Class 11th Physics - Kinetic Theory Case Study Questions and Answers 2022 - 2023
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Kinetic Theory Case Study Questions With Answer Key
11th Standard CBSE
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Reg.No. :
Physics
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The Kinetic theory was developed by Maxwell and Boltzmann to explain the behaviour of gases based on the idea that gases consist of rapidly moving atoms or molecules. The inter atomic forces binding the atoms are negligible and their size is. negligible. The theory is consistent with various gas laws and Avogadro's hypothesis. It gives molecular interpretation oftemperature pressure. Specific heat capacities. Compared to solids and liquids, it is easier to understand the properties of gases. This is because molecules in gases are far apart.
(i) Write the characteristics of ideal gas. Write the conditions at which real gases acquire ideal gas behaviour.
(ii) State Avogadro's hypothesis? What is Avogadro number?
(iii) Write the perfect gas equation by explaining Boyle's law and Charles's law.
(iv) What is meant by Boltzmann constant? Calculate its value in S.I units.
(v) The graph shows the variation of the product PV w.r.t the pressure P of given masses of three gases, A, Band C. The temperature is kept constant. State with proper arguments which of these gases is ideal.
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The molecules of a gas move in all directions with various speeds. The speeds of the molecules of a gas increase with rise in temperature. During its random motion, a fast molecule of ten strikes against the walls of the container of the gas. The collisions are assumed to be prefectly elastic i.e., the molecule bounces back with the same speed with which it strikes the wall. Since the number of molecules is very large, billions of molecules strike against the walls of the container every second. These molecules exert force on the wall. The force exerted per unit area is the pressure exerted by the gas on the walls. According to the kinetic theory, the pressure of a gas of density p at absolute temperature T is given by \(P=\frac{1}{3} \rho\) \(v_{r m s}^{2}, \) where vrms is the root mean square speed of the gas molecule and is given by \(v_{r m s}=\sqrt{\frac{3 K_{B} T}{M}}\) where M is the mass of a molecule and KB is Boltzmann constant.
(i) State the relation between pressure and kinetic energy of the gas.
(ii) On what factors does the average kinetic energy of translation per molecule of the gas depends?
(iii) State absolute zero temperature in terms of root mean square velocity of gas.
(iv) Define root mean square speed and establish its relation with temperature.
(v) The absolute temperature of a gas is made four times. How many times will its total kinetic energy pressure and r.m.s velocity become?
(vi) Two different gases have exactly the same temperature. Does this mean that their molecules have the same r.ms. speed?
(vii) On reducing the volume of the gas at constant temperature, the pressure of the gas increases. Explain on the basis of kinetic theory. -
The number of degrees of freedom of a dynamical system is the total number of co-ordinates or independent quantites required to describe completely the position and configuration of the system. For a dynamical system, the number of degrees of freedom is obtained by subtracting the number of independent relations from the total number of co-ordinates required to specify the position of constituebnt particles of the system.
N = 3A - R, where N - number of degrees of freedom of the system.
A - number of particles in the system
R - number of independent relations among the particles.
Each degree of freedom contributes equally In the distribution of the energy associated with each molecule. In thermal equilibrium the energy associated with each molecule per degree of freedom is \(\frac{1}{2} K_{B} T\) .
(i) Determine the number of degrees of freedom of a non-linear triatomic molecule.
(ii) State law of equipartition of energy.
(iii) If a gas has n degrees of freedom, determine the ratio of principal specific heat of the gas.
(iv) Determine the energy contributed by a vibrational mode in total energy.
(v) Determine the ratio of specific heat for monoatomic gas molecule. -
During their random motion, the molecules of a gas often come close to each other. Molecules are perfect elastic spheres and their size is very small compared to the distance between them. Gas molecules undergo elastic collision. Therefore, they cannot move straight unhindered. The paths of molecules keep on getting deflected incessantly. Path of a single gas molecule consists of a series of short zig zag paths of different lengths. These paths of different lengths are called free paths of the molecules and their mean is called mean free path. Mean free path of gas molecules depends on diameter (d) of gas molecule and molecular density (n) as follows \(\lambda=\frac{1}{n \pi d^{2}}\)
(i) Define mean free path of gas molecules.
(ii) On what factors do the mean free path of gas molecules depends?
(iii) What is significance of mean free path?
(iv) How many collisions per second does each molecule of a gas make, when the average speed of a molecule is 500 ms-1 and mean free path is 2.66 x 10- 7 m?
(v) Calculate the mean free path of molecules, if number of molecules per cm3 is 3 x 1019 and diameter of each molecule is 2A.
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Answers
Kinetic Theory Case Study Questions With Answer Key Answer Keys
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(i) In ideal gas
1. the size of the molecule of a gas is zero.
2. there is no force of attraction or repulsion amongst the molecules of the gas.
Real gases behave as ideal gas at extremely low pressure and high temperature.
(ii) It states that equal volumes of all gases under identical conditions of pressure and temperature would contain equal number of molecules.
Avogadro number NA is equal to 6.02 x 1023 number of molecules which any gas of amount 22.4 litre contains at N.T.P
(iii) Boyle's law: States that when temperature of a given mass of a gas is kept constant, its pressure varies inversely as the volume of the gas i.e., PV = constant.
Charles's law: When pressure of a given mass of a gas is kept constant, volume of gas varies directly as the temperature of the gas i.e \(\frac{V}{T}\) = constant from Boyle and Charles's law perfect gas equation is PV = nRT, where n-no. of moles of gas. R- Universal gas constant.
(iv) Boltzmann constant
\(K_{B}=\frac{R}{N_{A}}\)
Where R is universal gas constant and NA -Avogadro's constant.
\(K_{B}=\frac{8.31 \mathrm{~J} \mathrm{~mole}^{-1} \mathrm{~K}^{-1}}{6.023 \times 10^{23} \mathrm{~mol}^{-1}}\)
= 1.38 x 10- 23 JK-1
(v) Gas C is ideal, because PV is constant for this gas. It means the gas C obeys Boyle's law at all pressures. -
(i) The pressure exerted by an ideal gas is numerically equal to two third of the mean kinetic energy of translation per unit volume of gas.
\(P=\frac{2}{3} E\)
(ii) The average kinetic energy of a molecule is directly proportional to the absolute temperature of the gas. It is independent of pressure, volume or nature of the ideal gas.
(iii) Absolute zero temperature is that at which the root mean square velocity of the gas molecule reduces to zero i.e., molecular motion ceases at absolute zero.
(iv) It is defined as the square root of the mean of the squares of the random velocities of the individual molecules of a gas
\(v_{r m s} \text { or } C_{r m s}=\sqrt{\frac{C_{1}^{2}+C_{2}^{2}+\ldots C_{n}^{2}}{n}}=\sqrt{\frac{3 K_{B} T}{M}}\)
where KB - Boltz-mann constant, M - Mass of each molecule.
(v) Since K.E \(\propto\) Temperature, hence K.E will become four times \(v_{r m s} \propto \sqrt{T}\) hence vrms becomes twice and \(P \propto\left(v_{m s}\right)^{2}\) therefore pressure becomes 4 times.
(vi) When the two gases have exactly the same temperature, the average K.E per molecule for each gas is the same. But as the different gases may have molecules of different masses, the r.m.s speed (C) of molecules of different gases shall be different.
(vii) On reducing the volume, the number of molecules per unit volume increases. As a result of which more molecules collide with the walls of the vessel per second and hence a larger momentum is transferred to walls per second. Due to which the pressure of the gas increases. -
(i) In a non-linear triatomic molecule, the three atoms are present at the three vertices of a triangle,
N = 3A - R, Since A = 3, and R = 3
Number of degrees of freedom
N = 3 x 3 x 3 = 6
(ii) Law of equipartition of energy states that in thermal equilibrium, at temperature T each mode of energy: translational, rotational, and vibrational, contributes an average energy equal to \(\frac{1}{2} K_{B} T\) .
(iii) Ratio of Cp and Cv of the gas having n degrees of freedom \(\gamma=\frac{C_{P}}{C_{V}}=1+\frac{2}{n}\) .
(iv) Each vibrational mode contributes \(2 \times \frac{1}{2} K_{B} T\) total energy, because a vibrational mode has both kinetic and potential energy mode.
(v) \(\mathrm{C}_{\mathrm{V}}=\frac{3}{2} R, C_{P}=\frac{5}{2} R, \gamma=\frac{C_{P}}{C_{V}}=\frac{5}{3}=1.67\) -
(i) Mean free path of gas molecules is the average distance travelled by a molecule between two successive collisions. \(\lambda=\frac{\lambda_{1}+\lambda_{2}+\lambda_{3} \ldots \lambda_{n}}{n}=\frac{\overline{c t}}{n}\)
Where c is mean speed of the molecules.
(ii) Mean free path of gas molecules depends on number density (n) and area of molecules (d2) as follow
\(\lambda=\frac{1}{\sqrt{2} n \pi d^{2}}\)
(iii) The concept of mean free path is of great significance in understanding transport phenomena like diffusion, viscosity and thermal conduction.
(iv) Here \(\bar{\lambda}\) = 2.66 x 10- 7 m; v = 500 ms-1, Number of collision made per second by each molecule of the gas i.e., collision frequency,
\(f=\frac{v}{\bar{\lambda}}=\frac{500}{2.66 \times 10^{-7}}=1.88 \times 10^{9} \mathrm{~s}^{-1}\)
(v) Here n = 3 x 1019 cm-3 = 3 x 1025 m-3
d = 2A = 2 x 10- 10 m
\(\bar{\lambda}=\frac{1}{\sqrt{2} \pi d^{2} n}\)
\(=\frac{1}{\sqrt{2} \pi \times\left(2 \times 10^{-10}\right)^{2} \times 3 \times 10^{25}}\)
= 1.876 x 10- 7 m
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