept of Hess’s law, Gibb’s energy and Born Haber cycle. cbse class 11 chemistry ch 6 notes are a good source for revising the concepts just before the examination.

NCERT Class 11 Notes Chapter-Wise

Subject Wise NCERT Exemplar Solutions

Subject Wise NCERT Solutions

Frequently Asked Question (FAQs)

1. NCERT Class 11 Chemistry chapter 6 gives an example of a system being isolated?

Thermos flask with coffee is an example of an isolated system, where no exchange of matter or energy can be possible.

2. In thermodynamics Class 11 notes, when we studied about work is happening on the system then what is the internal energy of the system?

The internal energy of the system in this case will increase.

3. NCERT Class 11 Chemistry chapter 6 according to thermodynamic concern, which system do the animals and plants belong to?

Open system, For detailed understanding of the topic, students can refer to chemistry class 11 chapter 6 notes pdf . 

4. According to notes for Class 11 Chemistry chapter 6 and class 11 thermodynamics notes, what is the mathematical expression for enthalpy?

H=U+pv; where U belongs to internal energy.

5. What will be the entropy when two ideal gases are mixed together in an isolated system under the same condition of pressure and temperature according to NCERT notes for Class 11 Chemistry chapter 6?

Change in entropy is positive due to an increase in randomness. ch 6 chemistry class 11 notes can be referred for more understanding of the topic. 


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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg


An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)


Option 2)

\; K\;

Option 3)


Option 4)


In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)


Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)


Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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