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NCERT Solutions for Class 11 Physics Chapter 11 Thermodynamics

NCERT Solutions for Class 11 Physics Chapter 11 Thermodynamics

Edited By Vishal kumar | Updated on Apr 29, 2025 03:35 PM IST

Ever wondered how your geyser heats water or how steam powers a train? That is the magic of thermodynamics with our step by step Class 11 Physics Chapter 11 solutions, you will understand how heat turns into work and vice versa. These Thermodynamics Class 11 NCERT solutions make tough concepts simple and help boost your exam confidence!

This Story also Contains
  1. NCERT Solution for Class 11 Physics Chapter 11 Solutions: Download Solution PDF
  2. NCERT Solutions for Class 11 Physics Chapter 11 Thermodynamics Exercise
  3. Class 11 Physics NCERT Chapter 11: Higher Order Thinking Skills (HOTS) Questions
  4. Approach to Solve Questions of Thermodynamics Class 11
  5. What Extra Should Students Study Beyond NCERT for JEE/NEET?
  6. NCERT Solutions for Class 11 Physics Chapter Wise
  7. Importance of NCERT Solutions for Class 11 Physics Chapter 11 Thermodynamics
  8. NCERT Solutions for Class 11 Subject Wise
  9. Subject Wise NCERT Exemplar Solutions

The NCERT solution for thermodynamics class 11 physics is prepared by our highly experienced Subject Matter, which gives all the answers to the NCERT Book. Utilising the NCERT solutions for Class 11, specifically thermodynamics class 11 physics questions and answers, can aid students in practising questions and grasping important syllabus concepts before their exams. These solutions are available for free and can be accessed below.

NCERT Solution for Class 11 Physics Chapter 11 Solutions: Download Solution PDF

Free download NCERT solution for Class 11 Physics thermodynamics questions and answers PDF for CBSE exam

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NCERT Solutions for Class 11 Physics Chapter 11 Thermodynamics Exercise

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 o C -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, M N = 28 g

The number of moles is n

n=mMN
n=2028
n=0.714

As nitrogen is a diatomic gas, it's molar specific heat at constant pressure C P 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

(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 .

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 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 the 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

(d) 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 pressure be P 1 and P 2, respectively.

Let the initial and final volume be T 1 and T 2, respectively.

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

The pressure thus increases by a factor 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 P 1 = 1 atm

Initial Volume, V 1 = V

Final Volume, V 2 = 2V

Final Pressure P 2 will be

P2=P1V1V2P2=P12P2=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. (12.13)

screenshot123

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 change in Volume = 3 m 3

area(DEF)=12×DF×FE

area(DEF)=12×300×3

area(DEF)=450 J

The work done is, therefore, 450 J.

Class 11 Physics NCERT Chapter 11: Higher Order Thinking Skills (HOTS) Questions

Q.1 Three moles of an ideal gas being 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 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 the 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 joule) in both processes. It is given that y= 1.67.

Answer:

V1=7 m3,P1=2.4×105Nm2,V2=0.875 m3

In adiabatic process

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

Work done in adiabatic process,

W=P2V2P1V1γ1=(7.68×106×0.875)(2.4×105×7)1.671=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 the 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.

Approach to Solve Questions of Thermodynamics Class 11

  • 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.

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NCERT Solutions for Class 11 Physics Chapter Wise

Importance of NCERT Solutions for Class 11 Physics Chapter 11 Thermodynamics

  • As the Class 11 exams and competitive exams like NEET and JEE Main are considered, the NCERT solutions for Class 11 are important.
  • At least one question is expected for JEE Mains, and two questions are expected for NEET from the chapter on Thermodynamics.

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)

1. What is the weightage of Thermodynamics for JEE Main

One question from Class 11 chapter Thermodynamics is expected for JEE Main. Students should practice previous year papers of JEE Main to get a good score.

2. What is the weightage of thermodynamics for NEET

Two questions from class 11 chapter Thermodynamics is expected for NEET. The competition level of NEET exam is very high. So practicing more questions is important. To practice more questions refer to NCERT exemplar and NEET previous year papers

3. Is Thermodynamics helpful in higher studies

Yes, the concepts of thermodynamics are used in various fields of Engineering and Science. Not only Thermodynamics chapter, all the chapters in NCERT book are basics of some field of higher studies.

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2.45×10−3 kg

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