NCERT Solutions for Class 11 Physics Chapter 11 Thermodynamics

NCERT Solutions for Class 11 Physics Chapter 11 Thermodynamics

Vishal kumarUpdated on 24 Sep 2025, 01:17 AM IST

Have you ever asked yourself why it takes a long time to cool down a cup of hot tea or why the industrial revolution was run by steam engines? Such real-life phenomena were described in a clear and organised manner in the Thermodynamics Class 11 NCERT Solutions. The step-by-step solutions allow the students to learn about the transfer of energy, how heat can be converted into work, and how engines, refrigerators, and heat processes work.

This Story also Contains

  1. Class 11 Physics Chapter 11 - Thermodynamics: Download Solution PDF
  2. Thermodynamics NCERT Solutions: Exercise Questions
  3. Thermodynamics NCERT Solutions: Higher Order Thinking Skills (HOTS) Questions
  4. Class 11 Physics Chapter 11 - Thermodynamics: Topics
  5. Class 11 Physics Chapter 11 - Thermodynamics: Important Formulae
  6. Approach to Solve Questions of Class 11 Physics Chapter 11 - Thermodynamics
  7. What Extra Should Students Study Beyond NCERT for JEE/NEET?
  8. NCERT Solutions for Class 11 Physics Chapter Wise

These NCERT Solutions for Class 11 Physics Chapter 11 - Thermodynamics, as prepared by experts in the subject, according to the latest syllabus of CBSE Class 11 Physics, simplify difficult concepts, and they are easier to revise and remember. They are extremely helpful in school and board exams as well as in competitive exams such as JEE and NEET, in which a clear understanding of concepts and problem-solving skills are decisive. Through the NCERT solutions, students will be able to enhance their mastery of the laws of thermodynamics, energy conservation and power engines, as well as increase logic and confidence in exams. By using these elaborate NCERT Solutions for Class 11 Physics Chapter 11 - Thermodynamics, one will be able to master Thermodynamics with a lot of interest and concentrate on the exam.

Class 11 Physics Chapter 11 - Thermodynamics: Download Solution PDF

The Class 11 Physics Chapter 11 - Thermodynamics question answers give well-elaborated and step-wise Solutions to all questions in the textbook, thereby enabling the student to solidify his/her knowledge of the specific subject. The best part about this solution is that it is better suited when preparing to write exams, and it is in the form of an easy-to-download PDF file that can be revised quickly and at any time it is required.

Download PDF

Thermodynamics NCERT Solutions: Exercise Questions

The Thermodynamics class 11 question answers are step-by-step explanations to the exercise questions, and this makes the complex ideas easier to assimilate. These solutions equip students with the skills to master concepts such as heat, work, internal energy and the laws of thermodynamics to help them adequately prepare to face both the board examination and competitive examinations like JEE and NEET.

Q11.1 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 4.0×104 J/g?

Answer:

The volumetric flow of water is

dVdt=3 litres minute1

Density of water = 1000 g/litre

The mass flow rate of water is

dmdt=ρdVdt
dmdt=3000 g min1

Specific heat of water, c = 4.2 J g-1 oC-1

The rise in temperature is ΔT=7727=50 oC

The rate of energy consumption will be

dQdt=dmdtcΔT


dQdt=3000×4.2×50


dQdt=6.3×105 J min1

The heat of combustion of fuel =4.0×104J/g

The rate of consumption of fuel is

6.3×1054×104=15.75 g min1

Q11.2 What amount of heat must be supplied to 2.0×102 kg of nitrogen (at room temperature) to raise its temperature by 45 °C at constant pressure? (Molecular mass of N=28; R=8.3 J mol1K1 .)

Answer:

Mass of nitrogen, m=2.0×102kg=20g

Molar Mass of nitrogen, MN = 28 g

The number of moles is n

n=mMN


n=2028


n=0.714

As nitrogen is a diatomic gas, its molar specific heat at constant pressure CP is as follows

CP=7R2


CP=7×8.32


CP=29.05 J mol1 oC1

Rise in temperature, ΔT=45oC1

The amount of heat Q that must be supplied is

Q=nCPΔT


Q=0.714×29.05×45


Q=933.38 J

Q11.3 (a) Explain why two bodies at different temperatures T1 and T2 if brought in thermal contact do not necessarily settle to the mean temperature (T1+T2)/2 .

Answer:

As we know, heat will flow from the hotter body to the colder body till their temperatures become equal. That temperature will be equal to the mean of the initial temperatures of the bodies only if the two thermal capacities of the two bodies are equal.

Q11.3 (b) Explain why 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.

Answer:

The coolant should have a high specific heat so that it can absorb large amounts of heat without itself getting too hot, and its temperature lies in the permissible region. The higher the specific heat, the more heat will be absorbed by the same amount of material for a given increase in temperature.

Q11.3(c) Explain why air pressure in a car tyre increases during driving.

Answer:

As the car is driven, the air inside the tyre heats due to frictional forces. Thus, the temperature of the air inside the tyre increases and this, in turn, increases the pressure inside the tyre.

Q11.3 (d) Explain why the climate of a harbour town is more temperate than that of a town in a desert at the same latitude.

Answer:

The climate of a harbour town is more temperate than that of a town in a desert at the same latitude because of the formation of sea breezes.

Q11.4 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?

Answer:

As the walls of the cylinder and the piston are insulated, the process will be adiabatic. i.e. PVγ would be constant.

Hydrogen is a diatomic gas and therefore γ=1.4

Let the initial and final pressures be P1 and P2, respectively.

Let the initial and final volumes be V1 and V2, respectively.

P1V1γ=P2V2γ
P2P1=(V1V2)γ
P2P1=21.4
P2P1=2.639

The pressure thus increases by a factor of 2.639

Q11.5 In changing the state of a gas adiabatically from an equilibrium state A to another equilibrium state 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 1cal=4.19J )

Answer:

In the first case, the process is adiabatic, i.e. ΔQ=0

22.3 J work is done on the system, i.e. ΔW=22.3 J

ΔQ=ΔU+ΔW

0=ΔU22.3

ΔU=22.3 J

Since in the latter process as well,l the initial and final states are the same as those in the former process, ΔU will remain the same for the latter case.

In the latter case, the net heat absorbed by the system is 9.35 cal

ΔQ=9.35×4.2
ΔQ=39.3 J

ΔW=ΔQΔU

ΔW=39.322.3

ΔW=17.0 J

The network done by the system in the latter case is 17.0 J

Q11.6 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 the 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?

Answer:

As the entire system is thermally insulated and as free expansion will be taking place, the temperature of the gas remains the same. Therefore, PV is constant.

Initial Pressure P1 = 1 atm

Initial Volume, V1 = V

Final Volume, V2 = 2V

Final Pressure P2 will be

P2=P1V1V2
P2=P12
P2=0.5 atm

The final pressure of the gas in A and B is 0.5 atm.

b) Since the temperature of the gas does not change, its internal energy would also remain the same.

c) As the entire system is thermally insulated and as free expansion will be taking place, the temperature of the gas remains the same.

d) The intermediate states of the system do not lie on its P-V-T surface, as the process is a free expansion, it is rapid, and the intermediate states are non-equilibrium states.

Q11.7 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?

Answer:

The rate at which heat is supplied ΔQ=100 W

The rate at which work is done ΔW=75 Js1

The rate of change of internal energy is Δu

ΔU=ΔQΔWΔU=10075ΔU=25 J s1

The internal energy of the system is increasing at a rate of 25 J s-1

Q11.8 A thermodynamic system is taken from an original state to an intermediate state by the linear process shown in Fig.

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

Answer:

The work done by the gas as it goes from state D to E to F is equal to the area of triangle DEF

DF is change in pressure = 300 N m-2

FE is a change in Volume = 3 m3

area(DEF)=12×DF×FE

area(DEF)=12×300×3

area(DEF)=450 J

The work done is, therefore, 450 J.

Thermodynamics NCERT Solutions: Higher Order Thinking Skills (HOTS) Questions

The HOTS (Higher Order Thinking Skills) Questions for Class 11 Physics Chapter 11 - Thermodynamics are designed to push students beyond basic learning and encourage deeper conceptual understanding. These challenging problems strengthen analytical skills, logical reasoning, and application of thermodynamics principles, which are highly useful for JEE, NEET, and advanced exam preparation.

Q.1 Three moles of an ideal gas initially at a temperature T0=273K were isothermally expanded to n = 5.0 times its initial volume and then isochorically heated so that the pressure in the final state becomes equal to that in the initial state. The total amount of heat transferred to the gas during the process equals Q = 80 kJ. Find the ratio γ=CP/CV for this gas.

Answer:

Q=Q1+Q2

where
Q1 for Isothermal process
Q2 for Isochoric process

Q=nRT0lognV0V0+nCV(TCTB)

As TC=nT0 (If volume is made n times)

Q=nRT0logn+γ(n1)T0

γ=[1+n1QnRT0logn]

P-V graph

Putting the values

(Given n' = 3, n = 5)

γ=1.4

Hence, the answer is 1.4.

Q.2 In a pressure cooker, a sample of hydrogen gas was at a pressure of 1 atm. The volume of the cooker is 500 mL, and the mass of the cooker whistle is 100 g, and it has a cross-section area of 0.1cm2. Its initial temperature is 27°C. Select the correct option(s):

a) The final temperature of the gas if the gas can lift the whistle is 327°C
b) The final temperature of the gas if the gas can lift the whistle is 227°C
c) The pressure required to lift the whistle is 2×105Pa
d) The pressure required to lift the whistle is 105Pa

Answer:

The initial temperature of hydrogen, T1=273+27=300 K
Initial pressure of hydrogen, P1=1 atm=105 Pa

P2=P0+mgA=105+100×1030.01×104=2×105

As the volume is constant, we can use Gay Lussac's law:

P1T1=P2T2T2=T1×P2P1=300×21=600 K=600273=327C

Hence, the answer is option (a),(c).


Q.3 The volume of a gas is compressed adiabatically from state V1=7 m3 at a pressure of 2..4×105 N m2 to the volume V2=0.875 m3. If the same compression is performed isothermally, calculate the difference in work done (in 105 joules) in both processes. It is given that γ= 1.67.

Answer:

V1=7 m3,P1=2.4×105Nm2,V2=0.875 m3

In an adiabatic process

P1V1γ=P2V2γP2=2.4×105[70.875]1.6P2=2.4×105(8)53=7.68×106 N m2

Work done in an adiabatic process,

W=P2V2P1V1γ1W=(7.68×106×0.875)(2.4×105×7)1.671W=75.22×105 J

Work done in the isothermal process,

W=2.3026RTlogV2V1=2.3026P1V1logV2V1=2.3026×2.4×105×7×log[0.8757.0]=2.3026×2.4×105×7log(18)=34×105 J

Difference in work done =75.22×105 J+34×105 J=41.22×105 J

Q.4 1 mole of helium expands with temperature according to the relation V=KT2/3. If the temperature changes by 60 K, find the heat absorbed (in J) by the monoatomic gas in the above process.

a) 1080.30
b) 1120.32
c) 1200
d) 1150.20

Answer:

V=KT2/3V=K(PVnR)2/3PV1/2=αC=CV+R1x=3R2+R1+12=13R6Q=nCΔT=1×13R6×60=130R=1080.30 J

Hence, the answer is option (a).

Q.5. A diatomic ideal gas is heated at constant volume until the pressure is doubled and again heated at constant pressure until volume is doubled. The average molar heat capacity for the whole process is abR. Find the remainder when a is divided by b

Answer:

Let initial pressure, volume and temperature be P0,V0,T0 indicated by state A in the P-V diagram. The gas is isochorically taken to state B(2P0,V0,2T0) and then taken from state B to state C(2P0,2V0,4T0) isobarically.

P-V graph

Total heat absorbed by 1 mole of gas

ΔQ=Cv(2T0T0)+CP(4T02T0)=52RT0+72R×2T0=192RT0
Total change in temperature from state A to C is: ΔT=3T0
Molar heat capacity =ΔQΔT=192RT03T0=196R

Hence, the answer is 1.

Class 11 Physics Chapter 11 - Thermodynamics: Topics

The NCERT Class 11 Physics Chapter 11 Thermodynamics Topics provide a structured breakdown of all key concepts covered in the chapter. Since the laws of thermodynamics to the heat engines, refrigerators, and energy conservation, all concepts have been clearly explained to understand. Such topic-related information will assist the students in revising in a systematic manner and enhancing the preparation of board exams as well as competitive exams.

11.1 Introduction
11.2 Thermal Equilibrium
11.3 Zeroth Law Of Thermodynamics
11.4 Heat, Internal Energy And Work
11.5 First Law Of Thermodynamics
11.6 Specific Heat Capacity
11.7Thermodynamic State Variables And Equation Of State
11.8 Thermodynamic Processes
11.8.1 Quasi-Static Process
11.8.2 Isothermal Process
11.8.3 Adiabatic Process
11.8.4 Isochoric Process
11.8.5 Isobaric Process
11.8.6 Cyclic Process
11.9 Second Law Of Thermodynamics
11.10 Reversible And Irreversible Processes
11.11 Carnot Engine


Class 11 Physics Chapter 11 - Thermodynamics: Important Formulae

The Key Formulae in Class 11 Physics Chapter 11 - Thermodynamics serve as a quick revision guide to help solve the exercise questions and competitive examination problems. These formulas include relevant concepts such as heat, work, internal energy, and laws of thermodynamics, which students can use to enhance their ability to solve problems faster and accurately in exams such as JEE and NEET.

1. First Law of Thermodynamics

ΔQ=ΔU+ΔW
Where:
ΔQ= heat supplied to the system
ΔU= change in internal energy
ΔW= work done by the system

2. Work Done in an Isothermal Process

W=nRTln(VfVi)

(For ideal gas, T constant)

3. Work Done in Adiabatic Process

W=PiViPfVfγ1=nR(TiTf)γ1
Where γ=CpCv (adiabatic index)

4. Adiabatic Relation for Ideal Gas

PVγ= constant

5. Molar Specific Heats

For ideal gas:

CpCv=R

6. Internal Energy of Ideal Gas

U=f2nRT
Where f= degrees of freedom
For monoatomic gas: U=32nRT

7. Heat Capacity

C=dQdT,C=nCm
Where Cm= molar specific heat

9. Carnot Engine Efficiency

η=1T2T1
Where T1= temperature of source, T2= temperature of sink (in Kelvin)

Approach to Solve Questions of Class 11 Physics Chapter 11 - Thermodynamics

Solving questions from Class 11 Physics Chapter 11 - Thermodynamics requires a balance of conceptual clarity and practical application. Students should focus on understanding the laws of thermodynamics, interpreting heat, work, and internal energy relations, and applying formulas correctly. With a step-by-step approach, even complex numerical and reasoning-based questions become manageable and exam-ready.

  • Understand The Basic Thermodynamics Terms First
    Start by learning important terms like system, surroundings, internal energy, heat, and work. These are the basics that are required to solve thermodynamics questions.
  • Learn the Thermodynamic Laws
    Focus on the First Law of Thermodynamics and how energy is conserved. Know what happens to internal energy when heat is added or work is done.
  • Types of Processes
    Learn isothermal, adiabatic, isobaric, and isochoric processes. Know how temperature, volume, and pressure behave in each case.
  • Use PV Diagrams
    Practice sketching pressure-volume (PV) graphs they help you easily understand what kind of thermodynamic process is happening and how work is done.
  • Revise important formulas
    Revise formulas for work done in different processes, internal energy change (ΔU) and heat exchange (Q). These are very important for exam point of view.
  • Apply the first law in problems
    Use ΔQ = ΔU + W wisely. Take care sign conventions whether heat is gained or lost, and whether work is done by or on the system.
  • Solve NCERT and PYQs
    Practice a mix of conceptual and numerical questions. Focus on NCERT exercise questions as they directly appear in exams.

What Extra Should Students Study Beyond NCERT for JEE/NEET?

Although the foundation is made in NCERT Class 11 Physics Chapter 11 - Thermodynamics, the competitive examination such as JEE and NEET requires a deeper understanding and broad implementation of the concepts. Students should not only limit themselves to problems of the textbook, but they should also delve into advanced numerical problems, derivations and practical use of thermodynamic laws. The following table gives a comparison of the scope of NCERT vs. JEE/ NEET preparation to enable the students to prepare better.

NCERT Solutions for Class 11 Physics Chapter Wise

The NCERT Solutions for Class 11 Physics Chapter-wise links provide students with easy access to detailed, step-by-step answers for every chapter. These solutions help in building strong conceptual clarity, quick problem-solving skills, and effective exam preparation for both CBSE board exams and competitive tests like JEE and NEET.

NCERT Solutions for Class 11 Subject Wise

Also, check NCERT Books and NCERT Syllabus here

Subject Wise NCERT Exemplar Solutions

If you have any queries or uncertainties that are not addressed in the class 11 physics chapter 11 exercise solutions or any other chapter, you can contact us. You will get answers to your questions, which can aid you in achieving better scores in your exams.

Frequently Asked Questions (FAQs)

Q: Why is NCERT Solution Class 11 Physics important?
A:

They come in the form of step-by-step explanations to all NCERT textbook questions and thus are very useful in enabling students to comprehend the concepts properly and go into their exams well prepared.

Q: Are NCERT Solutions sufficient to prepare for JEE and NEET?
A:

Both JEE and NEET are based on NCERT, but students should also practise some reference books and solve higher-order problems to do well in the competitive exams.

Q: Is it possible to download the NCERT Solutions Class 11 Physics PDF file?
A:

Yes, NCERT Solutions are provided by us in an offline form in PDF format, chapter-wise.

Q: What are the benefits of chapter-wise solutions in exam preparation?
A:

They are prepared in such a way that they simplify revision since the concepts and questions are arranged chapter to chapter, making learning focused.

Q: Do NCERT Solutions contain major formulas?
A:

Yes, most solutions point to important formulas, and shortcuts, and they are very important in solving numerical problems in exams.

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