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.
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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.
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.
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.
Answer:
The volumetric flow of water is
Density of water = 1000 g/litre
The mass flow rate of water is
Specific heat of water, c = 4.2 J g-1 oC-1
The rise in temperature is
The rate of energy consumption will be
The heat of combustion of fuel
The rate of consumption of fuel is
Answer:
Mass of nitrogen,
Molar Mass of nitrogen, MN = 28 g
The number of moles is n
As nitrogen is a diatomic gas, its molar specific heat at constant pressure CP is as follows
Rise in temperature,
The amount of heat Q that must be supplied is
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.
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.
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.
Answer:
As the walls of the cylinder and the piston are insulated, the process will be adiabatic. i.e.
Hydrogen is a diatomic gas and therefore
Let the initial and final pressures be P1 and P2, respectively.
Let the initial and final volumes be V1 and V2, respectively.
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
Answer:
In the first case, the process is adiabatic, i.e.
22.3 J work is done on the system, i.e.
Since in the latter process as well,l the initial and final states are the same as those in the former process,
In the latter case, the net heat absorbed by the system is 9.35 cal
The network done by the system in the latter case is 17.0 J
(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?
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
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.
Answer:
The rate at which heat is supplied
The rate at which work is done
The rate of change of internal energy is
The internal energy of the system is increasing at a rate of 25 J s-1
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
The work done is, therefore, 450 J.
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
Answer:
where
As
Putting the values
(Given n' = 3, n = 5)
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
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
d) The pressure required to lift the whistle is
Answer:
The initial temperature of hydrogen,
Initial pressure of hydrogen,
As the volume is constant, we can use Gay Lussac's law:
Hence, the answer is option (a),(c).
Q.3 The volume of a gas is compressed adiabatically from state
Answer:
In an adiabatic process
Work done in an adiabatic process,
Work done in the isothermal process,
Difference in work done
Q.4 1 mole of helium expands with temperature according to the relation
a) 1080.30
b) 1120.32
c) 1200
d) 1150.20
Answer:
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
Answer:
Let initial pressure, volume and temperature be
Total heat absorbed by 1 mole of gas
Total change in temperature from state A to C is:
Hence, the answer is 1.
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
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.
Where:
(For ideal gas,
Where
For ideal gas:
Where
For monoatomic gas:
Where
Where
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.
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.
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.
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)
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.
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.
Yes, NCERT Solutions are provided by us in an offline form in PDF format, chapter-wise.
They are prepared in such a way that they simplify revision since the concepts and questions are arranged chapter to chapter, making learning focused.
Yes, most solutions point to important formulas, and shortcuts, and they are very important in solving numerical problems in exams.
This ebook serves as a valuable study guide for NEET 2025 exam.
This e-book offers NEET PYQ and serves as an indispensable NEET study material.
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