Welcome to the Chapter 13 - Kinetic Theory, Class 11 Physics NCERT Solutions page. Here, we provide detailed question answers for Chapter 13 - Kinetic Theory. The page is designed to help students gain a thorough understanding of the concepts related to natural resources, their classification, and sustainable development.
Our solutions explain each answer in a simple and comprehensive way, making it easier for students to grasp key topics Kinetic Theory and excel in their exams. By going through these Kinetic Theory question answers, you can strengthen your foundation and improve your performance in Class 11 Physics. Whether you’re revising or preparing for tests, this chapter-wise guide will serve as an invaluable resource.
Diameter of an oxygen molecule, d= 3Å
Radius, r = d /2 = 3/2 = 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 = 4/3 π r3 . N
Where, N is Avogadro's number = 6.023 × 1023 molecules/mole
∴ V = 4/3 x 3.14 x (1.5 × 10-8 )3 x 6.023 × 1023
= 8.51 cm3
Ratio of the molecular volume to the actual volume of oxygen = 8.51 / 22400 = 3.8 × 10-4
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 boreis 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 + hcm
And, length of the mercury column = 76 - hcm
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 = - 24 +- underroot [(24)2 + 4 x 1 x 1140] / 2 x 1
= 23.8 cm or -47.8 cm
Height cannot be negative.
Hence, 23.8 cm of mercurywill flow out from the boreand 52.2 cm of mercury will remain in it.
The length of the air column will be 24 + 23.8 = 47.8 cm.
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 = Underroot M2/M1
Where, M1 is the molecular mass of hydrogen = 2.020 g
M2 is the molecular mass of the unknown gas
∴ M2 =M1 (R1 / R2)2
=2.02 (28.7 / 7.2)2
= 32.09 g
32 g is the molecular mass of oxygen. Hence, the unknown gas is oxygen.
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 × 105 Nm-2
∴ V = nRT / P
= 1 x 8.314 x 273 / 1.013 x 105
= 0.0224 m3 = 22.4 litres
Hence, the molar volume of a gas at STP is 22.4 litres.
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 + dpg
Where, p 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: P1V1 / T1 = P2V2 / T2
Where, V2 is the volume of the air bubble when it reaches the surface
V2 = P1V1T2 / T1P2
= 493300 x (1.0 x 10-6) 308 / 285 x 1.013 x 105
= 5.263 × 10-6 m3 or 5.263 cm3
Therefore, when the air bubble reaches the surface, its volume becomes 5.263 cm3.
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 = kBNT
Where, KB is Boltzmann constant = 1.38 × 10-23 m2 kg s-2K-1
N is the number of air molecules in the room
∴ N = PV / KB T
= 1.013 x 105 x 25 / 1.38 x 10-23 x 300
= 6.11 × 1026 molecules
Therefore, the total number of air molecules in the given room is 6.11 × 1026.
(i) At room temperature, T= 27°C = 300 K
Average thermal energy = 3/2 kT
Where k is Boltzmann constant = 1.38 × 10-23m2 kg s-2K-1
∴ 3/2 kT = 3/2 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 × 10-21J.
(ii) On the surface of the sun, T= 6000 K
Average thermal energy = 3/2 kT
= 3/2 x 1.38 x 10-38 x 6000
= 1.241 × 10-19J
Hence, the average thermal energy of a helium atom on the surface of the sun is 1.241 × 10-19J .
(iii) At temperature, T= 107K
Average thermal energy = 3/2 kT
= 3/2 x 1.38 x 10-38 x 107
= 2.07 × 10-16J
Hence, the average thermal energy of a helium atom at the core of a star is 2.07 × 10-16J.
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 (vrms) of a gas of mass m, and temperature T, is given by the relation:
vrms = underroot 3kT / m
Where, k is Boltzmann constant
For the given gases, k and T are constants.
Hence vrmsdepends only on the mass of the atoms, i.e.,
vrms ∝ underroot 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.