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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.
Kinetic Theory Question Answers: NCERT Class 11 Physics
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 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.
Download PDF - Chapter 13 Kinetic Theory - Class 11 Physics
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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:- Give the location of the centre of mass of a (i) sphere, (ii) cylinder, (iii) ring, and (iv) cube, each of uniform mass density. Does the centre of mass of a body necessarily lie inside the body?
- Q:-
A shell of mass 0.020 kg is fired by a gun of mass 100 kg. If the muzzle speed of the shell is 80 ms-1, what is the recoil speed of the gun?
- Q:-
A constant force acting on a body of mass 3.0 kg changes its speed from 2.0 m s–1 to 3.5 m s–1 in 25 s. The direction of the motion of the body remains unchanged. What is the magnitude and direction of the force?
- Q:-
One end of a string of length l is connected to a particle of mass m and the other to a small peg on a smooth horizontal table. If the particle moves in a circle with speed v the net force on the particle (directed towards the centre) is:
(i) T, (ii) T - mv2 / l , (iii) T + mv2 / l , (iv) 0
T is the tension in the string. [Choose the correct alternative].
- Q:-
“It is more important to have beauty in the equations of physics than to have them agree with experiments”. The great British physicist P. A. M. Dirac held this view. Criticize this statement. Look out for some equations and results in this book which strike you as beautiful.
- Q:-
Textbooks on science may give you a wrong impression that studying science is dry and all too serious and that scientists are absent-minded introverts who never laugh or grin. This image of science and scientists is patently false. Scientists, like any other group of humans, have their share of humorists, and many have led their lives with a great sense of fun and adventure, even as they seriously pursued their scientific work. Two great physicists of this genre are Gamow and Feynman. You will enjoy reading their books listed in the Bibliography.
- Q:-
The shells of crabs found around a particular coastal location in Japan seem mostly to resemble the legendary face of a Samurai. Given below are two explanations of this observed fact. Which of these strikes you as a scientific explanation ?
(a) A tragic sea accident several centuries ago drowned a young Samurai. As a tribute to his bravery, nature through its inscrutable ways immortalised his face by imprinting it on the crab shells in that area.
(b) After the sea tragedy, fishermen in that area, in a gesture of honour to their dead hero, let free any crab shell caught by them which accidentally had a shape resembling the face of a Samurai. Consequently, the particular shape of the crab shell survived longer and therefore in course of time the shape was genetically propagated. This is an example of evolution by artificial selection.
[Note : This interesting illustration taken from Carl Sagan’s ‘The Cosmos’ highlights the fact that often strange and inexplicable facts which on the first sight appear ‘supernatural’ actually turn out to have simple scientific explanations. Try to think out other examples of this kind].
- 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:-
A stone tied to the end of a string 80 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolutions in 25 s, what is the magnitude and direction of acceleration of the stone?
- Q:-
Estimate the average mass density of a sodium atom assuming its size to be about 2.5 Å. (Use the known values of Avogadro’s number and the atomic mass of sodium). Compare it with the density of sodium in its crystalline phase : 970 kg m–3. Are the two densities of the same order of magnitude ? If so, why ?
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