# Class 11 Physics Chapter 13: Kinetic Theory - NCERT Solutions

Welcome to the complete NCERT solutions for Class 11 Physics Chapter 13: Kinetic Theory. In this section, we provide detailed, easy-to-understand solutions for all the questions from this chapter. Whether you're preparing for exams or seeking a deeper understanding of the subject, these Kinetic Theory question answers will offer you valuable insights and explanations. Each solution is crafted to ensure conceptual clarity and step-by-step problem-solving methods, enabling students to grasp the core themes and excel in their academics.

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### Exercise 1

•  Q1 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. Ans: 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 Q2 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: 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. Q5 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? Ans: 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. Q6 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. Ans: 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. Q7 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). Ans: (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. Q8 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 vrmsthe largest? Ans: 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. 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? Ans: 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. Q12 From a certain apparatus, the diffusion rate of hydrogen has an average value of 28.7 cm3s-1. The diffusion of another gas under the same conditions is measured to have an average rate of 7.2 cm3s-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.] Ans: 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.

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