Kinetic Theory of Gases - RCUB

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Kinetic Theory of Gases Prof. Avadhut Surekha Prakash Manage D.M.S. Mandal’s Bhaurao kakatkar College, Belgaum Kinetic Theory of Gases The kinetic theory of gases attempts to explain the microscopic properties of a gas in terms of the motion of its molecules. The gas is assumed to consist of a large number of identical, discrete particles called molecules, a molecule 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 theories 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 collision between any two gas molecules is very small. 5) The collision between gas molecules and the walls of container is perfectly elastic. It means kinetic energy and momentum 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.

Transcript of Kinetic Theory of Gases - RCUB

Page 1: Kinetic Theory of Gases - RCUB

Kinetic Theory of Gases

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

Kinetic Theory of Gases The kinetic theory of gases attempts to explain the

microscopic properties of a gas in terms of the motion of its

molecules. The gas is assumed to consist of a large number of

identical, discrete particles called molecules, a molecule 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

theories 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 collision between any two gas molecules

is very small.

5) The collision between gas molecules and the walls of

container is perfectly elastic. It means kinetic energy and

momentum 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.

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Kinetic Theory of Gases

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

Maxwell’s law of distribution of velocities

This law is to find the number of molecules which have a

velocity within small interval (ie. 𝑐 to 𝑐 + 𝑑𝑐).

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 probability theory, the probability of

molecules having velocity component 𝑒 to 𝑒 + 𝑑𝑒 is 𝑓(𝑒)𝑑𝑒

similarly the probability of molecules having velocity component

𝑣 to 𝑣 + 𝑑𝑣 is 𝑓(𝑣)𝑑𝑣 and the probability of molecules having

velocity component w to 𝑀 + 𝑑𝑀 is 𝑓(𝑀)𝑑𝑀. Thus the number of

molecules whose velocity lies between 𝑒 to +𝑑𝑒 , 𝑣 to 𝑣 + 𝑑𝑣 and

w to 𝑀 + 𝑑𝑀 is given by

𝑑𝑁 = 𝑁 𝑓(𝑒)𝑑𝑒 𝑓(𝑣)𝑑𝑣 𝑓(𝑀)𝑑𝑀

𝑑𝑁 = 𝑁 𝑓(𝑒) 𝑓(𝑣) 𝑓(𝑀) 𝑑𝑒 𝑑𝑣 𝑑𝑀 … … . . (1)

If velocity components 𝑒, 𝑣 and 𝑀

along the three axis. The space formed

is called as the velocity space. A

molecule having velocity components

𝑒, 𝑣 , 𝑀 be represented by a point 𝑝. Let

the small volume 𝑑𝑒 𝑑𝑣 𝑑𝑀 around 𝑝

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Kinetic Theory of Gases

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

contains 𝑑𝑁 number of molecules. Thus molecules have a resultant

velocity c is given by

𝑐2 = 𝑒2 + 𝑣2 + 𝑀2

Differentiating on both side, we get

0 = 2𝑒 𝑑𝑒 + 2𝑣 𝑑𝑣 + 2𝑀 𝑑𝑀 ∡ 𝑐 = π‘π‘œπ‘›π‘ π‘‘π‘Žπ‘›π‘‘

∴ 𝑒 𝑑𝑒 + 𝑣 𝑑𝑣 + 𝑀 𝑑𝑀 = 0 … … … … … (2)

The probability density 𝑓(𝑒) 𝑓(𝑣) 𝑓(𝑀) should depend only

on the magnitude but not the direction of 𝑐.

∴ 𝑓(𝑒) 𝑓(𝑣) 𝑓(𝑀) = 𝑓(𝑐2) (π‘€β„Žπ‘’π‘Ÿπ‘’ 𝑓(𝑐2) 𝑖𝑠 π‘“π‘’π‘›π‘π‘‘π‘–π‘œπ‘› π‘œπ‘“ 𝑐2)

Differentiating on both side, we get

𝑑[𝑓(𝑒) 𝑓(𝑣) 𝑓(𝑀)] = 0 ∡ 𝑐2 = π‘π‘œπ‘›π‘ π‘‘π‘Žπ‘›π‘‘

𝑓′(𝑒)𝑑𝑒𝑓(𝑣)𝑓(𝑀) + 𝑓′(𝑣) 𝑑𝑣𝑓(𝑒)𝑓(𝑀) + 𝑓′(𝑀)𝑑𝑀𝑓(𝑣)𝑓(𝑣) = 0

Dividing above equation by 𝑓(𝑒) 𝑓(𝑣) 𝑓(𝑀), we get

𝑓′(𝑒)

𝑓(𝑒)𝑑𝑒 +

𝑓′(𝑣)

𝑓(𝑣)𝑑𝑣 +

𝑓′(𝑀)

𝑓(𝑀)𝑑𝑀 = 0 . . . . . . . . . . (3)

Multiplying the equation (2) by an arbitrary constant (Ξ²)

and adding to equation (3), we get

[𝑓′(𝑒)

𝑓(𝑒)+ 𝛽𝑒] 𝑑𝑒 + [

𝑓′(𝑣)

𝑓(𝑣)+ 𝛽𝑣] 𝑑𝑣 + [

𝑓′(𝑀)

𝑓(𝑀)+ 𝛽𝑀] 𝑑𝑀 = 0

As 𝑑𝑒, 𝑑𝑣 & 𝑑𝑀 can not be zero independently, so we get

𝑓′(𝑒)

𝑓(𝑒)+ 𝛽𝑒 = 0

𝑓′(𝑣)

𝑓(𝑣)+ 𝛽𝑣 = 0

𝑓′(𝑀)

𝑓(𝑀)+ 𝛽𝑀 = 0

𝑓′(𝑒)

𝑓(𝑒)= βˆ’π›½π‘’

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Kinetic Theory of Gases

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

Integrating wrt u, we get

log 𝑓(𝑒) = βˆ’1

2 𝛽𝑒2 + log π‘Ž

(π‘Šβ„Žπ‘’π‘Ÿπ‘’ π‘™π‘œπ‘” π‘Ž 𝑖𝑠 π‘–π‘›π‘‘π‘’π‘”π‘Ÿπ‘Žπ‘‘π‘–π‘œπ‘› π‘π‘œπ‘›π‘ π‘‘π‘Žπ‘›π‘‘ )

Taking antilog on both side, we get

𝑓(𝑒) = π‘Žπ‘’βˆ’12

𝛽𝑒2

= π‘Žπ‘’βˆ’π‘π‘’2 (Where, 𝑏 = βˆ’

1

2 𝛽 = π‘π‘œπ‘›π‘ π‘‘π‘Žπ‘›π‘‘)

Similarly, 𝑓(𝑣) = π‘Žπ‘’βˆ’π‘π‘£2 & 𝑓(𝑀) = π‘Žπ‘’βˆ’π‘π‘€2

Substituting 𝑓(𝑒), 𝑓(𝑣) & 𝑓(𝑀) in eqn (1), we get

𝑑𝑁 = 𝑁 (π‘Žπ‘’βˆ’π‘π‘’2) (π‘Žπ‘’βˆ’π‘π‘£2

) (π‘Žπ‘’βˆ’π‘π‘€2) 𝑑𝑒 𝑑𝑣 𝑑𝑀

𝑑𝑁 = π‘π‘Ž3 π‘’βˆ’π‘(𝑒2+𝑣2+𝑀2)𝑑𝑒 𝑑𝑣 𝑑𝑀

∡ 𝑐2 = 𝑒2 + 𝑣2 + 𝑀2

𝑑𝑁 = π‘π‘Ž3 π‘’βˆ’π‘π‘2𝑑𝑒 𝑑𝑣 𝑑𝑀 … … … (4)

In this equation constant π‘Ž and 𝑏 is found to be

π‘Ž = βˆšπ‘š

2πœ‹π‘˜π‘‡ and 𝑏 =

π‘š

2π‘˜π‘‡ , where π‘š is the mass of the gas particle,

π‘˜ is Boltzmann constant and 𝑇 is temperature of gas

Then equation (4) becomes

𝑑𝑁 = 𝑁 (π‘š

2πœ‹π‘˜π‘‡)

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π‘’βˆ’ π‘šπ‘2

2π‘˜π‘‡ 𝑑𝑒 𝑑𝑣 𝑑𝑀 … … (5)

This is one form of Maxwell’s law of distribution of

velocities.

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Kinetic Theory of Gases

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

The number of molecules lies

within box of volume 𝑑𝑒 𝑑𝑣 𝑑𝑀 whose

velocity lies within 𝑐 to 𝑐 + 𝑑𝑐 is the

same as number of molecules within

the spherical shell of inner radius 𝑐 and

outer radius 𝑐 + 𝑑𝑐. Hence volume

𝑑𝑒 𝑑𝑣 𝑑𝑀 can be replaced volume of the

shell 4πœ‹π‘2𝑑𝑐. Therefore equation (5)

becomes

𝑑𝑁 = 𝑁 (π‘š

2πœ‹π‘˜π‘‡)

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π‘’βˆ’ π‘šπ‘2

2π‘˜π‘‡ 4πœ‹π‘2𝑑𝑐

𝑑𝑁 = 4πœ‹π‘ (π‘š

2πœ‹π‘˜π‘‡)

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π‘’βˆ’ π‘šπ‘2

2π‘˜π‘‡ 𝑐2𝑑𝑐

The above equation is called as Maxwell’s equation of

distribution of velocity which gives the value of number

molecules 𝑑𝑁 having velocity 𝑐 to 𝑐 + 𝑑𝑐.

Mean or Average velocity (π‘ͺ𝒂𝒗)

The average speed is the sum of all the velocities ranging from 0 to ∞ divided by total number of molecules N.

∴ πΆπ‘Žπ‘£ =∫ 𝑑𝑁 𝑋 𝑐

∞

0

𝑁

Substituting the value of 𝑑𝑁 in in above equation

πΆπ‘Žπ‘£ =

∫ (4πœ‹π‘ (π‘š

2πœ‹π‘˜π‘‡)

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π‘’βˆ’ π‘šπ‘2

2π‘˜π‘‡ 𝑐2𝑑𝑐) 𝑋 π‘βˆž

0

𝑁

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Kinetic Theory of Gases

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

πΆπ‘Žπ‘£ = 4πœ‹ (π‘š

2πœ‹π‘˜π‘‡)

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∫ π‘’βˆ’ π‘šπ‘2

2π‘˜π‘‡ 𝑐3𝑑𝑐

∞

0

πΆπ‘Žπ‘£ = 4πœ‹ (𝑏

πœ‹)

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∫ π‘’βˆ’π‘π‘2 𝑐3𝑑𝑐

∞

0

𝟏

∡ ∫ π‘’βˆ’π‘π‘2 𝑐3𝑑𝑐

∞

0

=1

2𝑏2

( Where 𝑏 =π‘š

2π‘˜π‘‡ )

πΆπ‘Žπ‘£ = 4πœ‹ (𝑏

πœ‹)

32 1

2𝑏2

πΆπ‘Žπ‘£ = √4

πœ‹π‘

∴ πΆπ‘Žπ‘£ = √4

πœ‹ (π‘š

2π‘˜π‘‡)

∴ πΆπ‘Žπ‘£ = √8π‘˜π‘‡

πœ‹π‘š

Where π‘š is the mass of gas molecule. If 𝑀 is molecular

weight which can be given by 𝑀 = π‘šπ‘π΄. Then π‘š =𝑀

𝑁𝐴 Therefore

the above equation becomes.

πΆπ‘Žπ‘£ = √8π‘˜π‘‡

πœ‹ (𝑀𝑁𝐴

)

Page 7: Kinetic Theory of Gases - RCUB

Kinetic Theory of Gases

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

∴ πΆπ‘Žπ‘£ = √8π‘˜π‘π΄π‘‡

πœ‹π‘€

∴ πΆπ‘Žπ‘£ = √8𝑅𝑇

πœ‹π‘€ ∡ π‘˜π‘π΄ = 𝑅

Root Mean Square Velocity (π‘ͺπ’“π’Žπ’”)

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.

∴ πΆπ‘Ÿπ‘šπ‘  = √∫ 𝑑𝑁 𝑋 𝑐2∞

0

𝑁

Substituting the value of 𝑑𝑁 in in above equation

πΆπ‘Ÿπ‘šπ‘  =√∫ (4πœ‹π‘ (

π‘š2πœ‹π‘˜π‘‡

)

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π‘’βˆ’ π‘šπ‘2

2π‘˜π‘‡ 𝑐2𝑑𝑐) 𝑋 𝑐2∞

0

𝑁

πΆπ‘Ÿπ‘šπ‘  = √4πœ‹ (π‘š

2πœ‹π‘˜π‘‡)

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∫ π‘’βˆ’ π‘šπ‘2

2π‘˜π‘‡ 𝑐4𝑑𝑐

∞

0

= √4πœ‹ (𝑏

πœ‹)

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∫ π‘’βˆ’π‘π‘2 𝑐4𝑑𝑐

∞

0

( Where 𝑏 =π‘š

2π‘˜π‘‡ )

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Kinetic Theory of Gases

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

= √4πœ‹ (𝑏

πœ‹)

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𝑋 3

8(

πœ‹

𝑏5)

12

= √3

2𝑏

= √3

2 (π‘š

2π‘˜π‘‡)

πΆπ‘Ÿπ‘šπ‘  = √3π‘˜π‘‡

π‘š

Substituting π‘š =𝑀

𝑁𝐴 in above equation. We get

πΆπ‘Žπ‘£ = √3π‘˜π‘‡

(𝑀𝑁𝐴

)

∴ πΆπ‘Žπ‘£ = √3π‘˜π‘π΄π‘‡

𝑀

∴ πΆπ‘Žπ‘£ = √3𝑅𝑇

𝑀 ∡ π‘˜π‘π΄ = 𝑅

Most Probable Velocity (π‘ͺ𝒑𝒓)

The most probable velocity is the velocity possessed by maximum number of molecules. For this condition is,

Page 9: Kinetic Theory of Gases - RCUB

Kinetic Theory of Gases

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

𝑑[𝑑𝑁]

𝑑𝑐= 0

βˆ΄π‘‘

𝑑𝑐(4πœ‹π‘ (

π‘š

2πœ‹π‘˜π‘‡)

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π‘’βˆ’ π‘šπ‘2

2π‘˜π‘‡ 𝑐2𝑑𝑐) = 0

4πœ‹π‘ (π‘š

2πœ‹π‘˜π‘‡)

32 𝑑

𝑑𝑐( π‘’βˆ’

π‘šπ‘2

2π‘˜π‘‡ 𝑐2𝑑𝑐) = 0

4πœ‹π‘ (π‘š

2πœ‹π‘˜π‘‡)

32

( π‘’βˆ’ π‘šπ‘2

2π‘˜π‘‡ [2𝑐] + [βˆ’ 2π‘šπ‘

2π‘˜π‘‡] π‘’βˆ’

π‘šπ‘2

2π‘˜π‘‡ 𝑐2) = 0

4πœ‹π‘ (π‘š

2πœ‹π‘˜π‘‡)

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π‘’βˆ’ π‘šπ‘2

2π‘˜π‘‡ ( 2𝑐 βˆ’ [ π‘šπ‘

π‘˜π‘‡] 𝑐2) = 0

2𝑐 βˆ’ [ π‘šπ‘

π‘˜π‘‡] 𝑐2 = 0

∴ π‘šπ‘3

π‘˜π‘‡= 2𝑐

∴ 𝑐2 =2π‘˜π‘‡

π‘š

∴ 𝑐 = √2π‘˜π‘‡

π‘š

Thus most probable velocity,

πΆπ‘π‘Ÿ = √2π‘˜π‘‡

π‘š

Substituting π‘š =𝑀

𝑁𝐴 in above equation. We get

Page 10: Kinetic Theory of Gases - RCUB

Kinetic Theory of Gases

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

πΆπ‘Žπ‘£ = √2π‘˜π‘‡

(𝑀𝑁𝐴

)

∴ πΆπ‘Žπ‘£ = √2π‘˜π‘π΄π‘‡

𝑀

∴ πΆπ‘Žπ‘£ = √2𝑅𝑇

𝑀 ∡ π‘˜π‘π΄ = 𝑅

Relation between π‘ͺ𝒑𝒓 , π‘ͺ𝒂𝒗 & π‘ͺπ’“π’Žπ’”

πΆπ‘π‘Ÿ ∢ πΆπ‘Žπ‘£ ∢ πΆπ‘Ÿπ‘šπ‘  = √2𝑅𝑇

𝑀: √

8𝑅𝑇

πœ‹π‘€βˆΆ √

3𝑅𝑇

𝑀

= √2: √8

πœ‹βˆΆ √3

= 1 ∢ 2

βˆšπœ‹βˆΆ √

3

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

collisions.

Page 11: Kinetic Theory of Gases - RCUB

Kinetic Theory of Gases

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

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 Ο€ 𝑑2𝐿.

No. of collisions suffered = No. molecules in the cylinder

by molecule A of volume Ο€ 𝑑2𝐿

= No. of molecules per unit volume

X Volume of cylinder

= 𝑛 𝑋 πœ‹ 𝑑2𝐿

= πœ‹ 𝑛 𝑑2𝐿

Now mean free path of molecule is given by Ξ»,

Ξ» =π‘‡π‘œπ‘‘π‘Žπ‘™ π‘‘π‘–π‘ π‘‘π‘Žπ‘›π‘π‘’ π‘‘π‘Ÿπ‘Žπ‘£π‘’π‘™π‘™π‘’π‘‘ 𝑖𝑛 π‘‘π‘–π‘šπ‘’ 𝑑

π‘π‘œ. π‘œπ‘“ π‘π‘œπ‘™π‘™π‘–π‘ π‘–π‘œπ‘›π‘  π‘ π‘’π‘“π‘“π‘’π‘Ÿπ‘‘

=𝐿

πœ‹ 𝑛 𝑑2𝐿

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Kinetic Theory of Gases

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

π‘Žπ‘™π‘ π‘œ, πœ† =𝑣 𝑑

πœ‹ 𝑛 𝑑2οΏ½Μ…οΏ½ 𝑑=

𝑣

πœ‹ 𝑛 𝑑2�̅�… . . (1)

where οΏ½Μ…οΏ½ 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 𝑣1 and 𝑣2 and ΞΈ

be the angle between them .Then

relative π‘£π‘Ÿ is given by,

π‘£π‘Ÿ = βˆšπ‘£12 + 𝑣2

2 βˆ’ 2𝑣1𝑣2 cos πœƒ … . (1)

If π‘‘π‘›πœƒ is the number of molecules per unit volume in

between ΞΈ to πœƒ + π‘‘πœƒ with a given direction the average relative

velocity 𝑣�̅� is given by

𝑣�̅� =∫ π‘£π‘Ÿπ‘‘π‘›πœƒ

𝑛 … … … . . (2)

From eqn (1), we get

𝑣�̅� =∫(βˆšπ‘£1

2 + 𝑣22 βˆ’ 2𝑣1𝑣2 cos πœƒ)π‘‘π‘›πœƒ

𝑛

Page 13: Kinetic Theory of Gases - RCUB

Kinetic Theory of Gases

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

To find the expression for

π‘‘π‘›πœƒ , imagine a sphere of radius r

and n be the number of molecules

lying in sphere again drawing two

cones with semi vertical angle ΞΈ &

πœƒ + π‘‘πœƒ then number of molecules

lying in this region is π‘‘π‘›πœƒ .

𝑛 𝛼 π‘Žπ‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘ π‘β„Žπ‘’π‘Ÿπ‘’

π‘‘π‘›πœƒ 𝛼 π‘Žπ‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘§π‘œπ‘›π‘’ 𝑏𝑒𝑑𝑀𝑒𝑒𝑛 πœƒ to πœƒ + π‘‘πœƒ

βˆ΄π‘‘π‘›πœƒ

𝑛=

π‘Žπ‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘§π‘œπ‘›π‘’ 𝑏𝑒𝑑𝑀𝑒𝑒𝑛 πœƒ to πœƒ + π‘‘πœƒ

π‘Žπ‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘ π‘β„Žπ‘’π‘Ÿπ‘’

=2πœ‹π‘Ÿ sin πœƒ π‘Ÿπ‘‘πœƒ

4πœ‹π‘Ÿ2

=1

2sin πœƒ π‘‘πœƒ

π‘‘π‘›πœƒ =1

2𝑛 sin πœƒ π‘‘πœƒ … … … … . . (3)

Substituting π‘‘π‘›πœƒ in equation (3), we get

𝑣�̅� =∫ βˆšπ‘£1

2 + 𝑣22 βˆ’ 2𝑣1𝑣2 cos πœƒ (

12 𝑛 sin πœƒ π‘‘πœƒ)

𝑛

𝑣�̅� =1

2𝑛

∫ βˆšπ‘£12 + 𝑣2

2 βˆ’ 2𝑣1𝑣2 cos πœƒ sin πœƒ π‘‘πœƒ

𝑛

Page 14: Kinetic Theory of Gases - RCUB

Kinetic Theory of Gases

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

𝑣�̅� =1

2∫ βˆšπ‘£1

2 + 𝑣22 βˆ’ 2𝑣1𝑣2 cos πœƒ sin πœƒ π‘‘πœƒ

πœ‹

0

𝑣�̅� =1

2(

1

3𝑣1𝑣2[(𝑣1

2 + 𝑣22 βˆ’ 2𝑣1𝑣2 cos πœƒ)

32]

0

πœ‹

)

𝑣�̅� =1

6𝑣1𝑣2[(𝑣1

2 + 𝑣22 + 2𝑣1𝑣2)

32 βˆ’ (𝑣1

2 + 𝑣22 βˆ’ 2𝑣1𝑣2)

32]

𝑣�̅� =1

6𝑣1𝑣2

[(𝑣1 + 𝑣2)3 βˆ’ (𝑣1 βˆ’ 𝑣2)3]

𝑣�̅� =1

6𝑣1𝑣2

[(𝑣13 + 𝑣2

3 + 3𝑣12𝑣2 + 3𝑣1𝑣2

2)

βˆ’ (𝑣13 βˆ’ 𝑣2

3 βˆ’ 3𝑣12𝑣2 + 3𝑣1𝑣2

2)]

𝑣�̅� =1

6𝑣1𝑣2

(2𝑣23 + 6𝑣1

2𝑣2)

𝑣�̅� =1

3𝑣1

(𝑣22 + 3𝑣1

2) … … . (4)

Since close gas molecules have almost same velocity.

Hence we can assume that 𝑣2 β‰ˆ 𝑣1 β‰ˆ 𝑣

∴ 𝑣�̅� =𝑣2 + 3𝑣2

3𝑣

∴ 𝑣�̅� =4

3𝑣

Substituting 𝑣�̅� in mean free path, πœ†. We get

πœ† =𝑣

πœ‹ 𝑛 𝑑2 (43 𝑣)

Page 15: Kinetic Theory of Gases - RCUB

Kinetic Theory of Gases

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

πœ† =3

4πœ‹ 𝑛 𝑑2

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 𝑣2, we get an

intermediate average relative velocity οΏ½Μ…οΏ½1

.

οΏ½Μ…οΏ½1 =∫ �̅�𝑑𝑁2

∞

0

𝑁

Where 𝑑𝑁2 is the number of molecules whose velocity ranging

from 𝑣2 to 𝑣2 + 𝑑𝑣2

∴ οΏ½Μ…οΏ½1 =∫ οΏ½Μ…οΏ½ 4πœ‹π‘ (

π‘πœ‹)

32

π‘’βˆ’π‘π‘£22 𝑣2

2𝑑𝑣2∞

0

𝑁

.

∴ οΏ½Μ…οΏ½1 =4πœ‹π‘ (

π‘πœ‹)

32

∫ οΏ½Μ…οΏ½ π‘’βˆ’π‘π‘£22 𝑣2

2𝑑𝑣2∞

0

𝑁

∴ οΏ½Μ…οΏ½1 = 4πœ‹ (𝑏

πœ‹)

32

∫ οΏ½Μ…οΏ½ π‘’βˆ’π‘π‘£22 𝑣2

2𝑑𝑣2

∞

0

Page 16: Kinetic Theory of Gases - RCUB

Kinetic Theory of Gases

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

Since οΏ½Μ…οΏ½ takes different values in for 𝑣1 > 𝑣2 (𝑣2

2+3𝑣12

3𝑣1)and 𝑣2 >

𝑣1 (𝑣1

2+3𝑣22

3𝑣2)

∴ οΏ½Μ…οΏ½1 = 4πœ‹ (𝑏

πœ‹)

32

{∫ (𝑣2

2 + 3𝑣12

3𝑣1) π‘’βˆ’π‘π‘£2

2 𝑣2

2𝑑𝑣2

𝑐1

0

+ ∫ (𝑣1

2 + 3𝑣22

3𝑣2) π‘’βˆ’π‘π‘£2

2 𝑣2

2𝑑𝑣2

∞

𝑐1

}

Introducing the probabilities of occurrence of 𝑣1 , we get the final

average οΏ½Μ…οΏ½2 for relative velocity

οΏ½Μ…οΏ½2 = 4πœ‹ (𝑏

πœ‹)

32

∫ οΏ½Μ…οΏ½1 π‘’βˆ’π‘π‘£12 𝑣1

2𝑑𝑣1

∞

0

∴ οΏ½Μ…οΏ½2 = 4πœ‹ (𝑏

πœ‹)

32

∫ [4πœ‹ (𝑏

πœ‹)

32

{∫ (𝑣2

2 + 3𝑣12

3𝑣1) π‘’βˆ’π‘π‘£2

2 𝑣2

2𝑑𝑣2

𝑐1

0

∞

0

+ ∫ (𝑣1

2 + 3𝑣22

3𝑣2) π‘’βˆ’π‘π‘£2

2 𝑣2

2𝑑𝑣2

∞

𝑐1

}] π‘’βˆ’π‘π‘£12 𝑣1

2𝑑𝑣1

∴ οΏ½Μ…οΏ½2 =16𝑏3

πœ‹βˆ« [{∫ (

𝑣22 + 3𝑣1

2

3𝑣1) π‘’βˆ’π‘π‘£2

2 𝑣2

2𝑑𝑣2

𝑐1

0

∞

0

+ ∫ (𝑣1

2 + 3𝑣22

3𝑣2) π‘’βˆ’π‘π‘£2

2 𝑣2

2𝑑𝑣2

∞

𝑐1

}] π‘’βˆ’π‘π‘£12 𝑣1

2𝑑𝑣1

∴ οΏ½Μ…οΏ½2 =16𝑏3

πœ‹[ 𝐽1 + 𝐽2]

Page 17: Kinetic Theory of Gases - RCUB

Kinetic Theory of Gases

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

Where

𝐽1 = ∫ {∫ (𝑣2

2 + 3𝑣12

3𝑣1) π‘’βˆ’π‘π‘£2

2 𝑣2

2𝑑𝑣2

𝑐1

0

} π‘’βˆ’π‘π‘£12 𝑣1

2𝑑𝑣1

∞

0

&

𝐽2 = ∫ { ∫ (𝑣1

2 + 3𝑣22

3𝑣2) π‘’βˆ’π‘π‘£2

2 𝑣2

2𝑑𝑣2

∞

𝑐1

} π‘’βˆ’π‘π‘£12 𝑣1

2𝑑𝑣1

∞

0

Integral 𝐽1& 𝐽2 can be calculated and given by

𝟐 𝐽1 = 𝐽2 =

1

8√2(

πœ‹

𝑏7)

12

∴ οΏ½Μ…οΏ½2 =16𝑏3

πœ‹[ 𝐽1 + 𝐽2]

∴ οΏ½Μ…οΏ½2 =16𝑏3

πœ‹[ 2 𝑋

1

8√2(

πœ‹

𝑏7)

12

]

∴ οΏ½Μ…οΏ½2 =4

√2πœ‹π‘ = √

8

πœ‹π‘

∴ οΏ½Μ…οΏ½2 = √8

πœ‹ (π‘š

2π‘˜π‘‡)

∴ οΏ½Μ…οΏ½2 = √2𝑋 8π‘˜π‘‡

πœ‹π‘š

Page 18: Kinetic Theory of Gases - RCUB

Kinetic Theory of Gases

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

∴ οΏ½Μ…οΏ½2 = √2√ 8π‘˜π‘‡

πœ‹π‘š

∴ οΏ½Μ…οΏ½2 = √2√ 8π‘˜π‘‡

πœ‹π‘š

∴ οΏ½Μ…οΏ½2 = √2𝑣

Where 𝑣 is the average velocity of molecule having mass π‘š

Therefore the mean free path, πœ† becomes

πœ† =𝑣

πœ‹ 𝑛 𝑑2(√2𝑣)

πœ† =1

√2πœ‹ 𝑛 𝑑2

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 Brownian

motion.

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.

Page 19: Kinetic Theory of Gases - RCUB

Kinetic Theory of Gases

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

iv) The lower is the viscosity 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 forces

1) The unbalanced force 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 pressure( 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 Ξ”( Ξ” is root mean square

distance). Let A be the cross section 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 = 𝑛1βˆ† 𝐴

Page 20: Kinetic Theory of Gases - RCUB

Kinetic Theory of Gases

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

The number of Brownian particles crossing at P side

=1

2𝑛1βˆ† 𝐴

The number of Brownian particles in the cylinder on the Q

side = 𝑛2βˆ† 𝐴

The number of Brownian particles crossing at Q side

=1

2𝑛2βˆ† 𝐴

So the total number of particles crossing the middle section R

from left to right =1

2(𝑛1 βˆ’ 𝑛2) βˆ† 𝐴 . . (1)

πŸ‘ By the definition of diffusion co-efficient, the number of

particles crossing the middle section R from left to right

=βˆ’π·π‘‘π‘›

𝑑π‘₯𝐴 𝑑 … … (2)

Where D= diffusion coefficient, -𝑑𝑛

𝑑π‘₯ concentration grdient and

t is the time taken for diffusion.

From equation (1)&(2)

1

2(𝑛1 βˆ’ 𝑛2) βˆ† 𝐴 = βˆ’π·

𝑑𝑛

𝑑π‘₯𝐴 𝑑

Where (𝑛1 βˆ’ 𝑛2) = βˆ’π‘‘π‘›

𝑑π‘₯ π›₯

∴ βˆ’1

2

𝑑𝑛

𝑑π‘₯ π›₯2 𝐴 = βˆ’π·

𝑑𝑛

𝑑π‘₯𝐴 𝑑

∴ 𝐷 =π›₯2

2𝑑… . (3)

Page 21: Kinetic Theory of Gases - RCUB

Kinetic Theory of Gases

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

πŸ’ Let P1 and P2 be the osmotic pressure acting on the surfaces P

and Q respectively.

Then 𝑃1 = 𝑛1π‘˜π‘‡ & 𝑃2 = 𝑛2π‘˜π‘‡

∴ π‘ƒπ‘Ÿπ‘’π‘ π‘ π‘’π‘Ÿπ‘’ π‘‘π‘–π‘“π‘“π‘’π‘Ÿπ‘’π‘›π‘π‘’ = 𝑃1 βˆ’ 𝑃2 = (𝑛1 βˆ’ 𝑛2)π‘˜π‘‡

The force exerted due to this pressure difference

= (𝑛1 βˆ’ 𝑛2)π‘˜π‘‡π΄

This force acts on all the particles in the cylinder PQ in which

the number of particles, where 𝑛 =𝑛1+𝑛2

2

∴ Force experience by each particle, 𝐹 = (𝑛1 βˆ’ 𝑛2)π‘˜π‘‡π΄

nΞ”A

∴ 𝐹 = (𝑛1 βˆ’ 𝑛2)π‘˜π‘‡

nΞ”

∡(𝑛1 βˆ’ 𝑛2)

Ξ”= βˆ’

𝑑𝑛

𝑑π‘₯

∴ 𝐹 = βˆ’π‘˜π‘‡

n

𝑑𝑛

𝑑π‘₯… . . (4)

πŸ“ Simultaneously particle also experience viscous drag (stokes

law) , which is given by

𝐹 = 6πœ‹πœ‚π‘Ÿπ‘£ … . . (5)

Where Ξ· is the coefficient of viscosity

π‘Ÿ is the radius of particles

𝑣 is the velocity of the particles

Page 22: Kinetic Theory of Gases - RCUB

Kinetic Theory of Gases

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

From equation (4) & (5)

𝐹 = βˆ’π‘˜π‘‡

n

𝑑𝑛

𝑑π‘₯= 6πœ‹πœ‚π‘Ÿπ‘£

𝑛𝑣

𝑑𝑛𝑑π‘₯

= βˆ’π‘˜π‘‡

6πœ‹πœ‚π‘Ÿ

Where 𝑛𝑣 is number of particles moving from left to right

throgh unit area in unit time and hence 𝑛𝑣

βˆ’π‘‘π‘›

𝑑π‘₯

gives diffusion

coefficient D

∴ 𝐷 =π‘˜π‘‡

6πœ‹πœ‚π‘Ÿβ€¦ . . (6)

From equation (3) & (6)

π›₯2

2𝑑=

π‘˜π‘‡

6πœ‹πœ‚π‘Ÿ

∡ π‘˜ =𝑅

𝑁𝐴

∴π›₯2

2𝑑=

𝑅𝑇

6πœ‹π‘πœ‚π‘Ÿ

∴ π›₯2 =𝑅𝑇

𝑁

𝑑

3πœ‹πœ‚π‘Ÿ

The above equation is Einstein’s equation which gives the

value of square of rms displacement of Brownian particles.