sign in sign up
<<
Home , Kinetic theory of gases

Kinetic of Kinetic Theory of Gases

The kinetic theory of gases attempts to explain the microscopic properties of a in terms of the of its . The gas is assumed to consist of a large number of identical, discrete called molecules, a being the smallest unit having the same chemical properties as the substance. Elements of kinetic theory were developed by Maxwell, Boltzmann and Clausius between 1860-1880’s. Kinetic are available for gas, solid as well as liquid. However this chapter deals with kinetic theory of gases only. Postulates of kinetic theory of gases

1) Any gas consist large number of molecules. These molecules are identical, perfectly elastic and hard sphere. 2) Gas molecules do not have preferred direction of motion, their motion is completely random. 3) Gas molecules travels in straight line. 4) The time interval of between any two gas molecules is very small. 5) The collision between gas molecules and the walls of container is perfectly elastic. It means and in such collision is conserved. 6) The motion of gas molecules is governed by Newton's laws of motion. 7) The effect of gravity on the motion of gas molecules is negligible.

Prof. Avadhut Surekha Prakash Manage D.M.S. Mandal’s Bhaurao kakatkar College, Belgaum Kinetic Theory of Gases Maxwell’s law of distribution of velocities

This law is to find the number of molecules which have a velocity within small interval (ie. 푐 to 푐 + 푑푐).

For deriving this equation Maxwell did following assumptions.

a) Speed of gas molecules ranges from zero to infinity. b) In the steady state, the density of gas remains constant. c) Though the speed of the individual molecule changes, definite number of gas molecules have speed between definite range.

Consider a gas containing N number of molecules having velocity c and its X, Y and Z components are u, v and w respectively. From the theory, the probability of molecules having velocity component 푢 to 푢 + 푑푢 is 푓(푢)푑푢 similarly the probability of molecules having velocity component 푣 to 푣 + 푑푣 is 푓(푣)푑푣 and the probability of molecules having velocity component w to 푤 + 푑푤 is 푓(푤)푑푤. Thus the number of molecules whose velocity lies between 푢 to +푑푢 , 푣 to 푣 + 푑푣 and w to 푤 + 푑푤 is given by

푑푁 = 푁 푓(푢)푑푢 푓(푣)푑푣 푓(푤)푑푤

푑푁 = 푁 푓(푢) 푓(푣) 푓(푤) 푑푢 푑푣 푑푤 … … . . (1) If velocity components 푢, 푣 and 푤 along the three axis. The space formed is called as the velocity space. A molecule having velocity components 푢, 푣 , 푤 be represented by a point 푝. Let the small 푑푢 푑푣 푑푤 around 푝

Prof. Avadhut Surekha Prakash Manage D.M.S. Mandal’s Bhaurao kakatkar College, Belgaum Kinetic Theory of Gases contains 푑푁 number of molecules. Thus molecules have a resultant velocity c is given by 푐2 = 푢2 + 푣2 + 푤2 Differentiating on both side, we get 0 = 2푢 푑푢 + 2푣 푑푣 + 2푤 푑푤 ∵ 푐 = 푐표푛푠푡푎푛푡 ∴ 푢 푑푢 + 푣 푑푣 + 푤 푑푤 = 0 … … … … … (2)

The probability density 푓(푢) 푓(푣) 푓(푤) should depend only on the magnitude but not the direction of 푐.

∴ 푓(푢) 푓(푣) 푓(푤) = 푓(푐2) (푤ℎ푒푟푒 푓(푐2) 𝑖푠 푓푢푛푐푡𝑖표푛 표푓 푐2) Differentiating on both side, we get

푑[푓(푢) 푓(푣) 푓(푤)] = 0 ∵ 푐2 = 푐표푛푠푡푎푛푡

푓′(푢)푑푢푓(푣)푓(푤) + 푓′(푣) 푑푣푓(푢)푓(푤) + 푓′(푤)푑푤푓(푣)푓(푣) = 0

Dividing above equation by 푓(푢) 푓(푣) 푓(푤), we get 푓′(푢) 푓′(푣) 푓′(푤) 푑푢 + 푑푣 + 푑푤 = 0 ...... (3) 푓(푢) 푓(푣) 푓(푤)

Multiplying the equation (2) by an arbitrary constant (β) and adding to equation (3), we get 푓′(푢) 푓′(푣) 푓′(푤) [ + 훽푢] 푑푢 + [ + 훽푣] 푑푣 + [ + 훽푤] 푑푤 = 0 푓(푢) 푓(푣) 푓(푤)

As 푑푢, 푑푣 & 푑푤 can not be zero independently, so we get 푓′(푢) 푓′(푣) 푓′(푤) + 훽푢 = 0 + 훽푣 = 0 + 훽푤 = 0 푓(푢) 푓(푣) 푓(푤) 푓′(푢) = −훽푢 푓(푢)

Prof. Avadhut Surekha Prakash Manage D.M.S. Mandal’s Bhaurao kakatkar College, Belgaum Kinetic Theory of Gases Integrating wrt u, we get 1 log 푓(푢) = − 훽푢2 + log 푎 2 (푊ℎ푒푟푒 푙표푔 푎 𝑖푠 𝑖푛푡푒푔푟푎푡𝑖표푛 푐표푛푠푡푎푛푡 ) Taking antilog on both side, we get 1 − 훽푢2 2 1 푓(푢) = 푎푒 2 = 푎푒−푏푢 (Where, 푏 = − 훽 = 푐표푛푠푡푎푛푡) 2

2 2 Similarly, 푓(푣) = 푎푒−푏푣 & 푓(푤) = 푎푒−푏푤

Substituting 푓(푢), 푓(푣) & 푓(푤) in eqn (1), we get

2 2 2 푑푁 = 푁 (푎푒−푏푢 ) (푎푒−푏푣 ) (푎푒−푏푤 ) 푑푢 푑푣 푑푤

2 2 2 푑푁 = 푁푎3 푒−푏(푢 +푣 +푤 )푑푢 푑푣 푑푤

∵ 푐2 = 푢2 + 푣2 + 푤2

2 푑푁 = 푁푎3 푒−푏푐 푑푢 푑푣 푑푤 … … … (4)

In this equation constant 푎 and 푏 is found to be

푚 푚 푎 = √ and 푏 = , where 푚 is the of the gas , 2휋푘푇 2푘푇 푘 is and 푇 is of gas

Then equation (4) becomes 3 푚푐2 푚 2 − 푑푁 = 푁 ( ) 푒 2푘푇 푑푢 푑푣 푑푤 … … (5) 2휋푘푇 This is one form of Maxwell’s law of distribution of velocities.

Prof. Avadhut Surekha Prakash Manage D.M.S. Mandal’s Bhaurao kakatkar College, Belgaum Kinetic Theory of Gases The number of molecules lies within box of volume 푑푢 푑푣 푑푤 whose velocity lies within 푐 to 푐 + 푑푐 is the same as number of molecules within the spherical shell of inner radius 푐 and outer radius 푐 + 푑푐. Hence volume 푑푢 푑푣 푑푤 can be replaced volume of the shell 4휋푐2푑푐. Therefore equation (5) becomes

3 푚푐2 푚 2 − 푑푁 = 푁 ( ) 푒 2푘푇 4휋푐2푑푐 2휋푘푇 3 푚푐2 푚 2 − 푑푁 = 4휋푁 ( ) 푒 2푘푇 푐2푑푐 2휋푘푇 The above equation is called as Maxwell’s equation of distribution of velocity which gives the value of number molecules 푑푁 having velocity 푐 to 푐 + 푑푐.

Mean or Average velocity (푪풂풗) The average speed is the sum of all the velocities ranging from 0 to ∞ divided by total number of molecules N. ∞ ∫ 푑푁 푋 푐 ∴ 퐶 = 0 푎푣 푁 Substituting the value of 푑푁 in in above equation 3 푚푐2 ∞ 푚 2 − ∫ (4휋푁 ( ) 푒 2푘푇 푐2푑푐) 푋 푐 0 2휋푘푇 퐶 = 푎푣 푁

Prof. Avadhut Surekha Prakash Manage D.M.S. Mandal’s Bhaurao kakatkar College, Belgaum Kinetic Theory of Gases 3 ∞ 푚푐2 푚 2 − 퐶 = 4휋 ( ) ∫ 푒 2푘푇 푐3푑푐 푎푣 2휋푘푇 0 3 ∞ 푏 2 2 퐶 = 4휋 ( ) ∫ 푒−푏푐 푐3푑푐 푎푣 휋 0 ∞ ퟏ 2 1 ∵ ∫ 푒−푏푐 푐3푑푐 = 2푏2 0 푚 ( Where 푏 = ) 2푘푇 3 푏 2 1 퐶 = 4휋 ( ) 푎푣 휋 2푏2

4 퐶 = √ 푎푣 휋푏

4 ∴ 퐶푎푣 = √ 푚 휋 ( ) 2푘푇

8푘푇 ∴ 퐶 = √ 푎푣 휋푚

Where 푚 is the mass of gas molecule. If 푀 is molecular 푀 weight which can be given by 푀 = 푚푁퐴. Then 푚 = Therefore 푁퐴 the above equation becomes.

8푘푇 퐶 = 푎푣 √ 푀 휋 ( ) 푁퐴

Prof. Avadhut Surekha Prakash Manage D.M.S. Mandal’s Bhaurao kakatkar College, Belgaum Kinetic Theory of Gases

8푘푁 푇 ∴ 퐶 = √ 퐴 푎푣 휋푀

8푅푇 ∴ 퐶 = √ ∵ 푘푁 = 푅 푎푣 휋푀 퐴

Root Mean Square Velocity (푪풓풎풔) The rms speed is the square root of the sum of all the squared velocities ranging from 0 to ∞ divided by total number of molecules N.

∞ ∫ 푑푁 푋 푐2 ∴ 퐶 = √ 0 푟푚푠 푁 Substituting the value of 푑푁 in in above equation

3 푚푐2 ∞ 푚 2 − ∫ (4휋푁 ( ) 푒 2푘푇 푐2푑푐) 푋 푐2 √ 0 2휋푘푇 퐶 = 푟푚푠 푁

3 ∞ 푚푐2 푚 2 − 퐶 = √4휋 ( ) ∫ 푒 2푘푇 푐4푑푐 푟푚푠 2휋푘푇 0

3 ∞ 푏 2 = √4휋 ( ) ∫ 푒−푏푐2 푐4푑푐 휋 0 푚 ( Where 푏 = ) 2푘푇

Prof. Avadhut Surekha Prakash Manage D.M.S. Mandal’s Bhaurao kakatkar College, Belgaum Kinetic Theory of Gases

3 1 푏 2 3 휋 2 = √4휋 ( ) 푋 ( ) 휋 8 푏5

3 = √ 2푏

3 = √ 푚 2 ( ) 2푘푇

3푘푇 퐶 = √ 푟푚푠 푚

푀 Substituting 푚 = in above equation. We get 푁퐴

3푘푇 퐶 = 푎푣 √ 푀 ( ) 푁퐴

3푘푁 푇 ∴ 퐶 = √ 퐴 푎푣 푀

3푅푇 ∴ 퐶 = √ ∵ 푘푁 = 푅 푎푣 푀 퐴

Most Probable Velocity (푪풑풓) The most probable velocity is the velocity possessed by maximum number of molecules. For this condition is,

Prof. Avadhut Surekha Prakash Manage D.M.S. Mandal’s Bhaurao kakatkar College, Belgaum Kinetic Theory of Gases 푑[푑푁] = 0 푑푐 3 푚푐2 푑 푚 2 − ∴ (4휋푁 ( ) 푒 2푘푇 푐2푑푐) = 0 푑푐 2휋푘푇

3 푚푐2 푚 2 푑 − 4휋푁 ( ) ( 푒 2푘푇 푐2푑푐) = 0 2휋푘푇 푑푐

3 푚푐2 푚푐2 푚 2 − 2푚푐 − 4휋푁 ( ) ( 푒 2푘푇 [2푐] + [− ] 푒 2푘푇 푐2) = 0 2휋푘푇 2푘푇

3 푚푐2 푚 2 − 푚푐 4휋푁 ( ) 푒 2푘푇 ( 2푐 − [ ] 푐2) = 0 2휋푘푇 푘푇 푚푐 2푐 − [ ] 푐2 = 0 푘푇 푚푐3 ∴ = 2푐 푘푇 2푘푇 ∴ 푐2 = 푚

2푘푇 ∴ 푐 = √ 푚

Thus most probable velocity,

2푘푇 퐶 = √ 푝푟 푚

푀 Substituting 푚 = in above equation. We get 푁퐴

Prof. Avadhut Surekha Prakash Manage D.M.S. Mandal’s Bhaurao kakatkar College, Belgaum Kinetic Theory of Gases

2푘푇 퐶 = 푎푣 √ 푀 ( ) 푁퐴

2푘푁 푇 ∴ 퐶 = √ 퐴 푎푣 푀

2푅푇 ∴ 퐶 = √ ∵ 푘푁 = 푅 푎푣 푀 퐴

Relation between 푪풑풓 , 푪풂풗 & 푪풓풎풔

2푅푇 8푅푇 3푅푇 퐶 ∶ 퐶 ∶ 퐶 = √ : √ ∶ √ 푝푟 푎푣 푟푚푠 푀 휋푀 푀

8 = √2: √ ∶ √3 휋

2 3 = 1 ∶ ∶ √ √휋 2

= 1: 1.128 ∶ 1.228

Mean free path

Mean free path of gas molecules is defined as the average distance travelled by a molecule between two successive . Prof. Avadhut Surekha Prakash Manage D.M.S. Mandal’s Bhaurao kakatkar College, Belgaum Kinetic Theory of Gases Expression for mean free path

Consider a gas in container having n molecules per unit volume. Let d be the diameter of molecule (A) which is assumed to be in motion, while all other molecules are at rest. The molecule A collides with other molecules like B and C whose centres are at distance d from the centre of molecule A as shown in the figure. If molecule moves a distance L with velocity v in time t, then this molecule collides with all molecules lying inside a cylinder of volume π 푑2퐿. No. of collisions suffered = No. molecules in the cylinder by molecule A of volume π 푑2퐿 = No. of molecules per unit volume

X Volume of cylinder

= 푛 푋 휋 푑2퐿

= 휋 푛 푑2퐿 Now mean free path of molecule is given by λ, 푇표푡푎푙 푑𝑖푠푡푎푛푐푒 푡푟푎푣푒푙푙푒푑 𝑖푛 푡𝑖푚푒 푡 λ = 푁표. 표푓 푐표푙푙𝑖푠𝑖표푛푠 푠푢푓푓푒푟푑 퐿 = 휋 푛 푑2퐿 Prof. Avadhut Surekha Prakash Manage D.M.S. Mandal’s Bhaurao kakatkar College, Belgaum Kinetic Theory of Gases 푣 푡 푣 푎푙푠표, 휆 = = … . . (1) 휋 푛 푑2푣̅ 푡 휋 푛 푑2푣̅ where 푣̅ is average velocity of all particles.

Clausius Method

Clausius derived the mean free path by considering the relative velocities of gas molecules to find average velocity.

Consider two molecules are

moving with velocities 푣1 and 푣2 and θ be the angle between them .Then

relative 푣푟 is given by,

2 2 푣푟 = √푣1 + 푣2 − 2푣1푣2 cos 휃 … . (1)

If 푑푛휃 is the number of molecules per unit volume in between θ to 휃 + 푑휃 with a given direction the average relative velocity 푣̅푟 is given by ∫ 푣 푑푛 푣̅ = 푟 휃 … … … . . (2) 푟 푛 From eqn (1), we get

2 2 ∫(√푣1 + 푣2 − 2푣1푣2 cos 휃)푑푛휃 푣̅ = 푟 푛

Prof. Avadhut Surekha Prakash Manage D.M.S. Mandal’s Bhaurao kakatkar College, Belgaum Kinetic Theory of Gases To find the expression for

푑푛휃 , imagine a sphere of radius r and n be the number of molecules lying in sphere again drawing two cones with semi vertical angle θ & 휃 + 푑휃 then number of molecules

lying in this region is 푑푛휃.

푛 훼 푎푟푒푎 표푓 푠푝ℎ푒푟푒

푑푛휃 훼 푎푟푒푎 표푓 푧표푛푒 푏푒푡푤푒푒푛 휃 to 휃 + 푑휃 푑푛 푎푟푒푎 표푓 푧표푛푒 푏푒푡푤푒푒푛 휃 to 휃 + 푑휃 ∴ 휃 = 푛 푎푟푒푎 표푓 푠푝ℎ푒푟푒 2휋푟 sin 휃 푟푑휃 = 4휋푟2 1 = sin 휃 푑휃 2 1 푑푛 = 푛 sin 휃 푑휃 … … … … . . (3) 휃 2

Substituting 푑푛휃 in equation (3), we get

2 2 1 ∫ √푣1 + 푣2 − 2푣1푣2 cos 휃 ( 푛 sin 휃 푑휃) 푣̅ = 2 푟 푛

1 ∫ √푣 2 + 푣 2 − 2푣 푣 cos 휃 sin 휃 푑휃 푣̅ = 푛 1 2 1 2 푟 2 푛

Prof. Avadhut Surekha Prakash Manage D.M.S. Mandal’s Bhaurao kakatkar College, Belgaum Kinetic Theory of Gases 휋 1 푣̅ = ∫ √푣 2 + 푣 2 − 2푣 푣 cos 휃 sin 휃 푑휃 푟 2 1 2 1 2 0

3 휋 1 1 2 2 푣̅푟 = ( [(푣1 + 푣2 − 2푣1푣2 cos 휃)2] ) 2 3푣1푣2 0

3 3 1 2 2 2 2 푣̅푟 = [(푣1 + 푣2 + 2푣1푣2)2 − (푣1 + 푣2 − 2푣1푣2)2] 6푣1푣2

1 3 3 푣̅푟 = [(푣1 + 푣2) − (푣1 − 푣2) ] 6푣1푣2

1 3 3 2 2 푣̅푟 = [(푣1 + 푣2 + 3푣1 푣2 + 3푣1푣2 ) 6푣1푣2 3 3 2 2 − (푣1 − 푣2 − 3푣1 푣2 + 3푣1푣2 )]

1 3 2 푣̅푟 = (2푣2 + 6푣1 푣2) 6푣1푣2

1 2 2 푣̅푟 = (푣2 + 3푣1 ) … … . (4) 3푣1 Since close gas molecules have almost same velocity.

Hence we can assume that 푣2 ≈ 푣1 ≈ 푣 푣2 + 3푣2 ∴ 푣̅ = 푟 3푣 4 ∴ 푣̅ = 푣 푟 3

Substituting 푣̅푟 in mean free path, 휆. We get 푣 휆 = 2 4 휋 푛 푑 (3 푣)

Prof. Avadhut Surekha Prakash Manage D.M.S. Mandal’s Bhaurao kakatkar College, Belgaum Kinetic Theory of Gases 3 휆 = 4휋 푛 푑2

Maxwell Expression

Maxwell has derived the mean free path by considering the probability of occurrence of the velocities gas molecules to find average velocity.

Introducing probability occurrence 푣2, we get an intermediate average relative velocity 푣1̅ . ∞ ∫ 푣̅푑푁2 푣̅ = 0 1 푁

Where 푑푁2 is the number of molecules whose velocity ranging from 푣2 to 푣2 + 푑푣2

3 ∞ 2 2 푏 −푏푣2 2 ∫ 푣̅ 4휋푁 ( ) 푒 푣2 푑푣2 ∴ 푣̅ = 0 휋 1 푁 . 3 2 ∞ 2 푏 −푏푣2 2 4휋푁 ( ) ∫ 푣̅ 푒 푣2 푑푣2 ∴ 푣̅ = 휋 0 1 푁

3 ∞ 푏 2 2 ∴ 푣̅ = 4휋 ( ) ∫ 푣̅ 푒−푏푣2 푣 2푑푣 1 휋 2 2 0

Prof. Avadhut Surekha Prakash Manage D.M.S. Mandal’s Bhaurao kakatkar College, Belgaum Kinetic Theory of Gases 2 2 푣2 +3푣1 Since 푣̅ takes different values in for 푣1 > 푣2 ( )and 푣2 > 3푣1 2 2 푣1 +3푣2 푣1 ( ) 3푣2

3 푐1 2 2 푏 2 푣2 + 3푣1 2 −푏푣2 2 ∴ 푣1̅ = 4휋 ( ) {∫ ( ) 푒 푣2 푑푣2 휋 3푣1 0 ∞ 2 2 푣1 + 3푣2 2 −푏푣2 2 + ∫ ( ) 푒 푣2 푑푣2} 3푣2 푐1

Introducing the of occurrence of 푣1 , we get the final average 푣̅2 for relative velocity

3 ∞ 푏 2 2 푣̅ = 4휋 ( ) ∫ 푣̅ 푒−푏푣1 푣 2푑푣 2 휋 1 1 1 0

3 ∞ 3 푐1 2 2 푏 2 푏 2 푣2 + 3푣1 2 −푏푣2 2 ∴ 푣̅2 = 4휋 ( ) ∫ [4휋 ( ) {∫ ( ) 푒 푣2 푑푣2 휋 휋 3푣1 0 0 ∞ 2 2 푣1 + 3푣2 2 2 −푏푣2 2 −푏푣1 2 + ∫ ( ) 푒 푣2 푑푣2}] 푒 푣1 푑푣1 3푣2 푐1

∞ 푐1 3 2 2 16푏 푣2 + 3푣1 2 −푏푣2 2 ∴ 푣̅2 = ∫ [{∫ ( ) 푒 푣2 푑푣2 휋 3푣1 0 0 ∞ 2 2 푣1 + 3푣2 2 2 −푏푣2 2 −푏푣1 2 + ∫ ( ) 푒 푣2 푑푣2}] 푒 푣1 푑푣1 3푣2 푐1 16푏3 ∴ 푣̅ = [ 퐽 + 퐽 ] 2 휋 1 2 Prof. Avadhut Surekha Prakash Manage D.M.S. Mandal’s Bhaurao kakatkar College, Belgaum Kinetic Theory of Gases Where ∞ 푐1 2 2 푣2 + 3푣1 2 2 −푏푣2 2 −푏푣1 2 퐽1 = ∫ {∫ ( ) 푒 푣2 푑푣2} 푒 푣1 푑푣1 3푣1 0 0

&

∞ ∞ 2 2 푣1 + 3푣2 2 2 −푏푣2 2 −푏푣1 2 퐽2 = ∫ { ∫ ( ) 푒 푣2 푑푣2} 푒 푣1 푑푣1 3푣2 0 푐1

Integral 퐽1& 퐽2 can be calculated and given by

1 ퟐ 1 휋 2 퐽1 = 퐽2 = ( ) 8√2 푏7 16푏3 ∴ 푣̅ = [ 퐽 + 퐽 ] 2 휋 1 2

3 1 16푏 1 휋 2 ∴ 푣̅2 = [ 2 푋 ( ) ] 휋 8√2 푏7

4 8 ∴ 푣̅2 = = √ √2휋푏 휋푏

8 ∴ 푣̅2 = √ 푚 휋 ( ) 2푘푇

2푋 8푘푇 ∴ 푣̅ = √ 2 휋푚

Prof. Avadhut Surekha Prakash Manage D.M.S. Mandal’s Bhaurao kakatkar College, Belgaum Kinetic Theory of Gases

8푘푇 ∴ 푣̅ = √2√ 2 휋푚

8푘푇 ∴ 푣̅ = √2√ 2 휋푚

∴ 푣̅2 = √2푣 Where 푣 is the average velocity of molecule having mass 푚

Therefore the mean free path, 휆 becomes 푣 휆 = 휋 푛 푑2(√2푣) 1 휆 = √2휋 푛 푑2

Brownian Motion

Botanist Brown observed under a high power microscope that when pollen grains suspended in liquid were moving rapidly and constantly in random fashion in all direction. This irregular motion of particles suspended in liquid is called .

Salient features of Brownian motion i) The motion is continuous , irregular and random, ii) The motion is independent of mechanical vibration. iii) The smaller the size of the particle, greater is the motion and vice-versa.

Prof. Avadhut Surekha Prakash Manage D.M.S. Mandal’s Bhaurao kakatkar College, Belgaum Kinetic Theory of Gases iv) The lower is the of the liquid, greater is the motion and vice-versa. v) The higher is the temperature, greater is the motion and vice- versa.

Einstein’s theory of Brownian motion

Brownian motion is acted upon three kind of

1) The unbalanced due to collision of unbalanced molecules in the opposite direction. 2) The viscous drag which opposes the Brownian motion. 3) Force due to difference in the osmotic ( ie. due to the difference in the concentration)

Imagine a cylinder with its axis along X axis whose surfaces P and Q are separated by a distance Δ( Δ is root mean square distance). Let A be the area of the cylinder. Let n1 and n2 be the concentration of the particles per unit volume at the surfaces P and Q respectively.

Imagine two more cylinders on both sides of P and Q as shown in the figure.

The number of Brownian particles in the cylinder on the P side = 푛1∆ 퐴 Prof. Avadhut Surekha Prakash Manage D.M.S. Mandal’s Bhaurao kakatkar College, Belgaum Kinetic Theory of Gases The number of Brownian particles crossing at P side

1 = 푛 ∆ 퐴 2 1 The number of Brownian particles in the cylinder on the Q side = 푛2∆ 퐴 The number of Brownian particles crossing at Q side 1 = 푛 ∆ 퐴 2 2 So the total number of particles crossing the middle section R 1 from left to right = (푛 − 푛 ) ∆ 퐴 . . (1) 2 1 2 ퟑ By the definition of co-efficient, the number of particles crossing the middle section R from left to right 푑푛 =−퐷 퐴 푡 … … (2) 푑푥

푑푛 Where D= diffusion coefficient, - concentration grdient and 푑푥 t is the time taken for diffusion.

From equation (1)&(2) 1 푑푛 (푛 − 푛 ) ∆ 퐴 = −퐷 퐴 푡 2 1 2 푑푥 푑푛 Where (푛 − 푛 ) = − 훥 1 2 푑푥 1 푑푛 푑푛 ∴ − 훥2 퐴 = −퐷 퐴 푡 2 푑푥 푑푥 훥2 ∴ 퐷 = … . (3) 2푡

Prof. Avadhut Surekha Prakash Manage D.M.S. Mandal’s Bhaurao kakatkar College, Belgaum Kinetic Theory of Gases ퟒ Let P1 and P2 be the osmotic pressure acting on the surfaces P and Q respectively.

Then 푃1 = 푛1푘푇 & 푃2 = 푛2푘푇

∴ 푃푟푒푠푠푢푟푒 푑𝑖푓푓푒푟푒푛푐푒 = 푃1 − 푃2 = (푛1 − 푛2)푘푇 The force exerted due to this pressure difference

= (푛1 − 푛2)푘푇퐴 This force acts on all the particles in the cylinder PQ in which 푛 +푛 the number of particles, where 푛 = 1 2 2 (푛 − 푛 )푘푇퐴 ∴ Force experience by each particle, 퐹 = 1 2 nΔA (푛 − 푛 )푘푇 ∴ 퐹 = 1 2 nΔ (푛 − 푛 ) 푑푛 ∵ 1 2 = − Δ 푑푥 푘푇 푑푛 ∴ 퐹 = − … . . (4) n 푑푥 ퟓ Simultaneously particle also experience viscous drag (stokes law) , which is given by

퐹 = 6휋휂푟푣 … . . (5)

Where η is the coefficient of viscosity

푟 is the radius of particles

푣 is the velocity of the particles

Prof. Avadhut Surekha Prakash Manage D.M.S. Mandal’s Bhaurao kakatkar College, Belgaum Kinetic Theory of Gases From equation (4) & (5) 푘푇 푑푛 퐹 = − = 6휋휂푟푣 n 푑푥 푛푣 푘푇 = − 푑푛 6휋휂푟 푑푥 Where 푛푣 is number of particles moving from left to right 푛푣 throgh unit area in unit time and hence 푑푛 gives diffusion − 푑푥 coefficient D 푘푇 ∴ 퐷 = … . . (6) 6휋휂푟

From equation (3) & (6)

훥2 푘푇 = 2푡 6휋휂푟 푅 ∵ 푘 = 푁퐴 훥2 푅푇 ∴ = 2푡 6휋푁휂푟 푅푇 푡 ∴ 훥2 = 푁 3휋휂푟 The above equation is Einstein’s equation which gives the value of square of rms displacement of Brownian particles.

Prof. Avadhut Surekha Prakash Manage D.M.S. Mandal’s Bhaurao kakatkar College, Belgaum