NCERT Exemplar Class 11 Physics Solutions Chapter 13 Kinetic Theory
Have you ever asked yourself why the scent of perfume can travel a long distance through a room, why air pressure is generated in a balloon or why gases expand when they are heated? Every one of these observations of our daily life is attributed to the Kinetic Theory of Gases that is the basis of NCERT Class 11 Physics Chapter 13: Kinetic Theory of Gases. The chapter enables the students to learn the microscopic behaviour of gas molecules and their connection with such macroscopically observable properties as pressure, temperature, and volume.
This Story also Contains
- NCERT Exemplar Class 11 Physics Solutions Chapter 13: MCQ I
- NCERT Exemplar Class 11 Physics Solutions Chapter 13: MCQII
- NCERT Exemplar Class 11 Physics Solutions Chapter 13: Very Short Answer
- NCERT Exemplar Class 11 Physics Solutions Chapter 13: Short Answer
- NCERT Exemplar Class 11 Physics Solutions Chapter 13: Long Answer
- Advantages of NCERT Exemplar Class 11 Physics Solutions Chapter 13: Kinetic Theory of Gases
- Approach to Solve Exemplar Questions of Chapter 13: Kinetic Theory of Gases
- NCERT Exemplar Class 11 Physics Solutions Chapter-Wise
- NCERT Exemplar Solutions Class 11 Subject-wise
- NCERT Solutions for Class 11 Physics Chapter-wise
The key concepts that NCERT Exemplar Class 11 Physics Solutions Chapter 13 Kinetic Theory of Gases discusses include the topics of molecular motion, kinetic definition of temperature, laws of gases, pressure of an ideal gas, root mean square speed, average and most probable speed and the law of equipartition of energy. These principles are essential in the explanation of the real-world gas behaviour and are usually sought in the CBSE board examination and in competitive exams such as JEE. The NCERT solutions of Exemplar are laid out in step by step format, and thus, numerical problems and questions based on theory would be easy to comprehend and answer appropriately. The NCERT Exemplar Solutions Class 11 Physics Chapter 13 are prepared by expert subject matter professionals and adhere strictly to the current CBSE syllabus and NCERT standards. All solutions emphasise the development of good conceptual understanding, minimisation of typical mistakes and a logical way of solving numerical problems involving gases. The consistent use of such questions will improve the level of critical thinking and increase confidence in using the skills to solve problems. Having a strong focus on real-life applications and systematic explanations, NCERT Exemplar Solutions Class 11 Physics on Kinetic Theory of Gases are an excellent study material for revision and exam preparation.
NCERT Exemplar Class 11 Physics Solutions Chapter 13: MCQ I
NCERT Exemplar Class 11 Physics Solutions Chapter 13: MCQ I are designed to test students' basic understanding of fundamental concepts of the kinetic theory of gases through single-correct objective questions. These MCQs assist in better clarity of concepts, accuracy, and speed, which makes it extremely convenient for fast revision and proper practice of exams.
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 the bottom wall of the vessel.
Answer:
The answer is option (b)In the rocket, the relative velocity of molecules does not change with respect to the 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 that 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 option (d)The rate of transfer of momentum on the wall exerts pressure on the wall due to the force exerted by the molecules. An elastic collision takes place, and hence, the molecules bounce back. The walls absorb the magnitude of momentum transferred to them 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 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 a 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 option (c)In the 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)The 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 option (d)The transfer of the rate of change of momentum by the particles to the walls is what causes the exertion of pressure by the gas. As the temperature rises and hydrogen molecules break into atoms, the mass of the particle is 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 the 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 $f_{1}(v)dv$ denote the fraction of molecules with speed between v and (v + dv) with $f_2(v)dv$, and 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 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 and 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 option (d)According to the 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
The NCERT Exemplar Class 11 Physics Solutions Chapter 13: MCQ II cover multiple-correct and reasoning-based objective tests that test greater conceptual knowledge of gas behaviour and the motion of the molecules. These questions also make students analyse statements carefully, eliminate possible misconceptions, and gain confidence in using higher-order MCQs in examinations.
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 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 makes 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 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 the 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 options (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 options (a) and (c)a) $PV = nRT$
where R and T are constant.
Since P is constant, $V \propto T$as given in the graph of option a.
Hence, the option is correct.
c) In the 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, and 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 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 is compressed. This, in turn, increases the temperature of the gas, which increases 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
Exemplar Class 11 Physics Solutions Chapter 13: Very Short Answer is filled with sharp and precise answers to the basic ideas regarding the kinetic theory of gases. These solutions bring the students to memorise the definitions and the formulas fast, which is why it is effective in the area of quick revision of the material and the pre-exam preparation, which is objective.
Question:13.14
Calculate the number of atoms in 39.4 g of 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, the 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
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
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
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
Answer:
The 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 the 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
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
NCERT Exemplar Class 11 Physics Chapter 13: Short Answer provides concise and clear answers to theory-based and numerical questions of the kinetic theory of gases. These solutions assist students to firm up on conceptual knowledge, exercise good presentation of answers and prepare well to score high in exams.
Question:13.20
| $V_{1}$ | $V_{2}$ |
| $\mu_{1}$ | $\mu_{2}$ |
| $p_{1}$ | $p_{2}$ |
Answer:
According to the ideal gas situation, we have PV = nRTFor 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
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
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
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 Ka) 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 $KE =\frac{3}{2}K_b$T 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 degrees of freedom + 2 rotational degrees 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
Answer:
Since the gas is monoatomic, only the 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
NCERT Exemplar Class 11 Physics Solutions Chapter 13: Long Answer have comprehensive step-wise explanation of in-depth questions concerning the kinetic theory of gases. Such solutions assist the students to possess good conceptual clarity, logical arguments and correct derivative abilities, and these are indispensable in delivering effectively in descriptive exams.
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 Earthescape 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 a larger intensity due to the lower density of the atmosphere. So, the molecules of sunlight have a larger RMS speed and hence can escape 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 a 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 provides a cooling effect.
Question:13.28
Consider an ideal gas with the following distribution of speeds.
| Speed | % of molecules |
| 200 | 10 |
| 400 | 20 |
| 600 | 40 |
| 800 | 20 |
| 1000 | 10 |
(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 a 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
Answer:
The motion of molecules in a confined space can be considered a plane. Mean free path λ can be considered as the 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
Answer:
Let v1 = volume of box = 1 m3Let 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 an 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
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 as 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 Exemplar Class 11 Physics Solutions Chapter 13 Kinetic Theory of Gases: Important Concepts and Formulas
The Important Concepts and Formulas part of NCERT Exemplar Class 11 Physics Chapter 13 - Kinetic Theory of Gases is the part that allows the student to revise the manner in which the macroscopically observable gas characteristics are explained by the microscopic movement of molecules. It plays a big role in enhancing conceptual skills and accuracy in solving numericals in board and competitive exams.
Important Concepts
1. Molecular Nature of Gases
Gases are made up of a great number of moving molecules that move around randomly. The distances between molecules are much larger than the size of the molecules.
2. Pressure of an Ideal Gas
The pressure of gases is a result of gas molecules colliding with the container walls. It is determined by the number concentration, mass, and the mean square velocity of molecules.
3. Kinetic Interpretation of Temperature
Temperature is a parameter of the mean kinetic energy of gas molecules. An increase in temperature implies an increase in the speed of molecules.
4. Root Mean Square (RMS), Average, and Most Probable Speeds
Molecules of gases travel at varying speeds. RMS speed is most applicable in calculation, whereas average and most probable speeds are used to characterise molecular distribution.
5. Law of Equipartition of Energy
In the Law of Equipartition of Energy, the energy is shared equally among all degrees of freedom of a molecule. The energies added per molecule are 1/2 kT with each degree of freedom.
6. Degrees of Freedom
Degrees of freedom are the independent methods by which a molecule is able to store energy, including translational, rotational, and vibrational energy.
7. Ideal Gas Behaviour
Gas laws hold true to an ideal gas at all temperatures and pressures, provided that there are no intermolecular forces and the collisions between the molecules are perfectly elastic.
Important Formulas
- Pressure of an ideal gas:
$
P=\frac{1}{3} \rho \overline{c^2}
$
- Mean kinetic energy per molecule:
$
\overline{E_k}=\frac{3}{2} k T
$
- RMS speed of gas molecules:
$
c_{r m s}=\sqrt{\frac{3 k T}{m}}=\sqrt{\frac{3 R T}{M}}
$
- Average speed:
$
c_{a v g}=\sqrt{\frac{8 R T}{\pi M}}
$
- Most probable speed:
$
c_{m p}=\sqrt{\frac{2 R T}{M}}
$
- Ideal gas equation:
$
P V=n R T
$
- Internal energy of an ideal gas:
$
U=\frac{f}{2} n R T
$
Advantages of NCERT Exemplar Class 11 Physics Solutions Chapter 13: Kinetic Theory of Gases
The Kinetic Theory of Gases describes the behaviour of gases which we encounter in everyday life by the underlying microscopic motion of their molecules. These abstract concepts can be easily comprehended and applied with the help of NCERT Exemplar Class 11 Physics Solutions Chapter 13 in order to support the student in developing a solid conceptual and numerical sense.
- The solutions are clear in explaining the mechanisms of molecular motion that explain pressure, temperature and volume, which will assist students in connecting what they study with everyday observations.
- Derivations like the pressure of an ideal gas and the RMS speed are described in a stepwise manner and are hence easy to learn and revise.
- The answers to calculation-based questions, in a systematic way, assist students in dealing with numericals that are related to temperature, speed, and energy with comfort.
- The Exemplar solutions fill the gap between the microscopic laws of molecular motion and the macroscopic gas laws and enhance the overall understanding of the concept.
- Important assumptions of ideal gases and conditions of applicability are clearly highlighted, reducing confusion and exam mistakes.
- These solutions are very useful in CBSE exams and competitive exams such as JEE, as the questions are concept-based and relevant to the exam.
- Well-organised explanations and key points help students revise efficiently before exams without feeling overwhelmed.
Approach to Solve Exemplar Questions of Chapter 13: Kinetic Theory of Gases
The chapter on the Kinetic Theory of Gases mostly includes questions which require the ability to comprehend the motion of the molecules and the ability to use the formulas properly. The well-structured and clear methodology prevents the confusion of students and allows them to answer Exemplar questions correctly and without doubts.
- First, determine whether the question is based on molecular speed, pressure, temperature, degrees of freedom, or equipartition of energy.
- Write down all the provided values, including temperature, pressure, molar mass, number of moles, or volume. Change temperature to Kelvin (when necessary).
- Decide whether to use the ideal gas equation, the RMS speed relation, or the energy formulas. Avoid mixing formulas unnecessarily.
- Make sure that there is uniformity in units and that the values of constants such as R and k have been used. The gas-related numericals tend to have unit errors.
- Check whether the question assumes ideal behaviour. It is important to remember that formulae of kinetic theory are applied to ideal conditions.
- Write each step clearly instead of jumping to the answer. This improves accuracy and helps in gaining full marks.
- Verify whether the final result is physically meaningful, such as speeds being positive and energy increasing with temperature.
NCERT Exemplar Class 11 Physics Solutions Chapter-Wise
NCERT Exemplar Class 11 Physics Solutions Chapter-Wise Links provide students with a structured and organised way to study physics concepts chapter by chapter. These links make it easy to access well-explained solutions for numericals, conceptual questions, and derivations as per the latest NCERT and CBSE guidelines. They help students revise efficiently, strengthen conceptual understanding, and prepare confidently for board and competitive examinations.
NCERT Exemplar Solutions Class 11 Subject-wise
NCERT Exemplar Solutions Class 11 Subject-Wise Links offer a convenient and organised way for students to access solutions for Physics, Chemistry, and Mathematics in one place. The links enable the students to study each topic in a systematic manner with correct step-by-step directions that strictly adhere to the latest NCERT syllabus. They are suitable for rapid revision, clarity of the concepts and proper preparation for exams in all science subjects in Class 11.
NCERT Solutions for Class 11 Physics Chapter-wise
NCERT Solutions for Class 11 Physics Chapter-Wise Links help students study physics in a well-organised and systematic manner. By accessing solutions chapter by chapter, learners can easily understand concepts, numericals, and derivations as prescribed by the latest NCERT syllabus. These solutions support effective revision, strengthen fundamentals, and assist students in preparing confidently for school exams and competitive examinations.
Also, Read NCERT Solution subject-wise -
- NCERT Solutions for Class 11 Maths
- NCERT Solutions for Class 11 Chemistry
- NCERT Solutions for Class 11 Biology
Check NCERT Notes subject-wise -
- NCERT Notes Class 11 Maths
- NCERT Notes for Class 11 Physics
- NCERT Notes for Class 11 Chemistry
- NCERT Notes for Class 11 Biology
Also, Check NCERT Books and NCERT Syllabus here
Frequently Asked Questions (FAQs)
Q: What are the benefits of these solutions?
A:
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.
Q: Are these solutions as per the CBSE pattern?
A:
Yes, these NCERT exemplar class 11 physics solutions chapter 13 are as per the CBSE pattern, as they are solved step by step and according to the marking scheme.
Q: Is this chapter important for exams like JEE and NEET?
A:
Yes, it builds strong basics about gases, the motion of molecules, and gas laws—topics that are often asked in competitive exams.
Q: What are the key assumptions of the kinetic theory?
A:
Some important ones include the following: gas is made of particles, they have negligible volume, there are no forces between them, and they move randomly in all directions.
Q: What is the kinetic theory of gases?
A:
It's a theory that explains the behaviour of gases by assuming they are made of tiny, fast-moving particles (molecules) that collide with each other and the container walls.
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