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Kinetic Theory Class 11th Notes - Free NCERT Class 11 Physics Chapter 13 Notes - Download PDF

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Kinetic Theory Class 11th Notes - Free NCERT Class 11 Physics Chapter 13 Notes - Download PDF

Edited By Safeer PP | Updated on Mar 16, 2022 06:11 PM IST

The Class 11 Physics chapter 13 notes (Kinetic theory) are notes of the NCERT chapter Kinetic theory. NCERT Class 11 Physics chapter 13 notes provide an overview of the chapter on Kinetic theories.

The main topics covered in Kinetic theory Class 11 notes are

  • Molecular nature of matter

  • Behavior of gases

  • Kinetic theory of ideal gases

  • Law of equipartition of energy

  • Specific heat capacity

  • Mean Free Path

In Class 11 Kinetic theory notes all these topics are important from the perspective of the CBSE exam and can be noted, all of these topics are relevant to the CBSE exam and can be obtained by downloading from Class 11 Physics chapter 13 notes pdf download. Kinetic theory Class 11 notes cover the summary of the chapter. The CBSE Class 11 Physics chapter 13 notes do not include the necessary derivations. The NCERT notes for Class 11 Physics chapter 13 or notes for class 11 physics chapter 13 are extremely beneficial for revision.

Additionally, students can refer to

NCERT Class 11 Physics Chapter 13 Notes

Some properties of gases

  • Gases can be compressed easily.

  • Gases have neither a definite volume nor a definite shape.

  • When a substance is in the gaseous state, its density is very small in comparison to the density in its solid or liquid state.

Boyle’s Law: At constant temperature, the volume V of a given mass of a gas is inversely proportional to its pressure P, according to this law.

1647002160714

{"type":"lalign*","id":"2","backgroundColor":"#ffffff","backgroundColorModified":false,"aid":null,"code":"\\begin{lalign*}\n&{V\\,\\propto\\,\\frac{1}{P}}\\\\\n&{PV\\,=\\,cons\\tan t}\t\n\\end{lalign*}","font":{"family":"Arial","size":11,"color":"#000000"},"ts":1638278566270,"cs":"jwvrFDPwD+l+nyqjqT3vvA==","size":{"width":124,"height":52}}

If P₁ and V₁ are the initial pressure and volume, and P₂ and V₂ are the final pressure and volume for a given mass of a gas at constant temperature, then

{"aid":null,"id":"3","backgroundColor":"#ffffff","type":"$$","code":"$$P_{1}V_{1}\\,=\\,P_{2}V_{2}$$","font":{"color":"#000000","size":11,"family":"Arial"},"backgroundColorModified":false,"ts":1638278831109,"cs":"9NHQuck97vM8PQr6Hvhr1g==","size":{"width":96,"height":13}}

A graph is drawn between P and V for a gas at different constant temperatures T₁, T₂, and T₃ (T₁> T₂>T₃).

The gases do not obey Boyle’s law strictly under all conditions of pressure and temperature, gas only obey Boyle’s law only at low pressures and high temperatures.

Charles’ Law:- At constant pressure the volume of the given mass of a gas increases by {"aid":null,"type":"$$","font":{"color":"#000000","family":"Arial","size":11},"id":"4","backgroundColor":"#ffffff","backgroundColorModified":false,"code":"$$\\frac{1}{273}th$$","ts":1638334680850,"cs":"YL2+z4UgHsyOxIkaPIXo1w==","size":{"width":46,"height":34}}part of its volume at 0℃ for 1℃ rise in temperature is called Charles's law.

{"backgroundColorModified":false,"font":{"size":11,"family":"Arial","color":"#000000"},"backgroundColor":"#ffffff","aid":null,"type":"$$","code":"$$V_{t}\\,=\\,V_{0}\\left(1+\\,\\frac{t}{273}\\right)$$","id":"5","ts":1638334827736,"cs":"fyKa1BSGpuzcMkFxrM5LMg==","size":{"width":146,"height":40}}

Charles’ Law in terms of absolute temperature:- Let at constant pressure, the volume of a given mass of a gas at 0℃, t₁℃ and t₂℃ are V₀, V₁ and V₂ respectively. T₁ and T₂ are the absolute temperatures corresponding to t₁℃ and t₂ ℃. By Charles' Law

{"backgroundColorModified":false,"backgroundColor":"#ffffff","id":"6","aid":null,"code":"\\begin{lalign*}\n&{V_{1}\\,=\\,V_{0}\\left(1\\,+\\,\\frac{t_{1}}{273}\\right)\\,=\\,V_{0}\\,\\left(\\frac{273+\\,t_{1}}{273}\\right)}\\\\\n&{V_{2}\\,=\\,V_{0}\\left(\\,1+\\,\\frac{t_{2}}{273}\\right)\\,=\\,V_{0}\\left(\\frac{273+t_{2}}{273}\\right)}\\\\\n&{\\therefore\\,\\frac{V_{1}}{V_{2}}\\,=\\,\\frac{273\\,+\\,t_{1}}{273+\\,t_{2}}\\,=\\,\\frac{T_{1}}{T_{2}}}\\\\\n&{\\frac{V_{1}}{T_{1}}\\,=\\,\\frac{V_{2}}{T_{2}}}\\\\\n&{\\frac{V}{T}\\,=\\,a\\,cons\\tan t}\\\\\n&{V\\,\\propto\\,T}\t\n\\end{lalign*}","type":"lalign*","font":{"color":"#000000","size":11,"family":"Arial"},"ts":1638336267686,"cs":"gMaIb4REonoseV0jACIrpA==","size":{"width":296,"height":232}}1647002161978

Therefore at constant pressure, the volume of a given mass of a gas is directly proportional to its absolute temperature. This is Charles’ Law in terms of absolute temperature.

A graph is drawn between T and V for a gas at three different constant pressures P₁, P₂, and P₃ ( P₁> P₂>P₃).

Perfect Gas ( or ideal gas):- A gas whose properties are similar to the properties of a real gas at infinitely low pressure such an imaginary gas is called perfect gas or ideal gas. There are following properties of perfect gas:

  • It strictly obeys Boyle's Law, Charles’ Law and the law of pressure under all conditions of temperature and pressure.

  • Its pressure coefficient and volume coefficient are exactly equal to each other.

  • Its molecules are infinitesimally small.

  • There is no force attraction between its molecules.

Equation of State for an Ideal Gas

Let initially pressure, volume, and temperature of a gas are P₁, V₁, and T₁respectively and finally change to P₂ V₂, and T₂ respectively. Now,

i. Let the temperature of the gas kept constant at T₁ and the pressure is changed from P₁ to P₂. If in this process the volume changes from V₁ to V’ then by Boyle’s law,

{"type":"lalign*","aid":null,"backgroundColor":"#ffffff","backgroundColorModified":false,"id":"7","font":{"size":11,"color":"#000000","family":"Arial"},"code":"\\begin{lalign*}\n&{P_{2}V^{\\prime}\\,=\\,P_{1}V_{1}}\\\\\n&{V^{\\prime}\\,=\\,\\frac{P_{1}V_{1}}{P_{2}}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,......\\left(i\\right)}\t\n\\end{lalign*}","ts":1638338107250,"cs":"zVYhfuqrCWFXC1tOqoerew==","size":{"width":206,"height":60}}

ii. Let the pressure of the gas is kept constant at P₂ and its absolute temperature is changed from T₁ to T₂. If in this process the volume of the gas changes from V’ to V₂ then by Charles’ law,

{"backgroundColor":"#ffffff","code":"\\begin{lalign*}\n&{\\frac{V^{\\prime}}{T_{1}}\\,=\\,\\frac{V_{2}}{T_{2}}}\\\\\n&{V^{\\prime}\\,=\\,\\frac{T_{1}V_{1}}{T_{2}}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,......\\left(ii\\right)}\t\n\\end{lalign*}","backgroundColorModified":false,"font":{"color":"#000000","size":11,"family":"Arial"},"id":"8","aid":null,"type":"lalign*","ts":1638338345179,"cs":"S+s1E1FLQCFbKEUhC8AokQ==","size":{"width":208,"height":81}}

From eqn (i) and (ii), we have

{"backgroundColor":"#ffffff","code":"\\begin{lalign*}\n&{\\frac{P_{1}V_{1}}{P_{2}}\\,=\\,\\frac{T_{1}V_{2}}{T_{2}}}\\\\\n&{\\frac{P_{1}V_{1}}{T_{1}}\\,=\\,\\frac{P_{2}V_{2}}{T_{2}}}\t\n\\end{lalign*}","backgroundColorModified":false,"id":"9","aid":null,"type":"lalign*","font":{"color":"#000000","family":"Arial","size":11},"ts":1638338489538,"cs":"ETfRPRy/VaR+d4gSE1DcUw==","size":{"width":108,"height":80}}

Hence for a gas, the volume of PV/T remains constant. Let this constant be equal to K, then

{"backgroundColorModified":false,"font":{"family":"Arial","color":"#000000","size":11},"aid":null,"backgroundColor":"#ffffff","type":"lalign*","code":"\\begin{lalign*}\n&{\\frac{PV}{T}\\,=\\,K}\\\\\n&{PV\\,=\\,KT}\t\n\\end{lalign*}","id":"10","ts":1638338606607,"cs":"Cag88HIzJv/AkTpaPUiaWg==","size":{"width":80,"height":52}}

We can write K = nk, k is the same for all gases. It is called Boltzmann constant.

{"type":"$$","id":"11","aid":null,"code":"$$PV\\,=\\,nkT\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,.....\\left(iii\\right)$$","backgroundColorModified":false,"font":{"family":"Arial","size":11,"color":"#000000"},"backgroundColor":"#ffffff","ts":1638338901041,"cs":"QAG8soS48kABZewqwio0oA==","size":{"width":189,"height":16}}

If P, V, and T are the same, then n is also the same for all gase.

If we take 1 gram-molecule (1 mole) of a gas-constant it will be the same for all gases. Then it is called the universal gas constant and it is represented by R. Then,

{"type":"$$","aid":null,"backgroundColorModified":false,"font":{"size":11,"color":"#000000","family":"Arial"},"id":"12","backgroundColor":"#ffffff","code":"$$PV\\,=\\,RT$$","ts":1638339110802,"cs":"ySWYBDo0K/4HhwUGjaA/Vg==","size":{"width":78,"height":12}}

Where R = Nk, where N = avogadro’s number and k = Boltzmann constant.

It is called the ideal gas equation or perfect gas equation.

The gas equation for μ moles of gas will be

{"backgroundColorModified":false,"code":"$$PV\\,=\\,\\mu RT\\,\\,\\,\\,\\,\\,\\,\\,.....\\left(iv\\right)$$","type":"$$","font":{"color":"#000000","family":"Arial","size":11},"backgroundColor":"#ffffff","id":"13","aid":null,"ts":1638340463351,"cs":"CIgzI2HOxTEvqmwi4pqLlw==","size":{"width":176,"height":16}}

Unit of R:- Unit of R is Jmol⁻¹K⁻¹

Dimension of R:- [ML²T⁻²?⁻¹].

Numerical value of R:- The value of R is 8.31Jmol⁻¹K⁻¹.

Kinetic Theory of Gases

Following are the fundamental assumptions of kinetic theory of gases:

  • A gas is composed of a large number of tiny invisible, perfectly elastic particles, called the molecules.

  • The molecules are always in a state of continuous motion with varying velocities in all possible directions.

  • The molecules traverse the straight path between any two collisions.

  • The size of the molecules is infinitely small compared to the average distance traversed by a molecule between any two consecutive collisions . The distance between any two consecutive collisions is called the free path and the average distance between any two consecutive collisions is called the mean free path.

  • The time of collisions is negligible as compared with the time taken to traverse the free path.

  • The collisions between molecules and with the walls are perfectly elastic so that there is no loss of kinetic energy in the collisions.

  • The molecules exert no force on each other except when they collide and the whole of the molecules energy is kinetic.

  • The volume of the molecules is negligible as compared to the volume of a vessel containing gas.

  • The inter molecular distances in a gas is much larger than that of a solid or liquid and the molecules of a gas are free to move in the entire space available to them.

Pressure Exerted by an Ideal Gas

When a certain mass of a perfect gas is confined with a vessel the molecule frequently collides with the walls of the vessel and is reflected from the walls.During the collision with the walls each molecule suffers a change in momentum .

1647002160107

According to Newton ‘s second law of motion the rate of momentum is equal to the force exerted .Thus,it exerts a pressure on the walls . This pressure may be calculated on the basis of postulates of kinetic theory of gases

Let us consider a perfect gas contained in a cubical vessel of side (l) and perfectly elastic walls. Let n be the total total number of molecules of a gas in the vessel and m is the mass of the molecule. Let the sides of the cube represent mutually perpendicular X, Y, and Z-axes.

Consider a molecule molecule P moving in a random direction with velocity v1. Let v1x, v1y, and v1z be the components of the velocity {"backgroundColorModified":false,"font":{"color":"#000000","size":11,"family":"Arial"},"aid":null,"code":"$$v_{1}$$","backgroundColor":"#ffffff","id":"76","type":"$$","ts":1638446014135,"cs":"1dP/3UYEegjcw8Y1CnEnww==","size":{"width":12,"height":9}} along three mutually perpendicular X, Y, and Z axes , respectively.

Then

{"font":{"family":"Arial","color":"#000000","size":11},"backgroundColor":"#ffffff","aid":null,"type":"$$","code":"$$v_{1}^{2}\\,=\\,v_{1x}^{2}\\,+\\,v_{1y}^{2}\\,+\\,v_{1z}^{2}$$","backgroundColorModified":null,"id":"17","ts":1638357915172,"cs":"ECv+gQ2fng1qw/pFvF7a/w==","size":{"width":160,"height":21}}

The component of velocity with which the molecule P will strike the face DEFA is v1x and the momentum of the molecule X- direction before collisions

{"font":{"family":"Arial","size":11,"color":"#000000"},"aid":null,"type":"$$","backgroundColorModified":false,"backgroundColor":"#ffffff","code":"$$=\\,mv_{1x}$$","id":"18","ts":1638358036696,"cs":"2QK4LLWT+Y9XIcdkxn5ZyQ==","size":{"width":56,"height":9}}

As the molecule and the wall are perfectly elastic, during collision speed of the molecule does not change and only the direction of velocity is changed.

Therefore, The momentum along X- axis after collisions

{"backgroundColor":"#ffffff","font":{"size":11,"family":"Arial","color":"#000000"},"code":"$$=\\,m\\left(-v_{1x}\\right)\\,=\\,-mv_{1x}$$","backgroundColorModified":false,"type":"$$","id":"19","aid":null,"ts":1638358165167,"cs":"nvFOOIQYdFXG5sDEyGrnvg==","size":{"width":168,"height":16}}

The change in the momentum during one collision along X-axis

{"id":"20","backgroundColor":"#ffffff","aid":null,"code":"$$=\\,mv_{1x}\\,-\\,\\left(-mv_{1x}\\right)\\,=\\,2mv_{1x}$$","type":"$$","font":{"family":"Arial","color":"#000000","size":11},"backgroundColorModified":false,"ts":1638358233462,"cs":"A8j+toCdS1Mo7AQh3CyWaQ==","size":{"width":220,"height":16}}

The molecule suffering reflection at face DEFA traverses a distance l and strikes the faceCGBO. It is equal to the force exerted on it. Hence, the force exerted by the molecule on the wall A will be mvx2/l Since there are n molecules, the total force exerted on the wall A is given by

{"type":"$$","font":{"size":11,"family":"Arial","color":"#000000"},"backgroundColorModified":false,"aid":null,"backgroundColor":"#ffffff","id":"22","code":"$$\\frac{m}{l}\\left(v_{x_{1}}^{2}\\,+\\,v_{x_{2}}^{2}+\\,......\\,v_{x_{n}}^{2}\\right)$$","ts":1638360030729,"cs":"s4d/Qf+ouizA+T8tk22N7g==","size":{"width":196,"height":30}}

Where vxn is the x-component of velocity of the molecule 1, vx2 of molecule 2,.... And so on. Since the area of the wall is l2, the pressure on the wall A is given by

{"backgroundColor":"#ffffff","type":"lalign*","code":"\\begin{lalign*}\n&{P\\,=\\,\\frac{m}{l^{3}}\\left(v_{x_{1}}^{2}\\,+\\,v_{x_{2}}^{2}+\\,......\\,v_{x_{n}}^{2}\\right)}\\\\\n&{\\,=\\,\\frac{mn}{l^{3}}\\left(\\frac{\\left(v_{x_{1}}^{2}\\,+\\,v_{x_{2}}^{2}+\\,......\\,v_{x_{n}}^{2}\\right)}{n}\\right)}\t\n\\end{lalign*}","aid":null,"font":{"color":"#000000","size":11,"family":"Arial"},"backgroundColorModified":false,"id":"26","ts":1638360414885,"cs":"p5N2CANfhZ7DcLu/C87LKw==","size":{"width":270,"height":88}}

But l3= V

{"code":"$$\\therefore\\,P\\,=\\,\\frac{mn}{V}\\left(\\frac{\\left(v_{x_{1}}^{2}\\,+\\,v_{x_{2}}^{2}+\\,......\\,v_{x_{n}}^{2}\\right)}{n}\\right)$$","type":"$$","aid":null,"backgroundColorModified":false,"backgroundColor":"#ffffff","font":{"color":"#000000","family":"Arial","size":11},"id":"27","ts":1638360452397,"cs":"hzslwuYktdZ+jWikrces+A==","size":{"width":302,"height":50}}

Now, {"backgroundColorModified":false,"type":"$$","code":"$$\\left(\\frac{\\left(v_{x_{1}}^{2}\\,+\\,v_{x_{2}}^{2}+\\,......\\,v_{x_{n}}^{2}\\right)}{n}\\right)$$","id":"28","font":{"size":11,"color":"#000000","family":"Arial"},"backgroundColor":"#ffffff","aid":null,"ts":1638360482540,"cs":"DeT7aLoCOlEMM53C2yjAqw==","size":{"width":205,"height":50}} is the average value of vx2 for all the n molecules. This can be written as

{"id":"30","backgroundColor":"#ffffff","type":"$$","aid":null,"font":{"color":"#000000","family":"Arial","size":11},"backgroundColorModified":false,"code":"$$\\bar{v_{x}^{2}}$$","ts":1638360586075,"cs":"2qNoRJlHqB7GZMl/S5OG6Q==","size":{"width":13,"height":18}}.

Therefore,

{"backgroundColorModified":false,"type":"$$","backgroundColor":"#ffffff","id":"31","font":{"color":"#000000","size":11,"family":"Arial"},"aid":null,"code":"$$P\\,=\\,\\frac{mn}{V}\\bar{v_{x}^{2}}\\,\\,\\,\\,\\,\\,......\\left(i\\right)$$","ts":1638360642515,"cs":"LoKWpa/ixOmQgwndUZtIww==","size":{"width":170,"height":30}}

We know that for any molecule, we have

{"backgroundColor":"#ffffff","aid":null,"font":{"color":"#000000","size":11,"family":"Arial"},"code":"$$v^{2}\\,=\\,\\bar{v_{x}^{2}}\\,+\\,\\bar{v_{y}^{2}\\,}+\\,\\bar{v_{z}^{2}}\\,\\,\\,\\,\\,\\,\\,\\,.....\\left(ii\\right)$$","type":"$$","backgroundColorModified":false,"id":"32","ts":1638360876728,"cs":"vv/SLrW7Wx1cqjMevaczhA==","size":{"width":228,"height":21}}

Because this equation is true for every molecule, this will also be true for the mean value of all the molecules.

{"aid":null,"id":"33","font":{"family":"Arial","size":11,"color":"#000000"},"backgroundColor":"#ffffff","type":"$$","backgroundColorModified":false,"code":"$$\\therefore\\,\\bar{v^{2}\\,}=\\,\\bar{v_{x}^{2}}\\,+\\,\\bar{v_{y}^{2}}\\,+\\,\\bar{v_{z}^{2}}\\,\\,\\,\\,\\,\\,.....\\left(iii\\right)$$","ts":1638361226100,"cs":"pfBrlKAPOai3+pZnh6XuZg==","size":{"width":248,"height":20}}

\text{ The directions of the molecules are quite at random, that is, all directions are equally possible. Hence, the values, of }\overline{v_x^2},\ \overline{v_y^2},\ \overline{v_z^2}\text{ will be equal, ie.}

{"backgroundColor":"#ffffff","type":"$$","id":"36","backgroundColorModified":false,"code":"$$\\bar{v_{x}^{2}}\\,=\\,\\bar{v_{y}^{2}}\\,=\\,\\bar{v_{z}^{2}}$$","aid":null,"font":{"size":11,"color":"#000000","family":"Arial"},"ts":1638361424065,"cs":"UUf+ekMw09vrsFPBgaNa2w==","size":{"width":101,"height":20}}

Therefore from eqn (ii), we have

{"id":"37","backgroundColor":"#ffffff","code":"\\begin{lalign*}\n&{\\bar{v^{2}}\\,=\\,3\\bar{v_{x}^{2}}}\\\\\n&{\\bar{v_{x}^{2}}\\,=\\,\\frac{1}{3}\\bar{v^{2}}}\t\n\\end{lalign*}","aid":null,"font":{"family":"Arial","size":11,"color":"#000000"},"type":"lalign*","backgroundColorModified":false,"ts":1638361498838,"cs":"oNR5ErJZgin0dJVNA7s5ng==","size":{"width":73,"height":58}}

or

{"type":"$$","id":"39","aid":null,"code":"$$P\\,=\\,\\frac{1}{3}\\,\\frac{mn}{V}\\,\\bar{v^{2}}\\,\\,\\,......\\left(iii\\right)$$","font":{"size":11,"family":"Arial","color":"#000000"},"backgroundColor":"#ffffff","backgroundColorModified":false,"ts":1638361598679,"cs":"+1o4bff0SAHSGjMWt+lkdQ==","size":{"width":194,"height":34}}

This is the equation for the pressure of the gas. {"backgroundColorModified":false,"id":"40","type":"$$","aid":null,"backgroundColor":"#ffffff","code":"$$\\bar{v^{2}}$$","font":{"size":11,"family":"Arial","color":"#000000"},"ts":1638361652175,"cs":"JEjeYbWDgsuganMagI+1Ag==","size":{"width":12,"height":14}}

is called the mean- square velocity of the molecules. Since, mn is the mass of the whole gas and V is the volume so, mn/V is the density ⍴ of the gas therefore, the above expression can be written as

{"type":"$$","font":{"family":"Arial","color":"#000000","size":11},"backgroundColor":"#ffffff","aid":null,"code":"$$P\\,=\\,\\frac{1}{3}\\rho \\bar{v^{2}}\\,\\,\\,\\,\\,\\,\\,\\,....\\left(iv\\right)$$","id":"41","backgroundColorModified":false,"ts":1638361828844,"cs":"PIzFxToCV6CX+xB4bOPMxw==","size":{"width":160,"height":34}}

Again,

{"aid":null,"font":{"color":"#000000","family":"Arial","size":11},"type":"$$","id":"42","code":"$$P\\,=\\,\\frac{1}{3}\\rho \\bar{v^{2}\\,}\\,=\\,\\frac{2}{3}\\left(\\frac{1}{2}\\rho \\bar{v^{2}}\\right)\\,=\\,\\frac{2}{3}E$$","backgroundColor":"#ffffff","backgroundColorModified":false,"ts":1638361919219,"cs":"PJaTcMoO9MmBpEUZpFjoXg==","size":{"width":252,"height":40}}

Where {"backgroundColorModified":false,"code":"$$E\\,=\\,\\left(\\frac{1}{2}\\rho \\bar{v^{2}}\\right)$$","font":{"color":"#000000","size":11,"family":"Arial"},"aid":null,"id":"43","backgroundColor":"#ffffff","type":"$$","ts":1638361959281,"cs":"j1gzaX2QRTsW+3owczYKjQ==","size":{"width":102,"height":40}} is the transnational kinetic energy of the gas per unit volume. Thus, the pressure is equal to two-thirds of its transnational kinetic energy per unit volume.

Law of Equipartition of Energy

It states that the total kinetic energy of a dynamical system consisting of a large number of particles in thermal equilibrium is equally distributed among all degrees of freedom 0.5KT, and the energy associated with each degree of freedom is where k is the Boltzmann constant and T is the absolute temperature of the system.

Thus, the average kinetic energy of a molecule per degree of freedom is 0.5KT which is the law of equipartition of energy.

Root mean-Square (rms) Speed of gas molecule

The speeds of the individual molecules of a gas vary over a wide range. Let v₁, v₂, and v₃…...vₙ. Be the speed of molecules of a gas. Then, the root-mean-square speed of the molecules is defined as the square root of the mean of the squares of the speeds of the all gas molecules, and it is denoted by rms. Thus,

{"backgroundColorModified":false,"type":"$$","code":"$$v_{rms}\\,=\\,{\\sqrt[]{\\bar{v^{2}}}}\\,{\\sqrt[]{\\frac{v_{1}^{2}\\,+\\,v_{2}^{2}\\,+\\,v_{3}^{2}}{n}}}$$","id":"44","font":{"color":"#000000","size":11,"family":"Arial"},"backgroundColor":"#ffffff","aid":null,"ts":1638362528327,"cs":"OM3yXR38I9YKfAjo2JWp2Q==","size":{"width":218,"height":50}}

We know that the pressure of a gas is given by

{"code":"\\begin{lalign*}\n&{P\\,=\\,\\frac{1}{3}\\rho \\bar{v^{2}}}\\\\\n&{v_{rms}\\,=\\,{\\sqrt[]{\\bar{v^{2}}}}\\,=\\,{\\sqrt[]{\\frac{3P}{\\rho}}}}\t\n\\end{lalign*}","backgroundColor":"#ffffff","aid":null,"id":"45","backgroundColorModified":false,"font":{"color":"#000000","size":11,"family":"Arial"},"type":"lalign*","ts":1638362644457,"cs":"Zi5w+zvx9pqOvzsc40phQg==","size":{"width":166,"height":92}}

Interpretation of temperature on the basis of Kinetic theory

Let us consider 1 gram molecule of a gas at absolute temperature, T, occupying a volume V. According to Kinetic theory, the pressure of the gas

{"font":{"color":"#000000","family":"Arial","size":11},"backgroundColorModified":false,"type":"lalign*","code":"\\begin{lalign*}\n&{P\\,=\\,\\frac{1}{3}.\\frac{mN}{V}\\bar{v^{2}\\,}\\,\\,\\,\\,\\,\\,\\,\\,\\,.....\\left(i\\right)}\\\\\n&{PV\\,=\\,\\frac{1}{3}mN\\bar{v^{2}}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,.....\\left(ii\\right)}\\\\\n&{\\because\\,PV\\,=\\,RT\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,.....\\left(iii\\right)}\t\n\\end{lalign*}","aid":null,"id":"46","backgroundColor":"#ffffff","ts":1638363603290,"cs":"aG+2LyJZ4AInyCk9kmA1zg==","size":{"width":206,"height":97}}

mN = M

Comparing eqn (ii) and (iii)

{"aid":null,"backgroundColorModified":false,"font":{"color":"#000000","size":11,"family":"Arial"},"type":"lalign*","code":"\\begin{lalign*}\n&{\\frac{1}{3}M\\bar{v^{2}}\\,=\\,RT}\\\\\n&{\\bar{v^{2}}\\,=\\,\\frac{3R}{M}T\\,}\\\\\n&{{\\sqrt[]{\\bar{v^{2}}}}\\,=\\,{\\sqrt[]{\\frac{3RT}{M}}}}\\\\\n&{v_{rms}\\,=\\,{\\sqrt[]{\\frac{3RT}{M}}}}\\\\\n&{v_{rms}\\,\\propto\\,{\\sqrt[]{T}}}\t\n\\end{lalign*}","id":"47","backgroundColor":"#ffffff","ts":1638436033134,"cs":"gaUS4f3mqsWIXdF6ZmmB7g==","size":{"width":120,"height":196}}

Thus, the root mean square speed of the molecules of a gas is directly proportional to the square root of the absolute temperature of the gas.

Relation between temperature and molecular kinetic energy:-

{"font":{"size":11,"color":"#000000","family":"Arial"},"backgroundColor":"#ffffff","type":"$$","code":"$$\\frac{1}{2}M\\bar{v^{2}}\\,=\\,\\frac{1}{2}M\\left(\\frac{3RT}{M}\\right)$$","id":"50","aid":null,"backgroundColorModified":false,"ts":1638436603700,"cs":"Xp2H+h/746NV0xeID1Vonw==","size":{"width":172,"height":40}}

\text{Average kinetic energy of 1 mole of gas} =\frac{3}{2}RT

Average kinetic energy of one molecule =

{"type":"$$","id":"52","code":"$$\\frac{\\left(3/2\\right)RT}{N}\\,=\\,\\frac{3}{2}\\left(\\frac{R}{N}\\right)T$$","backgroundColor":"#ffffff","aid":null,"backgroundColorModified":false,"font":{"size":11,"color":"#000000","family":"Arial"},"ts":1638436815230,"cs":"Id4TtkEmolA1q9QNMFjooQ==","size":{"width":174,"height":40}}

R/N = k (Boltzmann constant)

\text{Average kinetic energy of 1 molecule} =\frac{3}{2}KT

It means different gase at the same temperature have the same average kinetic energy per molecule and this energy is directly proportional to the absolute temperature of the gas.

Explanation of Gas Laws from Kinetic Theory

Boyle’s Law

We know

{"backgroundColorModified":false,"type":"$$","font":{"size":11,"family":"Arial","color":"#000000"},"id":"54","backgroundColor":"#ffffff","aid":null,"code":"$$PV\\,=\\,\\frac{1}{3}mN\\bar{v^{2}\\,\\,}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,....\\left(i\\right)$$","ts":1638437739824,"cs":"OT3dkV63V360Uoj7fSgqDg==","size":{"width":204,"height":34}}

According to kinetic theory, the average kinetic energy of translation of molecules of gas is directly proportional to its absolute temperature,

{"backgroundColor":"#ffffff","font":{"size":11,"family":"Arial","color":"#000000"},"backgroundColorModified":false,"id":"55","code":"$$\\frac{1}{2}m\\bar{v^{2}}\\,=\\,\\alpha T$$","aid":null,"type":"$$","ts":1638437936123,"cs":"JLNYkyuncVLjA//HXr9zRw==","size":{"width":96,"height":34}}

i.e., at a given temperature, average kinetic energy of a gas is constant and N is also constant, so eqn (i)

{"backgroundColor":"#ffffff","aid":null,"type":"$$","font":{"color":"#000000","family":"Arial","size":11},"id":"56","code":"$$PV\\,=\\,\\frac{2}{3}N.\\frac{1}{2}m\\bar{v^{2}}\\,=\\,\\frac{2}{3}N\\alpha T$$","backgroundColorModified":false,"ts":1638438112637,"cs":"RnR41LB6IWqnmFOfvc5U/g==","size":{"width":222,"height":34}}

= constant at a given temperature

This is Boyle's Law.

Charles’s law:-

From eqn(i)

{"code":"$$PV\\,=\\,\\frac{1}{3}mN\\bar{v^{2}}$$","aid":null,"id":"57","backgroundColor":"#ffffff","font":{"family":"Arial","size":11,"color":"#000000"},"type":"$$","backgroundColorModified":false,"ts":1638438250704,"cs":"TKSJtSiAiUxRtfJys+f16Q==","size":{"width":113,"height":34}}

Also the average K.E. of a molecule,

{"aid":null,"backgroundColorModified":false,"code":"\\begin{lalign*}\n&{\\frac{1}{2}m\\bar{v^{2}}\\,=\\,\\alpha T}\\\\\n&{PV\\,=\\,\\frac{2}{3}N.\\,\\alpha T}\\\\\n&{V\\,\\propto T}\t\n\\end{lalign*}","font":{"color":"#000000","size":11,"family":"Arial"},"type":"lalign*","backgroundColor":"#ffffff","id":"58","ts":1638438370506,"cs":"KlmD7HztlgybaOTqgz7/Dw==","size":{"width":118,"height":92}}

for constant value P.

Thus at a given pressure for a given mass of a gas, the volume is directly proportional to the absolute temperature. This is Charles' law.

Mean free Path:-The average distance traversed by a molecule between two successive collisions is defined as the mean free path.

Thus, if λ₁, λ₂, λ₃, ….λₙ are the successive free paths traversed in the total time t, then,

{"code":"$$\\lambda_{1}\\,+\\,\\lambda_{2}\\,+.....\\lambda_{n}\\,=\\,\\bar{v}t$$","font":{"color":"#000000","family":"Arial","size":11},"backgroundColorModified":false,"backgroundColor":"#ffffff","id":"59","type":"$$","aid":null,"ts":1638439036360,"cs":"OOqBw6kMEhIYOWKZp3gi3w==","size":{"width":176,"height":13}}

If N is the total no. of collisions suffered, then

{"backgroundColor":"#ffffff","backgroundColorModified":false,"aid":null,"id":"60","font":{"family":"Arial","size":11,"color":"#000000"},"code":"\\begin{lalign*}\n&{\\lambda=\\,\\frac{\\lambda_{1}\\,+\\,\\lambda_{2}\\,+\\,\\lambda_{3}\\,+.....\\lambda_{n}}{N}}\\\\\n&{\\,\\,\\,\\,\\,\\,=\\,\\frac{\\bar{v}t}{N}}\t\n\\end{lalign*}","type":"lalign*","ts":1638439150317,"cs":"a6DYysQKbmTzc+txoM+COQ==","size":{"width":216,"height":74}}

Expression for Mean Free Path

We know that,

{"backgroundColorModified":false,"font":{"color":"#000000","size":11,"family":"Arial"},"id":"61","code":"\\begin{lalign*}\n&{\\lambda=\\,\\frac{total\\,dis\\tan ce\\,covered\\,\\,\\,in\\,time\\,t}{number\\,of\\,collision\\,in\\,time\\,t}}\\\\\n&{\\,\\,\\,\\,\\,=\\,\\frac{\\bar{v}t}{n\\,\\times \\pi d^{2}\\bar{v}t}\\,=\\,\\frac{1}{\\pi nd^{2}}}\t\n\\end{lalign*}","type":"lalign*","backgroundColor":"#ffffff","aid":null,"ts":1638440765158,"cs":"6juKW+659prSD+AWghWemA==","size":{"width":278,"height":78}}

1647002162856

But by experiment Maxwell found that

{"type":"$$","font":{"color":"#000000","size":11,"family":"Arial"},"code":"$$\\lambda=\\,\\frac{1}{{\\sqrt[]{2}}\\left(\\pi \\sigma^{2}n\\right)}$$","aid":null,"backgroundColor":"#ffffff","backgroundColorModified":false,"id":"62","ts":1638440952626,"cs":"ln/J2O7rPYhl5v9il1kNGw==","size":{"width":113,"height":42}}

The mean free path is inversely proportional to the number of molecules per unit volume (n) which is proportional to the density of the gas.

For N molecule of an ideal gas, we have

{"backgroundColor":"#ffffff","aid":null,"font":{"family":"Arial","size":11,"color":"#000000"},"type":"lalign*","id":"63","backgroundColorModified":false,"code":"\\begin{lalign*}\n&{PV\\,=\\,NkT}\\\\\n&{\\therefore\\,n\\,=\\,\\frac{N}{V}\\,=\\,\\frac{P}{kT}}\\\\\n&{\\lambda=\\,\\frac{kT}{{\\sqrt[]{2}}\\pi d^{2}P}}\t\n\\end{lalign*}","ts":1638441177534,"cs":"JmHzNnjZ8FYKVBrjAm85hA==","size":{"width":136,"height":102}}

This is the mean free path of the molecules of a gas is directly proportional to its absolute temperature and inversely proportional to the pressure of the gas.

Avagadro’s Number: The Avagadro’s number is the number of molecules in 1 mole of a substance. It is denoted by N and its value is 6.0221 X 10 ²³.

Significance of NCERT Class 11 Physics Chapter 13 Notes

Kinetic theory Class 11 notes can be used to review the chapter and get a better understanding of the big topics can be used to go over the chapter again and get a better understanding of the major points covered. Also, these NCERT Class 11 Physics chapter 13 notes are useful for the Class 11 CBSE Physics Syllabus and competitive exams such as VITEEE, BITSAT, JEE Main, NEET, etc. Download all these notes from Class 11 Physics chapter 13 notes pdf download or Kinetic theory Class 11 notes pdf download

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Frequently Asked Question (FAQs)

1. What is the law of equipartition of energy?

It states that the total kinetic energy  of a dynamical system consisting of a large number of particles in thermal equilibrium is equally divided among all its degrees of freedom, and the energy contained with each degree of freedom is kT/2 where k signifies the Boltzmann constant and T signifies the system's absolute temperature

2. What is the significance of this chapter for the CBSE Board Exam?

From the CBSE Class 11 Physics chapter 13 notes, Students can expect 2 to 4 marks question. 

3. Write the Kinetic theory assumption of gases from Class 11 Physics chapter 2 notes and Class 11 Physics chapter 13 notes, Kinetic theory Class 11 notes.

Following are the fundamental assumptions of kinetic theory of gases:

  • A gas is composed of a large number of tiny invisible, perfectly elastic particles, called the molecules.

  • The molecules are always in a state of continuous motion with varying velocities in all possible directions.

  • The molecules traverse the straight path between any two collisions. 

  • The size of the molecules is infinitely small compared to the average distance traversed by a molecule between any two consecutive collisions . The distance between any two consecutive collisions is called the free path and the average distance between any two consecutive collisions is called the mean free path.

  • The time of collisions is negligible as compared with the time taken to traverse the free path.

  • The collisions between molecules and with the walls are perfectly elastic so that there is no loss of kinetic energy in the collisions.

  • The molecules exert no force on each other except when they collide and the whole of the molecules energy is kinetic.

  • The volume of the molecules is negligible as compared to the volume of a vessel containing gas.

  • The inter molecular distances in a gas is much larger than that of a solid or liquid and the molecules of a gas are free to move in the entire space available to them.

Articles

Get answers from students and experts

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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A multimedia specialist is a media professional who creates, audio, videos, graphic image files, computer animations for multimedia applications. He or she is responsible for planning, producing, and maintaining websites and applications. 

2 Jobs Available
Welding Engineer

Welding Engineer Job Description: A Welding Engineer work involves managing welding projects and supervising welding teams. He or she is responsible for reviewing welding procedures, processes and documentation. A career as Welding Engineer involves conducting failure analyses and causes on welding issues. 

5 Jobs Available
QA Manager
4 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 Manager
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
QA Lead

A QA Lead is in charge of the QA Team. The role of QA Lead comes with the responsibility of assessing services and products in order to determine that he or she meets the quality standards. He or she develops, implements and manages test plans. 

2 Jobs Available
Structural Engineer

A Structural Engineer designs buildings, bridges, and other related structures. He or she analyzes the structures and makes sure the structures are strong enough to be used by the people. A career as a Structural Engineer requires working in the construction process. It comes under the civil engineering discipline. A Structure Engineer creates structural models with the help of computer-aided design software. 

2 Jobs Available
Process Development Engineer

The Process Development Engineers design, implement, manufacture, mine, and other production systems using technical knowledge and expertise in the industry. They use computer modeling software to test technologies and machinery. An individual who is opting career as Process Development Engineer is responsible for developing cost-effective and efficient processes. They also monitor the production process and ensure it functions smoothly and efficiently.

2 Jobs Available
QA Manager
4 Jobs Available
AWS Solution Architect

An AWS Solution Architect is someone who specializes in developing and implementing cloud computing systems. He or she has a good understanding of the various aspects of cloud computing and can confidently deploy and manage their systems. He or she troubleshoots the issues and evaluates the risk from the third party. 

4 Jobs Available
Azure Administrator

An Azure Administrator is a professional responsible for implementing, monitoring, and maintaining Azure Solutions. He or she manages cloud infrastructure service instances and various cloud servers as well as sets up public and private cloud systems. 

4 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
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
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
ITSM Manager
3 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
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