# NCERT Solutions for Class 11 Physics Chapter 13 Kinetic Theory of Gases

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

NCERT solutions for class 11 physics chapter 13 Kinetic Theory is a crucial resource that helps you score good marks in 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.

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. This NCERT solution guides in getting deep insights through its answers to the textbook question, questions from previous papers and sample papers.

This solution also comprises of exemplary problems, worksheets, MCQ’s, HOTS, tips and tricks which will provide you with necessary details to help you in preparing notes and to prepare for the examination.

## 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.

### Important questions of Class 11 Physics Chapter 13 Kinetic Theory

Q.1: Calculate the ratio of molecular volume to the actual volume of oxygen gas at standard temperature and pressure. (Diameter of oxygen molecule = 3Å)

Sol:

Diameter of an oxygen molecule, d = 3Å
Radius, r =d/2 = 1.5 Å = 1.5 × 10 – 8 cm
We know:
Actual volume occupied by 1 mole of oxygen at STP = 22400 cm3
Molecular volume of oxygen, V = NA(4πr3/ 3)
Where, N is Avogadro’s number = 6.023 × 1023 molecules/mole
Therefore, molecular volume of oxygen ,V = 6.023 × 1023 x   3.14 (1.5 × 10 – 8)2 x (4/3) = 8.51 cm3
Thus, ratio of the molecular volume to the actual volume of oxygen = 8.51/22400 = 3.8 x 10-4

Q.2: The volume occupied by one mole of any (ideal) gas at STP is called molar volume (STP : 0 °C, 1 atmospheric pressure).  Prove that molar volume is 22.4 litres.

Ans:

We know that the ideal gas equation: PV = nRT
Where, R is the universal gas constant = 8.314 J mol-1 K-1
n = Number of moles = 1
T = Standard temperature = 273 K

P = Standard pressure = 1 atm = 1.013 × 105 Nm-2
Thus,  V = (nRT)/p
= (1 x 8.314 x 273)/( 1.013 × 105)
= 0.0224 m3
= 22.4 liters
Thus, it is proved that molar volume of a gas at standard temperature and pressure is 22.4 liters.

Q.3. The figure below is a graph of PV/T versus P for 1 x 10-3 kg of oxygen at two different temperatures.

(i) What is the significance of the dotted plot?
(ii) T1 < T2 or T1 > T2, which one is true?
(iii) Find the value of PV/T where the curves come together on the y axis.
(iv) In place of oxygen if we used 1 x 10-3 kg of hydrogen and we plotted similar graphs, would the value of PV/T be the same at the point where the curves come in contact with the y axis?
If the answer is no, find the mass of hydrogen that would give the same value (for high temperature low pressure region of the graph).
[Molecular mass of H2 = 2.02 u, of O2 = 32.0 u, R = 8.31 J mol-1 K-1]

Sol:

(i) The dotted plot signifies the ideal gas behavior  of oxygen  as it is parallel to P –axis and it says that the ratio PV/T remains constant even when P is changed.

(ii) The dotted line in the plot stands for  an ideal gas. At temperature T1 the curve of the gas is closer to the dotted plot than at temperature T2 . A real gas behaves more like an ideal gas when its temperature rises. Thus, T1 > T2 is true for the given graph.

(iii) At the point where the curves meet PV/T  = μR
Where μ = no. of moles = 1/32
R = 8.314 J mol-1 K-1
Thus, PV/T   = (1/32) x 8.314 = 0.26 J K-1
(iv) Even if we obtain  a similar curve for  1 x 10-3 kg of hydrogen, we will not get the same value for PV/T because the molar mass of H2 is 2.02 u and not 32u.
We have:
PV / T = 0.26
Given:
Molecular mass of hydrogen , M = 2.02 u
PV/T = μ R
where, μ = m/M
i.e. PV/T =  R (m/M)
i.e.       m = MVP/ TR
= 0.26 x (2.02/8.31) = 6.3 x 10-5 kg

Hence, 6.3 x 10-5 kg of H2 will give the value of PV/T = 0.26 J K-1
Q.4: A 30 liters oxygen cylinder has an initial temperature and gauge pressure of 27 0C and 20 atm respectively.  When a certain amount of oxygen escapes from the cylinder the temperature and gauge pressure drops to 17 0C and 22 atm, respectively. Find the mass of oxygen that escaped the cylinder.
[R = 8.31 J mol-1 K -1, molecular mass of O2 = 32 u]

Sol:

Given:
Initial volume of oxygen, V1 = 30 liters = 30 × 10-3 m3
Gauge pressure, P1 = 30,  atm = 30 × 1.013 × 105 Pa
Temperature, T1 = 27°C = 300 K
Universal gas constant, R = 8.314 J mole-1  K-1
Let the initial number of moles of oxygen in the cylinder be n1 .
We know:

P1 V1 = n1RT1
i.e. n1 = P1 V1 / RT1
= ( 30.39 × 105 × 30 × 10-3 ) / ( 8.314 × 300)
= 36.552
But, n1= m1/ M
Where,
m1 = initial mass of oxygen
M = molecular mass of oxygen = 32 g

i.e.  m1 = n1 x M = 36.552 x 32 = 1169.6 g
After some oxygen escapes:
Volume, V2 = 30 x 10-3 m3
Gauge pressure,
P2 = 22;  atm = 22 × 1.013 × 105 Pa
Temperature, T2 = 17°C = 290 K
Let the number of moles of oxygen  left in the cylinder be n2.
Now:
P2 V2 = n2 RT2

i.e.  n2 = P2 V2 / RT2
= (22.286 × 105 × 30 × 10-3 )/( 8.314 × 290) = 27.72
But, n2 = m2 / M
Where, m2 = remaining mass of oxygen
i.e. m2 = n2 x M = 27.72 x 32 = 906.2g
Therefore the mass of  oxygen that escaped the cylinder = m1 – m2 = 1169.6  – 906.2 = 263.4 g

Q.5:  An air bubble occupies a volume of 2 cm3 at the bottom of 20m deep lake. Assuming the bottom temperature of the lake is 12 0C , find the volume of this air bubble when it rises up to the lake surface which is at 35 0C ?

Sol:

Given:
Volume of the air bubble, V = 2.0 cm3= 2.0 × 10-6 m3
Bubble ascends a height of , d = 20 m
Temperature at a depth of 40 m, T = 12°C = 285 K
Temperature at the surface of the lake, T’ = 35°C = 308 K

The pressure on the surface of the lake: P’ = 1 atm = 1 ×1.013 × 105 Pa

And, The pressure at the bottom: P = 1atm + dρg
Where, ρ is the density of water = 103 kg/m3
g is the acceleration due to gravity = 9.8 m / s2
i.e. P = 1.013 × 105 + 20 × 103 × 9.8 = 297300 Pa
We know:

PV/T = P’V’/T’
Where, V’ is the volume of the bubble at the surface.
V’ = PVT’/P’T
= (297300 x 2 x 10-6 x 308) / (1.013 x 105 x 285) = 6.34 x 10-6 m3 or 6.34 cm3
Therefore, the volume of this bubble when it reaches the surface is 6.34 cm3.

Q.6:  In a 50 m3 room, at a pressure of 1 atm and temperature 270C, what is the number of air molecules (oxygen, nitrogen, water vapour and other constituents) present?

Sol:

Given:
Volume of the room, V = 50.0 m3

Temperature of the room, T = 27°C = 300 K

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

According to gas equation:

PV = kBNT

Where, kB is Boltzmann constant = 1.38 × 10 -23 m2 kg s -2 K -1
N is the number of air molecules in the room
Now, N = PV/ kBT

= (1.013 x 105 x 50) / (1.38 × 10 -23 x 300 ) = 1.22 x 1027
Therefore there is 1.22 × 10 27 molecules in the room.

Q.7: Calculate the average thermal energy of a helium atom at

(i) room temperature ( 27 0C),

(ii) the core of the earth  (6150 K),

(iii) at the core of the sun (10 million K)

Sol:

Given:
(i) At room temperature, T = 27°C = 300 K
Thus, average thermal energy = kT x ( 3/2 )
Where k is Boltzmann constant = 1.38 × 10-23 m2 kg s -2 K -1
Thus,
kT x ( 3/2 ) =  1.38 × 10-23 x 300 x 1.5 = 6.21 x 10 -21 J

(ii) In the core of the earth, T = 6150 K
Thus, average thermal energy  = kT x ( 3/2 )
i.e.  kT x ( 3/2 ) = 1.38 × 10-19 x 6150 x 1.5 = 1.27 x 10 -19 J
(iii) At core of the sun, T = 107
Thus, average thermal energy = kT x ( 3/2 )

i.e.  kT x ( 3/2 ) = 1.38 × 10-19 x 107 x 1.5 = 2.07 x 10 -16 J

Q.8: Three containers A, B and C , having the same capacity, contains neon ( monatomic), chlorine (diatomic) and uranium hexafluoride ( polyatomic) respectively at the same pressure and temperature .Do all the containers contain the same number of molecules? Also, do the molecules in the respective containers have the same root mean square value of speed? If not, molecules of which gas has the highest value of Vrms ?

Sol:

According to Avogadro’s principle, gases of the same volume at the same values of temperature and pressure will contain the same number of molecules. Thus, in the above case all the containers will contain equal number of molecules.
For a gas of mass (m) at temperature (T), its root mean square speed;
Vrms = $\sqrt{\frac{3kT}{m}}$
Where k is the Boltzmann constant.
As k and T are constants, we get:
Vrms = $\sqrt{\frac{1}{m}}$
Thus, Vrms is not the same for the molecules of the three gases.
As mass of neon is the least, it will have the highest Vrms .

Q.9: Calculate the temperature at which  the root mean square speed of an argon atom is the same as the root mean square speed of a helium gas atom at – 20 0C.
[Atomic mass of Ar = 39.9 u, of He = 4.0 u]

Sol:

Given:
Temperature of the helium atom, T’ = –20°C = 253 K
Atomic mass of argon, M = 39.9 u
Atomic mass of helium, M’ = 4.0 u
Let, (VRMS) Ar be the rms speed of argon and (VRMS) He be the rms speed of helium.
Now, we know:

(VRMS)Ar = $\sqrt{\frac{3RT}{M}}$. . . . . . . . . . . . ( 1 )
Where, R is the universal gas constant and T is temperature of argon gas
Now, (VRMS)He =  $\sqrt{\frac{3RT’}{M’}}$ . . . . . . . . . . ( 2 )
According to the question :
(VRMS)Ar = (VRMS)He

i.e. $\sqrt{\frac{3RT}{M}}$ = $\sqrt{\frac{3RT’}{M’}}$
i.e. T/M = T’ / M’
T = M × ( T’/M’ )
Therefore the temperature of argon, T = 39.9 × 253/4 = 2.52 × 103 K

Q.10: A cylinder contains nitrogen at 2 atm and 17 0C, find the collision frequency and the mean free path of a nitrogen molecule inside it. Considering the nitrogen molecule to have a radius of 1Å compare the time between two consecutive collisions and the collision time. [Molecular mass of N2 = 28]

Sol:

Given:
Pressure inside the cylinder containing nitrogen, P = 1.0 atm = 1 × 1.013 × 105 Pa

Temperature inside the cylinder, T = 17°C = 290 K
Radius of a nitrogen molecule, r = 1.0 Å = 1 × 1010 m
Diameter, d = 2 × 1 × 1010 = 2 × 10-10 m
Molecular mass of nitrogen, M = 28.0 g = 28 × 10-3 kg
We know, the root mean square speed ,VRMS = $\sqrt{\frac{3RT}{M}}$
VRMS = $\sqrt{\frac{3\times 8.314\times 290}{28 \times 10^{-3}}}$ = 508.26 m/s
For the mean free path (l) we have:
l = $\frac{kT}{\sqrt{2}\times \pi \times d^{2}\times P}$
Where, k is 1.38 x 10 -23 kg m2 s-2 K-1

Therefore, l = $\frac{1.38\times 10^{-23}\times 290}{\sqrt{2}\times 3.14\times (2\times 10^{-10})^{2}\times 1.013 \times 10^{5} }$ = 2.22 x 10-7 m
And, Collision frequency = VRMS / l = 2.29 x 10 9 s-1
Collision time T = d / VRMS

= 2 x 10-10 / 508.26 = 2.18 x 10-10 s = 3.93 x 10-13 s
Time between consecutive collisions:

T’ = l / VRMS

= 2.22 x 10-7 / 508.26 = 4.36 x 10-10 s
Thus, T’/ T = (4.36 x 10-10 )/ (2.22 x 10-7) = 1109.41.
Therefore the time between two consecutive collisions is 1109.41 times the collision time.

Q11: A narrow bore a meter long held horizontally contains a mercury thread of 70 cm, which traps air column of 20 cm. What will happen if tube is vertically held with bottom end open?

Sol:

Length of mercury thread, l = 70 cm

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

The air column length in between the closed end & mercury, la = 20 cm

Since the bottom end is open and the bore is vertically held in air, the air space occupied by the mercury length is: 100 – (70 + 20) = 10 cm

Hence, total air column length = 20 + 10 = 30 cm

Let, mercury out flow due to atmospheric pressure be ‘h’ cm

Therefore,

The air column length in the bore = (30 + h) cm

And, mercury column length = 80 – h cm

Initial pressure, P1 = 80 cm of mercury

Initial volume, V1 = 20 cm3

Final pressure, P2 = 80 – (80 – h) = h cm of mercury

Final volume is V2 = (30 + h) cm3

Throughout the process the temperature is constant.

$\\P_{1}V_{1} = P_{2}V_{2} \\ 70 \times 20 = h \left ( 30 + h \right ) \\ h^{2} + 30h – 1400 = 0$

Therefore, $h = \frac{-30\pm \sqrt{\left ( 30 \right )^{2} + 4 \times 1 \times 1400}}{2 \times 1} \\ = -55.3 \; cm \; or \; 25.3 \; cm$

Height is always positive. Hence, mercury that flow out from bore is 25.3 cm and mercury that remains in it is 54.7 cm. The air column length is 30 + 25.3 = 55.3 cm

Q12. Hydrogen gas’s diffusion rate from one certain apparatus has average value 30 cm2/ s. Under same condition the average diffusion rate of another gas is 8 cm2/s. What gas is it?

[Hint: Graham’s law of diffusion states that: (M2/ M1)1/2 = R1/ R2, where diffusion rates of gas 1 and gas 2 are given by R1, R2 and M1 and M2 are their molecular masses]

Sol:

Diffusion rate of hydrogen, R1 = 30 cm3/ s

Diffusion rate of the other gas, R2 = 8 cm3/ s

According to Graham’s Law of diffusion, we have:

$\\ \frac{R_{1}}{R_{2}} = \sqrt{\frac{M_{2}}{M_{1}}}$

Where, Molecular mass of hydrogen M1 = 2.020 g

Molecular mass of the unknown gas is M2

Therefore, $M_{2} = M_{1}\left ( \frac{R_{1}}{R_{2}} \right )^{2} \\ = 2.02\left ( \frac{30}{8} \right )^{2} \\ = 28.40 \; g$

Nitrogen has the molecular mass 28. Hence the other gas is nitrogen.

Q.13: Throughout the volume of a gas in equilibrium the density and pressure is uniform. It is true only if no external influences are used. Gas column because of gravity doesn’t have uniform density or pressure. Density of the gas decreases with height. The dependence precise is given by law of atmosphere n2 = n1 exp [ -mg(h2 – h1)/kBT ] Where n1, n2 are referred to density at h1 and h2 respectively. The sedimentation equilibrium equation of liquid column can be derived by using this relation: $n_{2} = n_{1} exp [-mgN_{A}(\rho – \rho’)(h_{2} – h_{1})/(\rho RT)]$ Where $\rho$ is the density of the particle suspended, and $\rho’$ is surrounding medium’s density. [NA = Avogadro’s number & R the universal gas constant.] [To find the suspended particle’s apparent weight use Archimedes principle]

Sol:

From law of atmosphere, we have:

n2 = n1 exp[-mg(h2 – h1)/kBT]…. (i)

Where, at height h1, number density is n1, and at height h2, number density is n2

Weight of suspended particle in gas column is mg

Medium density = $\rho’$

Suspended particle density = $\rho$

Suspended particle mass = m’

Displaced medium’s mass = m

Suspended particle’s volume = V

Archimedes’ principle states that the weight of the suspended particle in the liquid column is given by

Now, Displaced medium weight – suspended particle weight

$\\ = mg – m’g \\ = mg – V \rho ‘ g = mg – \left ( \frac{m}{\rho } \right )\rho ‘ g \\ = mg \left ( 1 – \frac{\rho ‘}{\rho } \right ) …. (ii) \\ Gas \; constant, \; R = k_{B}N\\ K_{B} = \frac{R}{N} …. (iii) \\ Substitution \; equation \; (ii) \; in \; place \;of \; mg \; in \; equation (i) \; and \; then \; using \; equation (iii), we ; get: \\ n_{2} = n_{1} \; exp [-mg(h_{2} – h_{1})/ k_{B}T]\\ = n_{1} \;exp [-mg\left ( 1 – \frac{\rho ‘}{\rho }(h_{2} – h_{1}) \frac{N}{RT\rho } \right )]$

Q14: Density of some of the solids and liquids are given below. Provide rough estimates of their atom sizes:

 Substance Atomic mass Density (103 kg m-3) Carbon (diamond) 12.01 2.22 Gold 197.00 19.32 Nitrogen 14.01 1.00 Lithium 6.94 0.53 Fluorine (liquid) 19.00 1.14

[Assume in solid and liquid phase the atoms are tightly packed, and use Avogadro’s number. Do not take actual numbers obtain for different atomic sizes? Because of tight packing approximation of the crudeness, the range of atomic size in between few Å]

Sol:

 Substance Radius (Å) Carbon (diamond) Gold Nitrogen (liquid) Lithium Fluorine (liquid) 1.29 1.59 1.77 1.73 1.88

Substance’s atomic mass = M

Substance’s density = $\rho$ $Avogadro’s \; number = N = 6.023 \times 10^{23} \\ Each \; atom’s \; volume = \frac{4}{3}\pi r^{3} \\ N \; number \; of \; molecules’ \; volume = \frac{4}{3}\pi r^{3} N …. (i)\\ One \; mole’s \; volume = \frac{M}{\rho } …. (iii)\\ \frac{4}{3}\pi r^{3} N = \frac{M}{\rho }$

Therefore, r = $\sqrt[3]{\frac{3M}{4\pi \rho N}}$

For carbon:

$\\ M = 12.01 \times 10^{-3} \; kg, \; \\ \rho = 2.22 \times 10^{3}kg m^{-3}$

Therefore, r = $\left ( \frac{3 \times 12.01 \times 10^{-3} }{4 \pi \times 2.22 \times 10^{3} \times 6.023 \times 10^{23}} \right )^{\frac{1}{3}} = 1.29$

Hence, radius of carbon atom = 1.29 Å

For gold:

$\\ M = 197.01 \times 10^{-3} \; kg, \; \\ \rho = 19.32 \times 10^{3}kg m^{-3}$

Therefore, r = $\left ( \frac{3 \times 197 \times 10^{-3} }{4 \pi \times 19.32 \times 10^{3} \times 6.023 \times 10^{23}} \right )^{\frac{1}{3}} = 1.59$

Hence, radius of gold atom = 1.59 Å

For nitrogen (liquid):

$\\ M = 14.01 \times 10^{-3} \; kg, \; \\ \rho = 1.00 \times 10^{3}kg m^{-3}$

Therefore, r = $\left ( \frac{3 \times 14.01 \times 10^{-3} }{4 \pi \times 1.00 \times 10^{3} \times 6.23 \times 10^{23}} \right )^{\frac{1}{3}} = 1.77$

Hence, radius of nitrogen (liquid) atom = 1.77 Å

For lithium:

$\\ M = 6.94 \times 10^{-3} \; kg, \; \\ \rho = 0.53 \times 10^{3}kg m^{-3}$

Therefore, r = $\left ( \frac{3 \times 6.94 \times 10^{-3} }{4 \pi \times 0.53 \times 10^{3} \times 6.23 \times 10^{23}} \right )^{\frac{1}{3}} = 1.73$

Hence, radius of lithium atom = 1.73 Å

For fluorine (liquid):

$M = 19.00 \times 10^{-3} \; kg, \; \\ \rho = 1.14 \times 10^{3}kg m^{-3}\\$

Therefore, r = $\left ( \frac{3 \times 19 \times 10^{-3} }{4 \pi \times 1.14 \times 10^{3} \times 6.023 \times 10^{23}} \right )^{\frac{1}{3}} = 1.88$

Hence, radius of fluorine (liquid) atom = 1.88 Å

 Also Access NCERT Exemplar for class 11 Physics Chapter 13 CBSE Notes for class 11 Physics Chapter 13

The Central Board of Secondary Education is one of the oldest educational board in the country. The CBSE follow the NCERT syllabus to conducts its class 10th and class 12th board examinations. The NCERT solutions are prepared by subject experts according to the latest CBSE syllabus (2018-19) of 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.

Kinetic Theory is one of the most scoring sections in CBSE Class 11 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).
 Related Links Ncert Questions Of Maths Class 10 Cbse Class 10 Science Ncert Solutions Ncert Biology Class 12 Pdf Free Download Ncert Class 12 Chemistry Part 2 Ncert Physics Class 12 Part 1 Ncert Solution For 7 Maths Solution Of Ncert Class 8 Maths Ncert Guide For Class 9 Ncert Solutions For Class 9 Chemistry NCERT Solutions Class 6 Science NCERT Solutions for Class 4 Maths NCERT Solutions for Class 4 Science