Question

NCERT Solutions Class 11 Physics Chapter 13 Kinetic Theory discusses the concepts of specific heat capacity and the theories which explain the equipartition of energy. Our solutions have a complete discussion of all the concepts related to the Kinetic Theory of the chapter.

NCERT Solutions Class 11 Physics Chapter 13 Kinetic Theory deals with the molecular nature of matter gases and how they behave. NCERT Solutions discuss the concepts of Kinetic Theory holistically. We have discussed all the important topics of the subject.

• Kinetic Theory – the kinetic theory explains the behavior of gases and it states that the gas consists of rapidly moving atoms or molecules.
• Molecular nature of matter – atomic hypothesis states that everything is made up of small fundamental units called atoms. 
• Behaviour of gases – the behavior of gases states how when temperature increases, the volume of gas also increases because of the expansion of gas molecules.
• Kinetic theory of an ideal gas – molecules of gas are always in a random motion and those molecules are continuously colliding with each other and with the walls of the container. Energy is not lost or gained due to the collisions because all the collisions are elastic thus total energy and total momentum are always conserved. It is mathematically represented as PV = nRT. Where ‘P’ is pressure, ‘T’ is temperature, ‘n’ is the mole of the ideal gas, and R is the gas constant.
• Law of equipartition of energy – when a dynamic system is in thermal equilibrium total energy of a system is equally divided among the different degrees of freedom.
• Specific heat capacity - The amount of heat needed to change the heat content of a 1-mole substance due to 1 Kelvin or 1 °C.
• Mean free path – mean free path is the average distance an object will move between the collisions.

Our NCERT Solutions Class 11 Physics discusses important topics such as atomic theory, gas laws, Boltzmann constant, Avogadro's number, postulates of kinetic theory, and specific heat capacities. Our experts suggest students study our solution for complete understanding to score good marks in the exam.

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Answer 1

NCERT Solutions 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 3Å.

Solution:

Diameter of an oxygen molecule, d = 3Å 

Radius, r = \(\frac d2 =\frac 32\) = 1.5 Å = 1.5 × 10^–8 cm 

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

Molecular volume of oxygen gas, V = \(\frac 43\pi r^3.N\)

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

∴ V = \(\frac 43 \) x 3.14 x (1.5 x 10-8)³ x 6.023 x 10²³ = 8.51 cm³

Ratio of the molecular volume to the actual volume of oxygen = \(\frac {8.51}{22400}\) = 3.8 × 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^–1 K^–1 

n = Number of moles = 1 

T = Standard temperature = 273 K 

P = Standard pressure = 1 atm = 1.013 × 10⁵ Nm^–2

= 0.0224 m3 

= 22.4 litres 

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

3. Figure shows plot of PV/T versus P for 1.00×10^–3 kg of oxygen gas at two different temperatures.

What does the dotted plot signify? 

Which is true: T₁ > T₂ or T₁ < T₂? 

What is the value of PV/T where the curves meet on the y-axis? 

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:

The dotted plot in the graph signifies the ideal behaviour of the gas, i.e., the ratio PV/T is equal. μR (μ is the number of moles and R is the universal gas constant) is a constant quality. It is not dependent on the pressure of the gas. 

The dotted plot in the given graph represents an ideal gas. The curve of the gas at temperature T₁ is closer to the dotted plot than the curve of the gas at temperature T₂. A real gas approaches the behaviour of an ideal gas when its temperature increases. Therefore, T₁ > T₂ is true for the given plot. 

The value of the ratio PV/T, where the two curves meet, is μR. This is because the ideal gas equation is given as:

PV = μRT

\(\frac {PV}{T}\) = μR

Where, 

P is the pressure 

T is the temperature 

V is the volume 

μ is the number of moles 

R is the universal constant 

Molecular mass of oxygen = 32.0 g 

Mass of oxygen = 1 × 10^–3 kg = 1 g 

R = 8.314 J mole^–1 K^–1

∴ \(\frac {PV}{T}\) = \(\frac {1}{32}\) x 8.314

= 0.26 J K^–1 

Therefore, the value of the ratio PV/T, where the curves meet on the y-axis, is 0.26 J K^–1. 

If we obtain similar plots for 1.00 × 10^–3 kg of hydrogen, then we will not get the same value of PV/T at the point where the curves meet the y-axis. This is because the molecular mass of hydrogen (2.02 u) is different from that of oxygen (32.0 u). 

We have:

\(\frac {PV}{T}\) = 0.26 J K^-1

R = 8.314 J mole^–1 K^–1

Molecular mass (M) of H₂ = 2.02 u

 \(\frac {PV}{T}\) = μR at constant temperature

Where, μ = m/M

m = Mass of H₂

∴ \(\frac {PV}{T}\) x \(\frac MR = \frac {0.26\times2.02}{8.31} \) 

= 6.3 × 10^–2 g = 6.3 × 10^–5 kg 

Hence, 6.3 × 10^–5 kg of H₂ will yield the same value of PV/T.

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 oxygen, V₁ = 30 litres = 30 × 10–3 m³ 

Gauge pressure, P₁ = 15 atm = 15 × 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 gas in the cylinder be n₁. 

The gas equation is given as:

P₁V₁ = n₁RT₁

= 18.276

But, n₁ = \(\frac {m_1}{M}\) Where, m₁ =

Initial mass of oxygen 

M = Molecular mass of oxygen = 32 g 

m₁ = n₁M = 18.276 × 32 = 584.84 g 

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

Volume, V₂ = 30 litres = 30 × 10^–3 m³ 

Gauge pressure, P₂ = 11 atm = 11 × 1.013 × 105 Pa 

Temperature, T₂ = 17°C = 290 K 

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

The gas equation is given as:

P₂V₂ = n₂RT₂

= 13.86

But,   n₂ = \(\frac {m_2}{M}\) Where, m₂ is the mass of oxygen 

remaining in the cylinder 

∴ m₂ = n₂ M = 13.86 × 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 

= 131.74 g 

= 0.131 kg 

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

Answer 2

 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 × 10^–6 m³ 

Bubble rises to height, d = 40 m 

Temperature at a depth of 40 m, 

T₁ = 12°C = 285 K 

Temperature at the surface of the lake, 

T₂ = 35°C = 308 K 

The pressure on the surface of the lake: P₂ = 1 atm = 1 ×1.013 × 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 

P₁ = 1.013 × 105 + 40 × 103 × 9.8 = 493300 Pa

We have: \(\frac {P_1V_1}{T_1} = \frac {P_2V2}{T_2}\) 

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

= 5.263 × 10^–6 m³ or 5.263 cm³

Therefore, 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 = 27°C = 300 K 

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

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

PV = k_BNT 

Where, 

K_B is Boltzmann constant = 1.38 × 10^–23 m 2 kg s^–2 K^–1 

N is the number of air molecules in the room

= 6.11 × 10²⁶ molecules

Therefore, the total number of air molecules in the given room is 6.11 × 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 :

 At room temperature, T = 27°C = 300 K 

Average thermal energy = \(\frac 32kT\)

Where k is Boltzmann constant = 1.38 × 10^–23 m² kg s^–2 K^–1 

∴ \(\frac 32kT\) = \(\frac 32\) x 1.38 x 10^-38 x 300

= 6.21 × 10^–21J 

Hence, the average thermal energy of a helium atom at room temperature (27°C) is 6.21 × 10^–21 J. 

On the surface of the sun, T = 6000 K 

Average thermal energy = \(\frac 32kT\)

  \(\frac 32\) x 1.38 x 10^-38 x 6000

1.241 × 10^–19 J 

Hence, the average thermal energy of a helium atom on the surface of the sun is 1.241 × 10^–19 J. 

At temperature, T = 10⁷ K

Average thermal energy = \(\frac 32kT\)

  \(\frac 32\) x 1.38 x 10^-23 x 10⁷

2.07 × 10^–16 J 

Hence, the average thermal energy of a helium atom at the core of a star is 2.07 × 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 v_rms the largest? 

Solution:

Yes. All contain the same number of the respective molecules. 

No. The root mean square speed of neon is the largest. 

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

Hence, 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 × 10²³ . 

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

v_rms = \(\sqrt {\frac {3kT}{m}}\) 

Where, k is Boltzmann constant 

For the given gases, k and T are constants. 

Hence v_rms depends only on the mass of the atoms, i.e.,

v_rms ∝ \(\sqrt {\frac {1}{m}}\) 

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

Answer 3

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:

Temperature of the helium atom, T_He = –20°C= 253 K 

Atomic mass of argon, M_Ar = 39.9 u 

Atomic mass of helium, M_He = 4.0 u 

Let, (v_rms)Ar be the rms speed of argon. 

Let (v_rms)He be the rms speed of helium. 

The rms speed of argon is given by:

Where, 

R is the universal gas constant 

T_Ar is temperature of argon gas 

The rms speed of helium is given by:

It is given that: 

(v_rms)Ar = (v_rms)He

= 2523.675 = 2.52 × 10³ K 

Therefore, the temperature of the argon atom is 2.52 × 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 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). 

Solution:

Mean free path = 1.11 × 10^–7 m 

Collision frequency = 4.58 × 109 s^–1 

Successive collision time ≈ 500 × (Collision time) 

Pressure inside the cylinder containing nitrogen, P = 2.0 atm = 2.026 × 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 × 10¹⁰ m 

Molecular mass of nitrogen, M = 28.0 g = 28 × 10^–3 kg 

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

v_rms = \(\sqrt {\frac {3RT}{M}}\) 

Where, 

R is the universal gas constant = 8.314 J mole^–1 K^–1

= 508.26 m/s

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

l = \(\frac {kT}{\sqrt2 \times d^2\times P}\) 

Where, k is the Boltzmann constant = 1.38 × 10^–23 kg 

m² s^–2 K^–1

= 1.11 × 10^–7 m 

Collision frequency = \(\frac {V_{rms}}{l} = \frac {508.26}{1.11 \times 10^{-7}}\) = 4.58 × 10⁹ s^–1

Collision time is given as: 

\(T = \frac d{v_{rms}} = \frac {2\times10^{-10}}{508.26}\) = 3.93 × 10^–13 s

Time taken between successive collisions:

T'= \(\frac {l}{V_{rms}} \)

Hence, 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, l_a = 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 

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

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

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

And, length of the mercury column = 76 – h cm 

Initial pressure, P₁ = 76 cm of mercury 

Initial volume, 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.

P₁V₁ = P₂V₂ 

76 × 15 = h (24 + h) 

h² + 24h – 1140 = 0

∴ h = \(\frac {-24\pm \sqrt {(24)^2 + 4 \times 1\times 1140}}{2\times1}\)

= 23.8 cm or –47.8 cm 

Height cannot be negative. Hence, 23.8 cm of mercury will flow out from the bore and 52.2 cm of mercury will remain in it. The length of the air column will be 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. 

[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: 

Rate of diffusion of hydrogen, R₁ = 28.7 cm³ s^–1 

Rate of diffusion of another gas, R₂ = 7.2 cm³ s^–1 

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

\(\frac {R_1}{R_2} = \sqrt{\frac {M_2}{M_1}}\) 

Where, 

M₁ is the molecular mass of hydrogen = 2.020 g 

M₂ is the molecular mass of the unknown gas

∴ M2 = M1 \((\frac {R_1}{R_2})^2\)

= \(2.02 (\frac {28.7}{R_2})^2\) 

= 32.09 g

32 g is the molecular mass of oxygen. Hence, the unknown gas is oxygen.

Answer 4

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 N_A(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.] 

[Use Archimedes principle to find the apparent weight of the suspended particle.] 

Solution:

According to the law of atmospheres, we have: 

n₂ = n₁ exp [-mg (h₂ – h₁)/ kBT] … (i) 

Where, n₁ is the number density at height h₁, and n₂ is the number density at height h₂ mg is the weight of the particle suspended in the gas column 

Density of the medium = ρ' 

Density of the suspended particle = ρ 

Mass of one suspended particle = m' 

Mass of the medium displaced = m 

Volume of a suspended particle = V 

According to Archimedes’ principle for a particle suspended in a liquid column, the effective weight of the suspended particle is given as: 

Weight of the medium displaced – Weight of the suspended particle 

= mg – m' g 

= mg - V ρ'g = mg - \((\frac mρ)ρ'g\) 

= mg \((1 - \frac {ρ'}{ρ})\) .....(ii)

Gas constant, R = k_BN

k_B = \(\frac RN\) ...(iii)

Substituting equation (ii) in place of mg in equation (i) and then using equation (iii), 

we get: n₂ = n₁ exp [-mg (h₂ – h₁)/ k_BT]

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

[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].

Solution:

Substance	Radious (\(\mathring{A}\))
Carbon (diamond)	1.29
Gold	1.59
Nitrogen (liquid)	1.77
Lithium	1.73
Fluorine (liquid)	1.88

 Atomic mass of a substance = M 

Density of the substance = ρ 

Avogadro’s number = N = 6.023 × 10²³ 

Volume of each atom = \(\frac 43 \pi r^3\)  

Volume of N number of molecules  = \(\frac 43 \pi r^3\) N … (i)

Volume of one mole of a substance = \(\frac Mρ\) ...(ii)

\(\frac 43 \pi r^3\) N = \(\frac Mρ\) 

∴ r = \(\sqrt [3]{\frac {3M}{4 \pi ρN}}\) 

For carbon: 

M = 12.01 × 10^–3 kg, 

ρ = 2.22 × 10³ kg m^–3

∴ r = \((\frac {3\times 12.01\times10^{-3}}{4 \pi \times 2.22 \times 10^3 \times 6.023 \times 10^{23}})^{\frac 13}\)

= 1.29 Å

Hence, the radius of a carbon atom is 1.29 Å.

For gold: 

M = 197.00 × 10^–3 kg 

ρ = 19.32 × 10³ kg m^–3

 ∴ r = \((\frac {3\times 197\times10^{-3}}{4 \pi \times 19.32 \times 10^3 \times 6.023 \times 10^{23}})^{\frac 13}\) 

= 1.59 Å

Hence, the radius of a gold atom is 1.59 Å. 

For liquid nitrogen:

M = 14.01 × 10^–3 kg 

ρ = 1.00 × 10³ kg m^–3

 ∴ r = \((\frac {3\times 14.01\times10^{-3}}{4 \pi \times 1.00 \times 10^3 \times 6.23 \times 10^{23}})^{\frac 13}\) 

= 1.77 Å 

Hence, the radius of a liquid nitrogen atom is 1.77 Å.

For lithium:

M = 6.94 × 10^–3 kg ρ 

= 0.53 × 10³ kg m^–3

 ∴ r = \((\frac {3\times 6.94\times10^{-3}}{4 \pi \times 0.53 \times 10^3 \times 6.23 \times 10^{23}})^{\frac 13}\) 

= 1.73 Å

Hence, the radius of a lithium atom is 1.73 Å. 

For liquid fluorine: 

M = 19.00 × 10^–3 kg 

ρ = 1.14 × 10³ kg m^–3

 ∴ r = \((\frac {3\times 19\times10^{-3}}{4 \pi \times 1.14 \times 10^3 \times 6.023 \times 10^{23}})^{\frac 13}\) 

= 1.88 Å 

Hence, the radius of a liquid fluorine atom is 1.88 Å.