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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.
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
Key Features of NCERT Class 11 Physics Chapter 'Kinetic Theory' question answers :
- All chapter question answers with detailed explanations.
- Simple language for easy comprehension.
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Popular Questions of Class 11 Physics
- Q:-
State the number of significant figures in the following:
(a) 0.007 m2
(b) 2.64 x 1024 kg
(c) 0.2370 g cm-3
(d) 6.320 J
(e) 6.032 N m-2
(f) 0.0006032 m2
- Q:-
Fill in the blanks by suitable conversion of units:
(a) 1 kg m2s–2= ....g cm2 s–2
(b) 1 m =..... ly
(c) 3.0 m s–2=.... km h–2
(d) G = 6.67 × 10–11 N m2 (kg)–2=.... (cm)3s–2 g–1.
- Q:-
A physical quantity P is related to four observables a, b, c and d as follows :
The percentage errors of measurement in a, b, c and d are 1%, 3%, 4% and 2%, respectively. What is the percentage error in the quantity P ? If the value of P calculated using the above relation turns out to be 3.763, to what value should you round off the result ?
- Q:-
Rain is falling vertically with a speed of 30 m s–1. A woman rides a bicycle with a speed of 10 m s–1 in the north to south direction. What is the direction in which she should hold her umbrella?
- Q:- Give the magnitude and direction of the net force acting on
(a) a drop of rain falling down with a constant speed
(b) a cork of mass 10 g floating on water
(c) a kite skillfully held stationary in the sky
(d) a car moving with a constant velocity of 30 km/h on a rough road
(e) a high-speed electron in space far from all material objects, and free of electric and magnetic fields. - Q:-
The mass of a box measured by a grocer's balance is 2.300 kg. Two gold pieces of masses 20.15 g and 20.17 g are added to the box. What is
(a) the total mass of the box,
(b) the difference in the masses of the pieces to correct significant figures?
- Q:-
On an open ground, a motorist follows a track that turns to his left by an angle of 60° after every 500 m. Starting from a given turn, specify the displacement of the motorist at the third, sixth and eighth turn. Compare the magnitude of the displacement with the total path length covered by the motorist in each case.
- Q:-
What amount of heat must be supplied to 2.0 x 10-2 kg of nitrogen (at room temperature) to raise its temperature by 45 °C at constant pressure? (Molecular mass of N2 = 28; R = 8.3 J mol-1 K-1.)
- Q:- In which of the following examples of motion, can the body be considered approximately a point object:
(a) a railway carriage moving without jerks between two stations.
(b) a monkey sitting on top of a man cycling smoothly on a circular track.
(c) a spinning cricket ball that turns sharply on hitting the ground.
(d) a tumbling beaker that has slipped off the edge of a table. - Q:-
A transverse harmonic wave on a string is described by
y(x,t) = 3.0 sin [36t + 0.018x + π /4]
Where x and y are in cm and t in s. The positive direction of x is from left to right.
(a) Is this a travelling wave or a stationary wave? If it is travelling, what are the speed and direction of its propagation?
(b) What are its amplitude and frequency?
(c) What is the initial phase at the origin?
(d) What is the least distance between two successive crests in the wave?
Recently Viewed Questions of Class 11 Physics
- Q:-
Precise measurements of physical quantities are a need of science. For example, to ascertain the speed of an aircraft, one must have an accurate method to find its positions at closely separated instants of time. This was the actual motivation behind the discovery of radar in World War II. Think of different examples in modern science where precise measurements of length, time, mass etc. are needed. Also, wherever you can, give a quantitative idea of the precision needed.
- Q:-
A transverse harmonic wave on a string is described by
y(x,t) = 3.0 sin [36t + 0.018x + π /4]
Where x and y are in cm and t in s. The positive direction of x is from left to right.
(a) Is this a travelling wave or a stationary wave? If it is travelling, what are the speed and direction of its propagation?
(b) What are its amplitude and frequency?
(c) What is the initial phase at the origin?
(d) What is the least distance between two successive crests in the wave?
- Q:-
A passenger arriving in a new town wishes to go from the station to a hotel located 10 km away on a straight road from the station. A dishonest cabman takes him along a circuitous path 23 km long and reaches the hotel in 28 min. What is
(a) the average speed of the taxi,
(b) the magnitude of average velocity? Are the two equal?
- Q:-
Rain is falling vertically with a speed of 30 m s–1. A woman rides a bicycle with a speed of 10 m s–1 in the north to south direction. What is the direction in which she should hold her umbrella?
- Q:-
Figure 3.23 gives the x-t plot of a particle executing one-dimensional simple harmonic motion. (You will learn about this motion in more detail in Chapter14). Give the signs of position, velocity and acceleration variables of the particle at t = 0.3 s, 1.2 s, – 1.2 s.
(Fig 3.23)
- Q:-
State the number of significant figures in the following:
(a) 0.007 m2
(b) 2.64 x 1024 kg
(c) 0.2370 g cm-3
(d) 6.320 J
(e) 6.032 N m-2
(f) 0.0006032 m2
- Q:-
Use the formula v = √ γP/ρ to explain why the speed of sound in air (a) is independent of pressure, (b) increases with temperature, (c) increases with humidity.
- Q:-
Attempt to formulate your ‘moral’ views on the practice of science. Imagine yourself stumbling upon a discovery, which has great academic interest but is certain to have nothing but dangerous consequences for the human society. How, if at all, will you resolve your dilemma?
- Q:-
A particle is in linear simple harmonic motion between two points, A and B, 10 cm apart. Take the direction from A to B as the positive direction and give the signs of velocity, acceleration and force on the particle when it is
(a) at the end A,
(b) at the end B,
(c) at the mid-point of AB going towards A,
(d) at 2 cm away from B going towards A,
(e) at 3 cm away from A going towards B, and
(f) at 4 cm away from B going towards A.
- Q:-
A player throws a ball upwards with an initial speed of 29.4 m s–1. What is the direction of acceleration during the upward motion of the ball? What are the velocity and acceleration of the ball at the highest point of its motion?
Choose the x = 0 m and t = 0 s to be the location and time of the ball at its highest point, vertically downward direction to be the positive direction of x-axis, and give the signs of position, velocity and acceleration of the ball during its upward, and downward motion. To what height does the ball rise and after how long does the ball return to the player’s hands? (Take g = 9.8 m s–2 and neglect air resistance).
- NCERT Chapter