 # NCERT Solutions for Class 11 Physics Chapter 13 Kinetic Theory

## NCERT Solutions Class 11 Physics Chapter 13 – Free PDF Download

NCERT Solutions for Class 11 Physics Chapter 13 Kinetic Theory is a crucial resource that helps you score good marks in the second term exams and the entrance examination for graduate courses. Chapter 13 Kinetic Theory is an important Chapter as it introduces you to the basic concepts which will be often asked in the entrance examination. NCERT Solutions contain elaborate explanations and accurate answers to provide students with a solid understanding of concepts to clear their term-wise exams.

Chapter 13 constitutes a big portion of the term – II CBSE Syllabus 2021-22. It contains crucial concepts like atomic theory, gas laws, Boltzmann constant, Avogadro’s number,  Kinetic theory postulates and specific heat capacities. Learning these topics is very important as numerous problems might appear in the second term exam based on this. Students will get a strong base of fundamental concepts which would also help them in their higher levels of education. NCERT Solutions for Class 11 Physics can be accessed by the link given below.             ### Access answers of NCERT Solutions for Class 11 Physics Chapter 13 Kinetic Theory

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.

Solution:

Diameter of an oxygen molecule, d = 3 Å

Radius, r = d / 2

r = 3 / 2 = 1.5 Å = 1.5 x 10 -8 cm

Actual volume occupied by 1 mole of oxygen gas at STP = 22400 cm3

Molecular volume of oxygen gas, V = 4 / 3 πr3. N

Where, N is Avogadro’s number = 6.023 x 1023 molecules/ mole

Hence,

V = 4 / 3 x 3.14 x (1.5 x 10-8)3 x 6.023 x 1023

We get,

V = 8.51 cm3

Therefore, ratio of the molecular volume to the actual volume of oxygen = 8.51/ 22400 = 3. 8 x 10-4

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

Solution:

The ideal gas equation relating pressure (P), volume (V), and absolute temperature (T) is given as:

PV = nRT

Where, R is the universal gas constant = 8.314 J mol-1K-1

n = Number of moles = 1

T = Standard temperature = 273 K

P = Standard pressure = 1 atm = 1.013 x 105Nm-2

Hence,

V = nRT / P

= 1 x 8.314 x 273 / 1.013 x 105

= 0.0224 m3

= 22.4 litres

Therefore, the molar volume of a gas at STP is 22.4 litres

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 mo1–1 K–1.)

Solution:

(a) dotted plot is parallel to X-axis, signifying that nR [PV/T = nR] is independent of P. Thus it is representing ideal gas behaviour

(b) the graph at temperature T1 is closer to ideal behaviour (because closer to dotted line) hence, T1 > T2 (higher the temperature, ideal behaviour is the higher)

(c) use PV = nRT

PV/ T = nR

Mass of the gas = 1 x 10-3 kg = 1 g

Molecular mass of O2 = 32g/ mol

Hence,

Number of mole = given weight / molecular weight

= 1/ 32

So, nR = 1/ 32 x 8.314 = 0.26 J/ K

Hence,

Value of PV / T = 0.26 J/ K

(d) 1 g of H2 doesn’t represent the same number of mole

Eg. molecular mass of H2 = 2 g/mol

Hence, number of moles of H2 require is 1/32 (as per the question)

Therefore,

Mass of H2 required = no. of mole of H2 x molecular mass of H2

= 1/ 32 x 2

= 1 / 16 g

= 0.0625 g

= 6.3 x 10-5 kg

Hence, 6.3 x 10-5 kg of H2 would yield the same value

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 = 8.31 J mol-1 K-1, molecular mass of O2 = 32 u).

Solution:

Volume of gas, V1 = 30 litres = 30 x 10-3 m3

Gauge pressure, P1 = 15 atm = 15 x 1.013 x 105 P a

Temperature, T1 = 270 C = 300 K

Universal gas constant, R = 8.314 J mol-1 K-1

Let the initial number of moles of oxygen gas in the cylinder be n1

The gas equation is given as follows:

P1V1 = n1RT1

Hence,

n1 = P1V1 / RT1

= (15.195 x 105 x 30 x 10-3) / (8.314 x 300)

= 18.276

But n1 = m1 / M

Where,

m1 = Initial mass of oxygen

M = Molecular mass of oxygen = 32 g

Thus,

m1 = N1M = 18.276 x 32 = 584.84 g

After some oxygen is withdrawn from the cylinder, the pressure and temperature reduce.

Volume, V2 = 30 litres = 30 x 10-3 m3

Gauge pressure, P2 = 11 atm

= 11 x 1.013 x 105 P a

Temperature, T2 = 170 C = 290 K

Let n2 be the number of moles of oxygen left in the cylinder

The gas equation is given as:

P2V2 = n2RT2

Hence,

n2 = P2V2 / RT2

= (11.143 x 105 x 30 x 10-30) / (8.314 x 290)

= 13.86

But

n2 = m2 / M

Where,

m2 is the mass of oxygen remaining in the cylinder

Therefore,

m2 = n2 x M = 13.86 x 32 = 453.1 g

The mass of oxygen taken out of the cylinder is given by the relation:

Initial mass of oxygen in the cylinder – Final mass of oxygen in the cylinder

= m1 – m2

= 584.84 g – 453.1 g

We get,

= 131.74 g

= 0.131 kg

Hence, 0.131 kg of oxygen is taken out of the cylinder

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?

Solution:

Volume of the air bubble, V1 = 1.0 cm3

= 1.0 x 10-6 m3

Air bubble rises to height, d = 40 m

Temperature at a depth of 40 m, T1 = 120 C = 285 K

Temperature at the surface of the lake, T2 = 350 C = 308 K

The pressure on the surface of the lake:

P2 = 1 atm = 1 x 1.013 x 105 Pa

The pressure at the depth of 40 m:

P1= 1 atm + dρg

Where,

ρ is the density of water = 103 kg / m3

g is the acceleration due to gravity = 9.8 m/s2

Hence,

P1 = 1.013 x 105 + 40 x 103 x 9.8

We get,

= 493300 Pa

We have

P1V1 / T1 = P2V2 / T2

Where, V2 is the volume of the air bubble when it reaches the surface

V2 = P1V1T2 / T1P2

= 493300 x 1 x 10-6 x 308 / (285 x 1.013 x 105)

We get,

= 5.263 x 10-6 m3 or 5.263 cm3

Hence, when the air bubble reaches the surface, its volume becomes 5.263 cm3

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.

Solution:

Volume of the room, V = 25.0 m3

Temperature of the room, T = 270 C = 300 K

Pressure in the room, P = 1 atm = 1 x 1.013 x 105 Pa

The ideal gas equation relating pressure (P), Volume (V), and absolute temperature (T) can be written as:

PV = (kBNT)

Where,

KB is Boltzmann constant = (1.38 x 10-23) m2 kg s-2 K-1

N is the number of air molecules in the room

Therefore,

N = (PV / kBT)

= (1.013 x 105 x 25) / (1.38 x 10-23 x 300)

We get,

= 6.11 x 1026 molecules

Hence, the total number of air molecules in the given room is 6.11 x 1026

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).

Solution:

(i) At room temperature, T = 270 C = 300 K

Average thermal energy = (3 / 2) kT

Where,

k is the Boltzmann constant = 1.38 x 10-23 m2 kg s-2 K-1

Hence,

(3 / 2) kT = (3 / 2) x 1.38 x 10-23 x 300

On calculation, we get,

= 6.21 x 10-21 J

Therefore, the average thermal energy of a helium atom at room temperature of 270 C is 6.21 x 10-21 J

(ii) On the surface of the sun, T = 6000 K

Average thermal energy = (3 / 2) kT

= (3 / 2) x 1.38 x 10-23 x 6000

We get,

= 1.241 x 10-19 J

Therefore, the average thermal energy of a helium atom on the surface of the sun is 1.241 x 10-19 J

(iii) At temperature, T = 107 K

Average thermal energy = (3 / 2) kT

= (3 / 2) x 1.38 x 10-23 x 107

We get,

= 2.07 x 10-16 J

Therefore, the average thermal energy of a helium atom at the core of a star is 2.07 x 10-16 J

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?

Solution:

All the three vessels have the same capacity, they have the same volume.

So, each gas has the same pressure, volume and temperature

According to Avogadro’s law, the three vessels will contain an equal number of the respective molecules.

This number is equal to Avogadro’s number, N = 6.023 x 1023.

The root mean square speed (Vrms) of a gas of mass m and temperature T is given by the relation:

Vrms = 3kT / m

Where,

k is Boltzmann constant

For the given gases, k and T are constants

Therefore, Vrms depends only on the mass of the atoms, i.e., Vrms ∝ (1/m)1/2

Hence, the root mean square speed of the molecules in the three cases is not the same.

Among neon, chlorine and uranium hexafluoride, the mass of neon is the smallest.

Therefore, neon has the largest root mean square speed among the given gases.

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).

Solution:

Given

Temperature of the helium atom, THe = -200 C = 253 K

Atomic mass of argon, MAr = 39.9 u

Atomic mass of helium, MHe = 4.0 u

Let (Vrms)Ar be the rms speed of argon and

Let (Vrms)He be the rms speed of helium

The rms speed of argon is given by:

(Vrms)Ar = √3RTAr / MAr ………… (i)

Where,

R is the universal gas constant

TAr is temperature of argon gas

The rms speed of helium is given by:

(Vrms)He = √3RTHe / MHe ………… (ii)

Given that,

(Vrms)Ar = (Vrms)He

√3RTAr / MAr = √3RTHe / MHe

TAr / MAr = THe / MHe

TAr = THe / MHe x MAr

= (253 / 4) x 39.9

We get,

= 2523.675

= 2.52 x 103 K

Hence, the temperature of the argon atom is 2.52 x 103 K

10. Estimate the mean free path and collision frequency of a nitrogen molecule in a cylinder containing nitrogen at 2.0 atm and temperature 170 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).

Solution:

Mean free path = 1.11 x 10-7 m

Collision frequency = 4.58 x 109 s-1

Successive collision time 500 x (Collision time)

Pressure inside the cylinder containing nitrogen, P = 2.0 atm = 2.026 x 105 Pa

Temperature inside the cylinder, T = 170 C = 290 K

Radius of a nitrogen molecule, r = 1.0 Å = 1 x 1010 m

Diameter, d = 2 x 1 x 1010 = 2 x 1010 m

Molecular mass of nitrogen, M = 28.0 g = 28 x 10-3 kg

The root mean square speed of nitrogen is given by the relation:

Vrms= √3RT / M

Where,

R is the universal gas constant = 8.314 J mol-1 K-1

Hence,

Vrms= 3 x 8.314 x 290 / 28 x 10-3

On calculation, we get,

= 508.26 m/s

The mean free path (l) is given by relation:

l = KT / √2 x π x d2 x P

Where,

k is the Boltzmann constant = 1.38 x 10-23 kg m2 s-2 K-1

Hence,

l = (1.38 x 10-23 x 290) / (√2 x 3.14 x (2 x 10-10)2 x 2.026 x 105

We get,

= 1.11 x 10-7 m

Collision frequency = Vrms / l

= 508.26 / 1.11 x 10-7

On calculation, we get,

= 4.58 x 109 s-1

Collision time is given as:

T = d / Vrms

= 2 x 10-10 / 508.26

On further calculation, we get

= 3.93 x 10-13 s

Time taken between successive collisions:

T’ = l / Vrms = 1.11 x 10-7 / 508.26

We get,

= 2.18 x 10-10

Hence,

T’ / T = 2.18 x 10-10 / 3.93 x 10-13

On calculation, we get,

= 500

Therefore, the time taken between successive collisions is 500 times the time taken for a collision

11. A metre 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?

Solution:

Length of the narrow bore, L = 1 m = 100 cm

Length of the mercury thread, l = 76 cm

Length of the air column between mercury and the closed end, la = 15 cm

Since the bore is held vertically in air with the open end at the bottom, the mercury length that occupies the air space is:

= 100 – (76 + 15)

= 9 cm

Therefore,

The total length of the air column = 15 + 9 = 24 cm

Let h cm of mercury flow out as a result of atmospheric pressure

So,

Length of the air column in the bore = 24 + h cm

And,

Length of the mercury column = 76 – h cm

Initial pressure, V1 = 15 cm3

Final pressure, P2 = 76 – (76 – h)

= h cm of mercury

Final volume, V2 = (24 + h) cm3

Temperature remains constant throughout the process

Therefore,

P1V1 = P2V2

On substituting, we get,

76 x 15 = h (24 + h)

h2 + 24h – 11410 = 0

On solving further, we get,

= 23.8 cm or -47.8 cm

Since height cannot be negative. Hence, 23.8 cm of mercury will flow out from the bore

Length of the air column = 24 + 23.8 = 47.8 cm

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.]

Solution:

Given

Rate of diffusion of hydrogen, R1 = 28.7 cm3s-1

Rate of diffusion of another gas, R2 = 7.2 cm3s-1

According to Graham’s Law of diffusion,

We have,

R1 / R2 = √M2 / M1

Where,

M1 is the molecular mass of hydrogen = 2.020g

M2 is the molecular mass of the unknown gas

Hence,

M2 = M1 (R1 / R2)2

= 2.02 (28.7 / 7.2)2

We get,

= 32.09 g

32 g is the molecular mass of oxygen.

Therefore, the unknown gas is oxygen.

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 a 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, nrefer to number density at heights hand 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 – ρ′ ) (h2–h1)/ (ρ RT)]
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:

Law of atmosphere

n2= n1 exp [ -mg (h2 – h1)/ kBT] ———(1)

The suspended particle experiences an apparent weight because of the liquid displaced

According to Archimedes principle

Apparent weight = Weight of the water displaced – weight of the suspended particle

= mg – m’g

= mg – Vρ’g  = mg – (m/ρ) ρ’g

= mg (1 – (ρ’/ρ)) ——-(2)

ρ′= Density of the water
ρ = Density of the suspended particle
m′ = Mass of the suspended particle
m = Mass of the water displaced
V =  Volume of a suspended particle

Boltzmann’s constant (K) = R/NA ——(3)

Substituting equation (2) and equation (3) in equation (1)

n2= n1 exp [ -mg (h2 – h1)/ kBT]

n2 = n1 exp [ -mg (1 – ρ′/ ρ ) (h2–h1)NA / (RT)]

n2 = n1 exp [ -mg NA (ρ – ρ′ ) (h2–h1)/ (ρ RT)]

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 Å].

Ans:

If r is the radius of the atom then the volume of each atom = (4/3)πr3

Volume of all the substance = (4/3)πrx N = M/ρ

M is the atomic mass of the substance

ρ is the density of the substance

One mole of the substance has 6.023 x 1023 atoms

r = (3M/4πρ x 6.023 x 1023)1/3

For carbon, M = 12. 01 x 10-3 kg and ρ = 2.22 x 103 kg m-3

R = (3 x 12. 01 x 10-3/4 x 3.14 x 2.22 x 103  x 6.023 x 1023)1/3

= (36.03 x  10-3  /167.94 x 1026)1/3

1.29 x 10 -10 m = 1.29 Å

For gold, M = 197 x 10-3 kg and ρ = 19. 32 x 103 kg m-3

R = (3 x 197 x 10-3/4 x 3.14 x 19.32 x 103  x 6.023 x 1023)1/3

= 1.59 x 10 -10 m = 1.59 Å

For lithium, M = 6.94 x 10-3 kg  and ρ = 0.53 x 103 kg/m3

R = (3 x 6.94 x 10-3/4 x 3.14 x 0.53 x 103  x 6.023 x 1023)1/3

= 1.73 x 10 -10 m = 1.73 Å

For nitrogen (liquid), M = 14.01 x 10-3 kg  and ρ = 1.00 x 103 kg/m3

R = (3 x 14.01 x 10-3/4 x 3.14 x 1.00 x 103  x 6.023 x 1023)1/3

= 1.77 x 10 -10 m = 1.77 Å

For fluorine (liquid), M = 19.00 x 10-3 kg  and ρ = 1.14 x 103 kg/m3

R = (3 x 19 x 10-3/4 x 3.14 x 1.14 x 103  x 6.023 x 1023)1/3

= 1.88 x 10 -10 m = 1.88 Å

Also Access

NCERT Exemplar for Class 11 Physics Chapter 13

CBSE Notes for Class 11 Physics Chapter 13

It is very important for you to get well versed with the concepts given in Chapter 13 Kinetic Theory in order to avoid any difficulty in understanding the advanced topics you face in the future. These NCERT Solutions guides you in getting deep insights through its answers to the textbook questions, also questions from previous papers and sample papers.

NCERT Solutions also consists of previous year question papers that are very useful in preparing for the term-wise CBSE examinations for 2021-22.

## Subtopics of Chapter 13 Kinetic Theory

1. Introduction
2. Molecular nature of matter
3. Behaviour of gases
4. Kinetic theory of an ideal gas
5. Law of equipartition of energy
6. Specific heat capacity
7. Mean free path

The NCERT Solutions are prepared by subject experts according to the latest term – II CBSE Syllabus (2021-22) of Class 11 Physics so that students can learn the concepts more effectively. The NCERT Solutions Class 11 Physics Kinetic Theory is given to make students understand the concepts of this chapter in-depth. The exams conducted by the CBSE are based on the NCERT Syllabus for both 10th as well as 12th Classes.

Kinetic Theory is one of the most scoring sections in the CBSE Class 11 term – II examination. Properties of gases are easier to understand than those of solids and liquids. All things are made of atoms – little particles that move around in perpetual motion, attracting each other when they are a little distance apart but repelling upon being squeezed into one another. Some key points of the kinetic theory of gases are given below.

• We should not have an exaggerated idea of the intermolecular distance in a gas. At ordinary pressures and temperatures, this is only 10 times or so the interatomic distance in solids and liquids. What is different is the mean free path, which in gas is 100 times the interatomic distance and 1000 times the size of the molecule.
• The pressure of a fluid is not only exerted on the wall. Pressure exists everywhere in a fluid. Any layer of gas inside the volume of a container is in equilibrium because the pressure is the same on both sides of the layer.
• Molecules of air in a room do not fall and settle on the ground (due to gravity) because of their high speeds and incessant collisions. In equilibrium, there is a very slight increase in density at lower heights (like in the atmosphere).

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Kinetic theory of gases is a theoretical model that describes the molecular composition of the gas in terms of a large number of submicroscopic particles which include atoms and molecules. Students of Class 11 are advised to refer to the NCERT Solutions from BYJU’S to gain a grip on the important concepts. The solutions are prepared with the main aim of helping students with their second term exam preparation. The stepwise explanations in simple language boost the confidence among students to appear for the term – II exam without fear.

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