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    NCERT Exemplar Class 11 Physics Solutions Chapter 13 Kinetic Theory

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    NCERT Exemplar Class 11 Physics Solutions Chapter 13 Kinetic Theory

    Edited By Safeer PP | Updated on Aug 09, 2022 12:21 PM IST

    In chapter 13, the Kinetic Theory of Gases is dealt with. In this chapter, the learners will come to know about the gases, their molecular movement, and how these phenomenons can be understood at both macroscopic and microscopic levels. This chapter can seem to be a bit complicated for the students in the beginning. But if one focuses on the terminology, the processes, and their applications step by step alongside NCERT exemplar Class 11 Physics solutions chapter 13 it will become easier to retain. In this chapter, the students are taught about the kinetic theory gases, gas laws of ideal gases, postulates, assumptions, and the non-ideal gas behavior. This chapter and NCERT exemplar Class 11 Physics chapter 13 solutions encapsulate all the topics related to gas molecule movement in a container along with the factors of temperature, pressure, and energy.

    This Story also Contains
    1. NCERT Exemplar Class 11 Physics Solutions Chapter 13-MCQI
    2. NCERT Exemplar Class 11 Physics Solutions Chapter 13-MCQII
    3. NCERT Exemplar Class 11 Physics Solutions Chapter 13-Very Short Answer
    4. NCERT Exemplar Class 11 Physics Solutions Chapter 13-Short Answer
    5. NCERT Exemplar Class 11 Physics Solutions Chapter 13-Long Answer
    6. Topics in NCERT Exemplar Class 11 Physics Solutions Chapter 13
    7. What will The Student Learn from NCERT Exemplar Class 11 Physics Solutions Chapter 13 Kinetic Theory Of Gases?
    8. NCERT Exemplar Class 11 Physics Solutions Chapter-Wise
    9. Important Topics To Cover From NCERT Exemplar Class 11 Physics Solutions Chapter 13 Kinetic Theory Of Gases
    10. NCERT Exemplar Class 11 Solutions

    NCERT Exemplar Class 11 Physics Solutions Chapter 13-MCQI

    Question:13.1

    A cubic vessel (with face horizontal + vertical) contains an ideal gas at NTP. The vessel is being carried by a rocket which is moving at a speed of 500 ms^{-1} in the vertical direction. The pressure of the gas inside the vessel as observed by us on the ground
    (a) remains the same because 500 ms^{-1} is very much smaller than vrms of the gas.
    (b) remains the same because the motion of the vessel as a whole does not affect the relative motion of the gas molecules and the walls.
    (c) will increase by a factor equal to\frac{\left ( v^{2}rms +\left ( 500 \right )^{2} \right )}{v^{2}rms} where vrms was the original mean square velocity of the gas.
    (d) will be different on the top wall and bottom wall of the vessel.

    Answer:

    The answer is the option (b)
    In the rocket, the relative velocity of molecules does not change with respect to walls of a container as the mass of a molecule when compared to that of the whole system is negligible. So, the whole gas system moves as a single unit. The rocket moves at a constant speed which makes the acceleration zero, and hence the pressure remains the same as observed by us inside the gas vessel.

    Question:13.2

    1 mole of an ideal gas is contained in a cubical volume V, ABCDEFGH at 300 K (figure). One face of the cube (EFGH) is made up of a material which totally absorbs any gas molecule incident on it. At any given time,
    (a) the pressure on EFGH would be zero
    (b) the pressure on all the faces will be equal
    (c) the pressure of EFGH would be double the pressure on ABCD
    (d) the pressure on EFGH would be half that on ABCD

    Answer:

    The answer is the option (d)
    The rate of transfer of momentum on the wall exerts pressure on the wall due to the force exerted by the molecules. Elastic collision takes place, and hence the molecules bounce back. The walls absorb the magnitude of momentum transferred to it by each molecule which is equal to 2mv. So, the net rate of change of momentum is mv, and the pressure of EFGH is halved as compared to ABCD.

    Question:13.3

    Boyle’s law is applicable for an
    (a) adiabatic process
    (b) isothermal process
    (c) isobaric process
    (d) isochoric process

    Answer:

    The answer is the option (b)
    At constant temperature, Boyle’s law is applicable
    PV = nRT (n, R, T are constant in this case)
    So, PV = constant and P\propto \frac{1}{V}

    Question:13.4

    A cylinder containing an ideal gas is in a vertical position and has a piston of mass M that is able to move up or down without friction ( figure). If the temperature is increased
    (a) both P and V of the gas will change
    (b) the only P will increase according to Charles’ law
    (c) V will change but not P
    (d) P will change but not V

    Answer:

    The answer is the option (c)
    In case of an ideal gas, the pressure remains the same throughout irrespective of the initial or final position. Here, P= \frac{Mg}{A} and no friction acts on piston and the cylinder
    We have, PV = nRT
    Since, P, n and R, are constant here, V\propto T. So, the temperature increases as volume increases and vice-versa at the same time, when pressure remains constant.

    Question:13.5

    Volume versus temperature graphs for a given mass of an ideal gas is shown in the figure. At two different values of constant pressure. What can be inferred about the relation between P_{1} and P_{2}?

    (a)P_{1}>P_{2}
    (b) P_{1}=P_{2}
    (c) P_{1}<P_{2}
    (d) Data is insufficient

    Answer:

    The answer is the option (a)

    Pressure and quantity of gas are constant here.
    From PV = nRT, we can say that, V \propto T
    So, \frac{V_{1}}{T_{1}} = constant (slope = constant)
    V = \frac{nRT}{P}
    \frac{dV}{dT}=n \frac{R}{P}
    hence its value decreases with pressure, so, \frac{dV}{dT}\propto \frac{1}{P}
    so, slope of P_{1} is smaller than P_{2} and hence a is the right option.

    Question:13.6

    1 mole of H_{2} gas is contained in a box of volume V = 1.00 m^3 at T = 300 K. The gas is heated to a temperature of T= 3000 K and the gas gets converted to a gas of hydrogen atoms. The final pressure would be (considering all gases to be ideal)
    (a) same as the pressure initially
    (b) 2 times the pressure initially
    (c) 10 times the pressure initially
    (d) 20 times the pressure initially

    Answer:

    The answer is the option (d)
    The transfer of the rate of change of momentum by the particles to the walls is what causes exertion of pressure by the gas. As temperature rises and hydrogen molecules break into atoms, the mass of the particle becomes halved, and the number of particles becomes double. Since the velocity is strictly dependent on the temperature, at the same temperature, i.e. at 300K, the velocity of particles of H_{2} as well as H remains the same. Also, the pressure does not change as the atomic form changes since the rate of transfer of momentum is also same in the case of H_{2} as well as H.

    Question:13.7

    A vessel of volume V contains a mixture of 1 mole of hydrogen and 1 mole of oxygen (both considered as ideal). Let f_{1}(v)dv denotes the fraction of molecules with speed between v and (v + dv) with f_2(v)dv, similarly for oxygen. Then,
    (a) f_{1}(v) + f_{2}(v) = f (v) obeys the Maxwell’s distribution law
    (b) f_{1}(v) , f_{2}(v) will obey the Maxwell’s distribution law separately
    (c) neither f_{1}(v) nor f_{2}(v)will obey Maxwell’s distribution law
    (d) f_{1}(v) and f_{2}(v)will be the same

    Answer:

    The correct answer is the option (b)
    F1(v): speed of n molecules = (v+dv)
    The number of molecules remains the same, which is one mole for f_{1}(v) and f_{2}(v). since the molecules differ in mass, they will also differ in speed, and hence both gases will separately obey Maxwell’s distribution.

    Question:13.8

    An inflated rubber balloon contains one mole of an ideal gas, has a pressure P. volume V and temperature T. If the temperature rises to 1.1 T, and the volume is increased to 1.05 V, the final pressure will be
    (a) 1.1 P
    (b) P
    (c) less than P
    (d) between P and 1.1

    Answer:

    The correct answer is the option (d)
    According to ideal gas equation, PV = nRT
    Here, \frac{PV}{T} = constant.
    \frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}
    So, P_{2}= \frac{1.1}{1.05}(p)=(1.0476)(p)
    So, the value of P_{2} is between p and 1.1 p and hence, option d is correct.

    NCERT Exemplar Class 11 Physics Solutions Chapter 13-MCQII

    Question:13.9

    ABCDEFGH is a hollow cube made of an insulator (figure) face ABCD has a positive charge on it. Inside the cube, we have ionised hydrogen.
    The usual kinetic theory expression for pressure
    (a) will be valid
    (b) will not be valid, since the ions would experience forces other than due to collisions with the walls
    (c) will not be valid, since collisions with walls would not be elastic
    (d) will not be valid because isotropy is lost

    Answer:

    The correct answer is the option (b) and (d)
    The wall of ABCD is positively charged. The presence of hydrogen ions and the charge on the walls lead to an electrostatic force which acts in contrast to the collision and make the kinetic theory of gases invalid in this case. The isotropy also gets lost given the presence of hydrogen ions instead of hydrogen molecules. Hence options b and d are correct.

    Question:13.10

    Diatomic molecules like hydrogen have energies due to both translational as well as rotational motion. From the equation in kinetic theory PV = \frac{2}{3}E, E is
    (a) the total energy per unit volume
    (b) only the translational part of energy because rotational energy is very small compared to the translational energy
    (c) only the translational part of the energy because during collisions with the wall pressure relates to change in linear momentum
    (d) the translational part of the energy because rotational energies of molecules can be of either sign and its average over all the molecules is zero

    Answer:

    The correct answer is the option (c)
    As per the kinetic theory of gases, the perpendicular forces exerted by the molecules on the walls while in motion are only responsible for the pressure exerted due to the gas molecules. So, for molecules striking at angles other than 90 degrees, no pressure will be exerted. Hence, in this case, only translational motion change leads to pressure on the wall. So, PV = \frac{2}{3}E, which represents the translational motion part and hence c is the correct option.

    Question:13.11

    In a diatomic molecule, the rotational energy at a given temperature
    (a) obeys Maxwell’s distribution
    (b) have the same value for all molecules
    (c) equals the translational kinetic energy for each molecule
    (d) is \left (\frac{2}{3} \right ){rd} the translational kinetic energy for each molecule

    Answer:

    The correct answer is the option (a) and (d)
    If we assume a diatomic molecule along the z-axis, its energy along that axis will be zero. The total energy of a diatomic molecule can be expressed as: E = \frac{1}{2} mvx^{2} + \frac{1}{2} mvy^{2} + \frac{1}{2} mvz^{2} + \frac{1}{2} IxWx^{2} + \frac{1}{2} IxWy^{2}
    The number of independent terms in the expression =5. The above expression obeys Maxwell’s distribution as their velocities can be predicted with Maxwell’s findings. In this case, for each molecule 3 translational and 2 rotational energies are associated. So, at any temperature, rotational energy = \left (\frac{2}{3} \right ){rd} translational KE

    Question:13.12

    Which of the following diagrams (figure) depicts ideal gas behaviour?

    Answer:

    The correct answer is the option (a) and (c)
    a) PV = nRT
    where R and T are constant.
    Since P is constant, V \propto Tas given in the graph of option a.
    Hence, option is correct.
    c) In case where, V = constant
    P \propto T, which is a straight line. Hence option c is correct.

    Question:10.13

    When an ideal gas is compressed adiabatically, its temperature rises the molecules on the average have more kinetic energy than before. The kinetic energy increases,
    (a) because of collisions with moving parts of the wall only
    (b) because of collisions with the entire wall
    (c) because the molecules get accelerated in their motion inside the volume
    (d) because of the redistribution of energy amongst the molecules

    Answer:

    The correct answer is the option (a)
    The number of collisions per second between the molecules and walls increases and the mean free path becomes smaller as an ideal gas compressed. This, in turn, increases the temperature of the gas, which increased the overall kinetic energy of the gas molecules as KE depends on the temperature.

    NCERT Exemplar Class 11 Physics Solutions Chapter 13-Very Short Answer

    Question:13.14

    Calculate the number of atoms in 39.4 g gold. The molar mass of gold is 197 g mole-1.

    Answer:

    Number of atoms on 197 g gold = 6.023 \times 10^{23}
    Number of atoms in 1g of gold, n
    =6.023 \times \frac{10^{23}}{197}
    Hence, number of atoms in 34g of gold =
    34 \times \left (6.023 \times \frac{10^{23}}{197} \right ) = 1.2 \times 10^{23} atoms

    Question:13.15

    The volume of a given mass of a gas at 27^{\circ}C, 1 atm is 100 cc. What will be its volume at 327^{\circ}C?

    Answer:

    Let, P_{1} = 1 atm, P_{2} = 1 atm, V_{1} = 100cc, V_{2} = ?T_{1} = 300K, T_{2} = 600K
    From the ideal gas equation, we have
    \frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}
    Hence,
    V_{2} = \frac{100 \times 600}{300}= 200 cc

    Question:13.16

    The molecules of a given mass a gas have a root mean square speeds of 100 ms-1 at and 1.00 atmospheric pressure. What will be the root mean square speeds of the molecules of the gas at and 2.0 atmospheric pressure?

    Answer:

    V_{1} rms = 100m/s, T_{1}= 300K, T_{2} = 400K, V_{2} rms =?
    We know that,
    V_{rms}=\sqrt{\frac{3RT}{m}}
    V_{rms}\propto \sqrt{T}
    Hence,
    \frac{V_{1}rms}{V_{2}rms}=\sqrt{\frac{T_{1}}{T_{2}}}
    \frac{100}{V_{2}rms}=\sqrt{\frac{300}{400}}
    V_{2}rms = 115.4 \frac{m}{s}

    Question:13.17

    Two molecules of a gas have speeds of 9 \times 10^{16} ms^{-1} and 1 \times 10^{16} ms^{-1} respectively. What is the root mean square speed of these molecules?

    Answer:

    V_{rms}=\sqrt{\frac{V1^{2}+V2^{2}...........Vn^{2}}{n}}
    We know that, V_{1}=9 \times 10^{16} ms^{-1} and V_{2}=1 \times 10^{16} ms^{-1}
    Hence, V_{rms}=\sqrt{\frac{\left ( 9 \times 10^{6} \right )^{2}+\left ( 1\times 10^{6} \right )^{2}}{2}}
    V_{rms}=10^{6}\times \sqrt{\frac{82}{2}}=6.4 \times 10^{6} m/s

    Question:13.18

    A gas mixture consists of 2.0 moIes of oxygen and 4.0 moles of neon at temperature T. Neglecting all vibrational modes, calculate the total internal energy of the system. (Oxygen has two rotational modes.)

    Answer:

    Degree of freedom is necessary to be known if we need to find the total energy of a particular molecule.
    An oxygen molecule has 2 atoms, so the degree of freedom = 2 rotational + 3 translational = 5
    Total internal energy for 2 moles of oxygen = 2\times \frac{5}{2} RT = 5RT
    Neon gas has a degree of freedom equal to 3, since it is a monoatomic gas-only translational degree of freedom is present. So, the total internal energy = \frac{3}{2} RT per mole.
    For 4 moles of neon gas, total internal energy = 4\times \frac{3}{2} RT = 6RT
    Hence total internal energy of 2 moles oxygen + 4 moles Neon = 5 + 6 = 11 RT

    Question:13.19

    Calculate the ratio of the mean free paths of the molecules of two gases having molecular diameters 1 A and 2 A. The gases may be considered under identical conditions of temperature, pressure and volume.

    Answer:

    Mean free path formula, \lambda =\frac{1}{\sqrt{2}}\pi d^{2}n
    Here, n = number of molecules per unit volume. In our case, n will be constant.
    Here, d = diameter of molecules.
    Now, \lambda \propto \frac{1}{d^{2}}
    So,\frac{\lambda _{1}}{\lambda _{2}}=\frac{d_2^{2}}{d_1^{2}}=\frac{4}{1}
    Hence, \lambda _{1}:\lambda _{2}=4:1

    NCERT Exemplar Class 11 Physics Solutions Chapter 13-Short Answer

    Question:13.20

    The container shown in Figure has two chambers, separated by a partition, of volumes V1 = 2.0 litre and V2 = 3.0 litre. The chambers contain \mu _{1}=4.0 and \mu _{2}=5.0 moles of a gas at pressures p_{1} = 1.00 atm and p_{2} = 2.00 atm. Calculate the pressure after the partition is removed and the mixture attains equilibrium.

    V_{1}V_{2}
    \mu_{1}\mu_{2}
    p_{1}p_{2}

    Answer:

    According to the ideal gas situation, we have PV = nRT
    For chamber 1 and chamber 2,
    P_{1}V_{1} = n_{1}RT_{1} and P_{2}V_{2} = n_{2}RT_{2}
    P_{1}=1atm, P_{2}=2 atm, V_{1}=2L, V_{2}=3L
    T_{1}=T, T_{2}=T, n_{1}=4, n_{2}=5
    When we remove the partition, n = n_{1}+n_{2} and V = V_{1} + V_{2}
    According to kinetic theory, Translational gas kinetic energy=PV = \frac{2}{3}. E per mole
    So, the translational KE for both cases will be,
    P_{1}V_{1} = \frac{2}{3}. n_{1} E_{1} and P_{2}V_{2} = \frac{2}{3}. n_{2} E_{2}
    Adding, P_{1}V_{1}+P_{2}V_{2} =\frac{2}{3}. n_{1} E_{1}+ \frac{2}{3}. n_{2} E_{2}
    n_{1}E_{1}+n_{2}E_{2}=\frac{3}{2}\left ( P_{1}V_{1}+P_{2} V_{2} \right )
    hence by combining the results we get,
    P(V_{1}+V_{2}) = \frac{2}{3}\left [ \frac{3}{2}(P_{1}V_{1}+P_{2}V_{2}) \right ]
    P = \frac{P_1V_1 + P_{2}V_2}{V_{1} + V_{2}} = 1.6 atm

    Question:13.21

    A gas mixture consists of molecules of A, B and C with masses. Rank the three types of molecules in decreasing order of (a) average KE, (b) RMS speeds.

    Answer:

    Vav =\sqrt{\frac{8 K_bT}{\pi m}} = \sqrt{\frac{8RT}{\pi m}} = \sqrt{\frac{8PV}{\pi m}}
    as the temperature and pressure are equal,
    Vav\propto \frac{1}{\sqrt{m}}
    hence, Vc > Vb > Va
    now KE \propto V^2 and KE \propto m
    hence KE in decreasing order will be,
    KE c > KE b > KE a
    b)
    now, V_{rms} =\sqrt{\frac{3K_bT}{m}}
    since it is given that, m(a) > m(b) > m(c)
    V_ {rms} (c) > V_{ rms} (b) > V_{ rms} (a)

    Question:13.22

    We have 0.5 g of hydrogen gas in a cubic chamber of size 3cm kept at NTP. The gas in the chamber is compressed, keeping the temperature constant till a final pressure of 100 atm. Is one justified in assuming the ideal gas law, in the final state?
    (Hydrogen molecules can be considered as spheres of radius 1 A ).

    Answer:

    The volume of a sphere = the volume of one molecule
    =\frac{4}{3}\pi r^{3}
    Here r= 10^{-10} m
    So, volume = 4 \times 1.05 \times 10^{-30} = 4.2 \times 10^{-30} m^{3}
    For 0.5 g of hydrogen gas, no of moles = \frac{0.5}{2}=0.25 moles
    Hence the volume of hydrogen molecules = 0.25 \times 6.023 \times 10^{23} \times 4.2 \times 10^{-30} = 6.3 \times 10^{-7} m^{3}
    For a constant temperature, ideal gas à Pi Vi = Pf Vf
    Vf =\frac{P_i V_i}{P_f} = \frac{1}{100}\times (3\times 10^{-2})^{3}
    Hence, V_f = 2.7 \times 10^{-7} m^{3}
    The gas here will not obey the ideal gas behaviour as the kinetic energy of the molecules will not interact with each other due to compression.

    Question:13.23

    When air is pumped into a cycle tyre the volume and pressure of the air in the tyre, both are increased. What about Boyle’s law in this case?

    Answer:

    The volume of a gas is inversely proportional to the pressure of the gas at a constant temperature as stated by Boyle’s law. When air is continuously pumped inside the tyre, the number of moles of gas (i.e. the mass of the gas) increases. Since Boyle’s law is only applicable for a constant mass of gas, it will not hold true in this case.

    Question:13.24

    A balloon has 5.0 mole of helium at 7^{\circ}C. Calculate
    (a) the number of atoms of helium in the balloon.
    (b) the total internal energy of the system.

    Answer:

    Helium gas, n = 5 moles and T = 280 K
    a) number of atoms of He = 5 \times 6.023 \times 10^{23} = 3.0115 \times 10^{24} atoms of He
    b) the degree of freedom for a helium atom is 3 since it is monoatomic
    so, average KE =\frac{3}{2}K_bT per molecule
    average KE = \frac{3}{2} \times 1.38 \times 10^{-23} \times 80 \times 3.0115 \times 10^{24}
    hence total energy = 1.74 \times 10^{ 4 }J

    Question:13.25

    Calculate the number of degrees of freedom of molecules of hydrogen in 1 cc of hydrogen gas at NTP.

    Answer:

    Degree of freedom of H2 molecule = 3 translational degree of freedom + 2 rotational degree of freedom = 5.
    Number of molecules in 1cc of hydrogen gas at NTP à 22.4 L = 22400 cc H2 gasà it has 6.023 \times 10^{23} molecules
    So, 1cc of H2 gas at STP has,\frac{ 6.023}{22400}\times 10^{23} = 2.688 \times 10^{19} molecules
    Hence the total degree of freedom =5 \times 2.688 \times 10^{19 }= 1.344 \times 10^{20}

    Question:13.26

    An insulated container containing monoatomic gas of molar mass m is moving with a velocity v_{0}. If the container is suddenly stopped, find the change in temperature.

    Answer:

    Since the gas is monoatomic, only translational degree of freedom will be applicable. KE per molecule = \frac{3}{2} R T. when the container which is insulated has suddenly stopped, a transfer of kinetic energy takes place to the gas molecules. Let \Delta T be the slight increase in temperature and n be the number of moles of gas.
    The resultant increase in translational KE can be presented as,
    KE = \frac{3}{2} n. R. \Delta T
    Increased KE due to velocity,
    v_{0} = \frac{1}{2} (m n) v_{0}^{2}
    Hence,
    \frac{mn}{2} v_{0}^{2}=\frac{3}{2} n. R. \Delta T
    So,
    \Delta T = \frac{m n v_{0}^{2} \times 2}{2 \times 3 \times n R} =\frac{ m v_{0}^{2}}{3R}

    NCERT Exemplar Class 11 Physics Solutions Chapter 13-Long Answer

    Question:13.27

    Explain why
    (a) there is no atmosphere on the moon.
    (b) there is a fall in temperature with altitude.

    Answer:

    a) The acceleration on the moon due to gravity is one-sixth that of Earth
    escape velocity on the moon can be written as: V es = \sqrt{2gR} = 2.38 km/s
    mass of hydrogen, m = 1.67 x 10-24 kg
    v (rms) =\sqrt{ \frac{3 Kb T}{m}} = \sqrt{\frac{3 \times 1.38 \times 10^{-23}\times 300}{1.67 \times 10^{-24}}} = 2.72 km/s
    v(RMS) is larger than the escape velocity as the gravitational force is small. The sun-moon distance is fairly equal to that of the earth. The energy of the sun reaches in a larger intensity to the moon due to its lower density of the atmosphere. So, the molecules of sunlight have a larger RMS speed and hence can escape out fairly easily. Hence, the moon has lost its atmosphere over time.
    b) The kinetic energy of the air molecule gives the characteristic temperature to the atmosphere. At greater heights, atmospheric pressure is low, which makes air molecules rise higher, which in turn leads to an increase in their potential energy and decrease in their kinetic energy. Since the overall kinetic energy decreases, the temperature also decreases. Also, the lower atmospheric pressure causes expansion of the gas, which also provide a cooling effect.

    Question:13.28

    Consider an ideal gas with the following distribution of speeds.

    Speed% of molecules
    20010
    40020
    60040
    80020
    100010

    (i) Calculate V rms and hence T. (m = 3.0 \times 10^{-26} kg)
    (ii) If all the molecules with speed 1000 m/s escape from the system, calculate new Vrms and hence T.

    Answer:

    (i)
    \\v^{2} (rms) = \frac{10 \times (200)^{2} + 20(400)^{2} + 40(600)^{2} + 20(800)^{2} + 10(1000)^{2}}{10+20+40+20+10}\\=4.08\times10^{5}

    \frac{1}{2}m v^{2} (rms)=\frac{3}{2} Kb T
    T = \frac{m v^{2} (rms)}{3 Kb} = \frac{3 \times 10^{-26} \times 10^{5} \times 4.08}{3\times 1.38\times 10^{-23}} = 296K
    (ii)
    Molecules escape at rate of 1000m/s, we can calculate the v2 (rms) as follows
    v^{2} (rms) = \frac{10 \times (200)^{2} + 20(400)^{2} + 40(600)^{2} + 20(800)^{2}}{10+20+40+20}
    v^2 (rms) = \frac{105[1\times 4 + 2 \times 16 + 4 \times 36 + 2 \times 64]}{90}
    v(rms) = \frac{100}{3}\sqrt{ 308} \sim 585 m/s
    T = \frac{1}{3} (\frac{m v^{2} (rms)}{Kb}) = \frac{3 \times 10-26 \times (585)2}{3 \times 1.28 \times 10^{-23}} = 284.04 K

    Question:13.29

    Ten small planes are flying at a speed of 150 km/h in total darkness in an air space that is 20 × 20 × 1.5 km3 in volume. You are in one of the planes, flying at random within this space with no way of knowing where the other planes are. On the average about how long a time will elapse between near collision with your plane. Assume for this rough computation that a safety region around the plane can be approximated by a sphere of radius 10m.

    Answer:

    The motion of molecules in a confined space can be considered a plane. Mean free path λ can be considered as distance travelled by molecules between two planes to avoid any collision, time = distance/speed
    =\frac{\lambda }{v}=\frac{1}{\sqrt{2n}\pi d^{2}.v}
    Number of particles per unit volume V = N/volume
    N =\frac{10}{20\times 20 \times 1.5} km^{3} = \frac{0.0167}{km^{3}}
    d = 2 \times 10 = 20m = 150 km/hr
    so, time
    =\frac{1}{\sqrt{2n}\pi d^{2}. V} \\= \frac{1}{1.414 \times 0.0167\times 3.14\times 20 \times 20\times 10^{-6}\times 150 }\\\\= 225 hours.

    Question:13.3

    A box of 1.00m3 is filled with nitrogen at 1.50 atm at 300K. The box has a hole of an area 0.010 mm2 . How much time is required for the pressure to reduce by 0.10 atm, if the pressure outside is 1 atm.

    Answer:

    Let v1 = volume of box = 1 m3
    Let p1 be the initial pressure = 1.5 atm
    Let p2’ be the final pressure =1.5 - 0.1 = 1.4 atm
    t1 = initial temperature = 300K, t2 = final temperature = 300 K
    let a be the area of hole = 10 ^{-8} m^{2}
    The pressure difference initially between the tyre and atmosphere, ΔP = (1.5 -1) = 0.5 atm
    Mass of N2 molecule
    =\frac{0.028}{6.023 \times 10^{23}} = 46.5 \times 10^{-27} kg
    Kb = 1.38 \times 10^{-23}
    Now since
    |v_{ix}|= | v_{iy} |= \left |v_{iz} \right |
    Hence, v^2_{rms} = 3v_{ix}^2
    For a gas molecule, Kinetic energy = \frac{3}{2}KbT
    Now,
    \frac{1}{2}mv^{2} = 3v_{ix}^{2}
    v_{ix }= \sqrt{\frac{Kb T}{m}} (A)
    The N2 molecules striking the wall in time \Delta t outwards
    \frac{1}{2} P_{n1} \sqrt{\frac{Kb T}{m}} (A) outwards.
    The temperature of the air and inside the box is equal to T.
    The N2 molecules striking the wall in time \Delta t inwards = \frac{1}{2} P_{n2} \sqrt{\frac{Kb T}{m}} (A) inwards.
    Here, Pn2 = density of air molecules
    So, we can calculate the net number of molecules going outward as:
    \frac{1}{2} \left [ P_{n1}-P_{n2} \right ] \sqrt{\frac{K_b T}{m}}(A)\Delta t
    We know that,PV = nRT, n= \frac{PV}{RT}
    Pn1 = total number of molecules in the box/ volume of the box
    =\frac{nNa}{V}= \frac{PNa}{RT} per unit volume.
    After a time, T, the pressure reduces to (1.5-0.1) = 1.4 atm = P’2
    The new density of the Na molecules =P^{'}n1 = \frac{P2^{'}Na}{RT} per unit volume.
    The number of molecules going out from volume V = (P1 - P^{'}1) v = \frac{P1Na}{RT} (V) -\frac{ P2Na}{RT}(V)
    = \frac{Na V}{RT} [P1 - P^{'}2] where P’2 = final pressure of the box
    The total number of molecules which exit the hole in time t’ is:
    \frac{1}{2}[Pn1 - Pn2] \sqrt{\frac{Kb T}{m t^{'}}} (A)
    Pn1 - Pn2 = \frac{P1Na}{RT} -\frac{ P2Na}{RT}= \frac{Na}{RT} [P1-P2]
    The net number of molecules exiting in time t' is
    \frac{Na}{2RT} [P1 - P2] \sqrt{\frac{Kb T}{m t^{'}}} (A)
    Now we can write from the above results,
    t^{'} = 2\frac{(P_1 - P^{'}_2)}{(P_1-P_2)}. \left ( \frac{V}{a} \right ).\sqrt{ \frac{m}{Kb T}}
    t^{'} = \frac{2[1.5-1.4]}{ [1.5-1]}.\frac{1}{10^{-8}}. \sqrt{\frac{46.5 \times 10^{-27}}{1.38 \times 10^{-23}\times 300} }
    t^{'}= 0.4 \times 10^{8}\sqrt{ \frac{775 \times 10^{-6}}{69}}
    t^{'} = 0.4 \times 10^{5} \times 3.35 = 1.34 \times 10^{5}seconds.

    Question:13.31

    Consider a rectangular block of wood moving with a velocity v0 in gas at temperature T and mass density \rho . Assume the velocity is along the x-axis and the area of cross-section of the block perpendicular to v_{0} is A. Show that the drag force on the block is 4\rho Av_{0}\sqrt{\frac{KT}{m}}, where m is the mass of the gas molecule.

    Answer:

    We assume ‘p’ to be the number of molecules per unit volume. Hence ‘p’ is the per unit volume molecular density. Let v be the velocity of gas molecules.
    The molecules of the gas strike the front face and the back face of the box when it moves. The back face relative velocity = (v-v_0)
    Change in momentum on the front face of the box = 2m(v+v_{0}) and the change in momentum on the back face of the box = 2m(v-v _0).
    Nf =Total number of molecules striking the box’s front face = \frac{1}{2}[A.(v+v_{0}) \Delta t] p
    Nf = \frac{1}{2}(v+v_{0}) A.p. \Delta t
    Nb = Total number of molecules striking the box’s back face
    =\frac{1}{2}[A.(v+v_{0}) \Delta t] p
    Nb
    =\frac{1}{2}(v+v_{0}) A.p. \Delta t
    Total change in momentum at the front face can be calculated s below:
    Pf = 2m(v+v_{0}) \times Nf = 2m(v+v_{0}) \times \frac{1}{2}(v+v_{o}) A.p. \Delta t
    Pf = - m(v+v_0)2 A.p. \Delta t in the backward direction.
    Force on front end = Ff = Pf = - m(v+v_0)2 A. p in the backward direction
    In the same way, the force on the back face, Fb = + m(v-v_{0})2 A. p
    So, the net force can be calculated as:
    - m(v+v_{0})^{2} A. p+ m(v-v_{0})^{2} A. p
    It can be expanded and simplified as,
    = -m A p 4v.v_{0}
    The magnitude of dragging force= 4m v.v _{0}.A.p
    Kinetic energy for a molecule of gas can be calculated as:
    KE = \frac{1}{2}mv^{2} = \frac{3}{2} Kb T
    Hence, v = \sqrt{\frac{Kb T}{m}}
    So, we can write the drag force as 4m.A. p. v_o \sqrt{\frac{Kb T}{m}}

    NCERT questions are one of the most crucial ones if one wants to score high in their exams and also want to learn better for entrance exams. We have solved all the questions given in the main exercise as per the CBSE pattern and marking scheme. This will help the student learn better keeping in mind how to answer for better scoring. Our NCERT exemplar Class 11 Physics solutions chapter 13 pdf download are also downloadable so that one can access them offline in the PDF format. Simply, click on the given link and it is done.

    Also, Read NCERT Solution subject wise -

    Also, check NCERT Notes subject wise -

    Topics in NCERT Exemplar Class 11 Physics Solutions Chapter 13

    • 1. Introduction
    • 2. Molecular nature of matter
    • 3. Behaviour of gases
    • 4. Kinetic theory of ideal gases
    • 5. Law of equipartition of energy
    • 6. Specific heat capacity
    • 7. Mean free path

    What will The Student Learn from NCERT Exemplar Class 11 Physics Solutions Chapter 13 Kinetic Theory Of Gases?

    Kinetic theory of gases is the chapter that derives all its topics from the idea that in the gas, the molecules are smaller than the distance between the gas molecules. They move around the container and collide with each other on the walls. Now, this movement of the molecules with the effect of temperature, pressure, and other factors will help in describing the physical properties of the gas. The learner will learn about kinetic theory, its assumptions, and postulates. There are around 9 assumptions like the gas is made of particles, particles have zero volume, there is no interaction between particles, etc.

    Students will also learn about the ideal gas’s gas laws like Avogadro’s law, Amonton’s law, Boyles’ law, Charles law, and Graham's law of diffusion. At the end of the chapter and NCERT exemplar Class 11 Physics chapter 13 solutions the learner will come across the Maxwell and Boltzmann energy and velocity distribution law, which has a lot of significance from an exam point of view.

    NCERT Exemplar Class 11 Physics Solutions Chapter-Wise


    Important Topics To Cover From NCERT Exemplar Class 11 Physics Solutions Chapter 13 Kinetic Theory Of Gases

    · For a better understanding of the kinetic theory of gases, the first thing that one should pay attention to is the gas laws. 5 gas laws explain how an ideal gas behaves when one factor is kept constant. The formulas of these laws will help in solving numerical based questions in the exam.

    · Secondly, a student should pay attention to non-ideal gas behaviour. In this topic, the real gases are discussed who do not follow the ideal gas behaviour under low pressure and high-temperature conditions as per NCERT exemplar Class 11 Physics solutions chapter 13.

    · Lastly, learners should pay attention to the Maxwell and Boltzmann law of velocity and energy distribution. In this, the root means square and average velocity is discussed, of the gaseous particles which act under proportional kinetic energy and temperature.

    NCERT Exemplar Class 11 Solutions

    Check Class 11 Physics Chapter-wise Solutions

    Chapter 1Physical world
    Chapter 2Units and Measurement
    Chapter 3Motion in a straight line
    Chapter 4Motion in a Plane
    Chapter 5Laws of Motion
    Chapter 6Work, Energy and Power
    Chapter 7System of Particles and Rotational motion
    Chapter 8Gravitation
    Chapter 9Mechanical Properties of Solids
    Chapter 10Mechanical Properties of Fluids
    Chapter 11Thermal Properties of Matter
    Chapter 12Thermodynamics
    Chapter 13Kinetic Theory
    Chapter 14Oscillations
    Chapter 15Waves

    Also Check NCERT Books and NCERT Syllabus here

    Frequently Asked Question (FAQs)

    1. What are the benefits of these solutions?

    Our solutions are detailed and created most exhaustively, and this will help in understanding the topic, solving questions, and is also learning how to answer questions in the exam.

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    Radio Jockey is an exciting, promising career and a great challenge for music lovers. If you are really interested in a career as radio jockey, then it is very important for an RJ to have an automatic, fun, and friendly personality. If you want to get a job done in this field, a strong command of the language and a good voice are always good things. Apart from this, in order to be a good radio jockey, you will also listen to good radio jockeys so that you can understand their style and later make your own by practicing.

    A career as radio jockey has a lot to offer to deserving candidates. If you want to know more about a career as radio jockey, and how to become a radio jockey then continue reading the article.

    3 Jobs Available
    Social Media Manager

    A career as social media manager involves implementing the company’s or brand’s marketing plan across all social media channels. Social media managers help in building or improving a brand’s or a company’s website traffic, build brand awareness, create and implement marketing and brand strategy. Social media managers are key to important social communication as well.

    2 Jobs Available
    Choreographer

    The word “choreography" actually comes from Greek words that mean “dance writing." Individuals who opt for a career as a choreographer create and direct original dances, in addition to developing interpretations of existing dances. A Choreographer dances and utilises his or her creativity in other aspects of dance performance. For example, he or she may work with the music director to select music or collaborate with other famous choreographers to enhance such performance elements as lighting, costume and set design.

    2 Jobs Available
    Talent Director

    Individuals who opt for a career as a talent director are professionals who work in the entertainment industry. He or she is responsible for finding out the right talent through auditions for films, theatre productions, or shows. A talented director possesses strong knowledge of computer software used in filmmaking, CGI and animation. A talent acquisition director keeps himself or herself updated on various technical aspects such as lighting, camera angles and shots. 

    2 Jobs Available
    Copy Writer

    In a career as a copywriter, one has to consult with the client and understand the brief well. A career as a copywriter has a lot to offer to deserving candidates. Several new mediums of advertising are opening therefore making it a lucrative career choice. Students can pursue various copywriter courses such as Journalism, Advertising, Marketing Management. Here, we have discussed how to become a freelance copywriter, copywriter career path, how to become a copywriter in India, and copywriting career outlook. 

    5 Jobs Available
    Journalist

    Careers in journalism are filled with excitement as well as responsibilities. One cannot afford to miss out on the details. As it is the small details that provide insights into a story. Depending on those insights a journalist goes about writing a news article. A journalism career can be stressful at times but if you are someone who is passionate about it then it is the right choice for you. If you want to know more about the media field and journalist career then continue reading this article.

    3 Jobs Available
    Publisher

    For publishing books, newspapers, magazines and digital material, editorial and commercial strategies are set by publishers. Individuals in publishing career paths make choices about the markets their businesses will reach and the type of content that their audience will be served. Individuals in book publisher careers collaborate with editorial staff, designers, authors, and freelance contributors who develop and manage the creation of content.

    3 Jobs Available
    Vlogger

    In a career as a vlogger, one generally works for himself or herself. However, once an individual has gained viewership there are several brands and companies that approach them for paid collaboration. It is one of those fields where an individual can earn well while following his or her passion. Ever since internet cost got reduced the viewership for these types of content has increased on a large scale. Therefore, the career as vlogger has a lot to offer. If you want to know more about the career as vlogger, how to become a vlogger, so on and so forth then continue reading the article. Students can visit Jamia Millia Islamia, Asian College of Journalism, Indian Institute of Mass Communication to pursue journalism degrees.

    3 Jobs Available
    Editor

    Individuals in the editor career path is an unsung hero of the news industry who polishes the language of the news stories provided by stringers, reporters, copywriters and content writers and also news agencies. Individuals who opt for a career as an editor make it more persuasive, concise and clear for readers. In this article, we will discuss the details of the editor's career path such as how to become an editor in India, editor salary in India and editor skills and qualities.

    3 Jobs Available
    Content Writer

    Content writing is meant to speak directly with a particular audience, such as customers, potential customers, investors, employees, or other stakeholders. The main aim of professional content writers is to speak to their targeted audience and if it is not then it is not doing its job. There are numerous kinds of the content present on the website and each is different based on the service or the product it is used for.

    2 Jobs Available
    Reporter

    Individuals who opt for a career as a reporter may often be at work on national holidays and festivities. He or she pitches various story ideas and covers news stories in risky situations. Students can pursue a BMC (Bachelor of Mass Communication), B.M.M. (Bachelor of Mass Media), or MAJMC (MA in Journalism and Mass Communication) to become a reporter. While we sit at home reporters travel to locations to collect information that carries a news value.  

    2 Jobs Available
    Linguist

    Linguistic meaning is related to language or Linguistics which is the study of languages. A career as a linguistic meaning, a profession that is based on the scientific study of language, and it's a very broad field with many specialities. Famous linguists work in academia, researching and teaching different areas of language, such as phonetics (sounds), syntax (word order) and semantics (meaning). 

    Other researchers focus on specialities like computational linguistics, which seeks to better match human and computer language capacities, or applied linguistics, which is concerned with improving language education. Still, others work as language experts for the government, advertising companies, dictionary publishers and various other private enterprises. Some might work from home as freelance linguists. Philologist, phonologist, and dialectician are some of Linguist synonym. Linguists can study French, German, Italian. 

    2 Jobs Available
    Production Manager

    Production Manager Job Description: A Production Manager is responsible for ensuring smooth running of manufacturing processes in an efficient manner. He or she plans and organises production schedules. The role of Production Manager involves estimation, negotiation on budget and timescales with the clients and managers. 

    Resource Links for Online MBA 

    3 Jobs Available
    Product Manager

    A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.  

    3 Jobs Available
    Quality Controller

    A quality controller plays a crucial role in an organisation. He or she is responsible for performing quality checks on manufactured products. He or she identifies the defects in a product and rejects the product. 

    A quality controller records detailed information about products with defects and sends it to the supervisor or plant manager to take necessary actions to improve the production process.

    3 Jobs Available
    Production Engineer

    A career as Production Engineer is crucial in the manufacturing industry. He or she ensures the functionality of production equipment and machinery to improve productivity and minimize production costs in order to drive revenues and increase profitability. 

    2 Jobs Available
    Automation Test Engineer

    An Automation Test Engineer job involves executing automated test scripts. He or she identifies the project’s problems and troubleshoots them. The role involves documenting the defect using management tools. He or she works with the application team in order to resolve any issues arising during the testing process. 

    2 Jobs Available
    Product Designer

    Individuals who opt for a career as product designers are responsible for designing the components and overall product concerning its shape, size, and material used in manufacturing. They are responsible for the aesthetic appearance of the product. A product designer uses his or her creative skills to give a product its final outlook and ensures the functionality of the design. 

    Students can opt for various product design degrees such as B.Des and M.Des to become product designers. Industrial product designer prepares 3D models of designs for approval and discusses them with clients and other colleagues. Individuals who opt for a career as a product designer estimate the total cost involved in designing.

    2 Jobs Available
    R&D Personnel

    A career as R&D Personnel requires researching, planning, and implementing new programs and protocols into their organization and overseeing new products’ development. He or she uses his or her creative abilities to improve the existing products as per the requirements of the target market.

    2 Jobs Available
    Commercial Manager

    A Commercial Manager negotiates, advises and secures information about pricing for commercial contracts. He or she is responsible for developing financial plans in order to maximise the business's profitability.

    2 Jobs Available
    Product Manager

    A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.  

    3 Jobs Available
    ITSM Manager

    ITSM Manager is a professional responsible for heading the ITSM (Information Technology Service Management) or (Information Technology Infrastructure Library) processes. He or she ensures that operation management provides appropriate resource levels for problem resolutions. The ITSM Manager oversees the level of prioritisation for the problems, critical incidents, planned as well as proactive tasks. 

    3 Jobs Available
    Information Security Manager

    Individuals in the information security manager career path involves in overseeing and controlling all aspects of computer security. The IT security manager job description includes planning and carrying out security measures to protect the business data and information from corruption, theft, unauthorised access, and deliberate attack 

    3 Jobs Available
    Computer Programmer

    Careers in computer programming primarily refer to the systematic act of writing code and moreover include wider computer science areas. The word 'programmer' or 'coder' has entered into practice with the growing number of newly self-taught tech enthusiasts. Computer programming careers involve the use of designs created by software developers and engineers and transforming them into commands that can be implemented by computers. These commands result in regular usage of social media sites, word-processing applications and browsers.

    3 Jobs Available
    Computer System Analyst

    Individuals in the computer systems analyst career path study the hardware and applications that are part of an organization's computer systems, as well as how they are used. They collaborate closely with managers and end-users to identify system specifications and business priorities, as well as to assess the efficiency of computer systems and create techniques to boost IT efficiency. Individuals who opt for a career as a computer system analyst support the implementation, modification, and debugging of new systems after they have been installed.

    2 Jobs Available
    Test Manager

    A Test Manager is a professional responsible for planning, coordinating and controlling test activities. He or she develops test processes and strategies to analyse and determine test methods and tools for test activities. The test manager jobs involve documenting tests that have been carried out, analysing and evaluating software quality to determine further recommended procedures. 

    2 Jobs Available
    Azure Developer

    A career as Azure Developer comes with the responsibility of designing and developing cloud-based applications and maintaining software components. He or she possesses an in-depth knowledge of cloud computing and Azure app service. 

    2 Jobs Available
    Deep Learning Engineer

    A Deep Learning Engineer is an IT professional who is responsible for developing and managing data pipelines. He or she is knowledgeable about analyzing and storing data collected from various sources.  A Career as a Deep Learning Engineer needs to help the  data scientists and analysts to create effective data sets.

    2 Jobs Available

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