# NCERT Solutions for Class 11 Physics Chapter 12

## NCERT Solutions for Class 11 Physics Chapter 12: Thermodynamics

Physics, which is the study of matter, motion, energy, and the behaviour of space and time, necessitates extensive reading and understanding. Thermal equilibrium, state variables, quasistatic process, sign convention, adiabatic, and other essential subtopics are included in the Thermodynamics Class 11 Physics topic.

Students are often confused about what to study and how to understand the laws underlying each Thermodynamics calculation. To address this issue, NCERT Solutions for Class 11 Physics Chap 12 are written in an engaging and easy manner to make learning more enjoyable.

The class 11th Physics Thermodynamics NCERT solutions are written in such a way that the notion of its various parameters, such as temperature, volume, and internal energy, is clear to students.

NCERT Solutions for Class 11 Physics Chapter 12 includes examples to assist students in revising and improving their grades. Furthermore, the answers are written in accordance with the latest CBSE pattern r to ensure good exam results.

### Access NCERT Solutions for Class 11 Physics, Thermodynamics.

Thermodynamics is one of the most scoring sections in Class 11 Physics. To do well on the term exam, students must read this chapter thoroughly. This is where referring to Thermodynamics Physics Class 11 Ncert Solutions can be of great help. The solutions have answers to all the questions that are listed at the end of Chapter 12 of the NCERT textbook. Students can access the solutions on the Extramarks website.

#### NCERT Solutions Class 11 Physics: Thermodynamics

NCERT Solutions for  Thermodynamics covers all of the major topics in detail so that students are able to grasp the concepts easily. Going over the solutions can assist you in understanding how to approach and answer difficult questions.

Subtopics of Class 11 Physics Chapter 12 Thermodynamics

1. Introduction
2. Thermal Equilibrium
3. Zeroth Law of Thermodynamics
4. Heat, Internal energy and work
5. The first law of thermodynamics
6. Specific heat capacity
7. Thermodynamic state variables and the equation of state
8. Thermodynamic processes
9. Heat engines
10. Refrigerators and heat pumps
11. The second law of thermodynamics
12. Reversible and irreversible processes
13. Carnot Engine

Extramarks is committed to providing you with study materials like notes, previous year question papers, sample papers, MCQs, short answer questions, long answer questions, and entrance exam textbooks. Extramarks interactive videos, animations, and infographics assist students in retaining information for a long time.

#### Benefits of Practicing from NCERT Solutions for Class 11 Physics:  Thermodynamics

A lack of necessary study materials can impair one’s ability to write answers and analyze critically. As a result,  NCERT answers are prepared in a way that will aid in the learning process.

• Solutions have answers to all the practise questions given at the end of the NCERT textbook. Students can refer to these solutions instead of looking around for answers elsewhere and wasting their valuable time.
• To make complex subjects easier to understand, the answers are explained in a simple and easy manner.

#### NCERT Solutions for Class 11 Physics Chapter 12

NCERT Solutions for Class 11 Physics Chapter 12 are prepared by subject matter experts at Extramarks, while ensuring that every answer meets the guidelines of CBSE. The answers are written in simple language and relevant examples, as well as diagrams, are used, wherever required, to explain the concept.

Q.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 Jg-1?

Ans.

$Here,initial temperature of water, T 1 = 27°C Final temperature of water, T 2 = 77°C ∴ Rise in temperature, ΔT = T 2 –T 1 ΔT= 77°C – 27°C = 50°C Heat of combustion = 4×10 4 Jg -1 Specific heat of water, c = 4 .2 J g –1 °C –1 Mass of water heated, m = 3 .0 litremin -1 = 3000 gmin -1 Total amount of heat used is given as: ΔQ = mc ΔT ΔQ = 3000 gmin -1 × 4 .2 J g –1 °C –1 × 50°C = 6 .3×10 5 Jmin -1 ∴Rate of consumption of fuel = 6 .3 × 10 5 Jmin -1 4 × 10 4 Jg -1 ∴Rate of consumption of fuel = 15 .75 gmin -1$

Q.2 What amount of heat must be supplied to 2.0 × 10–2 kg of nitrogen (at room temperature) to raise its temperature by 45°C at constant pressure? (Molecular mass of N2 = 28; R = 8.3 J mol–1 K–1.)

Ans.

$Here,mass of nitrogen, m = 2 .0 ×10 –2 kg = 20 g Increase in the temperature, ΔT = 45°C Molecular mass of nitrogen, M = 28 g Universal gas constant, R = 8 .3 J mol –1 K –1 Number of moles is given as: n = m M = 20 g 28 g = 0.714 As nitrogen is a diatomic gas, At constant pressure, molar specific heat for nitrogen, C P = 7 2 R C P = 7 2 × 8 .3 J mol –1 K –1 = 29.05 J mol -1 K -1 Total amount of heat to be supplied is given as: ΔQ = nC P ΔT ΔQ = 0.714 × 29.05 J mol -1 K -1 × 45 K ΔQ = 933.38 J ∴Total amount of heat to be supplied = 933.38 J$

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

Ans.

(a) When we bring two bodies at different temperatures and in thermal contact, heat starts flowing from the body at the higher temperature to the body at the lower temperature. This process goes on till the temperatures of both the bodies become equal. This is due to the fact the fact that heat flows from higher temperature to lower temperature. The equilibrium temperature of any two bodies can be equal to the mean temperature only if they have the same thermal capacities.

(b) Since the heat-absorbing capacity of the substance is directly proportional to its specific heat, therefore, the specific heat of the coolant in a chemical or nuclear plant must be high.

(c) While driving, the motion of air molecules inside the car increases. Due to the increased motion of air molecules inside the car, the temperature inside the car increases. According to Gay-Lussac’s law, at constant volume, temperature is directly proportional to pressure. Therefore, due to the increased temperature inside the car, pressure inside it also increases.

(d) Since the relative humidity in a harbour town is more than the relative humidity in a desert town, therefore, the climate of a harbour town is more temperate than that of a town in the desert at the same latitude.

Q.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?

Ans.

$Since the cylinder is insulated from its surroundings, the exchange of heat does not take place between the system ( cylinder ) and its surroundings. Therefore, it is an adiabatic process . Consider P i and P f be the initial and final pressure of cylinder respectively. Let V i and V f respectively are the initial and final volume of the cylinder. Ratio of specific heats, γ = 1.4 In an adiabatic process, P i V i γ = P f V f γ (i) Since the final volume is compressed to half of its initial volume, ∴ V f = V i 2 (ii) Substituting the value of V f from equation (ii) in equation (i), we obtain: P i ( V i ) γ = P f ( V i 2 ) γ P f P i = ( V i ) γ ( V i 2 ) γ = ( 2 ) γ = ( 2 ) 1.4 = 2.639 ∴Pressure increases by a factor of 2.639.$

Q.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 1 cal = 4.19 J)

Ans.

$Here, work done ( W ) on the system = 22.3 J As it is an adiabatic process, ∴ΔQ = 0 Since the work is done on the system, ∴ΔW = –22.3 J According to first law of thermodynamics, ΔQ = ΔU + ΔW Here, ΔU = Change in the internal energy of the gas ∴ΔU = ΔQ – ΔW ΔU = 0 – ( – 22.3 J ) ΔU = +22.3 J For the movement of gas from state A to state B via a process, ΔQ = 9.35 cal = 9.35 × 4.19 J = 39.1765 J Heat absorbed is given as: ΔQ = ΔU + ΔW ∴ΔW = ΔQ – ΔU ΔW = 39.1765 J – 22.3 J = 16.8765 J ∴Total work done by the system = 16.88 J$

Q.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 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?

Ans.

(a) As we remove the stopcock between cylinders A and B is suddenly, the volume available to the gas is doubled. According to Boyle’s law, at constant temperature, volume is inversely proportional to pressure. Therefore, as the volume available to the gas is doubled, the pressure will lowered to half of its original value. Since the initial pressure of the gas is 1 atm, therefore, the final pressure in each cylinder will be 0.5 atm.

(b) Since in this case, no work is done by or on the gas and no heat is supplied by an external source, therefore, the internal energy of the gas will not change at all.

(c) Since the internal energy of the gas does not change at all, therefore, the temperature of the gas will not change.

(d) In this process, the free and rapid expansion of gas takes place and it cannot be controlled. In this case, the intermediate states of the gas are non-equilibrium states and they do not satisfy the ideal gas equation. It shows that the given process is not quasi-static. Therefore, the intermediate states of the system do not lie on its P-V-T surface.

Q.7 A steam engine delivers 5.4 × 108 J of work per minute and services 3.6 × 109 J of heat per minute from its boiler. What is the efficiency of the engine? How much heat is wasted per minute?

Ans.

$Here, work done by steam engine per minute, W = 5.4 × 10 8 J Heat supplied from hot reservoir, H 1 = 3 .6 ×10 9 J Efficiency of the steam engine = Output energy Input energy ∴η = W H η = 5.4 × 10 8 J 3.6 × 10 9 J = 0.15 ∴The percentage efficiency of the engine is 15 %. Heat wasted per minute, H 2 = H 1 – W H 2 = 3 .6 × 10 9 J – 5 .4 ×10 8 J H 2 = 30 .6 × 10 8 = 3 .06 ×10 9 J ∴The amount of heat wasted per minute is 3.06 × 10 9 J.$

Q.8 An electric heater supplies heat to a system at a rate of 100 W. If system performs work at a rate of 75 Joules per second. At what rate is the internal energy increasing?

Ans.

$Here, rate of heat supplied, ΔQ = 100 W Work done by the system, ΔW = 75 Js -1 According to first law of thermodynamics, we have: ΔQ = ΔU + ΔW Here, ΔU = Change in internal energy ∴ΔU = ΔQ – ΔW ∴ ΔU =100 Js -1 – 75 Js -1 = 25 W Therefore, change in the internal energy of the given electric heater = 25 W$

Q.9 A thermodynamic system is taken from an original state to an intermediate state by the linear process shown in Fig. (12.13)

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 Ans.

$From the given figure, Change in pressure from E to F, dP = EF = 5.0 – 2 .0 Nm -2 = 3 .0 Nm -2 Change in volume from F to D, dV = FD = 600 – 300 = 300 m 3 From the given figure, it can be observed that Total work done by the gas from D to E to F, W = Area of ΔDEF W = Area of ΔDEF = 1 2 × FD × EF W = 1 2 × 300 m 3 × 3 Nm -2 = 450 J ∴Total work done by the gas from D to E to F = 450 J$

Q.10 A refrigerator is to maintain eatables kept inside at 9°C. If room temperature is 36°C, calculate the coefficient of performance.

Ans.

$Given, temperature inside the refrigerator, T 1 = 9°C = 282 K Room temperature, T 2 = 36 °C = 309 K Coefficient of performance of refrigerator, α = T 1 T 2 – T 1 α = 282 K (309 – 282) K = 10.44 ∴Coefficient of performance of the given refrigerator = 10.44$

## 1. How using the NCERT Solutions for Class 11 Physics Chapter 12 Thermodynamics will help you with your term – II exam preparation?

Students may save time and focus better on the important parts in the syllabus by using the NCERT Solutions for Class 11 Physics Chapter 12 Thermodynamics. The solutions are created by Extramarks professionals in accordance with the most recent term – II CBSE Syllabus.

## 2. How important is the chapter Thermodynamics?

The chapter is quite significant because it explains a variety of terms that are important in the field of physics. The chapter also aids in the development of a foundation that will aid in the study of subsequent chapters and allow you to prepare for your next academic session. It is critical that you comprehend the concepts and mathematical formulas discussed in this chapter. You will be better prepared to answer the questions in the examinations.

## 3. What is the chapter on thermodynamics about?

Thermodynamics, a branch of Physics, is introduced in this chapter. The rules of thermodynamics are also explained in this chapter. You’ll also learn the definitions of key concepts like heat, work, and energy. The numerous processes and their properties are also described in this chapter. You’ll acquire a general concept of several solids’ specific and molar heat capacities.

## 4. What is the meaning of thermal equilibrium?

According to NCERT Solutions for Class 11 Physics Chapter 12, when an object or a body comes into contact with another body of a different temperature, heat is transmitted from the high to the low temperature until both bodies reach thermal equilibrium. For example, when a hot cup of tea is held at room temperature, the heat is gradually transferred to the atmosphere, until the temperature of the tea and the atmosphere are equal.

## 5. What is the meaning of thermodynamics?

Thermodynamics is a field of Physics that deals with the study of the flow of heat. This subject of physics is concerned with the conversion of heat into other kinds of energy and vice versa. The link between heat, work, temperature, and energy is the focus. The entire chapter is based on the transfer of energy from one place to another, as well as from one form to another. Heat is a type of energy that correlates to a specific quantity of mechanical labour, according to the main principle.

## 6. What are the three laws of Thermodynamics?

Three laws of thermodynamics which facilitate the flow of heat and energy are:

First Law – The first law of thermodynamics is known as the law of conservation of energy which states that energy cannot be created or destroyed. Instead, this energy can be transferred or changed from one state to another. A way of expressing this law is that change in internal energy (∆E) is given by the sum of heat (q) which flows across the surrounding or boundaries and work done (w).

Second Law – This law states that an isolated system always increases as they evolve toward thermal equilibrium which is the state of maximum entropy.

Third Law – This law states that the entropy of a system approaches a constant value with the temperature reaching zero.