NCERT Solutions for Class 11 Physics Chapter 11 Thermodynamics

NCERT Solutions for Class 11 Physics Chapter 11 Thermodynamics: If you are a Class 11 CBSE student and searching for Thermodynamics Class 11 Physics NCERT solutions, this is where it comes in handy. In the Thermodynamics Class 11 chapter, you will study how heat is converted to work and vice versa. For example, electrical energy can be utilized to boil water, and similarly, steam can be utilized to produce electricity. The CBSE NCERT solutions for Class 11 Physics Chapter 11 Thermodynamics carry detailed explanations to exercise questions. Try to solve all these questions yourself before looking at the solutions. NCERT solutions help to improve conceptual knowledge.

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. Utilizing 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 Solutions for Class 11 Physics Chapter 11 Thermodynamics

Free download class 11 physics thermodynamics questions and answers pdf for CBSE exam

Download PDF

NCERT Solutions for Class 11 Physics Chapter 11 Thermodynamics Exercise

The class 11 physics questions and answers for Thermodynamics in Class 11 Physics are highly beneficial for students as they provide a solid foundation in understanding the principles of heat, work, energy, and the laws of thermodynamics. This knowledge not only aids in performing well in exams but also simplifies the comprehension of subsequent chapters in physics, particularly those related to heat and energy transfer. Additionally, problem-solving skills developed through these thermodynamics class 11 physics ncert solutions are transferable and valuable in a wide range of academic and practical scenarios.

NCERT Solutions for Class 11 Physics Chapter Wise

Have you ever wondered why gases appear to expand or how they produce pressure? The Kinetic Theory of Gases explains it by imagining that gases consist of small particles moving about rapidly. Scientists such as Boyle, Newton, Maxwell, and Boltzmann contributed significantly to this theory, which explains gas behavior, their response to changes in temperature and pressure, and even how they flow and mix together.

NCERT solutions for important chapter kinetic theory class 11 chapter 12 physics are created by subject matter experts to offer accurate and clear answers to each and every NCERT exercise problem. Students can build a strong concept by practicing these Kinetic Theory Class 11 Physics solutions. Step-by-step solutions make learning convenient and are provided below.

Also Read

• Kinetic Theory Class 11th Notes - Free NCERT Class 11 Physics Chapter 12 Notes Download PDF
• NCERT Exemplar Class 11 Physics Solutions Chapter 12 Kinetic Theory

NCERT Solutions for Class 11 Physics Chapter 12: Download PDF

Free download of Kinetic Theory of Gases Class 11 Solutions PDF for CBSE exams and strengthen your understanding with step-by-step explanations and solved exercises.

Download PDF

NCERT Solutions for Class 11 Physics Chapter 12 Kinetic Theory - Exercise Questions

Q12.1 Estimate the fraction of molecular volume to the actual volume occupied by oxygen gas at STP. Take the diameter of an oxygen molecule to be 3 Å.

Answer:

The diameter of an oxygen molecule, d = 3 Å.

The actual volume of a mole of oxygen molecules V _actual is

${V}_{actual}=N_A\frac{4}{3}\pi {\left(\frac{d}{2}\right)}^3$ $$
$V_{actual}=6.023\times {10}^{23}\times \frac{4}{3}\pi \times {\left(\frac{3\times {10}^{-10}}{2}\right)}^3$
$V_{actual}=8.51\times {10}^{-6}{m}^3$
$V_{actual}=8.51\times {10}^{-3}litres$

The volume occupied by a mole of oxygen gas at STP is V _molar = 22.4 litres

$\frac{V_{actual}}{V_{molar}}=\frac{8.51\times {10}^{-3}}{22.4}$

$\frac{V_{actual}}{V_{molar}}=3.8\times {10}^{-4}$

Q12.2 Molar volume is the volume occupied by 1 mol of any (ideal) gas at standard temperature and pressure (STP: 1 atmospheric pressure, $0^{0}C$ ). Show that it is $22.4Litres$ .

Answer:

As per the ideal gas equation

$$\begin{gathered} PV=nRT \\ V=\frac{nRT}{P} \end{gathered}$$

For one mole of a gas at STP we have

$V=\frac{1\times 8.314\times 273}{1.013\times {10}^5}$
$V=0.0224m^3$

$V=22.4litres$

Q12.3 Figure 13.8 shows plot of $PV/T$ versus P for $1.00\times {10}^{-3}$ kg of oxygen gas at two different temperatures.

(a) What does the dotted plot signify?
(b) Which is true: $T_1>T_{2}or{T}_1<T_2$ ?
(c) What is the value of $PV/T$ where the curves meet on the y-axis?
(d) If we obtained similar plots for $1.00\times {10}^{-3}$ kg of hydrogen, would we get the same
value of $PV/T$ at the point where the curves meet on the y-axis? If not, what mass
of hydrogen yields the same value of $PV/T$ (for low pressure high temperature
region of the plot) ? (Molecular mass of $H_2=2.02\mu$ , of $O_2=32.0\mu$ ,
$R=8.31Jmo{l}^{-1}{K}^{-1}$ .)

Answer:

(a) The dotted plot corresponds to the ideal gas behaviour.

(b) We know the behaviour of a real gas tends close to that of ideal gas as its temperature increases and since the plot corresponding to temperature T ₁ is closer to the horizontal line that the one corresponding to T ₂ we conclude T ₁ is greater than T ₂.

(c) As per the ideal gas equation

$\frac{PV}{T}=nR$

The molar mass of oxygen = 32 g

$n=\frac{1}{32}$

R = 8.314

$$\begin{gathered} nR=\frac{1}{32}\times 8.314 \\ nR=0.256J{K}^{-1} \end{gathered}$$

(d) If we obtained similar plots for $1.00\times {10}^{-3}$ kg of hydrogen we would not get the same
value of $PV/T$ at the point where the curves meet on the y-axis as 1 g of Hydrogen would contain more moles than 1 g of Oxygen because of having smaller molar mass.

Molar Mass of Hydrogen M = 2 g

mass of hydrogen

$m=\frac{PV}{T}\frac{M}{R}=0.256\times \frac{2}{8.314}=5.48\times {10}^{-5}Kg$

Q12.4 An oxygen cylinder of volume $30$ litres has an initial gauge pressure of $15$ atm and a temperature of ${27}^{0}C$ . After some oxygen is withdrawn from the cylinder, the gauge pressure drops to $11$ atm and its temperature drops to ${17}^{0}C$ . Estimate the mass of
oxygen taken out of the cylinder ( $R=8.31Jmo{l}^{-1}{K}^{-1}$ , molecular mass of $O_2=32\mu$ ).

Answer:

Initial volume, V ₁ = Volume of Cylinder = 30 l

Initial Pressure P ₁ = 15 atm

Initial Temperature T ₁ = 27 ^o C = 300 K

The initial number of moles n ₁ inside the cylinder is

$$\begin{gathered} {n}_1=\frac{P_1{V}_1}{R{T}_1} \\ {n}_1=\frac{15\times 1.013\times {10}^5\times 30\times {10}^{-3}}{8.314\times 300} \\ {n}_1=18.28 \end{gathered}$$

Final volume, V ₂ = Volume of Cylinder = 30 l

Final Pressure P ₂ = 11 atm

Final Temperature T ₂ = 17 ^o C = 290 K

The final number of moles n ₂ inside the cylinder is

$$\begin{gathered} {n}_2=\frac{P_2{V}_2}{R{T}_2} \\ {n}_2=\frac{11\times 1.013\times {10}^5\times 30\times {10}^{-3}}{8.314\times 290} \\ {n}_2=13.86 \end{gathered}$$

Moles of oxygen taken out of the cylinder = n ₂ -n ₁ = 18.28 - 13.86 = 4.42

Mass of oxygen taken out of the cylinder m is

$$\begin{gathered} m=4.42\times 32 \\ m=141.44g \end{gathered}$$

Q12.5 An air bubble of volume $1.0c{m}^3$ rises from the bottom of a lake $40m$ deep at a temperature of ${12}^{0}C$ . To what volume does it grow when it reaches the surface, which is at a temperature of ${35}^{0}C$ ?

Answer:

Initial Volume of the bubble, V ₁ = 1.0 cm ³

Initial temperature, T ₁ = 12 ^o C = 273 + 12 = 285 K

The density of water is ${\rho }_w={10}^{3}kg{m}^{-3}$

Initial Pressure is P ₁

Depth of the bottom of the lake = 40 m

$$\begin{gathered} {P}_1=AtmosphericPressure+Pressureduetowater \\ {P}_1=P_{atm}+{\rho }_{w}gh \\ {P}_1=1.013\times {10}^5+{10}^3\times 9.8\times 40 \\ {P}_1=4.93\times {10}^{5}Pa \end{gathered}$$

Final Temperature, T ₂ = 35 ^o C = 35 + 273 = 308 K

Final Pressure = Atmospheric Pressure $=1.013\times {10}^{5}Pa$

Let the final volume be V ₂

As the number of moles inside the bubble remains constant, we have

$\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}$

$V_2=\frac{P_1{T}_2{V}_1}{P_2{T}_1}$

$V_2=\frac{4.93\times {10}^5\times 308\times 1}{1.013\times {10}^5\times 285}$

$V_2=5.26c{m}^3$

Q12.6 Estimate the total number of air molecules (inclusive of oxygen, nitrogen, water vapour and other constituents) in a room of capacity $25.0m^3$ at a temperature of ${27}^{0}C$ and $1atm$ pressure.

Answer:

The volume of the room, V = 25.0 m ³

Temperature of the room, T = 27 ^o C = 300 K

The pressure inside the room, P = 1 atm

Let the number of moles of air molecules inside the room be n

$$\begin{gathered} n=\frac{PV}{RT} \\ n=\frac{1.013\times {10}^5\times 25}{8.314\times 300} \\ n=1015.35 \end{gathered}$$

Avogadro's Number, $N_A=6.022\times {10}^{23}$

The number of molecules inside the room is N

$$\begin{gathered} N=n{N}_A \\ N=1015.35\times 6.022\times {10}^{23} \\ N=6.114\times {10}^{26} \end{gathered}$$

Q12.7 Estimate the average thermal energy of a helium atom at (i) room temperature ( ${27}^{0}C$ ), (ii) the temperature on the surface of the Sun ( $6000K$ ), (iii) the temperature of $10$ million kelvin (the typical core temperature in the case of a star).

Answer:

The average energy of a Helium atom is given as $\frac{3kT}{2}$ since it is monoatomic

(i)

$E=\frac{3kT}{2}$

$E=\frac{3\times 1.38\times {10}^{-23}\times 300}{2}$

$E=6.21\times {10}^{-21}J$

(ii)

$E=\frac{3kT}{2}$

$E=\frac{3\times 1.38\times {10}^{-23}\times 6000}{2}$

$E=1.242\times {10}^{-19}J$

(iii)

$E=\frac{3kT}{2}$

$E=\frac{3\times 1.38\times {10}^{-23}\times {10}^7}{2}$

$E=2.07\times {10}^{-16}J$

Q12.8 Three vessels of equal capacity have gases at the same temperature and pressure. The first vessel contains neon (monatomic), the second contains chlorine (diatomic), and the third contains uranium hexafluoride (polyatomic). Do the vessels contain an equal number of respective molecules? Is the root mean square speed of molecules the same in the three cases? If not, in which case is v _rms the largest?

Answer:

As per Avogadro's Hypothesis, under similar conditions of temperature and pressure, equal volumes of gases contain an equal number of molecules. Since the volume of the vessels is the same and all vessels are kept at the same conditions of pressure and temperature, they would contain an equal number of molecules.

Root mean square velocity is given as

$v_{rms}=\sqrt{\frac{3kT}{m}}$

As we can see, v_rms is inversely proportional to the square root of the molar mass; the root mean square velocity will be maximum in the case of Neon as its molar mass is the least.

Q12.9 At what temperature is the root mean square speed of an atom in an argon gas cylinder equal to the rms speed of a helium gas atom at $-{20}^{0}C$ ? (atomic mass of $Ar=39.9\mu$ , of $He=4.0\mu$ ).

Answer:

As we know root mean square velocity is given as $v_{rms}=\sqrt{\frac{3RT}{M}}$

Let at temperature T the root mean square speed of an atom in an argon cylinder equal to the rms speed of a helium gas atom at $-{20}^{0}C$

$$\begin{gathered} \sqrt{\frac{3R\times T}{39.9}}=\sqrt{\frac{3R\times 253}{4}} \\ T=2523.7K \end{gathered}$$

Q12.10 Estimate the mean free path and collision frequency of a nitrogen molecule in a cylinder containing nitrogen at $2.0atm$ and temperature ${17}^{0}C$ . Take the radius of a nitrogen molecule to be roughly $1.0A$ . Compare the collision time with the time the
molecule moves freely between two successive collisions (Molecular mass of $N_2=28.0\mu$ ).

Answer:

Pressure, P = 2atm

Temperature, T = 17 ^o C

The radius of the Nitrogen molecule, r=1 Å.

The molecular mass of N ₂ = 28 u

The molar mass of N ₂ = 28 g

From the ideal gas equation

$PV=nRT$

$\frac{n}{V}=\frac{P}{RT}$

The above tells us about the number of moles per unit volume; the number of molecules per unit volume would be given as

${n}^{\prime }=\frac{N_{A}n}{V}=\frac{6.022\times {10}^{23}\times 2\times 1.013\times {10}^5}{8.314\times (17+273)}$

$n^{\prime }=5.06\times {10}^{25}$

The mean free path $\lambda$ is given as

$\lambda =\frac{1}{\sqrt{2}\pi n^{\prime }{d}^2}$

$\lambda =\frac{1}{\sqrt{2}\times \pi \times 5.06\times {10}^{25}\times (2\times 1\times {10}^{-10}{)}^2}$

$\lambda =1.11\times {10}^{-7}m$

The root mean square velocity v _rms is given as

${v}_{rms}=\sqrt{\frac{3RT}{M}}$

$v_{rms}=\sqrt{\frac{3\times 8.314\times 290}{28\times {10}^{-3}}}$

$v_{rms}=508.26m{s}^{-1}$

The time between collisions T is given as

$T=\frac{1}{CollisionFrequency}$

$T=\frac{1}{\nu }$

$T=\frac{\lambda }{v_{rms}}$

$T=\frac{1.11\times {10}^{-7}}{508.26}$

$T=2.18\times {10}^{-10}s$

Collision time T' is equal to the average time taken by a molecule to travel a distance equal to its diameter

${T}^{\prime }=\frac{d}{v_{rms}}$

$T^{\prime }=\frac{2\times 1\times {10}^{-10}}{508.26}$

$T^{\prime }=3.935\times {10}^{-13}s$

The ratio of the average time between collisions to the collision time is

$\begin{array}{rl}\frac{T}{T^{\prime }}& =\frac{2.18\times {10}^{-10}}{3.935\times {10}^{-13}}\\ \frac{T}{T^{\prime }}& =554\end{array}$
Thus, we can see that the time between collisions is much larger than the collision time.

Additional Questions

Q 1) A metre-long narrow bore held horizontally (and closed at one end) contains a $76cm$ long mercury thread, which traps a $15cm$ column of air. What happens if the tube is held vertically with the open end at the bottom?

Answer:

Initially, the pressure of the 15 cm long air column is equal to the atmospheric pressure, P ₁ = 1 atm = 76 cm of Mercury

Let the crossectional area of the tube be x cm ²

The initial volume of the air column, V ₁ = 15x cm ³

Let's assume that once the tube is held vertical, y cm of Mercury flows out of it.

The pressure of the air column after y cm of Mercury has flown out of the column P ₂ = 76 - (76 - y) cm of Mercury = y cm of mercury

Final volume of air column V ₂ = (24 + y)x cm ³

Since the temperature of the air column does not change

$$\begin{gathered} {P}_1{V}_1=P_2{V}_2 \\ 76\times 15x=y\times (24+y)x \\ 1140=y^2+24y \\ {y}^2+24y-1140=0 \end{gathered}$$

Solving the above quadratic equation, we get y = 23.8 cm or y = -47.8 cm

Since a negative amount of mercury cannot flow out of the column, y cannot be negative. Therefore, y = 23.8 cm.

Length of the air column = y + 24 = 47.8 cm.

Therefore, once the tube is held vertically, 23.8 cm of Mercury flows out of it, and the length of the air column becomes 47.8 cm

Q 2) From a certain apparatus, the diffusion rate of hydrogen has an average value of $28.7c{m}^3{s}^{-1}$. The diffusion of another gas under the same conditions is measured to have an average rate of $7.2c{m}^3{s}^{-1}$. Identify the gas.

Answer:

As per Graham's Law of diffusion, if two gases of Molar Mass M ₁ and M ₂ diffuse with rates R ₁ and R ₂ respectively, their diffusion rates are related by the following equation

$\frac{R_1}{R_2}=\sqrt{\frac{M_2}{M_1}}$

In the given question

R ₁ = 28.7 cm ³ s ^-1

R ₂ = 7.2 cm ³ s ^-1

M ₁ = 2 g

$$\begin{gathered} {M}_2=M_1{\left(\frac{R_1}{R_2}\right)}^2 \\ {M}_2=2\times {\left(\frac{28.7}{7.2}\right)}^2 \\ {M}_2=31.78g \end{gathered}$$

The above Molar Mass is close to 32, therefore, the gas is Oxygen.

Q 3) A gas in equilibrium has uniform density and pressure throughout its volume. This is strictly true only if there are no external influences. A gas column under gravity, for example, does not have uniform density (and pressure). As you might expect, its density decreases with height. The precise dependence is given by the so-called law of atmospheres $n_2=n_{1}exp[-mg(h_2-h_1)/k_{b}T]$

where n2, n1 refer to number density at heights h2 and h1 respectively. Use this relation to derive the equation for the sedimentation equilibrium of a suspension in a liquid column:

$n_2=n_{1}exp[-mg{N}_A(\rho -{\rho }^{{}^{\prime }})(h_2-h_1)/(\rho RT)]$
where $\rho$ is the density of the suspended particle, and ${\rho }^{{}^{\prime }}$ , that of the surrounding medium. [ $N_A$ s Avogadro’s number, and R the universal gas constant.] [Hint: Use Archimedes' principle to find the apparent weight of the suspended particle.]

Answer:

$n_2=n_{1}exp[-mg(h_2-h_1)/k_{b}T]$ $(i)$

Let the suspended particles be spherical and have a radius r

The gravitational force acting on the suspended particles would be

$F_G=\frac{4}{3}\pi r^3\rho g$

The buoyant force acting on them would be

$F_B=\frac{4}{3}\pi r^3{\rho }^{\prime }g$

The net force acting on the particles become

$$\begin{gathered} {F}_{net}=F_G-F_B \\ {F}_{net}=\frac{4}{3}\pi r^3\rho g-\frac{4}{3}\pi r^3{\rho }^{\prime }g \\ {F}_{net}=\frac{4}{3}\pi r^{3}g(\rho -{\rho }^{\prime }) \end{gathered}$$

Replacing mg in equation (i) with the above equation, we get

$$\begin{gathered} {n}_2=n_{1}exp[-\frac{4}{3}\pi r^{3}g(\rho -{\rho }^{\prime })(h_2-h_1)/k_{b}T] \\ {n}_2=n_{1}exp\left[\frac{-\frac{4}{3}\pi r^{3}g(\rho -{\rho }^{\prime })(h_2-h_1)}{\frac{RT}{N_A}}\right] \\ {n}_2=n_{1}exp\left[\frac{-mg{N}_A(\rho -{\rho }^{\prime })(h_2-h_1)}{RT{\rho }^{\prime }}\right] \end{gathered}$$

The above is the equation to be derived

Q 4) Given below are densities of some solids and liquids. Give rough estimates of the size of their atoms :

Answer:

Let one mole of a substance of atomic radius r and density $\rho$ have molar mass M

Let us assume the atoms to be spherical

Avogadro's number is $N_A=6.022\times {10}^{23}$

$$\begin{gathered} {N}_A\frac{4}{3}\pi r^3\rho =M \\ r={\left(\frac{3M}{4N_A\pi \rho }\right)}^{\frac{1}{3}} \end{gathered}$$

For Carbon

$$\begin{gathered} {N}_A\frac{4}{3}\pi r^3\rho =M \\ r={\left(\frac{3\times 12.01}{4\times 6.022\times {10}^{23}\times \pi \times 2.22\times {10}^3}\right)}^{\frac{1}{3}} \\ r=1.29 \end{gathered}$$ Å.

For gold

$$\begin{gathered} {N}_A\frac{4}{3}\pi r^3\rho =M \\ r={\left(\frac{3\times 197.00}{4\times 6.022\times {10}^{23}\times \pi \times 19.32\times {10}^3}\right)}^{\frac{1}{3}} \\ r=1.59 \end{gathered}$$ Å.

For Nitrogen

$$\begin{gathered} {N}_A\frac{4}{3}\pi r^3\rho =M \\ r={\left(\frac{3\times 14.01}{4\times 6.022\times {10}^{23}\times \pi \times 1.00\times {10}^3}\right)}^{\frac{1}{3}} \\ r=1.77 \end{gathered}$$ Å.

For Lithium

$$\begin{gathered} {N}_A\frac{4}{3}\pi r^3\rho =M \\ r={\left(\frac{3\times 6.94}{4\times 6.022\times {10}^{23}\times \pi \times 0.53\times {10}^3}\right)}^{\frac{1}{3}} \\ r=1.73 \end{gathered}$$ Å.

For Fluorine

$$\begin{gathered} {N}_A\frac{4}{3}\pi r^3\rho =M \\ r={\left(\frac{3\times 19.00}{4\times 6.022\times {10}^{23}\times \pi \times 1.14\times {10}^3}\right)}^{\frac{1}{3}} \\ r=1.88 \end{gathered}$$ Å.

Class 11 Physics Chapter 12 NCERT solutions help students understand how tiny particles in matter behave. This chapter explains gases and their thermal properties, linking to thermodynamics. Studying it helps in exams and builds a strong foundation in physics.

Kinetic Theory Chapter 12 Physics NCERT Solutions: Concepts and Important Formulas

12.1 Introduction

The kinetic theory explains the behaviour of gases by assuming they consist of a large number of tiny particles (atoms or molecules) in constant random motion. This theory helps derive gas laws and explain properties like pressure, temperature, and specific heat.

12.2 Molecular Nature of Matter

• Matter is made up of tiny molecules that are in continuous motion.
• The molecular interactions vary in solids, liquids, and gases.
• Gases have negligible interatomic forces, allowing them to expand and fill any container.

12.3 Behavior of Gases

• Gases obey different laws like Boyle's Law and Charles's Law at low pressure and high temperature.
• The Ideal Gas Equation is given by:

$PV=nRT$

where $\mathbf{P}=$ pressure, $\mathbf{V}=$ volume, $\mathbf{n}=$ number of moles, $\mathbf{R}=$ universal gas constant, $\mathbf{T}=$ temperature (Kelvin).

12.4 Kinetic Theory of an Ideal Gas

• Gas molecules move in random directions, colliding elastically with each other and with the walls of the container.
• Pressure of an Ideal Gas:

$P=\frac{1}{3}\rho v_{\mathrm{rms}}^2$

where $\rho =$ density of gas, ${\mathrm{vamx}}_{\mathrm{am}}=$ root mean square velocity of gas molecules.

• Root Mean Square Speed ( ${\mathrm{v}}_{\mathrm{mas}}$ ):

$v_{\mathrm{rms}}=\sqrt{\frac{3kT}{m}}$

where $\mathbf{k}=$ Boltzmann constant, $\mathbf{T}=$ absolute temperature, $\mathbf{m}=$ mass of a gas molecule.

12.5 Law of Equipartition of Energy

• Energy is equally distributed among all degrees of freedom in thermal equilibrium.
• Energy per molecule in an ideal gas:

$E=\frac{f}{2}kT$

where $f=$ degrees of freedom.

12.6 Specific Heat Capacity

• Monoatomic gas: $C_v=\frac{3}{2}R,C_p=\frac{5}{2}R$
• Diatomic gas: $C_v=\frac{5}{2}R,C_p=\frac{7}{2}R$
• Relation between Cp and Cv :

$C_p-C_v=R$

12.7 Mean Free Path

• The mean free path $(\lambda )$ is the average distance a molecule travels before colliding with another molecule.

$\lambda =\frac{kT}{\sqrt{2}\pi d^{2}P}$

where $\mathbf{d}=$ diameter of the gas molecule, $\mathbf{P}=$ pressure .

Importance of NCERT Solutions for Class 11 Physics Chapter 12 – Kinetic Theory

• Helps in understanding key concepts and equations in the chapter.
• Useful for class exams and competitive exams like NEET & JEE Mains (1 question is usually asked).
• Covers important formulas, making problem-solving easier.
• This chapter holds 2-3% weightage in competitive exams, so mastering it is beneficial.

Kinetic Theory Class 11 Physics NCERT Topics

NCERT Solutions for Class 11 Physics Chapter Wise

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

NCERT Solutions of Thermodynamics Physics 11: Important Formulas

• Thermodynamics First Law

$\mathrm{\Delta }\mathbf{Q}=\mathrm{\Delta }\mathbf{U}+\mathrm{\Delta }\mathbf{W}$

Where:

ΔQ is 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

• Thermodynamics Second Law

ds >= 0

Where: S is entropy

• Thermodynamics Third Law

S →0 as T →0

Where: s is entropy and T is the temperature

• Molar Specific Heat Capacity

$C=S/\mu =\mathrm{\Delta }Q/\mu \mathrm{\Delta }T$

Here: μ= the mass of 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.

These thermodynamics physics class 11 formulas are valuable tools that students can utilize to calculate and analyze various thermodynamic parameters, making it easier to solve complex problems and gain a deeper understanding of heat, energy, and the laws of thermodynamics.

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.

How to Solve Class 11 Physics Chapter 11 NCERT Questions

• Read the question carefully and list the given values (P, V, T, Q, W, U).
• Identify the type of process (Isothermal, Adiabatic, Isobaric, Isochoric or Cyclic).
• Use the First Law of Thermodynamics to relate Heat, Work, and Energy.
• Learn Important Formulas
• Practice with Example Problems
• Check your units and dimensions in every answer

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.