Chapter 11 - Thermodynamics

NCERT Solutions Class 11 Physics Chapter 11 – Thermodynamics

The NCERT Solutions for Class 11 Physics Chapter 11 (Thermodynamics) explains the macroscopic world of heat, work and internal energy. The previous chapters dealt with the properties of matter. Thermodynamics shifts our focus to the laws of energy transformation and thermal equilibrium. This chapter is fundamental to classical physics. It is the theoretical foundation for the operation of engines, refrigerators and heat pumps .

NCERT Solutions for Class 11 Physics Chapter 11 by ALLEN are designed with the help of expert faculties to bridge the gap between the law and practical use of numericals. These solutions are prepared systematically according to the CBSE, JEE and NEET requirements so that students can learn everything from the sign conventions to the complex cyclic processes. These solutions range from the theoretical elegance of the Carnot cycle to the practicalities of the First Law of Thermodynamics. They provide a rigorous basis for understanding why energy can be neither created nor destroyed, only transformed—and why some of it is always "lost" to the environment.

1.0Class 11 Physics Chapter 11 Thermodynamics: Key Concepts

This chapter examines the transformation of heat into work and the governing laws of thermal equilibrium. Key lessons include:

Thermal Equilibrium and Zeroth Law: If two systems are each in thermal equilibrium with a third system, then they are in thermal equilibrium with each other. This law determines temperature.
Internal Energy (U): The sum of molecular kinetic and potential energies. It is a state function, meaning it depends only on the current state, not the path taken.
First Law of Thermodynamics: The principle of conservation of energy applied to thermal systems:
$Δ Q = Δ U + Δ W$ .
Thermodynamic Processes:
Isothermal: Constant temperature ( $Δ T = 0$ ).
Adiabatic: No heat exchange (Q = 0).
Isobaric: Constant pressure ( $Δ P = 0$ ).
Isochoric: Constant volume ( $Δ V = 0$ ).
Heat Engines: Devices that convert heat into work. Understanding Efficiency ( $η$ ):
$η = 1 - \frac{Q _{2}}{Q _{1}}$ .
Refrigerators and Heat Pumps: Devices that transfer heat from a cold reservoir to a hot one using external work. Understanding the Coefficient of Performance ( $α$ ).
Second Law of Thermodynamics:
Kelvin-Planck Statement: No engine can be 100% efficient.
Clausius Statement: Heat cannot of itself pass from a colder to a hotter body.
Reversible and Irreversible Processes: Understanding why real-world processes are never perfectly reversible.
Carnot Engine: The theoretical ideal engine. Mastering the Carnot Cycle and its efficiency based on temperature:
$η = 1 - \frac{T _{2}}{T _{1}}$ .

2.0Detailed NCERT Textbook Solutions : Class 11 Physics Chapter 11

Note : No Solved examples available in this chapter.

EXERCISE QUESTIONS WITH SOLUTIONS

A geyser heats water flowing at the rate of 3.0 litres per minute from 27 °C to 77 °C. If the geyser operates on a gas burner, what is the rate of consumption of the fuel if its heat of combustion is $4.0 \times 1 0^{4} J / g$ ? Sol. Water is flowing at a rate of 3.0 litre/min. The geyser heats the water, raising the temperature from 27°C to 77°C Initial temperature, $T_{1} = 2 7^{\circ} C$ Final temperature, $T_{2} = 7 7^{\circ} C$ ∴ Rise in temperature, $Δ T = T_{2} - T_{1} a$ $= 77 - 27 = 5 0^{\circ} C$ Heat of combustion $= 4 \times 1 0^{4} J / g$ Specific heat of water $= 4 \times 1 0^{4} J / g$ Mass of flowing water, $m = 3.0$ litre/min = 3000 g/min Total heat used, $Δ Q = mc Δ T$

$= 3000 \times 4.2 \times 50 = 6.3 \times 1 0^{5} J / min ∴ Rate of consumption = \frac{6.3 \times 1 0 ^{5}}{( 4 \times 1 0 ^{4} )} = 15.75 g / min$

What amount of heat must be supplied to $2.0 \times 1 0^{- 2} kg$ of nitrogen (at room temperature) to raise its temperature by 45°C at constant pressure? (molecular mass of $N_{2} = 28; R = 8.3 J$ $mol^{- 1} K^{- 1}$ ) Sol. Mass of nitrogen, $m = 2.0 \times 1 0^{- 2} kg = 20 g$ Rise in temperature, $Δ T = 4 5^{\circ} C$ Molecular mass of $N_{2}, M = 28$ Universal gas constant, $R = 8.3 J mol^{- 1} K^{- 1}$ . Number of moles, $n = \frac{m}{M} = \frac{( 2 \times 1 0 ^{- 2} \times 1 0 ^{3} )}{28}$ $= 0.714$ Molar specific heat at constant pressure for nitrogen, $C_{P} = (\frac{7}{2}) R = (\frac{7}{2}) \times 8.3$ $= 29.05 J mol^{- 1} K^{- 1}$ The total amount of heat to be supplied is given by the relation: $Δ Q = nC_{p} Δ T = 0.714 \times 29.05 \times 45$ $= 933.38 J$ Therefore, the amount of heat to be supplied is 933.38J.
Explain why (a) Two bodies at different temperatures $T_{1}$ and $T_{2}$ if brought in thermal contact do not necessarily settle to the mean temperature $\frac{( T _{1} + T _{2} )}{2}$ . (b) The coolant in a chemical or a nuclear plant (i.e., the liquid used to prevent the different parts of a plant from getting too hot) should have high specific heat. (c) Air pressure in a car tyre increases during driving. (d) The climate of a harbour town is more temperate than that of a town in a desert at the same latitude. Sol. (a) In thermal contact, heat flows from the body at higher temperature to the body at lower temperature till temperatures becomes equal. The final temperature can be the mean temperature $\frac{( T _{1} + T _{2} )}{2}$ only when thermal capacities of the two bodies are equal. (b) This is because heat absorbed by a substance is directly proportional to the specific heat of the substance.

(c) During driving, the temperature of air inside the tyre increase due to motion. According to Charle's law, $P \propto T$ . Therefore, air pressure inside the tyre increase. (d) This is because in a harbour town, the relative humidity is more than in a desert town, as well as sea breeze & land breeze are there. Hence, the climate of a harbour town is without extremes of hot and cold.

A cylinder with a movable piston contains 3 moles of hydrogen at standard temperature and pressure. The walls of the cylinder are made of a heat insulator, and the piston is insulated by having a pile of sand on it. By what factor does the pressure of the gas increase if the gas is compressed to half its original volume? Sol. The cylinder is completely insulated from its surrounding. As a result, no heat is exchanged between the system (cylinder) and its surroundings. Thus, the process is adiabatic. Initial pressure inside the cylinder $= P_{1}$ Final pressure inside the cylinder $= P_{2}$ Initial volume inside the cylinder $= V_{1}$ Final volume inside the cylinder $= V_{2}$ Ratio of specific heats, $γ = 1.4$ For an adiabatic process, we have

$P_{1} V_{1}^{γ} = P_{2} V_{2}^{γ}$

The final volume is compressed to half of its initial volume

$∴ V_{2} = \frac{V _{1}}{2} P_{1} V_{1}^{γ} = P_{2} (\frac{V _{1}}{2})^{γ} \frac{P _{2}}{P _{1}} = \frac{V _{1}^{γ}}{( \frac{V _{1}}{2} ) ^{γ}} = 2^{γ} = 2^{1.4} = 2.639$

Hence, the pressure increases by a factor of 2.639.

In changing the state of a gas adiabatically from an equilibrium state A to another equilibrium ate B, an amount of work equal to 22.3 J is done on the system. If the gas is taken from state A to B via a process in which the net heat absorbed by the system is 9.35 cal, how much is the net work done by the system in the latter case? (Take 1 cal = 4.19 J)

Sol. The work done (W) on the system while the gas changes from state (A) to state B is 22.3J. This is an adiabatic process. Hence, change in heat is zero.

$∴ Δ Q = 0$

$Δ W = - 22.3 J$ (Since the work is done on the system) From the first law of thermodynamics, we have: $Δ Q = Δ U + Δ W$ Where, $Δ U =$ Change in the internal energy of the gas

$∴ Δ U = Δ Q - Δ W = - (- 22.3 J) Δ U = + 22.3 J$

When the gas goes from state A to state B via a process, the net heat absorbed by the system is :

$Δ Q = 9.35 cal = 9.35 \times 4.19 = 39.1765 J$

Heat absorbed, $Δ Q = Δ U + Δ Q$

$∴ Δ W = Δ Q - Δ U = 39.1765 - 22.3 = 16.8765 J$

Therefore, 16.88J of work is done by the system.

Two cylinders A and B of equal capacity are connected to each other via a stopcock. A contains a gas at standard temperature and pressure. B is completely evacuated. The entire system is thermally insulated. The stopcock is suddenly opened. Answer the following: (a) What is the final pressure of the gas in A and B? (b) What is the change in internal energy of the gas? (c) What is the change in the temperature of the gas? (d) Do the intermediate states of the system (before settling to the final equilibrium state) lie on its P-V-T surface? Sol. (a) When the stopcock is suddenly opened, the volume available to the gas at 1 atmospheric pressure will become two times. Therefore, pressure will decrease to one-half, i.e., 0.5 atmosphere. (b) There will be no change in the internal energy of the gas as no work is done on/by the gas. (c) Since no work is being done by the gas during the expansion of the gas, the temperature of the gas will not change at all. (d) No, because the process called free expansion is rapid and cannot be controlled. the intermediate states are non-equilibrium states and do not satisfy the gas equation. In due course, the gas does return to an equilibrium state.
An electric heater supplies heat to a system at a rate of 100W. If system performs work at a rate of 75 joules per second. At what rate is the internal energy increasing? Sol. Heat is supplied to the system at a rate of 100W. $∴$ Heat supplied, $Q = 100 J / s$ The system performs at a rate of 75 J/s $∴$ Work done, $W = 75 J / s$ From the first law of thermodynamics, we have: $Q = U + W$ Where U = Internal energy

$∴ U = Q - W = 100 - 75 = 25 J / s = 25 W$

Therefore, the internal energy of the given system increases at a rate of 25W.

A thermodynamic system is taken from an original state to an intermediate state by the linear process shown in Figure

class-11-physics-chap-11-exercise-ques-7

Its volume is then reduced to the original value from E to F by an isobaric process. Calculate the total work done by the gas from D to E to F. Sol. Total work done by the gas from D to E to $F =$ Area of $△ DEF$ Area of $△ DEF = (\frac{1}{2}) DE \times EF$ Where

$DF FE = Change in pressure = 600 N / m^{2} - 300 N / m^{2} = 300 N / m^{2} = Change in volume = 5.0 m^{3} - 2.0 m^{3} = 3.0 m^{3}$

Area of $△ DEF = (\frac{1}{2}) \times 300 \times 3 = 450 J$ Therefore, the total work done by the gas from D to E to F is 450J.

3.0NCERT Class 11 Physics – Chapter-Wise Solutions & Explanations

Get NCERT Solutions for Class 11 Physics chapter wise with detailed answers and explanations for all the questions from the textbook. Build a strong foundation in Physics . Understand the basics. Learn problem solving step wise.

Chapter Number	Chapterwise NCERT Solutions
Chapter 1	Units and Measurements
Chapter 2	Motion in a Straight Line
Chapter 3	Motion in a Plane
Chapter 4	Laws of Motion
Chapter 5	Work, Energy and Power
Chapter 6	System of Particles and Rotational Motion
Chapter 7	Gravitation
Chapter 8	Mechanical Properties of Solids
Chapter 9	Mechanical Properties of Fluids
Chapter 10	Thermal Properties of Matter
Chapter 12	Kinetic Theory
Chapter 13	Oscillations
Chapter 14	Waves

4.0Key Features of NCERT Solutions for Class 11 Physics Chapter 11

Clear Work Sign Convention: Understand the P-V work sign convention, including work done by and on a thermodynamic system.
P-V Diagram Analysis: Learn to interpret P-V diagrams and calculate work done in cyclic and non-cyclic thermodynamic processes.
Heat Engine and Refrigerator Problems: Solve questions based on heat engines, refrigerators, heat pumps, efficiency, work output, and heat rejection.
Adiabatic Process Relations: Apply PVγ=constantPV^\gamma = \text{constant} to calculate changes in pressure, volume, and temperature during adiabatic processes.
Prepared by ALLEN Subject Experts: Get accurate and detailed NCERT Solutions for Class 11 Physics Chapter 11, prepared according to the latest NCERT syllabus.
Simple and Concept-Based Explanations: Understand the thermodynamic processes, first law of thermodynamics, heat transfer, work, internal energy and thermal efficiency with simple explanations and step-by-step solutions.
Complete NCERT Exercise Coverage: Get detailed solutions to NCERT Class 11 Physics Chapter 11 examples and exercise questions, covering important concepts of Thermodynamics for CBSE, JEE, and NEET preparation.

NCERT Solutions Class 11 Physics Chapter 11 – Thermodynamics

The NCERT Solutions for Class 11 Physics Chapter 11 (Thermodynamics) explains the macroscopic world of heat, work and internal energy. The previous chapters dealt with the properties of matter. Thermodynamics shifts our focus to the laws of energy transformation and thermal equilibrium. This chapter is fundamental to classical physics. It is the theoretical foundation for the operation of engines, refrigerators and heat pumps .

NCERT Solutions for Class 11 Physics Chapter 11 by ALLEN are designed with the help of expert faculties to bridge the gap between the law and practical use of numericals. These solutions are prepared systematically according to the CBSE, JEE and NEET requirements so that students can learn everything from the sign conventions to the complex cyclic processes. These solutions range from the theoretical elegance of the Carnot cycle to the practicalities of the First Law of Thermodynamics. They provide a rigorous basis for understanding why energy can be neither created nor destroyed, only transformed—and why some of it is always "lost" to the environment.

1.0Class 11 Physics Chapter 11 Thermodynamics: Key Concepts

This chapter examines the transformation of heat into work and the governing laws of thermal equilibrium. Key lessons include:

Thermal Equilibrium and Zeroth Law: If two systems are each in thermal equilibrium with a third system, then they are in thermal equilibrium with each other. This law determines temperature.
Internal Energy (U): The sum of molecular kinetic and potential energies. It is a state function, meaning it depends only on the current state, not the path taken.
First Law of Thermodynamics: The principle of conservation of energy applied to thermal systems:
$Δ Q = Δ U + Δ W$ .
Thermodynamic Processes:
Isothermal: Constant temperature ( $Δ T = 0$ ).
Adiabatic: No heat exchange (Q = 0).
Isobaric: Constant pressure ( $Δ P = 0$ ).
Isochoric: Constant volume ( $Δ V = 0$ ).
Heat Engines: Devices that convert heat into work. Understanding Efficiency ( $η$ ):
$η = 1 - \frac{Q _{2}}{Q _{1}}$ .
Refrigerators and Heat Pumps: Devices that transfer heat from a cold reservoir to a hot one using external work. Understanding the Coefficient of Performance ( $α$ ).
Second Law of Thermodynamics:
Kelvin-Planck Statement: No engine can be 100% efficient.
Clausius Statement: Heat cannot of itself pass from a colder to a hotter body.
Reversible and Irreversible Processes: Understanding why real-world processes are never perfectly reversible.
Carnot Engine: The theoretical ideal engine. Mastering the Carnot Cycle and its efficiency based on temperature:
$η = 1 - \frac{T _{2}}{T _{1}}$ .

2.0Detailed NCERT Textbook Solutions : Class 11 Physics Chapter 11

Note : No Solved examples available in this chapter.

EXERCISE QUESTIONS WITH SOLUTIONS

A geyser heats water flowing at the rate of 3.0 litres per minute from 27 °C to 77 °C. If the geyser operates on a gas burner, what is the rate of consumption of the fuel if its heat of combustion is $4.0 \times 1 0^{4} J / g$ ? Sol. Water is flowing at a rate of 3.0 litre/min. The geyser heats the water, raising the temperature from 27°C to 77°C Initial temperature, $T_{1} = 2 7^{\circ} C$ Final temperature, $T_{2} = 7 7^{\circ} C$ ∴ Rise in temperature, $Δ T = T_{2} - T_{1} a$ $= 77 - 27 = 5 0^{\circ} C$ Heat of combustion $= 4 \times 1 0^{4} J / g$ Specific heat of water $= 4 \times 1 0^{4} J / g$ Mass of flowing water, $m = 3.0$ litre/min = 3000 g/min Total heat used, $Δ Q = mc Δ T$

$= 3000 \times 4.2 \times 50 = 6.3 \times 1 0^{5} J / min ∴ Rate of consumption = \frac{6.3 \times 1 0 ^{5}}{( 4 \times 1 0 ^{4} )} = 15.75 g / min$

What amount of heat must be supplied to $2.0 \times 1 0^{- 2} kg$ of nitrogen (at room temperature) to raise its temperature by 45°C at constant pressure? (molecular mass of $N_{2} = 28; R = 8.3 J$ $mol^{- 1} K^{- 1}$ ) Sol. Mass of nitrogen, $m = 2.0 \times 1 0^{- 2} kg = 20 g$ Rise in temperature, $Δ T = 4 5^{\circ} C$ Molecular mass of $N_{2}, M = 28$ Universal gas constant, $R = 8.3 J mol^{- 1} K^{- 1}$ . Number of moles, $n = \frac{m}{M} = \frac{( 2 \times 1 0 ^{- 2} \times 1 0 ^{3} )}{28}$ $= 0.714$ Molar specific heat at constant pressure for nitrogen, $C_{P} = (\frac{7}{2}) R = (\frac{7}{2}) \times 8.3$ $= 29.05 J mol^{- 1} K^{- 1}$ The total amount of heat to be supplied is given by the relation: $Δ Q = nC_{p} Δ T = 0.714 \times 29.05 \times 45$ $= 933.38 J$ Therefore, the amount of heat to be supplied is 933.38J.
Explain why (a) Two bodies at different temperatures $T_{1}$ and $T_{2}$ if brought in thermal contact do not necessarily settle to the mean temperature $\frac{( T _{1} + T _{2} )}{2}$ . (b) The coolant in a chemical or a nuclear plant (i.e., the liquid used to prevent the different parts of a plant from getting too hot) should have high specific heat. (c) Air pressure in a car tyre increases during driving. (d) The climate of a harbour town is more temperate than that of a town in a desert at the same latitude. Sol. (a) In thermal contact, heat flows from the body at higher temperature to the body at lower temperature till temperatures becomes equal. The final temperature can be the mean temperature $\frac{( T _{1} + T _{2} )}{2}$ only when thermal capacities of the two bodies are equal. (b) This is because heat absorbed by a substance is directly proportional to the specific heat of the substance.

(c) During driving, the temperature of air inside the tyre increase due to motion. According to Charle's law, $P \propto T$ . Therefore, air pressure inside the tyre increase. (d) This is because in a harbour town, the relative humidity is more than in a desert town, as well as sea breeze & land breeze are there. Hence, the climate of a harbour town is without extremes of hot and cold.

A cylinder with a movable piston contains 3 moles of hydrogen at standard temperature and pressure. The walls of the cylinder are made of a heat insulator, and the piston is insulated by having a pile of sand on it. By what factor does the pressure of the gas increase if the gas is compressed to half its original volume? Sol. The cylinder is completely insulated from its surrounding. As a result, no heat is exchanged between the system (cylinder) and its surroundings. Thus, the process is adiabatic. Initial pressure inside the cylinder $= P_{1}$ Final pressure inside the cylinder $= P_{2}$ Initial volume inside the cylinder $= V_{1}$ Final volume inside the cylinder $= V_{2}$ Ratio of specific heats, $γ = 1.4$ For an adiabatic process, we have

$P_{1} V_{1}^{γ} = P_{2} V_{2}^{γ}$

The final volume is compressed to half of its initial volume

$∴ V_{2} = \frac{V _{1}}{2} P_{1} V_{1}^{γ} = P_{2} (\frac{V _{1}}{2})^{γ} \frac{P _{2}}{P _{1}} = \frac{V _{1}^{γ}}{( \frac{V _{1}}{2} ) ^{γ}} = 2^{γ} = 2^{1.4} = 2.639$

Hence, the pressure increases by a factor of 2.639.

In changing the state of a gas adiabatically from an equilibrium state A to another equilibrium ate B, an amount of work equal to 22.3 J is done on the system. If the gas is taken from state A to B via a process in which the net heat absorbed by the system is 9.35 cal, how much is the net work done by the system in the latter case? (Take 1 cal = 4.19 J)

Sol. The work done (W) on the system while the gas changes from state (A) to state B is 22.3J. This is an adiabatic process. Hence, change in heat is zero.

$∴ Δ Q = 0$

$Δ W = - 22.3 J$ (Since the work is done on the system) From the first law of thermodynamics, we have: $Δ Q = Δ U + Δ W$ Where, $Δ U =$ Change in the internal energy of the gas

$∴ Δ U = Δ Q - Δ W = - (- 22.3 J) Δ U = + 22.3 J$

When the gas goes from state A to state B via a process, the net heat absorbed by the system is :

$Δ Q = 9.35 cal = 9.35 \times 4.19 = 39.1765 J$

Heat absorbed, $Δ Q = Δ U + Δ Q$

$∴ Δ W = Δ Q - Δ U = 39.1765 - 22.3 = 16.8765 J$

Therefore, 16.88J of work is done by the system.

Two cylinders A and B of equal capacity are connected to each other via a stopcock. A contains a gas at standard temperature and pressure. B is completely evacuated. The entire system is thermally insulated. The stopcock is suddenly opened. Answer the following: (a) What is the final pressure of the gas in A and B? (b) What is the change in internal energy of the gas? (c) What is the change in the temperature of the gas? (d) Do the intermediate states of the system (before settling to the final equilibrium state) lie on its P-V-T surface? Sol. (a) When the stopcock is suddenly opened, the volume available to the gas at 1 atmospheric pressure will become two times. Therefore, pressure will decrease to one-half, i.e., 0.5 atmosphere. (b) There will be no change in the internal energy of the gas as no work is done on/by the gas. (c) Since no work is being done by the gas during the expansion of the gas, the temperature of the gas will not change at all. (d) No, because the process called free expansion is rapid and cannot be controlled. the intermediate states are non-equilibrium states and do not satisfy the gas equation. In due course, the gas does return to an equilibrium state.
An electric heater supplies heat to a system at a rate of 100W. If system performs work at a rate of 75 joules per second. At what rate is the internal energy increasing? Sol. Heat is supplied to the system at a rate of 100W. $∴$ Heat supplied, $Q = 100 J / s$ The system performs at a rate of 75 J/s $∴$ Work done, $W = 75 J / s$ From the first law of thermodynamics, we have: $Q = U + W$ Where U = Internal energy

$∴ U = Q - W = 100 - 75 = 25 J / s = 25 W$

Therefore, the internal energy of the given system increases at a rate of 25W.

A thermodynamic system is taken from an original state to an intermediate state by the linear process shown in Figure

Its volume is then reduced to the original value from E to F by an isobaric process. Calculate the total work done by the gas from D to E to F. Sol. Total work done by the gas from D to E to $F =$ Area of $△ DEF$ Area of $△ DEF = (\frac{1}{2}) DE \times EF$ Where

$DF FE = Change in pressure = 600 N / m^{2} - 300 N / m^{2} = 300 N / m^{2} = Change in volume = 5.0 m^{3} - 2.0 m^{3} = 3.0 m^{3}$

Area of $△ DEF = (\frac{1}{2}) \times 300 \times 3 = 450 J$ Therefore, the total work done by the gas from D to E to F is 450J.

3.0NCERT Class 11 Physics – Chapter-Wise Solutions & Explanations

Get NCERT Solutions for Class 11 Physics chapter wise with detailed answers and explanations for all the questions from the textbook. Build a strong foundation in Physics . Understand the basics. Learn problem solving step wise.

Chapter Number	Chapterwise NCERT Solutions
Chapter 1	Units and Measurements
Chapter 2	Motion in a Straight Line
Chapter 3	Motion in a Plane
Chapter 4	Laws of Motion
Chapter 5	Work, Energy and Power
Chapter 6	System of Particles and Rotational Motion
Chapter 7	Gravitation
Chapter 8	Mechanical Properties of Solids
Chapter 9	Mechanical Properties of Fluids
Chapter 10	Thermal Properties of Matter
Chapter 12	Kinetic Theory
Chapter 13	Oscillations
Chapter 14	Waves

4.0Key Features of NCERT Solutions for Class 11 Physics Chapter 11

Clear Work Sign Convention: Understand the P-V work sign convention, including work done by and on a thermodynamic system.
P-V Diagram Analysis: Learn to interpret P-V diagrams and calculate work done in cyclic and non-cyclic thermodynamic processes.
Heat Engine and Refrigerator Problems: Solve questions based on heat engines, refrigerators, heat pumps, efficiency, work output, and heat rejection.
Adiabatic Process Relations: Apply PVγ=constantPV^\gamma = \text{constant} to calculate changes in pressure, volume, and temperature during adiabatic processes.
Prepared by ALLEN Subject Experts: Get accurate and detailed NCERT Solutions for Class 11 Physics Chapter 11, prepared according to the latest NCERT syllabus.
Simple and Concept-Based Explanations: Understand the thermodynamic processes, first law of thermodynamics, heat transfer, work, internal energy and thermal efficiency with simple explanations and step-by-step solutions.
Complete NCERT Exercise Coverage: Get detailed solutions to NCERT Class 11 Physics Chapter 11 examples and exercise questions, covering important concepts of Thermodynamics for CBSE, JEE, and NEET preparation.

Frequently Asked Questions

Why are the NCERT Solutions for Class 11 Physics Chapter 11 Thermodynamics important for students?

What concepts are covered in NCERT Solutions for Class 11 Physics Chapter 11 Thermodynamics?

Is NCERT Solutions for Class 11 Physics Chapter 11 by ALLEN useful for competitive examinations?

What is the difference between Isothermal and Adiabatic process?

What is the First Law of Thermodynamics in Class 11 Physics?

Join ALLEN!

NCERT Solutions Class 11 Physics Chapter 11 – Thermodynamics

1.0Class 11 Physics Chapter 11 Thermodynamics: Key Concepts

2.0Detailed NCERT Textbook Solutions : Class 11 Physics Chapter 11

Note : No Solved examples available in this chapter.

EXERCISE QUESTIONS WITH SOLUTIONS

3.0NCERT Class 11 Physics – Chapter-Wise Solutions & Explanations

4.0Key Features of NCERT Solutions for Class 11 Physics Chapter 11

Table of Contents

Frequently Asked Questions

Why are the NCERT Solutions for Class 11 Physics Chapter 11 Thermodynamics important for students?

What concepts are covered in NCERT Solutions for Class 11 Physics Chapter 11 Thermodynamics?

Is NCERT Solutions for Class 11 Physics Chapter 11 by ALLEN useful for competitive examinations?

What is the difference between Isothermal and Adiabatic process?

What is the First Law of Thermodynamics in Class 11 Physics?

Join ALLEN!

NCERT Solutions Class 11 Physics Chapter 11 – Thermodynamics

1.0Class 11 Physics Chapter 11 Thermodynamics: Key Concepts

2.0Detailed NCERT Textbook Solutions : Class 11 Physics Chapter 11

Note : No Solved examples available in this chapter.

EXERCISE QUESTIONS WITH SOLUTIONS

3.0NCERT Class 11 Physics – Chapter-Wise Solutions & Explanations

4.0Key Features of NCERT Solutions for Class 11 Physics Chapter 11

Table of Contents