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

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.

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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 consists of previous year questions papers that are very useful in preparing for the board examinations.

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: 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

 

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: 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:

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. 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.)Untitled

 

Sol:

(i) The dotted plot signifies the ideal gas behaviour 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 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 ?

 

Sol:

Given:
Volume of the air bubble, V = 2.0 cm3= 1.0 × 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:  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.

 

Sol:

Given:
Volume of the room, V = 25.0 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: 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). 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 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 an 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?

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 = 3kTm\sqrt{\frac{3kT}{m}}
Where k is the Boltzmann constant.
As k and T are constants, we get:
Vrms = 1m\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: 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)

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

i.e. 3RTM\sqrt{\frac{3RT}{M}} = 3RTM\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: Estimate the mean free path and collision frequency of a nitrogen molecule in a cylinder containing nitrogen at 2.0 atm and temperature 17 0C. 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)

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 = 3RTM\sqrt{\frac{3RT}{M}}
VRMS = 3×8.314×29028×103\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 = kT2×π×d2×P\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 = 1.38×1023×2902×3.14×(2×1010)2×1.013×105\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 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?

Sol:

Length of mercury thread, l = 76 cm

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

The air column length in between the closed end & mercury, la = 15 cm

Since the bottom end is open and the bore is vertically held in the air, the air space occupied by the mercury length is: 100 – (76 + 15) = 9 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.

P1V1=P2V270×20=h(30+h)h2+30h1400=0\\P_{1}V_{1} = P_{2}V_{2} \\ 70 \times 20 = h \left ( 30 + h \right ) \\ h^{2} + 30h – 1400 = 0

Therefore, h=30±(30)2+4×1×14002×1=55.3  cm  or  25.3  cmh = \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. 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: 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:

R1R2=M2M1\\ \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, M2=M1(R1R2)2=2.02(308)2=28.40  gM_{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: n2=n1exp[mgNA(ρρ)(h2h1)/(ρRT)]n_{2} = n_{1} exp [-mgN_{A}(\rho – \rho’)(h_{2} – h_{1})/(\rho RT)] Where ρ \rho is the density of the particle suspended, and ρ \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 = ρ \rho’

Suspended particle density = ρ \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

=mgmg=mgVρg=mg(mρ)ρg=mg(1ρρ).(ii)Gas  constant,  R=kBNKB=RN.(iii)Substitution  equation  (ii)  in  place  of  mg  in  equation(i)  and  then  using  equation(iii),we;get:n2=n1  exp[mg(h2h1)/kBT]=n1  exp[mg(1ρρ(h2h1)NRTρ)]\\ = 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 = ρ \rho Avogadros  number=N=6.023×1023Each  atoms  volume=43πr3N  number  of  molecules  volume=43πr3N.(i)One  moles  volume=Mρ.(iii)43πr3N=MρAvogadro’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 = 3M4πρN3\sqrt[3]{\frac{3M}{4\pi \rho N}}

For carbon:

M=12.01×103  kg,   ρ = 2.22×103kgm3\\ M = 12.01 \times 10^{-3} \; kg, \;  \\ \rho  =  2.22 \times 10^{3}kg m^{-3}

Therefore, r = (3×12.01×1034π×2.22×103×6.023×1023)13=1.29\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=197.01×103  kg,  ρ =19.32×103kgm3\\ M = 197.01 \times 10^{-3} \; kg, \; \\ \rho  = 19.32 \times 10^{3}kg m^{-3}

Therefore, r = (3×197×1034π×19.32×103×6.023×1023)13=1.59\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=14.01×103  kg,  ρ=1.00×103kgm3\\ M = 14.01 \times 10^{-3} \; kg, \; \\ \rho = 1.00 \times 10^{3}kg m^{-3}

Therefore, r = (3×14.01×1034π×1.00×103×6.23×1023)13=1.77\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=6.94×103  kg,  ρ =0.53×103kgm3\\ M = 6.94 \times 10^{-3} \; kg, \; \\ \rho  = 0.53 \times 10^{3}kg m^{-3}

Therefore, r = (3×6.94×1034π×0.53×103×6.23×1023)13=1.73\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=19.00×103  kg,  ρ=1.14×103kgm3M = 19.00 \times 10^{-3} \; kg, \; \\ \rho = 1.14 \times 10^{3}kg m^{-3}\\

Therefore, r = (3×19×1034π×1.14×103×6.023×1023)13=1.88\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

 

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. The exams conducted by the CBSE is based on the NCERT syllabus for both 10th as well as 12th classes.

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).

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