Suppose there is a cup of tea on the table that is hot--you see it cools down slowly till it reaches the temperature of the room. This simple observation is a beautiful explanation of the principles of thermodynamics, the science of physics, which is concerned with the interrelation of heat, work, temperature, and energy. Class 11 Physics Chapter 11 - Thermodynamics is a very important chapter that addresses these ideas in detail and is important in understanding how energy transformation takes place in many physical processes.
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To simplify complex topics, the NCERT Notes for Class 11 Physics Chapter 11 Thermodynamics are well drafted by experts in the subject to ensure that students can learn and understand the handy materials. These notes are clear explanations of important concepts such as thermal equilibrium, the zero law, internal energy, the first and second law, and such processes as the isothermal and adiabatic changes. These NCERT notes are prepared in line with the new CBSE syllabus and are invaluable to students who are about to take CBSE board exams, JEE, NEET, and other competitive exams. These NCERT Notes for Class 11 Physics Chapter 11 Thermodynamics are easy to revise due to simplified derivations, significant formulas, definitions, clearly labelled diagrams, and solved examples. Not only do they enhance conceptual knowledge, but they also enable students to solve real-life as well as exam-level problems with the confidence of applying thermodynamic concepts.
Also, students can refer,
The Thermodynamics Class 11 Notes can be downloaded easily by the students in order to get well-structured and exam-oriented study material. The notes are in structured form with brief explanations, major formulas, and derivations, which one can revise within a short period. Created by professionals according to the newest CBSE standards, they assist students in enhancing their conceptual understanding and train well to sit CBSE, JEE, and NEET exams.
The NCERT Notes for Class 11 Physics Chapter 11 Thermodynamics assist students in knowing the connections between heat and work as well as how these entities relate to energy within a physical system. These notes include all the significant concepts, such as thermal equilibrium, laws of thermodynamics, and heat transfer, in a simple and sequential manner. They are ready-made and based on the latest CBSE syllabus; hence, they are ideal in terms of revision within a short time and performing highly in board and competitive exams.
The equation of state establishes a link between key variables such as pressure, mass, volume, and density, describing the state of matter when certain physical conditions are met. In the case of an ideal gas, this equation provides information about its behaviour under various conditions, making it a useful tool for understanding and predicting its state.
Consider an ideal gas, whose state equation is
$P V=\mu R T$
Here, P, V, and T are state variables, with μ = no. of moles.
$\mathrm{dW}=\mathrm{PdV}$
Where P is the pressure of the gas in the cylinder.
$\Delta Q=\Delta U+\Delta W$
$\sum \Delta Q=\sum \Delta W$
Where: ΔQ is the heat provided to the system by the environment.
ΔW denotes the amount of work done by the system in relation to the environment.
And ΔU is the change in the system's internal energy.
CP - CV = R
$S=\frac{\Delta Q}{m \Delta T}$
Where,
m is the body's mass, ΔQ is the quantity of heat that a substance absorbs or rejects, and ΔT stands for temperature change.
$C=S / \mu=\Delta Q / \mu \Delta T$
Where: μ= the mass of a material in moles, C is the substance's molar specific heat capacity, ΔQ is the quantity of heat that a substance absorbs or rejects and ΔT stands for temperature change.
PV=constant
This means that the product of pressure and volume for an ideal gas remains constant as long as the process takes place at the same temperature. A hyperbolic curve is commonly used to represent the behaviour of an isothermal process on a pressure-volume (PV) diagram.
Points in the Graph of the Isothermal Process:
$\tan \theta=\frac{d P}{d V}=\frac{-P}{V}$
$\begin{aligned} & W=n R T \log _e\left(\frac{V_f}{V_i}\right)=2.303 n R T \log _{10}\left(\frac{V_f}{V_i}\right) \\ & W=n R T \log _e\left(\frac{P_i}{P_f}\right)=2.303 n R T \log _{10}\left(\frac{P_i}{P_f}\right)\end{aligned}$
$\begin{gathered}P V^\gamma=\text { constant; where } \gamma=\frac{C_P}{C_V} \\ T V^{\gamma-1}=\text { constant } \Rightarrow T_1 V_1^{\gamma-1}=T_2 V_2^{\gamma-1} \text { or } T \propto V^{1-\gamma} \\ \frac{T^\gamma}{P^{\gamma-1}}=\text { constant } \\ \Rightarrow T_1^\gamma P_1^{1-\gamma}=T_2^\gamma P_2^{1-\gamma} \text { or } T \propto P^{\frac{\gamma-1}{\gamma}} \text { or } P \propto T^{\frac{\gamma}{\gamma-1}}\end{gathered}$
ΔW =0
$\begin{aligned} & W=\mu R\left(V_2-V_1\right) \\ & \Delta Q=\Delta U+\mu R\left(T_2-T_1\right)\end{aligned}$
ΔU = 0
This means that the entire amount of heat absorbed is equal to the amount of work done by the system.
Any heat engine's basic function is to transfer heat Q1 from a hot reservoir at temperature T1 to a cool reservoir at temperature T2.
It is isothermal expansion because the system absorbs heat. At temperature T1, the engine absorbs heat Q1.
Inside the engine, an adiabatic process occurs, causing the temperature of the engine to rise from T1 to T2, but with no heat flow.
Isothermal contraction occurs as the system releases heat. At temperature T2, the engine produces heat Q2.
An adiabatic process occurs once more, changing the system's temperature from T2 to T1.
Isothermal expansion, followed by the adiabatic process, and then isothermal contraction, followed by the adiabatic process, make up one cycle of the Carnot engine.
This will continue to happen.
$\eta=\frac{W}{Q_1-Q_2}=1-\frac{T_2}{T_1}$
The Class 11 Physics Chapter 11 - Thermodynamics Previous Year Questions and Answers can guide the students to know the kind of questions that are asked in board and competitive exams. The questions are to enhance the conceptual knowledge and accuracy of problem-solving in terms of heat, energy, and laws of thermodynamics. By regularly practising them, the students can learn to recognise critical subject matter, can become faster, and gain confidence in tests such as CBSE, JEE and NEET.
Q1: An ideal gas undergoes an isothermal process from some initial state I to final state f. Choose the correct alternatives
a) dU = 0
b) dQ = 0
c) dQ = dU
d) dQ = dW
Answer:
Since the process is isothermal $\Delta T=0$ or T is constant For an ideal gas, dU=Change in internal energy=nCvdT $\mathrm{dT}=0$; Thus $\mathrm{dU}=0$
$
\begin{gathered}
d Q=d U+d W \\
d Q=d W
\end{gathered}
$
Hence, the answers are options (a) and (d) only.
Q2: An ideal gas undergoes a cyclic process ABCDA as shown in the given P-V diagram (Figure). The amount of work done by the gas is:
Answer:
Let us take the $P-V$ diagram. In this figure, we have work done in the process of $A B C D$ is equal to the area of the rectangle in ABCDA.
$\Rightarrow$ work done in the process of $A B C D=$ area of the rectangle in $A B C D A$.
$\Rightarrow$ work done in the process of $A B C D=A B \times B C$
$\Rightarrow$ work done in the process of $\mathrm{ABCD}=\left(3 V_o-V_o\right) \times\left(2 P_o-P_o\right)$
$\Rightarrow$ work done in the process of $\mathrm{ABCD}=2 V_o \times P_o$
$\Rightarrow$ work done in the process of $\mathrm{ABCD}=2 V_o P_o$
Here we can see that the process is in an anti-clockwise direction, therefore, work done in the process of $\mathrm{ABCD}=-2 V_o P_o$
Q3: An ideal gas undergoes four different processes from the same initial state (Figure). Four processes are adiabatic, isothermal, isobaric and isochoric. Out of 1, 2, 3 and 4, which one is adiabatic
.
Answer:
For curve 4, pressure is constant, so this is an isobaric process.
For curve 1, volume is constant, so it is an isochoric process. Between curves 3 and 2, curve 2 is steeper, so it is adiabatic, and 3 curve is isothermal
The Thermodynamics Class 11 Notes are essential to understand one of the most basic aspects of physics, the connection between heat, work, and energy. These notes make complex laws and derivations less challenging, thereby assisting students in establishing a solid conceptual foundation. Not only are they required to score well in CBSE board examinations, but they are also the basis
These Class 11 Physics chapter-wise notes of NCERT offer students a simplified and structured study guide on all aspects of the syllabus. These notes are designed by experts in the subject, comprising important formulas, important definitions, solved examples, and practice questions, to ensure that the student can revise at an efficient pace. They are also perfect for preparing for exams in a short time, whether they are CBSE board exams or competitive exams, such as the JEE and NEET. These notes can be read online or downloaded by the students as PDF files so that they can study them at any time and place.
Frequently Asked Questions (FAQs)
Thermodynamics is a part of physics that concerns the correlation between heat, work and energy. it discusses the transfer of energy in a system and surrounding.
The NCERT notes provide well-written explanations, essential formulas, derivations and diagram-based revision for last-minute revision
Devices such as ACs, engines, other power plants, and even biological metabolism in organisms are based on thermodynamics.
The Second Law of Thermodynamics Asserts that a hot object will naturally give heat to a cold object until both are at the same temperature.
It is, in fact, very important chapter as far as JEE and NEET are concerned because it acts as a foundation to several conceptual and numerical questions on heat, energy, and the efficiency of a system.
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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