# NCERT Solutions for Class 11 Physics Chapter 13 Kinetic Theory

NCERT Solutions for Class 11 Physics Chapter 13 Kinetic Theory: Maxwell, Boltzmann, and others developed the kinetic theory of gas in the 19th century. Before going to solutions of NCERT for class 11 physics chapter 13 kinetic theory one must be familiar with ideal gas equations and gas laws. The kinetic theory interprets the pressure and temperature at the molecular level. Kinetic theory is consistent with gas laws and it correctly explains specific heat and relates measurable properties of gases such as viscosity, diffusion, etc. The CBSE NCERT solutions for class 11 physics chapter 13 kinetic theory gives an idea about how to apply the formulas studied in the chapter in numerical problems. NCERT solutions explain all the exercise and additional exercise questions.

Exercise

A brief summary of formulas required for NCERT solutions for class 11 physics chapter 13 kinetic theory are given below.

• The ideal gas equation is given by

$PV=\mu RT=K_BNT$

where $\mu$ is the number of moles

N is the number of molecules

R and kB are universal constants.

• The pressure of an ideal gas

$P=\frac{1}{3}nm\bar v^2$

where n is the number density, m is the mass of the molecule and $\bar v^2$ is the mean of the squared speed.

• The RMS velocity of the gas molecule

$v_{rms}=\sqrt{\bar v^2}=\sqrt{\frac{3K_BT}{m}}$ Where T is the temperature.

• The translational kinetic energy is

$\frac{3K_BNT}{2}$

• The mean free path

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

where n is the number density and d is the diameter of the molecule

## The main topics of  NCERT Class 11 Physics Chapter 13 are listed below.

13.1 Introduction

13.2 Molecular nature of matter

13.3 Behaviour of gases

13.4 Kinetic theory of an ideal gas

13.5 Law of equipartition of energy

13.6 Specific heat capacity

13.7 Mean free path

## Q13.1 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.

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

The actual volume of a mole of oxygen molecules Vactual 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 Vmolar = 22.4 litres

$\\\frac{V_{actual}}{V_{molar}}=\frac{8.51\times 10^{-3}}{22.4}\\ \frac{V_{actual}}{V_{molar}}=3.8\times 10^{-4}$

As per the ideal gas equation

$\\PV=nRT\\ V=\frac{nRT}{P}$

For one mole of a gas at STP we have

$\\V=\frac{1\times 8.314\times 273}{1.013\times 10^{5}}\\ V=0.0224 m^{3}\\ V=22.4\ litres$

## Q13.3 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: $T_{1}> T_{2}\: \: orT_{1}< T_{2}$ ?
(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 $O_{2}=32.0\mu$,
$R=8.31J mol^{-1}K^{-1}$.)

(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 T1 is closer to the horizontal line that the one corresponding to T2 we conclude T1 is greater than T2.

(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

$\\nR=\frac{1}{32}\times 8.314\\ nR=0.256JK^{-1}$

(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

$m=\frac{PV}{T} \frac{M}{R}={0.256}\times \frac{2}{8.314}=5.48\times10^{-5}Kg$

Initial volume, V1 = Volume of Cylinder = 30 l

Initial Pressure P1 = 15 atm

Initial Temperature T1 = 27 oC = 300 K

The initial number of moles n1 inside the cylinder is

$\\n_{1}=\frac{P_{1}V_{1}}{RT_{1}}\\ n_{1}=\frac{15\times 1.013\times 10^{5}\times 30\times 10^{-3}}{8.314\times 300}\\ n_{1}=18.28$

Final volume, V2 = Volume of Cylinder = 30 l

Final Pressure P2 = 11 atm

Final Temperature T2 = 17 oC = 290 K

Final number of moles n2 inside the cylinder is

$\\n_{2}=\frac{P_{2}V_{2}}{RT_{2}}\\ n_{2}=\frac{11\times 1.013\times 10^{5}\times 30\times 10^{-3}}{8.314\times 290}\\ n_{2}=13.86$

Moles of oxygen taken out of the cylinder = n2 -n1 = 18.28 - 13.86 = 4.42

Mass of oxygen taken out of the cylinder m is

$\\m=4.42\times 32\\ m=141.44g$

Initial Volume of the bubble, V1 = 1.0 cm3

Initial temperature, T1 = 12 oC = 273 + 12 = 285 K

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

Initial Pressure is P1

Depth of the bottom of the lake = 40 m

$\\P_{1}=Atmospheric\ Pressure+Pressure\ due\ to\ water\\ P_{1}=P_{atm}+\rho _{w}gh\\ P_{1}=1.013\times 10^{5}+10^{3}\times 9.8\times 40\\ P_{1}=4.93\times 10^{5}Pa$

Final Temperature, T2 = 35 oC = 35 + 273 = 308 K

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

Let the final volume be V2

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

$\\\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}\\ V_{2}=\frac{P_{1}T_{2}V_{1}}{P_{2}T_{1}}\\ V_{2}=\frac{4.93\times 10^{5}\times 308\times 1}{1.013\times 10^{5}\times 285}\\ V_{2}=5.26\ cm^{3}$

The volume of the room, V = 25.0 m3

Temperature of the room, T = 27oC = 300 K

The pressure inside the room, P  = 1 atm

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

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

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

Number of molecules inside the room is N

$\\N=nN_{A}\\ N=1015.35\times 6.022\times 10^{23}\\ N=6.114\times 10^{26}$

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)

$\\E=\frac{3kT}{2}\\ E=\frac{3\times 1.38\times 10^{-23}\times10^{7}}{2}\\ E=2.07\times 10^{-16}\ J$

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

Root mean square velocity is given as

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

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

### 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 $-20^{0}C$

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

Pressure, P = 2atm

Temperature, T = 17 oC

The radius of the Nitrogen molecule , $r=1\ \AA$

The molecular mass of  N= 28 u

The molar mass of N= 28 g

From 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

$\\n'=\frac{N_{A}n}{V}=\frac{6.022\times 10^{23}\times 2\times 1.013\times 10^{5}}{8.314\times (17+273)}\\ n'=5.06\times 10^{25}$

The mean free path $\lambda$ is given as

$\\\lambda =\frac{1}{\sqrt{2}\pi n'd^{2}}\\ \lambda =\frac{1}{\sqrt{2}\times \pi \times 5.06\times 10^{25}\times (2\times 1\times 10^{-10})^{2}}\\ \lambda =1.11\times 10^{-7}\ m$

The root mean square velocity vrms is given as

$\\v_{rms}=\sqrt{\frac{3RT}{M}}\\ v_{rms}=\sqrt{\frac{3\times 8.314\times 290}{28\times 10^{-3}}}\\ v_{rms}=508.26\ m\ s^{-1}$

The time between collisions T is given as

$\\T=\frac{1}{Collision\ Frequency}\\ T=\frac{1}{\nu }\\ T=\frac{\lambda }{v_{rms}}\\ T=\frac{1.11\times 10^{-7}}{508.26}\\ T=2.18\times 10^{-10}s$

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

$\\T'=\frac{d}{v_{rms}}\\ T'=\frac{2\times 1\times 10^{-10}}{508.26}\\ T'=3.935\times 10^{-13}s$

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

$\\\frac{T}{T'}=\frac{2.18\times 10^{-10}}{3.935\times 10^{-13}}\\ \frac{T}{T'}=554$

Thus we can see time between collisions is much larger than the collision time.

## NCERT solutions for class 11 physics chapter 13 kinetic theory additional exercise

Initially, the pressure of the 15 cm long air column is equal to the atmospheric pressure, P1 = 1 atm = 76 cm of Mercury

Let the crossectional area of the tube be x cm2

The initial volume of the air column, V1 = 15x cm3

Let's assume 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= 76 - (76 - y) cm of Mercury = y cm of mercury

Final volume of air column V2 = (24 + y)x cm3

Since the temperature of the air column does not change

$\\P_{1}V_{1}=P_{2}V_{2}\\ 76\times 15x=y\times (24+y)x\\ 1140=y^{2}+24y\\ y^{2}+24y-1140=0$

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

As per Graham's Law of diffusion if two gases of Molar Mass M1 and M2 diffuse with rates R1 and R2 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

R1 = 28.7 cm3 s-1

R2 = 7.2 cm3 s-1

M1 = 2 g

$\\M_{2}=M_{1}\left ( \frac{R_{1}}{R_{2}} \right )^{2}\\ M_{2}=2\times \left ( \frac{28.7}{7.2} \right )^{2}\\ M_{2}=31.78g$

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

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:

$n_{2}=n_{1}exp\left [ -mg N_{A}(\rho -\rho ^{'})(h_{2}-h_{1})/ (\rho RT)\right ]$
where   $\rho$ is the density of the suspended particle, and $\rho ^{'}$ , 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.]

$n_{2}=n_{1}exp\left [ -mg(h_{2}-h_{1}) /k_{b}T\right ]$                                   $(i)$

Let the suspended particles be spherical and have 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

$F_{B}=\frac{4}{3}\pi r^{3}\rho' g$

The net force acting on the particles become

$\\F_{net}=F_{G}-F_{B}\\ F_{net}=\frac{4}{3}\pi r^{3}\rho g-\frac{4}{3}\pi r^{3}\rho' g\\F_{net}=\frac{4}{3}\pi r^{3}g(\rho -\rho ')$

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

$\\n_{2}=n_{1}exp\left [ -\frac{4}{3}\pi r^{3}g(\rho -\rho ')(h_{2}-h_{1}) /k_{b}T\right ]\\ n_{2}=n_{1}exp\left [ \frac{-\frac{4}{3}\pi r^{3}g(\rho -\rho ')(h_{2}-h_{1})}{\frac{RT}{N_{A}}} \right ]\\ n_{2}=n_{1}exp\left [ \frac{-mgN_{A}(\rho -\rho ')(h_{2}-h_{1})}{RT\rho '} \right ]$

The above is the equation to be derived

 Substance Atomic Mass (u) Density (103 Kg m3) Carbon (diamond) 12.01 2.22 Gold 197 19.32 Nitrogen (liquid) 14.01 1 Lithium 6.94 0.53 Fluorine 19 1.14

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 $N_{A}=6.022\times 10^{23}$

$\\N_{A}\frac{4}{3}\pi r^{3}\rho =M\\ r=\left ( \frac{3M}{4N_{A}\pi \rho } \right )^{\frac{1}{3}}$

For Carbon

$\\N_{A}\frac{4}{3}\pi r^{3}\rho =M\\ r=\left ( \frac{3\times 12.01}{4\times 6.022\times 10^{23}\times \pi\times 2.22\times 10^{3} } \right )^{\frac{1}{3}}\\ r=1.29\AA$

For gold

$\\N_{A}\frac{4}{3}\pi r^{3}\rho =M\\ r=\left ( \frac{3\times 197.00}{4\times 6.022\times 10^{23}\times \pi\times 19.32\times 10^{3} } \right )^{\frac{1}{3}}\\ r=1.59\AA$

For Nitrogen

$\\N_{A}\frac{4}{3}\pi r^{3}\rho =M\\ r=\left ( \frac{3\times 14.01}{4\times 6.022\times 10^{23}\times \pi\times 1.00\times 10^{3} } \right )^{\frac{1}{3}}\\ r=1.77\AA$

For Lithium

$\\N_{A}\frac{4}{3}\pi r^{3}\rho =M\\ r=\left ( \frac{3\times 6.94}{4\times 6.022\times 10^{23}\times \pi\times 0.53\times 10^{3} } \right )^{\frac{1}{3}}\\ r=1.73\AA$

For Fluorine

$\\N_{A}\frac{4}{3}\pi r^{3}\rho =M\\ r=\left ( \frac{3\times19.00}{4\times 6.022\times 10^{23}\times \pi\times 1.14\times 10^{3} } \right )^{\frac{1}{3}}\\ r=1.88\AA$

## NCERT solutions for class 11 physics chapter wise

 Chapter 1 NCERT solutions for class 11 physics chapter 1 Physical world Chapter 2 Solutions of NCERT for class 11 physics chapter 2 Units and Measurement Chapter 3 CBSE NCERT solutions for class 11 physics chapter 3 physics Motion in a straight line Chapter 4 NCERT solutions for class 11 physics chapter 4 Motion in a Plane Chapter 5 Solutions of NCERT for class 11 physics chapter 5 Laws of Motion Chapter 6 CBSE NCERT solutions for class 11 physics chapter 6 Work, Energy and Power Chapter 7 NCERT solutions for class 11 physics chapter 7 System of Particles and Rotational motion Chapter 8 Solutions of NCERT for class 11 physics chapter 8 Gravitation Chapter 9 CBSE NCERT solutions for class 11 physics chapter 9 Mechanical Properties of Solids Chapter 10 NCERT solutions for class 11 physics chapter 10 Mechanical Properties of Fluids Chapter 11 Solutions of NCERT for class 11 physics chapter 11 Thermal Properties of Matter Chapter 12 CBSE NCERT solutions for class 11 physics chapter 12 Thermodynamics Chapter 13 NCERT solutions for class 11 physics chapter 13 Kinetic Theory Chapter 14 Solutions of NCERT for class 11 physics chapter 14 Oscillations Chapter 15 CBSE NCERT solutions for class 11 physics chapter 15 Waves

## NCERT Solutions for Class 11 Subject wise

 CBSE NCERT solutions for class 11 biology NCERT solutions for class 11 maths CBSE NCERT solutions for class 11 chemistry NCERT solutions for class 11 physics

## Use of NCERT solutions for class 11 physics chapter 13 kinetic theory:

• To familiarise with the concepts studied in the chapter, it is important to study the CBSE NCERT solutions for class 11 physics chapter 13 kinetic theory.
• All the equations studied in the chapter are utilized in the solutions of NCERT class 11 physics chapter 13 kinetic theory.
• The chapter is important for class exams and competitive exams. For exams like NEET and JEE Mains one question can be expected from the chapter and the questions can be solved easily if you are familiar with the concepts and formulas of the chapter.