# NCERT Solutions for Class 11 Physics Chapter 13

## NCERT Solutions for Class 11: Physics Chapter 13 – Kinetic Theory

NCERT Solutions for Class 11:  Physics Chapter 13 provide detailed explanations and accurate solutions to key concepts that students must refer to, in order to ace their term tests, final exams as well as competitive exams at a later stage.

NCERT Class 11:  Physics Chapter 13 : Kinetic Theory covers some of the most important ideas, such as different scientists’ gas laws, atomic theory, the importance and applications of constant values like the Boltzmann constant and Avogadro’s number, kinetic theory postulates, and specific heat capacities.

Learning these subjects is crucial because the second term exam may have questions based on these concepts. Additionally, with these subjects, students will have a strong foundation that will  definitely help them in their next academic session. NCERT Solutions, which are available on Extramarks, provide  access to in-text and end-text exercises as well as additional study resources.

### NCERT Solutions For Class 11 Physics Chapter 13

Solutions will not only help in preparing notes and framing excellent answers but will also boost their confidence. These solutions follow the latest CBSE guidelines  and also include suggested questions for your final and entrance test exams. Access NCERT Solutions for Class 11 Physics Chapter 13 from Extramarks to find these study resources offline and enjoy  the hassle free  and non-stop learning experience.

### NCERT  Class 11: Kinetic Theory

In Class 11, Kinetic Theory is one of the highest scoring chapters. Gas properties are simpler to comprehend than those of solids and liquids. All the objects are made up of atoms, which are tiny particles that move around in a constant state of motion, attracting one another when they are separated by a little distance but repelling when pushed together. The following are some key points from the kinetic theory of gases.

• In a gas, we should not have an exaggerated notion of intermolecular distance. This is only around 10 times the interatomic distance in solids and liquids at ordinary pressures and temperatures.  This means, that the free path, which is 100 times the interatomic distance and 1000 times the molecule’s size in gas, is different.
• A fluid’s pressure isn’t just exerted on the wall. In a fluid, pressure exists everywhere. Because the pressure on both sides of the layer is equal, any layer of gas inside the volume of a container is in equilibrium.
• Because of their rapid speed and constant collisions, air molecules in a room do not fall and settle on the ground (due to gravity). At lower heights, there is a very modest rise in density in equilibrium (like in the atmosphere).

#### NCERT Solutions For Class 11 Physics Chapter 13

It is crucial  that you understand the principles presented in Chapter 13 Kinetic Theory so that you do not have any difficulty understanding the advanced topics you will come across in higher classes. NCERT Solutions will help you gain a deeper understanding of the study material by providing answers to textbook problems as well as questions from sample papers and model question papers.

NCERT Solutions also include past years’ questions, which are quite helpful in preparing for the CBSE examinations for the years 2022-23.

#### NCERT Solutions for Class 11 Physics Chapter 13 Sub – Topics

You can have a clear picture of the principles taught in Kinetic Theory of Gases Class 11 by reading the summary of all seven topics covered in the chapter provided below.

Part 1: Introduction

Before  getting into the details, the first section lays out the fundamentals of the Kinetic Theory of Gases and how it was developed. Students will study the structure and contents of gas molecules, as well as their characteristics and thermodynamic behaviour when contrasted with liquids and solids. Questions about terminology like conduction, viscosity, diffusion, and specific heat capacities have been thoroughly addressed in NCERT Solutions Class 11 Physics Chapter 13 Kinetic Theory.

Part 2: Molecular Nature of Matter

This section goes into the molecular behaviour of gases in greater depth. It includes the Atomic Theory as well as its principles. Interatomic spaces, mean free paths, and dynamic equilibrium of gases have all been emphasised. Students will be expected to comprehend Gay Loussac’s Law and Avogadro’s Law, as well as their relationship.

Part 3: Behaviour of Gases

The title concept of Kinetic Theory of Gases, as well as its postulates, are introduced here. This section covers key terms and ideas such as the Boltzmann constant and Avogadro’s number, as well as its applications in determining temperature, pressure, and volume of gases. You’ll also have to work out difficulties with mole number and molar mass.

Part 4: Kinetic Theory of an Ideal Gas

This section is further divided into two subsections, each of which expands on the previously established concepts and relationships in light of specific circumstances.

• Pressure of an Ideal Gas

The momentum and, finally, the pressure of gas particles may be calculated using their fixed velocities. There is a full explanation of Pascal’s Law.

• Kinetic Interpretation of Temperature

With the help of the Boltzmann constant, the link between kinetic energy, pressure, and temperature is calculated using the ideal gas equation and preceding derivations.

Examples and activities relating to the above-mentioned ideas are included in Chapter 13 Physics Class 11 NCERT Solutions to reinforce the equations and derivations for quick and efficient learning.

Part 5: Law of Equipartition of Energy

Before defining the Law of Equipartition of Energy, this portion covers some of the key concepts of vibrational energy of molecules, rotational vibration, the moment of inertia, and degrees of movement of gas particles.

Part 6: Specific Heat Capacity

In the computations of molar-specific heat capacity of gases, solids, and water at constant volume (cv) and constant pressure, the Law of Equipartition Energy is utilised (cp). There are also suggestions for avoiding gaps between projected and experimental specific heat capacity values. The relevance of quantum physics and its use in numerical problems is conveyed in physics chapter 13 class 11 NCERT Solutions.

Part 7: Mean Free Path

The final subtopic in Kinetic Theory of Gases Class 11 Physics defines the term Mean Free Path and illustrates how it works with real-life examples. It is also calculated using a formula based on the size, density, and the number of gas molecules. In Kinetic Theory of Gases Class 11 Physics NCERT Solutions, students will discover a well-explained numerical based on the formula that will help them improve their problem-solving skills.

Q.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 3A.

Ans.

$\begin{array}{l}\text{Given,}\text{\hspace{0.17em}}\text{diameter of the oxygen molecule,}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{d}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\text{3A}\\ \therefore \text{Radius}\text{\hspace{0.17em}}\text{of the oxygen molecule,}\text{\hspace{0.17em}}\text{r}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\frac{\text{d}}{\text{2}}\\ \text{=}\frac{3}{2}\\ \text{=1}.\text{5}\stackrel{\text{o}}{\text{A}}\text{}\\ =\text{1}.\text{5}×\text{1}{0}^{–\text{8}}\text{cm}\\ \text{The}\text{\hspace{0.17em}}\text{actual volume occupied by 1 mole of oxygen}\text{\hspace{0.17em}}\text{gas at}\text{\hspace{0.17em}}\text{STP,}\text{\hspace{0.17em}}{\text{V}}_{\text{A}}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}{\text{22400 cm}}^{\text{3}}\\ \text{Molecular volume of oxygen gas}\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\text{given}\text{\hspace{0.17em}}\text{by}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{relation:}\\ {\text{V}}_{\text{m}}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\frac{4}{3}\pi {\text{r}}^{3}\text{N}\\ \text{Here,}\text{\hspace{0.17em}}\text{N}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\text{Avogadro’s number}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\text{6}.0\text{23}\text{\hspace{0.17em}}×\text{\hspace{0.17em}}\text{1}{0}^{\text{23}}\text{\hspace{0.17em}}\text{molecules}\text{\hspace{0.17em}}{\text{mole}}^{\text{-1}}\\ \therefore {\text{V}}_{\text{m}}=\frac{4}{3}×3.14×{\left(1.5×{10}^{-8}\right)}^{3}×\text{6}.0\text{23}×\text{1}{0}^{\text{23}}\\ \therefore {\text{V}}_{\text{m}}=8.51\text{\hspace{0.17em}}c{m}^{3}\\ \therefore \text{Ratio of molecular volume to the actual volume of}\text{\hspace{0.17em}}\text{oxygen}\text{\hspace{0.17em}}\text{gas}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\frac{{\text{V}}_{\text{m}}}{{\text{V}}_{\text{A}}}\\ =\frac{8.51}{\text{224}00}\\ \text{=3}.\text{8}×\text{1}{0}^{–\text{4}}\end{array}$

Q.2 Molar volume is the volume occupied by 1 mol of any (ideal) gas at standard temperature and pressure (STP: 1 atmospheric pressure, 0°C). Show that it is 22.4 litres.

Ans.

$\begin{array}{l}\text{The ideal gas equation}\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\text{given}\text{\hspace{0.17em}}\text{as:}\\ \text{PV}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\text{nRT}\\ \text{Here,}\\ \text{R}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\text{Universal gas constant}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{8}.{\text{314 Jmol}}^{–\text{1}}{\text{K}}^{–\text{1}}\\ \text{n}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\text{Number of moles}\\ \text{For}\text{\hspace{0.17em}}\text{one}\text{\hspace{0.17em}}\text{mole}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\text{gas,}\text{\hspace{0.17em}}\text{n}=1\\ \text{Standard temperature,}\text{\hspace{0.17em}}\text{T}=\text{273 K}\\ \text{Standard pressure,}\text{\hspace{0.17em}}\text{P}=\text{1 atm}\\ =\text{1}.0\text{13}×\text{1}{0}^{\text{5}}{\text{Nm}}^{–\text{2}}\\ \therefore \text{Molar}\text{\hspace{0.17em}}\text{volume}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\text{an}\text{\hspace{0.17em}}\text{ideal}\text{\hspace{0.17em}}\text{gas}\text{\hspace{0.17em}}\text{at}\text{\hspace{0.17em}}\text{STP}\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\text{given}\text{\hspace{0.17em}}\text{as:}\text{\hspace{0.17em}}\\ \text{V}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\frac{\text{nRT}}{\text{P}}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\frac{1×\text{8}.\text{314}×\text{273}}{\text{1}.0\text{13}×\text{1}{0}^{\text{5}}}=\text{}0.0{\text{224 m}}^{\text{3}}\\ \text{V}=\text{22}.\text{4 litres}\\ \therefore \text{Molar volume of a gas at STP}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\text{22}.\text{4 litres}.\end{array}$

Q.3 Figure 13.8 shows plot of PV/T versus P for 1.00×10–3 kg of oxygen gas at two different temperatures.
(a) What does the dotted plot signify?
(b) Which is true: T1 > T2 or T1 < T2?
(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 ×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 H2 = 2.02 u, of O2 = 32.0 u, R = 8.31 J mo1–1K–1.)

Ans.

$\begin{array}{l}\left(\text{a}\right)\text{The dotted plot is a straight line parallel to pressure axis}\text{. It shows the ideal behavior of the gas}\text{.}\\ \text{It signifies that the quantity}\text{\hspace{0.17em}}\frac{\text{PV}}{\text{T}}\text{\hspace{0.17em}}\text{is constant and it does not depend upon the pressure of the gas}\text{.}\\ \text{Thus, the dotted plot is for an ideal gas}\text{.}\\ \left(\text{b}\right)\text{\hspace{0.17em}}\text{As the temperature increases, a real gas approaches the behavior of an ideal gas}\text{. In the given}\\ {\text{plot, the curve of the gas at temperature T}}_{\text{1}}\text{\hspace{0.17em}}\text{is closer to the dotted plot than the curve of the gas at}\\ {\text{temperature T}}_{\text{2}}{\text{. Thus, we conclude that T}}_{\text{1}}\text{\hspace{0.17em}}\text{>}\text{\hspace{0.17em}}{\text{T}}_{\text{2}}\text{\hspace{0.17em}}\text{for the given plot}\text{.}\\ \text{(c)}\text{\hspace{0.17em}}\text{Where the two curves}\text{\hspace{0.17em}}\text{meet,}\text{\hspace{0.17em}}\text{the value of}\text{\hspace{0.17em}}\frac{\text{PV}}{\text{T}}\text{is}\text{\hspace{0.17em}}\mu \text{R}.\text{}\\ \text{Since ideal gas equation is given as:}\\ \text{PV}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\mu \text{RT}\\ \therefore \frac{\text{PV}}{\text{T}}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\mu \text{R}\to \text{(i)}\\ \text{Here},\text{\hspace{0.17em}}\mu \text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}\text{R represent}\text{\hspace{0.17em}}\text{number of moles}\text{\hspace{0.17em}}\text{and Universal}\text{\hspace{0.17em}}\text{gas}\text{\hspace{0.17em}}\text{constant}\text{\hspace{0.17em}}\text{respectively}\text{.}\\ \text{Since}\text{\hspace{0.17em}}\text{mass of oxygen}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\text{1}\text{\hspace{0.17em}}×\text{\hspace{0.17em}}\text{1}{0}^{–\text{3}}\text{ }\text{kg}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\text{1 g}\\ \text{Molecular mass of oxygen}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\text{32}.0\text{g}\\ \therefore \mu \text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\frac{1}{32}\to \text{(ii)}\\ \text{R}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\text{8}.{\text{314 J mole}}^{–\text{1}}{\text{K}}^{–\text{1}}\to \text{(iii)}\\ \text{From}\text{\hspace{0.17em}}\text{equation}\text{\hspace{0.17em}}\text{(i),}\text{\hspace{0.17em}}\text{(ii)}\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}\text{(iii),}\text{\hspace{0.17em}}\text{we}\text{\hspace{0.17em}}\text{obtain:}\\ \frac{\text{PV}}{\text{T}}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\frac{1}{32}\text{\hspace{0.17em}}×\text{\hspace{0.17em}}\text{8}.\text{314}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}0.26\text{\hspace{0.17em}}{\text{JK}}^{-1}\\ \therefore \text{The value of}\text{\hspace{0.17em}}\frac{\text{PV}}{\text{T}}\text{\hspace{0.17em}}\text{where the curves meet on}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{y}-\text{axis}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}0.{\text{26 JK}}^{\text{-1}}\end{array}$ $\begin{array}{l}\text{(d)}\text{\hspace{0.17em}}\text{Since}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{molecular masses of hydrogen}\left(\text{2}\text{.02 u}\right)\text{and}\text{\hspace{0.17em}}\text{oxygen}\text{\hspace{0.17em}}\text{}\left(\text{32}\text{.0 u}\right)\text{\hspace{0.17em}}\text{are different,}\text{\hspace{0.17em}}\text{therefore,}\text{\hspace{0.17em}}\text{if we obtained}\\ \text{similar graphs for 1}.00\text{\hspace{0.17em}}×\text{\hspace{0.17em}}\text{1}{0}^{–\text{3}}\text{kg of}\text{\hspace{0.17em}}\text{hydrogen,}\text{\hspace{0.17em}}\text{then}\text{\hspace{0.17em}}\text{we}\text{\hspace{0.17em}}\text{will not get}\text{\hspace{0.17em}}\text{the same value of}\text{\hspace{0.17em}}\frac{\text{PV}}{\text{T}}\text{\hspace{0.17em}}\text{at the point where the}\\ \text{curves}\text{\hspace{0.17em}}\text{meet the}\text{\hspace{0.17em}}\text{y}-\text{axis}.\text{}\\ \text{Here,}\text{\hspace{0.17em}}\frac{\text{PV}}{\text{T}}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}0.26\text{\hspace{0.17em}}{\text{JK}}^{-1}\\ \text{R}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\text{8}.{\text{314 Jmole}}^{–\text{1}}{\text{K}}^{–\text{1}}\\ \text{Molecular mass of Hydrogen,}\text{\hspace{0.17em}}\text{M}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\text{2}.0\text{2 u}\\ \text{At}\text{\hspace{0.17em}}\text{constant}\text{\hspace{0.17em}}\text{temperature},\text{\hspace{0.17em}}\frac{\text{PV}}{\text{T}}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\mu R\to \text{(i)}\\ \text{Let}\text{\hspace{0.17em}}\text{mass}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}{\text{H}}_{2}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}m\\ \therefore \text{Number}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\text{moles,}\text{\hspace{0.17em}}\mu \text{\hspace{0.17em}}=\text{\hspace{0.17em}}\frac{\text{m}}{\text{M}}\text{\hspace{0.17em}}\to \text{(ii)}\\ \text{From}\text{\hspace{0.17em}}\text{equation}\text{\hspace{0.17em}}\text{(i)}\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}\text{(ii),}\text{\hspace{0.17em}}\text{we}\text{\hspace{0.17em}}\text{have:}\\ \frac{\text{PV}}{\text{T}}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\frac{\text{m}}{\text{M}}\text{\hspace{0.17em}}\text{R}\\ \text{Substituting}\text{\hspace{0.17em}}\text{values}\text{\hspace{0.17em}}\text{in}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{above}\text{\hspace{0.17em}}\text{relation,}\text{\hspace{0.17em}}\text{we}\text{\hspace{0.17em}}\text{obtain:}\\ 0.26=\frac{m}{2.02}×\text{8}.\text{314}\\ \therefore \text{m}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\frac{0.26×2.02}{\text{8}.\text{314}}=\text{6}.\text{3}\text{\hspace{0.17em}}×\text{\hspace{0.17em}}\text{1}{0}^{–\text{2}}\text{g}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}6.3\text{\hspace{0.17em}}×\text{\hspace{0.17em}}{10}^{-5}\text{\hspace{0.17em}}\text{kg}\\ \therefore \text{Same value of}\text{\hspace{0.17em}}\frac{\text{PV}}{\text{T}}\text{\hspace{0.17em}}\text{will}\text{\hspace{0.17em}}\text{be}\text{\hspace{0.17em}}\text{obtained}\text{\hspace{0.17em}}\text{from}\text{\hspace{0.17em}}\text{6}.\text{3}\text{\hspace{0.17em}}×\text{\hspace{0.17em}}\text{1}{0}^{–\text{5}}\text{\hspace{0.17em}}{\text{kg of H}}_{\text{2}}.\end{array}$

Q.4 An oxygen cylinder of volume 30 litres has an initial gauge pressure of 15 atm and a temperature of 27°C. After some oxygen is withdrawn from the cylinder, the gauge pressure drops to 11 atm and its temperature drops to 17°C. Estimate the mass of oxygen taken out of the cylinder (R = 8.31J mol–1 K–1, molecular mass of O2 = 32 u).

Ans.

$\begin{array}{l}\text{Here,}\text{\hspace{0.17em}}\text{initial}\text{\hspace{0.17em}}\text{volume of oxygen,}\text{\hspace{0.17em}}{\text{V}}_{\text{i}}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\text{30 litres}\\ =\text{3}0×\text{1}{0}^{–\text{3}}{\text{m}}^{\text{3}}\\ \text{Initial}\text{\hspace{0.17em}}\text{gauge pressure,}\text{\hspace{0.17em}}{\text{P}}_{\text{i}}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\text{15 atm}\\ =\text{\hspace{0.17em}}\text{15}\text{\hspace{0.17em}}×\text{\hspace{0.17em}}\text{1}.0\text{13}\text{\hspace{0.17em}}×\text{\hspace{0.17em}}\text{1}{0}^{\text{5}}\text{Pa}\\ \text{=}\text{\hspace{0.17em}}\text{15}\text{.195}\text{\hspace{0.17em}}×\text{\hspace{0.17em}}\text{1}{0}^{\text{5}}\text{Pa}\\ \text{Initial}\text{\hspace{0.17em}}\text{temperature},\text{\hspace{0.17em}}{\text{T}}_{i}=\text{27}°\text{C}\\ =\text{3}00\text{K}\\ \text{Universal gas constant,}\text{\hspace{0.17em}}\text{R}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\text{8}.{\text{314 Jmole}}^{–\text{1}}{\text{K}}^{–\text{1}}\\ \text{Molecular}\text{\hspace{0.17em}}\text{mass}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}{\text{O}}_{\text{2}}\text{,}\text{\hspace{0.17em}}\text{M}=32\text{\hspace{0.17em}}\text{g}\\ \text{Let}\text{\hspace{0.17em}}\text{initial mass of oxygen}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}{\text{m}}_{\text{i}}\\ \text{Let the initial number of moles of oxygen gas in the}\text{\hspace{0.17em}}\text{cylinder}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}{\text{n}}_{\text{i}}\\ \text{The gas equation is given by:}\\ {\text{P}}_{i}{\text{V}}_{i}={\text{n}}_{i}{\text{RT}}_{i}\\ \therefore {\text{n}}_{i}=\frac{{\text{P}}_{i}{\text{V}}_{i}}{{\text{RT}}_{i}}\\ =\frac{\text{15}\text{.195}×\text{1}{0}^{\text{5}}×\text{3}0×\text{1}{0}^{–\text{3}}}{\text{8}.\text{314}×\text{3}00}\\ =\text{18}.\text{276}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\\ \text{Since,}\text{\hspace{0.17em}}{\text{n}}_{i}=\frac{{m}_{i}}{M}\\ \therefore {\text{m}}_{i}={\text{n}}_{i}\text{M}\\ =\text{18}.\text{276}×\text{32}\\ =\text{584}.\text{84 g}\\ \text{Since}\text{\hspace{0.17em}}\text{initial}\text{\hspace{0.17em}}\text{values}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\text{pressure}\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}\text{temperature}\text{\hspace{0.17em}}\text{decrease,}\text{\hspace{0.17em}}\\ \text{after withdrawing}\text{\hspace{0.17em}}\text{some oxygen from the cylinder,}\\ \text{Given,}\text{\hspace{0.17em}}\text{final}\text{\hspace{0.17em}}\text{volume}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\text{oxygen,}\text{\hspace{0.17em}}{\text{V}}_{\text{f}}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\text{30 litres}\\ =\text{\hspace{0.17em}}\text{3}0×\text{1}{0}^{–\text{3}}{\text{m}}^{\text{3}}\\ \text{Final}\text{\hspace{0.17em}}\text{gauge pressure,}\text{\hspace{0.17em}}{\text{P}}_{\text{f}}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\text{11 atm}\\ =\text{\hspace{0.17em}}\text{11}\text{\hspace{0.17em}}×\text{\hspace{0.17em}}\text{1}.0\text{13}\text{\hspace{0.17em}}×\text{\hspace{0.17em}}\text{1}{0}^{\text{5}}\text{\hspace{0.17em}}\text{Pa}\\ \text{=}\text{\hspace{0.17em}}\text{11}\text{.143}×\text{1}{0}^{\text{5}}\text{Pa}\\ \text{Final}\text{\hspace{0.17em}}{\text{temperature,T}}_{f}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\text{17}°\text{C}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\text{29}0\text{K}\\ \text{Let the number of moles of oxygen left in the cylinder}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}{\text{n}}_{\text{f}}\\ \text{The gas equation is given as:}\\ {\text{P}}_{f}{\text{V}}_{f}={\text{n}}_{f}{\text{RT}}_{f}\\ \therefore {\text{n}}_{f}=\frac{{\text{P}}_{f}{\text{V}}_{f}}{{\text{RT}}_{f}}\\ =\frac{\text{11}\text{.143}×\text{1}{0}^{\text{5}}×\text{3}0×\text{1}{0}^{–\text{3}}}{\text{8}.\text{314}×290}\\ =\text{13}.\text{86}\\ \text{But,}\text{\hspace{0.17em}}\text{number}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\text{moles}\text{\hspace{0.17em}}\text{left},\text{\hspace{0.17em}}{\text{n}}_{f}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\frac{{m}_{f}}{M}\\ \text{Here,}\text{\hspace{0.17em}}{\text{m}}_{\text{f}}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\text{Mass of oxygen left in the cylinder}\\ \therefore {\text{m}}_{\text{f}}={\text{n}}_{f}\text{M}\\ =\text{13}.\text{86}×\text{32}\\ =\text{453}.\text{1 g}\\ \text{Mass of oxygen withdrawn from the cylinder,}\\ \Delta \text{M}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\text{Initial mass of oxygen–Final mass of oxygen}\\ \Delta \text{M}={\text{m}}_{i}–{\text{m}}_{f}\\ =\text{584}.\text{84 g}–\text{453}.\text{1 g}\\ \text{=131}\text{.74}\text{\hspace{0.17em}}\text{g}\\ \therefore \Delta \text{M=0}\text{.131}\text{\hspace{0.17em}}\text{kg}\\ \therefore \text{0}\text{.131}\text{\hspace{0.17em}}\text{kg}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\text{oxygen is}\text{\hspace{0.17em}}\text{withdrawn}\text{\hspace{0.17em}}\text{from the cylinder}\text{.}\end{array}$

Q.5 An air bubble of volume 1.0 cm3 rises from the bottom of a lake 40 m deep at a temperature of 12 °C. To what volume does it grow when it reaches the surface, which is at a temperature of 35°C?

Ans.

$\begin{array}{l}\text{Here,}\text{\hspace{0.17em}}\text{initial}\text{\hspace{0.17em}}\text{volume of the air bubble,}\text{\hspace{0.17em}}\text{V}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\text{1}.0{\text{cm}}^{\text{3}}\text{}\\ =\text{1}.0\text{\hspace{0.17em}}×\text{\hspace{0.17em}}\text{1}{0}^{–\text{6}}{\text{m}}^{\text{3}}\\ \text{Height}\text{\hspace{0.17em}}\text{upto}\text{\hspace{0.17em}}\text{which}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{air}\text{\hspace{0.17em}}\text{bubble}\text{\hspace{0.17em}}\text{rises}\text{\hspace{0.17em}}\text{,}\text{\hspace{0.17em}}\text{d}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\text{40 m}\\ \text{Temperature at the}\text{\hspace{0.17em}}\text{bottom}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}{\text{lake,T}}_{\text{1}}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\text{12°C}\\ =\text{\hspace{0.17em}}\text{285 K}\\ {\text{Temperature at the surface of the lake,T}}_{\text{s}}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\text{35}°\text{C}\\ =\text{\hspace{0.17em}}\text{3}0\text{8 K}\\ \text{Pressure on the surface of the lake,}\text{\hspace{0.17em}}{\text{P}}_{s}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\text{1 atm}\\ =\text{\hspace{0.17em}}\text{1}\text{\hspace{0.17em}}×\text{\hspace{0.17em}}\text{1}.0\text{13}\text{\hspace{0.17em}}×\text{\hspace{0.17em}}\text{1}{0}^{\text{5}}\text{Pa}\\ \text{Pressure at the bottom}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{lake,}\text{\hspace{0.17em}}{\text{P}}_{\text{b}}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\text{1 atm}\text{\hspace{0.17em}}+\text{\hspace{0.17em}}\text{d}\rho \text{g}\\ \text{Here,}\\ \rho =\text{Density of water}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{1}{0}^{\text{3}}{\text{kgm}}^{\text{-3}}\\ \text{g =Acceleration due to gravity}\\ =\text{9}.{\text{8 ms}}^{\text{-2}}\\ \therefore {\text{P}}_{b}=\text{1}.0\text{13}×\text{1}{0}^{\text{5}}+\text{4}0×\text{1}{0}^{\text{3}}×\text{9}.\text{8}\\ =\text{4933}00\text{Pa}\\ \text{From}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{relation,}\\ \frac{{P}_{b}{V}_{b}}{{T}_{b}}=\frac{{P}_{s}{V}_{s}}{{T}_{s}}\\ \text{Here,}\text{\hspace{0.17em}}{\text{V}}_{\text{s}}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\text{Volume of the air bubble at the surface}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\text{lake}\\ {\text{V}}_{s}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\frac{{P}_{b}{V}_{b}{T}_{s}}{{T}_{b}{P}_{s}}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\frac{\text{4933}00\text{\hspace{0.17em}}×\text{\hspace{0.17em}}\text{1}.0\text{\hspace{0.17em}}×\text{\hspace{0.17em}}\text{1}{0}^{–\text{6}}\text{\hspace{0.17em}}×\text{\hspace{0.17em}}\text{3}0\text{8}}{\text{285}\text{\hspace{0.17em}}×\text{\hspace{0.17em}}\text{1}.0\text{13}\text{\hspace{0.17em}}×\text{\hspace{0.17em}}\text{1}{0}^{\text{5}}}\\ =\text{\hspace{0.17em}}\text{5}.\text{263}\text{\hspace{0.17em}}×\text{\hspace{0.17em}}\text{1}{0}^{–\text{6}}{\text{m}}^{\text{3}}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\text{5}.{\text{263 cm}}^{\text{3}}\\ \therefore \text{When the air bubble reaches the surface,}\text{\hspace{0.17em}}\text{its volume = 5}.{\text{263 cm}}^{\text{3}}\end{array}$

Q.6 Estimate the total number of air molecules (inclusive of oxygen, nitrogen, water vapour and other constituents) in a room of capacity 25.0 m3 at a temperature of 27 °C and 1 atm pressure.

Ans.

$\begin{array}{l}\text{Here,}\text{\hspace{0.17em}}\text{volume of room,V}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\text{25}{\text{.0 m}}^{\text{3}}\\ \text{Temperature of room,T}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\text{27}°\text{C}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\text{300 K}\\ \text{Pressure inside the room,}\text{\hspace{0.17em}}\text{P}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\text{1 atm}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\text{1}\text{\hspace{0.17em}}\text{×}\text{\hspace{0.17em}}\text{1}\text{.013}\text{\hspace{0.17em}}\text{×}\text{\hspace{0.17em}}{\text{10}}^{\text{5}}\text{Pa}\\ \text{The}\text{\hspace{0.17em}}\text{ideal}\text{\hspace{0.17em}}\text{gas}\text{\hspace{0.17em}}\text{equation}\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\text{given}\text{\hspace{0.17em}}\text{as:}\\ \text{PV}={\text{k}}_{\text{B}}\text{NT}\\ \text{Here},\text{\hspace{0.17em}}{\text{K}}_{\text{B}}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\text{Boltzmann constant}\\ =\text{1}.\text{38}\text{\hspace{0.17em}}×\text{\hspace{0.17em}}\text{1}{0}^{–\text{23}}{\text{m}}^{\text{2}}\text{\hspace{0.17em}}\text{kg}\text{\hspace{0.17em}}{\text{s}}^{\text{–2}}\text{\hspace{0.17em}}{\text{K}}^{\text{–1}}\\ \text{N}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\text{Number}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\text{air}\text{\hspace{0.17em}}\text{molecules}\text{\hspace{0.17em}}\text{inside}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{room}\\ \therefore \text{N}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\frac{\text{PV}}{{\text{k}}_{\text{B}}\text{T}}=\frac{\text{1}.0\text{13}×\text{1}{0}^{\text{5}}×\text{25}}{\text{1}.\text{38}\text{\hspace{0.17em}}×\text{\hspace{0.17em}}\text{1}{0}^{–\text{23}}×\text{3}00}\\ =\text{6}.\text{11}×\text{1}{0}^{\text{26}}\text{molecules}\\ \therefore \text{Total number of air molecules in the given room}\\ \text{= 6}.\text{11}×\text{1}{0}^{\text{26}}\end{array}$

Q.7 Estimate the average thermal energy of a helium atom at
(i) room temperature (27°C),
(ii) the temperature on the surface of the Sun (6000 K),
(iii) the temperature of 10 million Kelvin (the typical core temperature in the case of a star).

Ans.

$\begin{array}{l}\left(\text{i}\right)\text{\hspace{0.17em}}\text{At room temperature,}\\ \text{Temperature,}\text{\hspace{0.17em}}\text{T}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\text{27}°\text{C}=\text{3}00\text{K}\\ \text{Average thermal energy}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{3}{2}\text{kT}\\ \text{Where}\text{\hspace{0.17em}}\text{k=Boltzmann constant}\\ =\text{1}.\text{38}\text{\hspace{0.17em}}×\text{\hspace{0.17em}}\text{1}{0}^{–\text{23}}{\text{m}}^{\text{2}}\text{\hspace{0.17em}}\text{kg}\text{\hspace{0.17em}}{\text{s}}^{–\text{2}}\text{\hspace{0.17em}}{\text{K}}^{–\text{1}}\\ \therefore \text{Average}\text{\hspace{0.17em}}\text{thermal}\text{\hspace{0.17em}}\text{energy}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\frac{3}{2}\text{\hspace{0.17em}}\text{kT}\\ =\frac{3}{2}\text{\hspace{0.17em}}×\text{\hspace{0.17em}}\text{1}.\text{38}\text{\hspace{0.17em}}×\text{\hspace{0.17em}}\text{1}{0}^{–\text{23}}\text{\hspace{0.17em}}×\text{\hspace{0.17em}}\text{3}00\\ =\text{6}.\text{21}\text{\hspace{0.17em}}×\text{\hspace{0.17em}}\text{1}{0}^{–\text{21}}\text{\hspace{0.17em}}\text{J}\\ \therefore \text{Average thermal energy of a helium atom at room temperature}\left(\text{27}°\text{C}\right)\text{is}\text{\hspace{0.17em}}\text{6}.\text{21}×\text{1}{0}^{–\text{21}}\text{J}.\\ \left(\text{ii}\right)\text{On the surface of the sun,}\text{\hspace{0.17em}}\\ \text{Temperature,}\text{\hspace{0.17em}}\text{T}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\text{6000 K}\\ \text{Average thermal energy}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\frac{\text{3}}{\text{2}}\text{kT}\\ =\text{\hspace{0.17em}}\frac{3}{2}\text{\hspace{0.17em}}×\text{\hspace{0.17em}}\text{1}.\text{38}\text{\hspace{0.17em}}×\text{\hspace{0.17em}}\text{1}{0}^{–\text{23}}\text{\hspace{0.17em}}×\text{\hspace{0.17em}}6000\text{}\\ \text{=}\text{\hspace{0.17em}}\text{1}.\text{241}\text{\hspace{0.17em}}\text{\hspace{0.17em}}×\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{1}{0}^{–\text{19}}\text{J}\\ \therefore \text{Average thermal energy of a helium atom on the}\text{\hspace{0.17em}}\text{surface of the sun is}\text{\hspace{0.17em}}\text{1}.\text{241}\text{\hspace{0.17em}}×\text{\hspace{0.17em}}\text{1}{0}^{–\text{19}}\text{J}.\\ \left(\text{iii}\right)\text{\hspace{0.17em}}\text{At temperature,T}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\text{1}{0}^{\text{7}}\text{K}\\ \text{Average thermal energy}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\frac{3}{2}\text{\hspace{0.17em}}\text{kT}\\ =\frac{3}{2}\text{\hspace{0.17em}}×\text{\hspace{0.17em}}\text{1}.\text{38}\text{\hspace{0.17em}}×\text{\hspace{0.17em}}\text{1}{0}^{–\text{23}}\text{\hspace{0.17em}}×\text{\hspace{0.17em}}\text{1}{0}^{\text{7}}\\ =\text{\hspace{0.17em}}\text{2}.0\text{7}\text{\hspace{0.17em}}×\text{\hspace{0.17em}}\text{1}{0}^{–\text{16}}\text{J}\\ \therefore \text{Average thermal energy of a helium atom at the core of a star}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\text{2}.0\text{7}\text{\hspace{0.17em}}×\text{\hspace{0.17em}}\text{1}{0}^{–\text{16}}\text{J}\end{array}$

Q.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 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 vrms the largest?

Ans.

$\begin{array}{l}\text{Since all the three vessels have same volume at same temperature and pressure,}\\ \text{Therefore, according to Avogadro’s law, all the three vessels contain the same number of the respective molecules}\text{.}\\ {\text{The root mean square speed (v}}_{\text{rms}}\text{) of a gas of mass}\text{\hspace{0.17em}}\text{m,}\text{\hspace{0.17em}}\text{and temperature}\text{\hspace{0.17em}}\text{T, is given as:}\\ {\text{v}}_{\text{rms}}=\sqrt{\frac{3kT}{m}}\\ \text{Where,}\text{\hspace{0.17em}}\text{k}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\text{Boltzmann constant}\\ \text{Since}\text{\hspace{0.17em}}\text{for}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{given}\text{\hspace{0.17em}}\text{gases,}\text{\hspace{0.17em}}\text{k}\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}\text{T}\text{\hspace{0.17em}}\text{are}\text{\hspace{0.17em}}\text{constants,}\\ \therefore {\text{v}}_{\text{rms}}\propto \sqrt{\frac{1}{m}}\\ \therefore \text{The}\text{\hspace{0.17em}}\text{root}\text{\hspace{0.17em}}\text{mean}\text{\hspace{0.17em}}\text{square}\text{\hspace{0.17em}}\text{speed}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{molecules}\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\text{not}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{same}\text{\hspace{0.17em}}\text{in}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{three}\text{\hspace{0.17em}}\text{cases}\text{.}\\ \text{Out of the given gases neon has the smallest mass, therefore, root mean square}\text{\hspace{0.17em}}\text{speed of neon is the largest}\text{.}\end{array}$

Q.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°C? (Atomic mass of Ar = 39.9 u, of He = 4.0 u).

Ans.

$\begin{array}{l}\text{Here,}\text{\hspace{0.17em}}\text{Temperature of helium atom,}\text{\hspace{0.17em}}{\text{T}}_{\text{He}}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}–\text{2}0°\text{C}\\ =\text{\hspace{0.17em}}\text{253 K}\\ \text{Atomic mass of the}\text{\hspace{0.17em}}\text{argon}\text{\hspace{0.17em}}\text{gas,}\text{\hspace{0.17em}}{\text{M}}_{\text{Ar}}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\text{39}.\text{9 u}\\ \text{Atomic mass of the}\text{\hspace{0.17em}}\text{helium}\text{\hspace{0.17em}}\text{gas},\text{\hspace{0.17em}}{\text{M}}_{\text{He}}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\text{4}.0\text{u}\\ \text{Let}\text{\hspace{0.17em}}\text{rms speed of argon}\text{\hspace{0.17em}}\text{gas}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}{\left({\text{v}}_{\text{rms}}\right)}_{\text{Ar}}\text{}\\ \text{Let rms speed of helium}\text{\hspace{0.17em}}\text{gas}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}{\left({\text{v}}_{\text{rms}}\right)}_{\text{He}}\text{}\\ \text{Rms speed of the}\text{\hspace{0.17em}}\text{argon gas}\text{\hspace{0.17em}}\text{is given as:}\\ {\left({\text{v}}_{\text{rms}}\right)}_{\text{Ar}}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\sqrt{\frac{3R{T}_{Ar}}{{M}_{Ar}}}\text{}\to \text{\hspace{0.17em}}\left(\text{i}\right)\\ \text{Where,}\text{\hspace{0.17em}}\text{R}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\text{Universal gas constant}\text{\hspace{0.17em}}\\ {\text{T}}_{\text{Ar}}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\text{Temperature of argon gas}\\ \text{Rms speed of the}\text{\hspace{0.17em}}\text{helium gas}\text{\hspace{0.17em}}\text{is given as:}\\ {\left({\text{v}}_{\text{rms}}\right)}_{\text{He}}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\sqrt{\frac{3R{T}_{He}}{{M}_{He}}}\to \text{\hspace{0.17em}}\left(\text{ii}\right)\\ \text{As}\text{\hspace{0.17em}}\text{per}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{question:}\\ {\left({\text{v}}_{\text{rms}}\right)}_{\text{Ar}}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}{\left({\text{v}}_{\text{rms}}\right)}_{\text{He}}\text{\hspace{0.17em}}\\ \sqrt{\frac{3R{T}_{Ar}}{{M}_{Ar}}}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\sqrt{\frac{3R{T}_{He}}{{M}_{He}}}\\ \frac{{T}_{Ar}}{{M}_{Ar}}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\frac{{T}_{He}}{{M}_{He}}\\ {T}_{Ar}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\frac{{T}_{He}}{{M}_{He}}\text{\hspace{0.17em}}×\text{\hspace{0.17em}}{M}_{Ar}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\frac{253}{4}\text{\hspace{0.17em}}×\text{\hspace{0.17em}}39.9\\ =\text{\hspace{0.17em}}\text{2523}.\text{675}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\text{2}.\text{52}\text{\hspace{0.17em}}×\text{\hspace{0.17em}}\text{1}{0}^{\text{3}}\text{\hspace{0.17em}}\text{K}\\ \therefore \text{Temperature of the argon gas}\text{\hspace{0.17em}}\text{atom}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\text{2}.\text{52}\text{\hspace{0.17em}}×\text{\hspace{0.17em}}\text{1}{0}^{\text{3}}\text{K}\end{array}$

Q.10 Estimate the mean free path and collision frequency of a nitrogen molecule in a cylinder containing nitrogen at 2.0 atm and temperature 17 °C. Take the radius of a nitrogen molecule to be roughly 1.0 Å.
Compare the collision time with the time the molecule moves freely between two successive collisions (Molecular mass of N2 = 28.0 u).

Ans.

$Here, pressure inside the cylinder, P = 2.0 atm = 2 .026×10 5 Pa Temperature inside the cylinder, T = 17°C = 290 K Radius of a nitrogen molecule, r = 1 .0 Å = 1×10 10 m Diameter, d = 2r = 2×1×10 10 m = 2×10 10 m Molecular mass of nitrogen, M N = 28 .0 g = 28×10 –3 kg The root mean square speed of nitrogen is given as: v rms = 3RT M Where,R = Universal gas constant = 8 .314 Jmol –1 K –1 ∴ v rms = 3×8.314×290 28×10 -3 = 508 .26 ms -1 Mean free path, l = kT 2 ×d 2 ×P Here, k=Boltzmann constant = 1 .38×10 –23 kgm 2 s –2 K –1 ∴l = 1 .38×10 –23 kgm 2 s –2 K –1 ×290 K 2 ×3.14× 2×10 -10 m 2 ×2 .026×10 5 Pa = 1 .11×10 –7 m Collision frequency = v rms l = 508.26 1 .11×10 –7 = 4 .58×10 9 s –1 Collision time is given by the relation: T = d v rms T = 2×10 -10 508.26 = 3 .93×10 -13 s Time taken between successive collisions is given as: T’ = l v rms = 1 .11×10 –7 508.26 = 2 .18×10 –10 s ∴ T’ T = 2 .18×10 –10 3 .93×10 -13 ≈500 ∴Time taken between successive collisions = 500 times the time taken for a collision.$

Q.11 A meter long narrow bore held horizontally (and closed at one end) contains a 76 cm long mercury thread, which traps a 15 cm column of air. What happens if the tube is held vertically with the open end at the bottom?

Ans.

$\begin{array}{l}\text{Here,}\text{\hspace{0.17em}}\text{length of narrow bore,}\text{\hspace{0.17em}}\text{L}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\text{1 m}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\text{1}00\text{cm}\\ \text{Length of mercury thread,}\text{\hspace{0.17em}}\text{l}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\text{76 cm}\\ \text{Length of air column between mercury and the}\text{\hspace{0.17em}}\text{closed end,}\text{\hspace{0.17em}}{\text{l}}_{\text{a}}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\text{15 cm}\\ \text{As the bore is held vertically in air with the open}\text{\hspace{0.17em}}\text{end}\text{\hspace{0.17em}}\text{at the bottom,}\\ \text{The air space}\text{\hspace{0.17em}}\text{occupied}\text{\hspace{0.17em}}\text{by}\text{\hspace{0.17em}}\text{the mercury}\text{\hspace{0.17em}}\text{length}\\ \text{=1}00\text{\hspace{0.17em}}–\text{\hspace{0.17em}}\left(\text{76}\text{\hspace{0.17em}}+\text{\hspace{0.17em}}\text{15}\right)\\ =\text{9 cm}\\ \therefore \text{Total length of the air column}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\text{15}\text{\hspace{0.17em}}\text{+}\text{\hspace{0.17em}}\text{9}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\text{24 cm}\\ \text{Let mercury flow out due}\text{\hspace{0.17em}}\text{to}\text{\hspace{0.17em}}\text{atmospheric pressure}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\text{h}\text{\hspace{0.17em}}\text{cm}\\ \therefore \text{Length of air column inside the bore}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\text{24}\text{\hspace{0.17em}}+\text{\hspace{0.17em}}\text{h}\text{\hspace{0.17em}}\text{cm}\\ \text{Length of mercury column}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\text{76}\text{\hspace{0.17em}}–\text{\hspace{0.17em}}\text{h}\text{\hspace{0.17em}}\text{cm}\\ \text{Initial pressure,}\text{\hspace{0.17em}}{\text{P}}_{\text{i}}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\text{76 cm of mercury}\\ \text{Initial volume,}\text{\hspace{0.17em}}{\text{V}}_{\text{i}}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}{\text{15 cm}}^{\text{3}}\\ \text{Final pressure,}\text{\hspace{0.17em}}{\text{P}}_{\text{f}}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\text{76}\text{\hspace{0.17em}}–\text{\hspace{0.17em}}\left(\text{76}\text{\hspace{0.17em}}–\text{\hspace{0.17em}}\text{h}\right)\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\text{h}\text{\hspace{0.17em}}\text{cm of mercury}\\ \text{Final volume,}\text{\hspace{0.17em}}{\text{V}}_{\text{f}}=\left(\text{24}\text{\hspace{0.17em}}+\text{\hspace{0.17em}}\text{h}\right){\text{cm}}^{\text{3}}\\ \text{As}\text{\hspace{0.17em}}\text{temperature remains constant,}\\ \therefore \text{From}\text{\hspace{0.17em}}\text{Boyle’s}\text{\hspace{0.17em}}\text{law}\\ {\text{P}}_{\text{i}}{\text{V}}_{\text{i}}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}{\text{P}}_{\text{f}}{\text{V}}_{\text{f}}\\ \text{76}\text{\hspace{0.17em}}×\text{\hspace{0.17em}}\text{15}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\text{h}\left(\text{24}\text{\hspace{0.17em}}+\text{\hspace{0.17em}}\text{h}\right)\\ {\text{h}}^{\text{2}}\text{\hspace{0.17em}}+\text{\hspace{0.17em}}\text{24h}\text{\hspace{0.17em}}–\text{\hspace{0.17em}}\text{114}0=0\\ \therefore \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{h}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\frac{-24±\sqrt{{\left(\text{24}\right)}^{2}+4×1×1140}}{2×1}\\ =\text{23}.\text{8 cm or}–\text{47}.\text{8 cm}\\ \text{Since}\text{\hspace{0.17em}}\text{height cannot be negative,}\\ \therefore \text{23}.\text{8 cm of mercury will flow out from the bore}\text{\hspace{0.17em}}\text{and}\\ \left(76\text{\hspace{0.17em}}\text{cm}-\text{23}.\text{8}\text{\hspace{0.17em}}\text{cm}\right)\text{\hspace{0.17em}}\text{52}.\text{2 cm of mercury will be}\text{\hspace{0.17em}}\text{left in it}\text{.}\\ \therefore \text{Length of the air column}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\text{24}\text{\hspace{0.17em}}\text{+}\text{\hspace{0.17em}}\text{23}\text{.8}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\text{47}\text{.8 cm}\text{.}\end{array}$

Q.12 From a certain apparatus, the diffusion rate of hydrogen has an average value of 28.7 cm3 s–1. The diffusion of another gas under the same conditions is measured to have an average rate of 7.2 cm3 s–1. Identify the gas.
[Hint: Use Graham’s law of diffusion: R1/R2 = (M2/M1)1/2, where R1, R2 are diffusion rates of gases 1 and 2, and M1 and M2 their respective molecular masses. The law is a simple consequence of kinetic theory.]

Ans.

$\begin{array}{l}\text{Here,}\text{\hspace{0.17em}}\text{average rate of diffusion of hydrogen},\text{\hspace{0.17em}}{\text{R}}_{\text{1}}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\text{28}.{\text{7 cm}}^{\text{3}}{\text{s}}^{–\text{1}}\\ \text{Average}\text{\hspace{0.17em}}\text{Rate of diffusion of another gas,}\text{\hspace{0.17em}}{\text{R}}_{\text{2}}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\text{7}.{\text{2 cm}}^{\text{3}}{\text{s}}^{–\text{1}}\\ \text{Molecular mass of hydrogen,}\text{\hspace{0.17em}}{\text{M}}_{\text{1}}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\text{2}.0\text{2}0\text{g}\\ \text{Let}\text{\hspace{0.17em}}\text{molecular mass of the unknown gas}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}{\text{M}}_{\text{2}}\\ \text{As}\text{\hspace{0.17em}}\text{per Graham’s Law of diffusion,}\\ \frac{{\text{R}}_{\text{1}}}{{\text{R}}_{\text{2}}}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\sqrt{\frac{{\text{M}}_{\text{2}}}{{\text{M}}_{\text{1}}}}\\ \therefore \text{\hspace{0.17em}}{\text{M}}_{\text{2}}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}{\text{M}}_{\text{1}}{\left(\frac{{\text{R}}_{\text{1}}}{{\text{R}}_{\text{2}}}\right)}^{2}\\ =\text{\hspace{0.17em}}\text{2}.0\text{2}0{\left(\frac{28.7}{7.2}\right)}^{2}\\ =\text{\hspace{0.17em}}32.09\text{\hspace{0.17em}}\text{g}\\ \text{Since molecular mass of oxygen is 32, therefore, the unknown gas is oxygen}\text{.}\end{array}$

Q.13 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
n2 = n1 exp [-mg (h2 – h1)/ kBT]
Where n2, n1 refer to number density at heights h2 and h1 respectively. Use this relation to derive the equation for sedimentation equilibrium of a suspension in a liquid column:
n2 = n1 exp [-mg NA(ρ – P′) (h2 –h1)/ (ρRT)]
Where ρ is the density of the suspended particle, and ρ’ that of surrounding medium.
[NA is Avogadro’s number, and R the universal gas constant.] [Hint: Use Archimedes principle to find the apparent weight of the suspended particle.]

Ans.

$\begin{array}{l}\text{Given}\text{\hspace{0.17em}}\text{that,}\text{\hspace{0.17em}}\text{as}\text{\hspace{0.17em}}\text{per}\text{\hspace{0.17em}}\text{the law of atmospheres,}\\ {\text{n}}_{\text{2}}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}{\text{n}}_{\text{1}}\text{exp}\text{\hspace{0.17em}}\left[-\text{mg}\left({\text{h}}_{\text{2}}\text{\hspace{0.17em}}–\text{\hspace{0.17em}}{\text{h}}_{\text{1}}\right)/{\text{k}}_{\text{B}}\text{T}\right]\text{}\to \text{}\left(\text{i}\right)\\ {\text{Here,n}}_{\text{1}}\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}{\text{n}}_{\text{2}}\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\text{the number density of}\text{\hspace{0.17em}}\text{particles}\text{\hspace{0.17em}}\text{at height}\text{\hspace{0.17em}}{\text{h}}_{\text{1}}\text{and}\text{\hspace{0.17em}}{\text{h}}_{\text{2}}\text{\hspace{0.17em}}\text{respectively}\text{.}\\ \text{mg}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\text{Weight of the particle hanged in the gas}\text{\hspace{0.17em}}\text{column}\\ \text{Let}\text{\hspace{0.17em}}{\rho }_{m}\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}{\rho }_{s}\text{\hspace{0.17em}}be\text{\hspace{0.17em}}the\text{\hspace{0.17em}}d\text{ensity of medium}\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}d\text{ensity of}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{suspended particle}\text{\hspace{0.17em}}\text{respectively}\text{.}\\ \text{Let}\text{\hspace{0.17em}}\text{mass of one suspended particle}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}{\text{m}}_{\text{s}}\\ \text{Let}\text{\hspace{0.17em}}\text{mass of the medium displaced}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}{\text{m}}_{\text{m}}\\ \text{Let}\text{\hspace{0.17em}}\text{volume of the suspended particle}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\text{V}\\ \text{As}\text{\hspace{0.17em}}\text{per Archimedes’ principle for a particle suspended}\text{\hspace{0.17em}}\text{in a liquid column,}\text{\hspace{0.17em}}\text{we}\text{\hspace{0.17em}}\text{have:}\\ \text{Effective weight of the suspended particle}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\text{Weight of the medium displaced}\text{\hspace{0.17em}}–\text{\hspace{0.17em}}\text{Weight of the}\text{\hspace{0.17em}}\text{suspended particle}\\ =\text{\hspace{0.17em}}\text{mg}\text{\hspace{0.17em}}–\text{\hspace{0.17em}}\text{m}‘\text{g}\\ \text{=}\text{\hspace{0.17em}}\text{mg}\text{\hspace{0.17em}}–\text{\hspace{0.17em}}\text{V}\rho ‘\text{g}\\ =\text{mg}\text{\hspace{0.17em}}–\text{\hspace{0.17em}}\left(\frac{\text{m}}{\rho }\right)\rho ‘\text{g}\\ =\text{mg}\left(1-\frac{\rho ‘}{\rho }\right)\to \text{(ii)}\\ \text{Gas constant}\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\text{given}\text{\hspace{0.17em}}\text{as:}\\ \text{R}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}{\text{k}}_{\text{B}}\text{N}\\ {\text{k}}_{\text{B}}=\frac{\text{R}}{\text{N}}\text{}\to \left(\text{iii}\right)\\ \text{Putting equation}\text{\hspace{0.17em}}\left(\text{ii}\right)\text{in place of}\text{\hspace{0.17em}}\text{mg in equation}\left(\text{i}\right)\text{and}\text{\hspace{0.17em}}\text{then using equation}\left(\text{iii}\right)\text{,we obtain:}\\ {\text{n}}_{\text{2}}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}{\text{n}}_{\text{1}}\text{exp}\left[\frac{-\text{mg}\left({\text{h}}_{\text{2}}\text{\hspace{0.17em}}–\text{\hspace{0.17em}}{\text{h}}_{\text{1}}\right)}{{\text{k}}_{\text{B}}\text{T}}\right]\\ ={\text{n}}_{\text{1}}\text{exp}\left[-\text{mg}\left(1\text{\hspace{0.17em}}-\text{\hspace{0.17em}}\frac{\rho ‘}{\rho }\right)\left({\text{h}}_{\text{2}}\text{\hspace{0.17em}}–\text{\hspace{0.17em}}{\text{h}}_{\text{1}}\right)\text{\hspace{0.17em}}\frac{\text{N}}{\text{RT}}\text{}\right]\\ ={\text{n}}_{\text{1}}\text{exp}\left[-\text{mg}\left(\rho -\rho ‘\right)\left({\text{h}}_{\text{2}}\text{\hspace{0.17em}}–\text{\hspace{0.17em}}{\text{h}}_{\text{1}}\right)\frac{\text{N}}{\text{RT}\rho }\right]\end{array}$

Q.14 Given below are densities of some solids and liquids. Give rough estimates of the size of their atoms:

 Substance Atomic Mass(u) Density (103 Kg m–3) Carbon (diamond) 12.01 2.22 Gold 197.00 19.32 Nitrogen (liquid) 14.01 1.00 Lithium 6.94 0.53 Fluorine (liquid) 19.00 1.14

[Hint: Assume the atoms to be ‘tightly packed’ in a solid or liquid phase, and use the known value of Avogadro’s number. You should, however, not take the actual numbers you obtain for various atomic sizes too literally. Because of the crudeness of the tight packing approximation, the results only indicate that atomic sizes are in the range of a few A].

Ans.

$\begin{array}{l}\text{Let}\text{\hspace{0.17em}}\text{radius}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\text{an}\text{\hspace{0.17em}}\text{atom}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\text{r}\\ \therefore \text{Volume of an atom}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\frac{\text{4}}{\text{3}}\pi {\text{r}}^{\text{3}}\\ \text{Avogadro’s number,}\text{\hspace{0.17em}}\text{N}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\text{6}\text{.023}\text{\hspace{0.17em}}\text{×}\text{\hspace{0.17em}}{\text{10}}^{\text{23}}\\ \text{Volume of}\text{\hspace{0.17em}}\text{N}\text{\hspace{0.17em}}\text{number of molecules}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\frac{\text{4}}{\text{3}}\pi {\text{r}}^{\text{3}}\text{N}\to \text{(i)}\\ \text{Volume of one mole of the substance}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\frac{\text{M}}{\rho }\to \text{(ii)}\\ \frac{\text{4}}{\text{3}}\pi {\text{r}}^{\text{3}}\text{N}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\frac{\text{M}}{\rho }\\ \therefore \text{Radius}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\text{an}\text{\hspace{0.17em}}\text{atom,}\text{\hspace{0.17em}}\text{r}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\sqrt[3]{\frac{3M}{4\pi \rho N}}\\ \text{For carbon}\text{\hspace{0.17em}}\text{(diamond)}:\\ \text{Atomic}\text{\hspace{0.17em}}\text{mass,}\text{\hspace{0.17em}}\text{M}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\text{12}.0\text{1}\text{\hspace{0.17em}}×\text{\hspace{0.17em}}\text{1}{0}^{–\text{3}}\text{kg},\\ \text{Density,}\text{\hspace{0.17em}}\rho \text{\hspace{0.17em}}=\text{\hspace{0.17em}}\text{2}.\text{22}\text{\hspace{0.17em}}×\text{\hspace{0.17em}}\text{1}{0}^{\text{3}}{\text{kgm}}^{–\text{3}}\\ \therefore \text{Radius},\text{\hspace{0.17em}}r={\left(\frac{3\text{\hspace{0.17em}}×\text{\hspace{0.17em}}\text{12}.0\text{1}\text{\hspace{0.17em}}×\text{\hspace{0.17em}}\text{1}{0}^{–\text{3}}}{4\pi \text{\hspace{0.17em}}×\text{\hspace{0.17em}}\text{2}.\text{22}\text{\hspace{0.17em}}×\text{\hspace{0.17em}}\text{1}{0}^{\text{3}}\text{\hspace{0.17em}}×\text{\hspace{0.17em}}\text{6}.0\text{23}\text{\hspace{0.17em}}×\text{\hspace{0.17em}}\text{1}{0}^{\text{23}}}\right)}^{\frac{1}{3}}\\ =\text{1}\text{.77}\text{\hspace{0.17em}}\stackrel{\text{o}}{\text{A}}\\ =\text{1}.\text{29 A}\\ \therefore \text{The radius of a carbon atom is 1}\text{.29}\stackrel{\text{o}}{\text{A}}\text{.}\\ \text{For gold:}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\\ \text{M}=\text{197}.00\text{\hspace{0.17em}}×\text{\hspace{0.17em}}\text{1}{0}^{–\text{3}}\text{kg}\\ \text{Density},\text{\hspace{0.17em}}\rho \text{\hspace{0.17em}}=\text{\hspace{0.17em}}\text{19}.\text{32}×\text{1}{0}^{\text{3}}{\text{kg m}}^{\text{–3}}\\ =\text{1}\text{.59}\stackrel{\text{o}}{\text{A}}\\ \therefore \text{Radius,}\text{\hspace{0.17em}}\text{r}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}{\left(\frac{3\text{\hspace{0.17em}}×\text{\hspace{0.17em}}\text{197}\text{\hspace{0.17em}}×\text{\hspace{0.17em}}\text{1}{0}^{–\text{3}}}{4\pi \text{\hspace{0.17em}}×\text{\hspace{0.17em}}19.32\text{\hspace{0.17em}}×\text{\hspace{0.17em}}\text{1}{0}^{\text{3}}\text{\hspace{0.17em}}×\text{\hspace{0.17em}}\text{6}.0\text{23}\text{\hspace{0.17em}}×\text{\hspace{0.17em}}\text{1}{0}^{\text{23}}}\right)}^{\frac{1}{3}}\\ =\text{1}\text{.59}\text{\hspace{0.17em}}\stackrel{\text{o}}{\text{A}}\\ \therefore \text{The radius of a gold atom is 1}\text{.59 A}\text{.}\\ \text{For liquid nitrogen:}\\ \text{Atomic}\text{\hspace{0.17em}}\text{mass,}\text{\hspace{0.17em}}\text{M}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\text{14}.0\text{1}×\text{1}{0}^{–\text{3}}\text{kg}\\ \text{Density,}\text{\hspace{0.17em}}\rho =\text{1}.00×\text{1}{0}^{\text{3}}{\text{kg m}}^{–\text{3}}\\ \therefore \text{Radius,}\text{\hspace{0.17em}}\text{r}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}{\left(\frac{3\text{\hspace{0.17em}}×\text{\hspace{0.17em}}\text{14}\text{.01}\text{\hspace{0.17em}}×\text{\hspace{0.17em}}\text{1}{0}^{–\text{3}}}{4\pi \text{\hspace{0.17em}}×\text{\hspace{0.17em}}1.00\text{\hspace{0.17em}}×\text{\hspace{0.17em}}\text{1}{0}^{\text{3}}\text{\hspace{0.17em}}×\text{\hspace{0.17em}}\text{6}.0\text{23}\text{\hspace{0.17em}}×\text{\hspace{0.17em}}\text{1}{0}^{\text{23}}}\right)}^{\frac{1}{3}}\\ =1.77\text{\hspace{0.17em}}\stackrel{\text{o}}{\text{A}}\\ \therefore \text{The radius of a liquid nitrogen atom is 1}\text{.77}\stackrel{\text{o}}{\text{A}}\text{.}\\ \text{For lithium:}\\ \text{Atomic}\text{\hspace{0.17em}}\text{mass,}\text{\hspace{0.17em}}\text{M}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\text{6}.\text{94}\text{\hspace{0.17em}}×\text{\hspace{0.17em}}\text{1}{0}^{–\text{3}}\text{kg}\\ \text{Density},\text{\hspace{0.17em}}\rho =0.\text{53}×\text{1}{0}^{\text{3}}{\text{kgm}}^{–\text{3}}\text{}\\ \therefore \text{Radius,}\text{\hspace{0.17em}}\text{r}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}{\left(\frac{3\text{\hspace{0.17em}}×\text{\hspace{0.17em}}6.\text{94}\text{\hspace{0.17em}}×\text{\hspace{0.17em}}\text{1}{0}^{–\text{3}}}{4\pi \text{\hspace{0.17em}}×\text{\hspace{0.17em}}0.53\text{\hspace{0.17em}}×\text{\hspace{0.17em}}\text{1}{0}^{\text{3}}\text{\hspace{0.17em}}×\text{\hspace{0.17em}}\text{6}.0\text{23}\text{\hspace{0.17em}}×\text{\hspace{0.17em}}\text{1}{0}^{\text{23}}}\right)}^{\frac{1}{3}}\\ =1.73\text{\hspace{0.17em}}\stackrel{\text{o}}{\text{A}}\\ \therefore \text{The radius of a lithium atom is 1}\text{.73}\stackrel{\text{o}}{\text{A}}\text{.}\\ \text{For}\text{\hspace{0.17em}}\text{fluorine}\text{\hspace{0.17em}}\text{(liquid):}\\ \text{Atomic}\text{\hspace{0.17em}}\text{mass,}\text{\hspace{0.17em}}\text{M}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\text{19}\text{.00}\text{\hspace{0.17em}}×\text{\hspace{0.17em}}\text{1}{0}^{–\text{3}}\text{kg}\end{array}$ $\begin{array}{l}\text{Density,}\text{\hspace{0.17em}}\rho =\text{1}.\text{14}×\text{1}{0}^{\text{3}}{\text{kg m}}^{\text{–3}}\\ \therefore \text{Radius,}\text{\hspace{0.17em}}\text{r}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}{\left(\frac{3\text{\hspace{0.17em}}×\text{\hspace{0.17em}}19\text{\hspace{0.17em}}×\text{\hspace{0.17em}}\text{1}{0}^{–\text{3}}}{4\pi \text{\hspace{0.17em}}×\text{\hspace{0.17em}}1.14\text{\hspace{0.17em}}×\text{\hspace{0.17em}}\text{1}{0}^{\text{3}}\text{\hspace{0.17em}}×\text{\hspace{0.17em}}\text{6}.0\text{23}\text{\hspace{0.17em}}×\text{\hspace{0.17em}}\text{1}{0}^{\text{23}}}\right)}^{\frac{1}{3}}\\ =\text{1}\text{.88}\text{\hspace{0.17em}}\stackrel{\text{o}}{\text{A}}\\ \therefore \text{Radius of a liquid fluorine atom}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\text{1}\text{.88}\text{\hspace{0.17em}}\stackrel{\text{o}}{\text{A}}\end{array}$

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### 1. What are the most significant topics to cover in Class 11 Physics Chapter 13?

Students can refer to the website of Extramarks or its mobile application to get accurate answers to questions asked at the end of Chapter 13 Physics. Additional practise questions are also added in the solutions for students to assist them to understand the concepts. . The extra questions are derived from past years’ papers from various schools. All of the answers are provided in accordance with the CBSE board’s most recent guidelines and are available on the Extramarks website.

### 2. What is explained by the kinetic theory of gases?

The basic macroscopic properties of gases, such as volume, pressure, and temperature, as well as some transport properties, such as viscosity and thermal conductivity, are explained by the kinetic theory of gases. Because gases are made up of fast-moving atoms and molecules, it gives a basic notion of how they behave. This is conceivable because, in gases, the short-range interatomic forces that are so important in solids and liquids may be ignored.

### 3. What are the five assumptions of the kinetic theory of gases?

The following are the five assumptions:

• Gases are made up of an endless number of particles known as molecules that travel in different directions at random.
• The distance between the two molecules is larger than the molecules’ individual sizes.
• The collision between the two molecules and their environment has a very short time interval.
• The laws of motion apply to all the molecules in the gas.
• Between the molecules, only elastic collisions exist.

### 4. What is the Kinetic Theory of Gases? Class 11 Physics KInetic theory

Kinetic Theory provides a basic understanding of the behaviour of gases because they are made up of fast-moving atoms or molecules. In the nineteenth century, Maxwell, Boltzmann, and others proposed the kinetic theory of gases. An ideal gas is one that strictly follows gas laws such as Charles’, Boyle’s, and Gay Lussac’s law. There is no force of attraction or repulsion in the ideal gas molecule since it is a point mass.

### 5. How can I improve my score in Physics Class 11 chapter Kinetic Theory?

Our highly experienced faculty has curated NCERT Solutions for Class 11 Physics Chapter 13 Kinetic Theory after completing an extensive study on each concept. Students can use these solutions to acquire a general sense of the concepts covered in this chapter while working on the textbook questions. Students can access the solutions both online and offline, depending on their needs. At Extramarks, solutions are provided in an easy-to-understand and simple format, which students can easily access to clarify their doubts and understand how to write accurate, precise answers to the questions. Since Extramarks follows the NCERT syllabus, it also prepares the students for competitive exams.