NCERT Solutions for Class 11 Physics Chapter 12 Kinetic Theory

NCERT Solutions for Class 11 Physics Chapter 12 - Kinetic Theory: Download PDF

The Class 11 Physics Chapter 12 - Kinetic Theory question answers include step-by-step solutions of the textbook questions, which help the students to have a clear picture of the behaviour of gases at the molecular scale. One can access these class 11 physics chapter 12 Kinetic Theory questions answers online and download them as a PDF solution without any payment, making practice and revision convenient.

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

Kinetic Theory class 11 question answers provide all the detailed solutions to the exercises, which allow students to understand the behaviour of gases easily at the microscopic level. The class 11 physics chapter 12 Kinetic Theory questions answers make things easy to understand, such as explaining concepts such as pressure, mean free path, gas laws, and the kinetic interpretation of temperature, developing competitive fundamentals to prepare students for exams and other competitive tests like the JEE or NEET.

NCERT Solutions for Class 11 Physics Chapter 12 - Kinetic Theory: Higher Order Thinking Skills (HOTS) Questions

The HOTS questions of Class 11 Physics Chapter 12 - Kinetic Theory aim at engaging students in the real-life application of the concept and solving the problems that challenge them beyond the core of the topic. These higher-order questions develop thinking, analytical and good conceptual clarity, which are the requirements of board examination as well as competitive examinations, such as JEE and NEET.

Q1:

Two moles of an ideal gas with $\frac{C_p}{C_v}=\frac{5}{3}$ are mixed 3 moles of another ideal gas with $\frac{C_p}{C_v}=\frac{4}{3}$. The value of $\frac{C_p}{C_v}$ for the mixture is:-

Answer:

For ideal gas:- $C_p-C_v=R$
For the first case:-

$
\frac{C_{p 1}}{C_{v 1}}=\frac{5}{3} \text { and } C_{p 1}-C_{v 1}=R
$

$
C_{p 1}=\frac{5}{3} C_{v 1} \text { and } \frac{5}{3} C_{v 1}-C_{v 1}=R$

$\Rightarrow \frac{2}{3} C_{v 1}=R$

$\Rightarrow C_{v 1}=\frac{3}{2} R
$

${ }_{\text {So, }} C_{p 1}=\frac{5}{2} R$
For the second case:-

$
\frac{C_{p 2}}{C_{v 2}}=\frac{4}{3} \text { and } C_{p 2}-C_{v 2}=R \\$

$C_{p 2}=\frac{4}{3} C_{v 2} \text { and } \frac{4}{3} C_{v 2}-C_{v 2}=R$

$\Rightarrow C_{v 2}=3 R{ } \ {\text {and }} C_{p 2}=4 R \\$

$ Y_{\text {mix }}=\frac{n_1 C_{p 1}+n_2 C_{p 2}}{n_1 C_{v 1}+n_2 C_{v 2}}=\frac{2 \times \frac{5}{2} R+3 \times 4 R}{2 \times \frac{3}{2} R+3 \times 3 R}=1.417=1.42$

Q2:

If temperature of the atmosphere varies with height as $\mathrm{T}=\left(\mathrm{T}_0-\mathrm{ah}\right)$, where a and $\mathrm{T}_0$ are positive constants, then the pressure as a function of height $h$ is (assume atmospheric pressure at sea level ( $\mathrm{h}=0$ ) is ${P_0}$ and molecule mass M of the air and acceleration due to gravity g be constant)

Answer:

$
\frac{\mathrm{dP}}{\mathrm{dh}}=-\rho \mathrm{g}=-\left(\frac{\mathrm{PM}}{\mathrm{RT}}\right) \mathrm{g}=-\left[\frac{\mathrm{PM}}{\mathrm{R}\left(\mathrm{~T}_0-\mathrm{ah}\right)}\right] g
$

$
\frac{\mathrm{dP}}{\mathrm{P}}=-\left(\frac{\mathrm{Mg}}{\mathrm{R}}\right) \frac{\mathrm{dh}}{\mathrm{~T}_0-\mathrm{ah}}
$

$
\int_{P_0}^{\mathrm{P}} \frac{\mathrm{dP}}{\mathrm{P}}=-\left(\frac{\mathrm{Mg}}{\mathrm{R}}\right) \int_0^{\mathrm{h}} \frac{\mathrm{dh}}{\left(\mathrm{~T}_0-\mathrm{ah}\right)}
$

$
\mathrm{P}=\mathrm{P}_0\left(\frac{\mathrm{~T}_0-\mathrm{ah}}{\mathrm{~T}_0}\right)^{\frac{\mathrm{M}_{\mathrm{s}}}{\mathrm{Ra}}}
$

Q3:

In two jars A and B, the pressure, volume and temperature in jar A are respectively P, V, and T, and those of B are 2P, V/4 and 2T. Then the ratio of the number of molecules in jars A and B will be,

Answer:

$\begin{aligned} & \frac{N_A}{N_B}=\frac{P_A V_A}{P_B P_B} \times \frac{T_B}{T_A} \\ & \frac{N_A}{N_B}=\frac{P \times V \times(2 T)}{2 P \times \frac{V}{4} \times T}=\frac{4}{1}\end{aligned}$

Q4:

The temperature of an open room of volume 30 m3 increases from 17 °C to 27^oC due to the sunshine. The atmospheric pressure in the room remains 1×105 Pa. If $n_i$ and $n_f$ are the number of molecules in the room before and after heating, then $n_f-n_i$ will be :

Answer:

PV = nRT

$\begin{aligned} & \Rightarrow n_i=\frac{P V}{R T_i}, \quad n_f=\frac{P V}{R T_f} \\ & \Rightarrow n_f-n_i=\frac{P V}{R}\left(\frac{1}{T_f}-\frac{1}{T_i}\right)=\frac{10^5 \times 30}{8.31} \times\left(\frac{1}{300}-\frac{1}{290}\right) \\ & n_f-n_i=\frac{10^5 \times 30}{8.31} \times \frac{-10}{300 \times 290}=\frac{-10^5}{290 \times 8.31} \\ & \text { Change in the Number of molecules }=\frac{-10^5 \times 6.023 \times 10^{23}}{290 \times 8.31}=-2.5 \times 10^{25}\end{aligned}$

Q5:

A gas mixture contains 3 moles of oxygen and x moles of monoatomic gas at temperature T. Considering only translational and rotational but not vibrational modes, the total energy of the system is 15 RT, then the value of x is.

Answer:

$U=\frac{n f}{2} R T$

According to the question,

$\begin{aligned} 15 R T & =3 \times \frac{5}{2} R T+x \times \frac{3}{2} R T \\ 15 & =\frac{15}{2}+\frac{3 x}{2} \\ x & =5\end{aligned}$

Class 11 Physics Chapter 12 - Kinetic Theory: Topics

The Class 11 Physics Chapter 12 - Kinetic Theory examines how the behaviour of gases can be considered at the molecular level, in which the behaviour at the microscopic level is related to the macroscopic behaviour in the form of pressure, volume and temperature. The chapter addresses all the basic assumptions of kinetic theory, derivations and applications that describe real-life phenomena like diffusion and pressure of gases. Knowledge of these topics not only enhances clarity in concepts but also equips students to board examinations and other competitive exams such as JEE and NEET.

12.1 Introduction
12.2 Molecular Nature Of Matter
12.3 Behaviour Of Gases
12.4 Kinetic Theory Of An Ideal Gas
12.4.1 Pressure Of An Ideal Gas
12.4.2 Kinetic Interpretation Of Temperature
12.5 Law Of Equipartition Of Energy
12.6 Specific Heat Capacity
12.6.1 Monatomic Gases
12.6.2 Diatomic Gases
12.6.3 Polyatomic Gases
12.6.4 Specific Heat Capacity Of Solids
12.7 Mean Free Path

Class 11 Physics Chapter 12 - Kinetic Theory: Important Formulae

Kinetic Theory NCERT Solutions- Important Formulae contain important equations that describe the behaviour of the molecular motion in relation to the macroscopic properties of gas, such as pressure, temperature and volume. With the help of these formulas, students can solve numerical problems and see a correlation between the behaviour of microscopic particles and the observed physical phenomena.

1. Ideal Gas Equation

$
P V=n R T
$

Where:

$
\begin{aligned}
& P=\text { pressure }, \\
& V=\text { volume }, \\
& n=\text { number of moles }, \\
& R=\text { universal gas constant }, \\
& T=\text { temperature (in Kelvin) }
\end{aligned}
$

2. Average Kinetic Energy per Molecule

$
K E_{\mathrm{avg}}=\frac{3}{2} k T
$
Where $k=$ Boltzmann constant, $T=$ temperature in Kelvin

3. Total Kinetic Energy of Gas

$K E_{\text {total }}=\frac{3}{2} n R T$

4. Pressure of Ideal Gas from Kinetic Theory

$
P=\frac{1}{3} \rho \bar{v}^2
$
Where $\rho=$ mass per unit volume, $\bar{v}^2=$ mean square speed

5. Root Mean Square Speed

$
v_{\mathrm{rms}}=\sqrt{\frac{3 k T}{m}}=\sqrt{\frac{3 R T}{M}}
$
Where:
$m=$ mass of one molecule,
$M=$ molar mass

7. Relation Between Different Speeds

$
v_{\text {rms }}>v_{\text {avg }}>v_{\text {mp }}
$
Where:
- $v_{\mathrm{rms}}=\sqrt{\frac{3 R T}{M}}$
- $v_{\mathrm{avg}}=\sqrt{\frac{8 R T}{\pi M}}$
- $v_{\mathrm{mp}}=\sqrt{\frac{2 R T}{M}}$

8. Degrees of Freedom and Energy

$
E=\frac{f}{2} k T \text { per molecule }
$
Where $f=$ degrees of freedom

$
E_{\text {total }}=\frac{f}{2} n R T \text { for } n \text { moles }
$

Approach to Solve Questions of the Class 11 Physics Chapter 12 - Kinetic Theory

The Kinetic Theory of Gases has several assumptions, equations and derivations that must be understood clearly to solve some of its questions. Students are expected to learn formulas not by memorization, but to work out the logical application of formulas and interpret physical meaning underlying a formula. Organized planning assists in dealing with numerical and conceptual issues with ease.

• Start with the revision of the fundamental postulates of the kinetic theory, including the following: gases consist of small particles in a state of perpetual random motion, the volume is insignificant, and collisions are perfectly elastic. These assumptions assist you in relating concepts to formulas.

• Determine whether the problem is based on the derivation of pressure, mean free path, root mean square (rms) velocity or laws, such as Boyle and Charles. Saving time is achieved through categorizing and guides to the correct formula.

• Don’t just apply formulas like $P=\frac{1}{3} \rho c^2$ or $v_{r m s}=\sqrt{\frac{3 k T}{m}} ;$ consider what they are, relation of pressure with molecular movement or temperature with speed of particles.

• Begin with familiar values and change all the units into SI and then carefully replace them in the equation. Always verify dimensional consistency- it means that it is accurate.

• Reasoning is involved in many questions such as pressure-temperature or volume-temperature graphs. Explanations are better understood and remembered through graphs.

• Derivations such as the pressure exerted by the gas molecules, mean free path and degrees of freedom are common questions. The conceptual power is developed through writing them step by step.

How Can NCERT Solutions for Class 11 Physics Chapter 12 Kinetic Theory Help in Exam Preparation?

Kinetic Theory class 11 question answers help students build a clear understanding of how gases behave at the molecular level, which is essential for both conceptual and numerical questions in exams. Such Kinetic Theory class 11 question answers break down complex concepts such as pressure, temperature, mean free path, and the law of equipartition of energy into simple explanations so that students can be able to relate microscopic motion with macroscopic properties. With the help of these questions, students are well equipped to solve problems through derivations and numericals, which are often encountered in CBSE, JEE and NEET examinations. The solutions also focus on the connection between the kinetic energy and temperature to allow the students to imagine the way gas behaves in the real world. The frequent use of these NCERT solutions will boost student confidence, increase their ability to use the laws of gases and give them a solid background on other areas of physics and thermodynamics in higher classes.

What Extra Should Students Study Beyond NCERT for JEE/NEET?

NCERT textbooks are almost a must-have in studying and preparing competitive exams such as JEE and NEET, but are definitely lacking in preparing the numericals, tricky applications and deeper conceptual advancement. This is a list of what NCERT does not cover and what additional students are expected to learn in the chapter Kinetic Theory of Gases in preparation for JEE/NEET

NCERT Solutions for Class 11 Physics Chapter-Wise

Chapter-wise links of NCERT Solutions Class 11 provide an easy way to access all chapters at one stop. These solutions are aimed at allowing students to easily get explanations, solved exercises, vital formulas, and additional questions to each chapter and therefore making exam preparation and concept reviewing easier and organised.

NCERT Solutions for Class 11 Subject Wise

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• NCERT Books Class 11 Physics
• NCERT Syllabus Class 11 Physics
• NCERT Books Class 11
• NCERT Syllabus Class 11

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