NCERT Solutions for Class 11 Physics Chapter 11 Thermodynamics
Have you ever asked yourself why hot tea will cool down slowly or how the steam engines worked to drive the industrial revolution? The Thermodynamics in Class 11 Physics explains these real life situations in the most beautiful way possible. Our NCERT Solutions to Class 11 Physics Chapter 11 Thermodynamics will be step by step in nature to help the students to learn how the energy transfer takes place as well as how the heat is converted into work, and also what types of engines and refrigerators and how heating itself works.
This Story also Contains
- NCERT Solution for Class 11 Physics Chapter 11 Solutions: Download Solution PDF
- NCERT Solutions for Class 11 Physics Chapter 11: Exercise Solutions
- Class 11 Physics NCERT Chapter 11: Higher Order Thinking Skills (HOTS) Questions
- NCERT Chapter 11 Thermodynamics Topics
- NCERT Solutions for Class 11 Chapter 11: Important Formulae
- Approach to Solve Questions of Thermodynamics Class 11
- What Extra Should Students Study Beyond NCERT for JEE/NEET?
- NCERT Solutions for Class 11 Physics Chapter Wise
- Importance of NCERT Solutions for Class 11 Physics Chapter 11 Thermodynamics
Designed by professionals according to the new CBSE curriculum, these NCERT solutions do make even the most difficult things to learn simpler to revise and memorize. They not only aid in school examinations and board exams but also play an important role in preparation of competitive examinations such as JEE and NEET. Practicing them on a regular basis, the students will be able to improve their problem-solving skills, acquire logical thinking, and understand the laws of thermodynamics, heat engines and principles of energy conservation better. These NCERT Solutions for Class 11 Physics make learning Thermodynamics interesting and examination-driven.
NCERT Solution for Class 11 Physics Chapter 11 Solutions: Download Solution PDF
The NCERT Solutions for Class 11 Physics Chapter 11 Thermodynamics 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.
NCERT Solutions for Class 11 Physics Chapter 11: Exercise Solutions
The solutions of NCERT to Physics Class 11 Chapter 11 Thermodynamics 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
$\\\frac{dV}{dt}=3\ litres\ minute^{-1}\\$
Density of water = 1000 g/litre
The mass flow rate of water is
$\\\frac{dm}{dt}=\rho \frac{dV}{dt}\\ $
$\frac{dm}{dt}=3000\ g\ min^{-1}$
Specific heat of water, c = 4.2 J g-1 oC-1
The rise in temperature is $\Delta T=77-27=50\ ^{o}C$
The rate of energy consumption will be
$\\\frac{dQ}{dt}=\frac{dm}{dt}c\Delta T\\ $
$\frac{dQ}{dt}=3000\times 4.2\times 50\\$
$ \frac{dQ}{dt}=6.3\times 10^{5}\ J\ min^{-1}$
The heat of combustion of fuel $=4.0\times 10^{4}J/g$
The rate of consumption of fuel is
$\\\frac{6.3\times 10^{5}}{4\times 10^{4}}\\ =15.75\ g\ min^{-1}$
Answer:
Mass of nitrogen, $m=2.0\times 10^{-2}kg=20g$
Molar Mass of nitrogen, MN = 28 g
The number of moles is n
$\\n=\frac{m}{M_{N}}\\$
$ n=\frac{20}{28}\\$
$ n=0.714$
As nitrogen is a diatomic gas, its molar specific heat at constant pressure CP is as follows
$\\C_{P}=\frac{7R}{2}\\ $
$C_{P}=\frac{7\times 8.3}{2}\\$
$ C_{P}=29.05\ J\ mol^{-1}\ ^{o}C^{-1}$
Rise in temperature, $\\\Delta T=45 ^{o}C^{-1}$
The amount of heat Q that must be supplied is
$\\Q=nC_{P}\Delta T\\ $
$Q=0.714\times 29.05\times 45\\ $
$Q=933.38\ J$
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. $PV^{\gamma }$ would be constant.
Hydrogen is a diatomic gas and therefore $\gamma =1.4$
Let the initial and final pressures be P1 and P2, respectively.
Let the initial and final volumes be V1 and V2, respectively.
$\\P_{1}V_{1}^{\gamma }=P_{2}V_{2}^{\gamma }\\ $
$\frac{P_{2}}{P_{1}}=\left ( \frac{V_{1}}{V_{2}} \right )^{\gamma }\\ $
$\frac{P_{2}}{P_{1}}=2^{1.4}\\$
$ \frac{P_{2}}{P_{1}}=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 $1 cal = 4.19 J$ )
Answer:
In the first case, the process is adiabatic, i.e. $\Delta Q=0$
22.3 J work is done on the system, i.e. $\Delta W=-22.3\ J$
$\\\Delta Q=\Delta U+\Delta W$
$\\ 0=\Delta U-22.3\\ $
$\Delta U=22.3\ J$
Since in the latter process as well,l the initial and final states are the same as those in the former process, $\Delta U$ will remain the same for the latter case.
In the latter case, the net heat absorbed by the system is 9.35 cal
$\\\Delta Q=9.35\times 4.2\\$
$\Delta Q=39.3\ J$
$\\\Delta W=\Delta Q-\Delta U\\ $
$\Delta W=39.3-22.3\\ $
$\Delta W=17.0\ J$
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
$P_{2}=\frac{P_{1}V_{1}}{V_{2}}$
$ P_{2}=\frac{P_{1}}{2}$
$ P_{2}=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.
Answer:
The rate at which heat is supplied $\Delta Q=100\ W$
The rate at which work is done $\Delta W=75\ J s^{-1}$
The rate of change of internal energy is $\\\Delta u$
$\begin{aligned}
\Delta U & =\Delta Q-\Delta W \\
\Delta U & =100-75 \\
\Delta U & =25 \mathrm{~J} \mathrm{~s}^{-1}
\end{aligned}$
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
$\\area(DEF)=\frac{1}{2}\times DF\times FE\\$
$area(DEF)=\frac{1}{2}\times300\times 3\\$
$area(DEF)=450\ J$
The work done is, therefore, 450 J.
Class 11 Physics NCERT Chapter 11: 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 being initially at a temperature $T_0=273 K$ 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 $\gamma=C_P / C_V$ for this gas.
Answer:
$Q=Q_1+Q_2$
where
$Q_1 \Rightarrow$ for Isothermal process
$Q_2 \Rightarrow$ for Isochoric process
$Q=n^{\prime} \mathrm{R} T_0 \log \frac{n V_0}{V_0}+n^{\prime} C_V\left(T_C-T_B\right)$
As $T_C=n T_0 \quad$ (If volume is made n times)
$Q=n^{\prime} \mathrm{R} T_0 \log n+\gamma(n-1) T_0$
$\therefore \gamma=\left[1+\frac{n-1}{\frac{Q}{n^{\prime} R T_0}-\log n}\right]$
Putting the values
(Given n' = 3, n = 5)
$\gamma=1.4$
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 $0.1\, cm^{2}$. 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 $2\times 10^{5}$Pa
d) The pressure required to lift the whistle is $10^5$Pa
Answer:
The initial temperature of hydrogen, $T_1=273+27=300 \mathrm{~K}$
Initial pressure of hydrogen, $P_1=1 \mathrm{~atm}=10^5 \mathrm{~Pa}$
$
\begin{aligned}
& P_2=P_0+\frac{m g}{A} \\
& =10^5+\frac{100 \times 10^{-3}}{0.01 \times 10^{-4}} \\
& =2 \times 10^5
\end{aligned}
$
As the volume is constant, we can use Gay Lussac's law:
$
\frac{P_1}{T_1}=\frac{P_2}{T_2} \Rightarrow T_2=T_1 \times \frac{P_2}{P_1}=300 \times \frac{2}{1}=600 \mathrm{~K}=600-273=327^{\circ} \mathrm{C}
$
Hence, the answer is option (a),(c).
Q.3 The volume of a gas is compressed adiabatically from state $V_1=7 \mathrm{~m}^3$ at a pressure of $2. .4 \times 10^5 \mathrm{~N} \mathrm{~m}^{-2}$ to the volume $V_2=0.875 \mathrm{~m}^3$. If the same compression is performed isothermally, calculate the difference in work done (in $10^5$ joules) in both processes. It is given that $\gamma=$ 1.67.
Answer:
$
V_1=7 \mathrm{~m}^3, P_1=2.4 \times 10^5 \mathrm{Nm}^{-2}, V_2=0.875 \mathrm{~m}^3
$
In an adiabatic process
$
\begin{aligned}
& P_1 V_1^\gamma=P_2 V_2^\gamma \\
& \Rightarrow P_2=2.4 \times 10^5\left[\frac{7}{0.875}\right]^{1.6} \\
& \Rightarrow P_2=2.4 \times 10^5(8)^{\frac{5}{3}}=7.68 \times 10^6 \mathrm{~N} \mathrm{~m}^{-2}
\end{aligned}
$
Work done in an adiabatic process,
$
\begin{aligned}
& W=-\frac{P_2 V_2-P_1 V_1}{\gamma-1} \\
& W=-\frac{\left(7.68 \times 10^6 \times 0.875\right)-\left(2.4 \times 10^5 \times 7\right)}{1.67-1} \\
& W=-75.22 \times 10^5 \mathrm{~J}
\end{aligned}
$
Work done in the isothermal process,
$
\begin{aligned}
& W=2.3026 \mathrm{RT} \log \frac{V_2}{V_1}=2.3026 P_1 V_1 \log \frac{V_2}{V_1} \\
& =2.3026 \times 2.4 \times 10^5 \times 7 \times \log \left[\frac{0.875}{7.0}\right] \\
& =2.3026 \times 2.4 \times 10^5 \times 7 \log \left(\frac{1}{8}\right)=-34 \times 10^5 \mathrm{~J}
\end{aligned}
$
Difference in work done $=-75.22 \times 10^5 \mathrm{~J}+34 \times 10^5 \mathrm{~J}=-41.22 \times 10^5 \mathrm{~J}$
Q.4 1 mole of helium expands with temperature according to the relation $V=K T^{2 / 3}$. 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:
$\begin{aligned} & V=K T^{2 / 3} \\ \Rightarrow & V=K\left(\frac{P V}{n R}\right)^{2 / 3} \\ \Rightarrow & P V^{-1 / 2}=\alpha \\ \because & C=C_V+\frac{R}{1-x}=\frac{3 R}{2}+\frac{R}{1+\frac{1}{2}}=\frac{13 R}{6} \\ Q & =n C \Delta T=1 \times \frac{13 R}{6} \times 60=130 R=1080.30 \mathrm{~J}\end{aligned}$
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 $\frac{a}{b} R$. Find the remainder when $a$ is divided by $b$
Answer:
Let initial pressure, volume and temperature be $P_0, V_0, T_0$ indicated by state A in the P-V diagram. The gas is isochorically taken to state $B\left(2 P_0, V_0, 2 T_0\right)$ and then taken from state B to state $\mathrm{C}\left(2 P_0, 2 V_0, 4 T_0\right)$ isobarically.
Total heat absorbed by 1 mole of gas
$
\begin{aligned}
& \Delta Q=C_v\left(2 T_0-T_0\right)+C_P\left(4 T_0-2 T_0\right) \\
& =\frac{5}{2} R T_0+\frac{7}{2} R \times 2 T_0 \\
& =\frac{19}{2} R T_0
\end{aligned}
$
Total change in temperature from state A to C is: $\Delta T=3 T_0$
$\therefore$ Molar heat capacity $=\frac{\Delta Q}{\Delta T}=\frac{\frac{19}{2} R T_0}{3 T_0}=\frac{19}{6} R$
Hence, the answer is 1.
NCERT Chapter 11 Thermodynamics Topics
In Class 11 Physics, in Chapter 11, Thermodynamics, concepts of the motion of heat and the connection of heat, work and energy are given. It describes such concepts as internal energy, first law of thermodynamics, heat engines, and second law.
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
NCERT Solutions for Class 11 Chapter 11: Important Formulae
The Key Formulae in Chapter 11 of Class 11 Physics 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
$
\Delta Q=\Delta U+\Delta W
$
Where:
$\Delta Q=$ heat supplied to the system
$\Delta U=$ change in internal energy
$\Delta W=$ work done by the system
2. Work Done in an Isothermal Process
$
W=n R T \ln \left(\frac{V_f}{V_i}\right)
$
(For ideal gas, $T$ constant)
3. Work Done in Adiabatic Process
$
W=\frac{P_i V_i-P_f V_f}{\gamma-1}=\frac{n R\left(T_i-T_f\right)}{\gamma-1}
$
Where $\gamma=\frac{C_p}{C_v}$ (adiabatic index)
4. Adiabatic Relation for Ideal Gas
$P V^\gamma=\text { constant }$
5. Molar Specific Heats
For ideal gas:
$
C_p-C_v=R
$
6. Internal Energy of Ideal Gas
$
U=\frac{f}{2} n R T
$
Where $f=$ degrees of freedom
For monoatomic gas: $U=\frac{3}{2} n R T$
7. Heat Capacity
$
C=\frac{d Q}{d T}, \quad C=n C_m
$
Where $C_m=$ molar specific heat
9. Carnot Engine Efficiency
$
\eta=1-\frac{T_2}{T_1}
$
Where $T_1=$ temperature of source, $T_2=$ temperature of sink (in Kelvin)
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.
What Extra Should Students Study Beyond NCERT for JEE/NEET?
The following is a speedy comparison table of how to prepare only in NCERT and how it is supposed to also be in addition to entering JEE/NEET relevant to the understanding of the students of which additional efforts are necessary to do better in such exams which are very competitive in nature.
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.
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
- NCERT Books Class 11 Physics
- NCERT Syllabus Class 11 Physics
- NCERT Books Class 11
- NCERT Syllabus Class 11
Subject Wise NCERT Exemplar Solutions
- NCERT Exemplar Class 11th Solutions
- NCERT Exemplar Class 11th Maths
- NCERT Exemplar Class 11th Physics
- NCERT Exemplar Class 11th Chemistry
- NCERT Exemplar Class 11th Biology
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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