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Postulates Kinetic Theory of Gases, Lecture notes of Chemistry

Explain in kinetic theory of gases, maxwell's law of distribution of velocities and Brownian Motion.

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Kinetic Theory of Gases
Prof. Avadhut Surekha Prakash Manage
D.M.S. Mandals 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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Download Postulates Kinetic Theory of Gases and more Lecture notes Chemistry in PDF only on Docsity!

Prof. Avadhut Surekha Prakash Manage

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.

  1. Gas molecules do not have preferred direction of motion, their

motion is completely random.

  1. Gas molecules travels in straight line.

  2. The time interval of collision between any two gas molecules

is very small.

  1. The collision between gas molecules and the walls of

container is perfectly elastic. It means kinetic energy and

momentum in such collision is conserved.

  1. The motion of gas molecules is governed by Newton's laws of

motion.

  1. The effect of gravity on the motion of gas molecules is

negligible.

Prof. Avadhut Surekha Prakash Manage

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 𝑢 to 𝑢 + 𝑑𝑢 is 𝑓(𝑢)𝑑𝑢

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 𝑤 + 𝑑𝑤 is 𝑓(𝑤)𝑑𝑤. Thus the number of

molecules whose velocity lies between 𝑢 to +𝑑𝑢 , 𝑣 to 𝑣 + 𝑑𝑣 and

w to 𝑤 + 𝑑𝑤 is given by

If velocity components 𝑢, 𝑣 and 𝑤

along the three axis. The space formed

is called as the velocity space. A

molecule having velocity components

𝑢, 𝑣 , 𝑤 be represented by a point 𝑝. Let

the small volume 𝑑𝑢 𝑑𝑣 𝑑𝑤 around 𝑝

Prof. Avadhut Surekha Prakash Manage

Integrating wrt u, we get

log 𝑓

2

  • log 𝑎

(𝑊ℎ𝑒𝑟𝑒 𝑙𝑜𝑔 𝑎 𝑖𝑠 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑡𝑖𝑜𝑛 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 )

Taking antilog on both side, we get

1

2

𝛽𝑢

2

−𝑏𝑢

2

(Where, 𝑏 = −

Similarly, 𝑓(𝑣) = 𝑎𝑒

−𝑏𝑣

2

−𝑏𝑤

2

Substituting 𝑓(𝑢), 𝑓(𝑣) & 𝑓(𝑤) in eq

n

(1), we get

−𝑏𝑢

2

−𝑏𝑣

2

−𝑏𝑤

2

3

−𝑏(𝑢

2

+𝑣

2

+𝑤

2

)

2

2

2

2

3

−𝑏𝑐

2

In this equation constant 𝑎 and 𝑏 is found to be

𝑚

2 𝜋𝑘𝑇

and 𝑏 =

𝑚

2 𝑘𝑇

, where 𝑚 is the mass of the gas particle,

𝑘 is Boltzmann constant and 𝑇 is temperature of gas

Then equation (4) becomes

3

2

𝑚𝑐

2

2 𝑘𝑇 𝑑𝑢 𝑑𝑣 𝑑𝑤 … … ( 5 )

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

velocities.

Prof. Avadhut Surekha Prakash Manage

The number of molecules lies

within box of volume 𝑑𝑢 𝑑𝑣 𝑑𝑤 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

𝑑𝑢 𝑑𝑣 𝑑𝑤 can be replaced volume of the

shell 4 𝜋𝑐

2

𝑑𝑐. Therefore equation (5)

becomes

3

2

𝑚𝑐

2

2 𝑘𝑇 4 𝜋𝑐

2

3

2

𝑚𝑐

2

2 𝑘𝑇

𝑐

2

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 (𝑪

𝒂𝒗

The average speed is the sum of all the velocities ranging

from 0 to ∞ divided by total number of molecules N.

𝑎𝑣

0

Substituting the value of 𝑑𝑁 in in above equation

𝑎𝑣

3

2

𝑚𝑐

2

2 𝑘𝑇 𝑐

2

0

Prof. Avadhut Surekha Prakash Manage

𝑎𝑣

𝐴

𝑎𝑣

𝐴

Root Mean Square Velocity (𝑪 𝒓𝒎𝒔

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.

𝑟𝑚𝑠

2

0

Substituting the value of 𝑑𝑁 in in above equation

𝑟𝑚𝑠

3

2

𝑚𝑐

2

2 𝑘𝑇 𝑐

2

2

0

𝑟𝑚𝑠

3

2

𝑚𝑐

2

2 𝑘𝑇 𝑐

4

0

3

2

−𝑏𝑐

2

4

0

( Where 𝑏 =

Prof. Avadhut Surekha Prakash Manage

3

2

5

1

2

𝑟𝑚𝑠

Substituting 𝑚 =

𝑀

𝑁 𝐴

in above equation. We get

𝑎𝑣

𝐴

𝑎𝑣

𝐴

𝑎𝑣

𝐴

Most Probable Velocity (𝑪

𝒑𝒓

The most probable velocity is the velocity possessed by

maximum number of molecules. For this condition is,

Prof. Avadhut Surekha Prakash Manage

𝑎𝑣

𝐴

𝑎𝑣

𝐴

𝑎𝑣

𝐴

Relation between 𝑪

𝒑𝒓

𝒂𝒗

𝒓𝒎𝒔

𝑝𝑟

𝑎𝑣

𝑟𝑚𝑠

Mean free path

Mean free path of gas molecules is defined as the average

distance travelled by a molecule between two successive

collisions.

Prof. Avadhut Surekha Prakash Manage

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 𝑑

2

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

by molecule A of volume π 𝑑

2

= No. of molecules per unit volume

X Volume of cylinder

2

2

Now mean free path of molecule is given by λ,

λ =

2

Prof. Avadhut Surekha Prakash Manage

To find the expression for

𝜃

, 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 &

𝜃 + 𝑑𝜃 then number of molecules

lying in this region is 𝑑𝑛

𝜃

𝜃

𝛼 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑧𝑜𝑛𝑒 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝜃 to 𝜃 + 𝑑𝜃

𝜃

𝑎𝑟𝑒𝑎 𝑜𝑓 𝑧𝑜𝑛𝑒 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝜃 to 𝜃 + 𝑑𝜃

2 𝜋𝑟 sin 𝜃 𝑟𝑑𝜃

2

sin 𝜃 𝑑𝜃

𝜃

𝑛 sin 𝜃 𝑑𝜃 … … … …..

Substituting 𝑑𝑛

𝜃

in equation (3), we get

𝑟

1

2

2

2

1

2

cos 𝜃 (

𝑛 sin 𝜃 𝑑𝜃)

𝑟

1

2

2

2

1

2

cos 𝜃 sin 𝜃 𝑑𝜃

Prof. Avadhut Surekha Prakash Manage

𝑟

1

2

2

2

1

2

cos 𝜃 sin 𝜃 𝑑𝜃

𝜋

0

𝑟

1

2

[

1

2

2

2

1

2

cos 𝜃

3

2

]

0

𝜋

𝑟

1

2

[(𝑣

1

2

2

2

1

2

3

2 − (𝑣

1

2

2

2

1

2

3

2 ]

𝑟

1

2

[(𝑣

1

2

3

1

2

3

]

𝑟

1

2

[(𝑣

1

3

2

3

1

2

2

1

2

2

1

3

2

3

1

2

2

1

2

2

)]

𝑟

1

2

2

3

1

2

2

𝑟

1

2

2

1

2

Since close gas molecules have almost same velocity.

Hence we can assume that 𝑣 2

1

𝑟

2

2

𝑟

Substituting 𝑣

𝑟

in mean free path, 𝜆. We get

2

Prof. Avadhut Surekha Prakash Manage

Since 𝑣̅ takes different values in for 𝑣 1

2

𝑣

2

2

  • 3 𝑣

1

2

3 𝑣 1

)and 𝑣

2

1

𝑣

1

2

  • 3 𝑣

2

2

3 𝑣

2

1

3

2

2

2

1

2

1

−𝑏𝑣

2

2

2

2

2

𝑐

1

0

1

2

2

2

2

−𝑏𝑣

2

2

2

2

2

𝑐 1

Introducing the probabilities of occurrence of 𝑣 1

, we get the final

average 𝑣̅

2

for relative velocity

2

3

2

1

−𝑏𝑣

1

2

1

2

1

0

2

3

2

∫ [ 4 𝜋 (

3

2

2

2

1

2

1

−𝑏𝑣 2

2

2

2

2

𝑐

1

0

0

1

2

2

2

2

−𝑏𝑣 2

2

2

2

2

𝑐

1

}] 𝑒

−𝑏𝑣 1

2

1

2

1

2

3

∫ [{∫ (

2

2

1

2

1

−𝑏𝑣

2

2

2

2

2

𝑐

1

0

0

1

2

2

2

2

−𝑏𝑣

2

2

2

2

2

𝑐

1

}] 𝑒

−𝑏𝑣

1

2

1

2

1

2

3

[

1

2

]

Prof. Avadhut Surekha Prakash Manage

Where

1

2

2

1

2

1

−𝑏𝑣

2

2

2

2

2

𝑐

1

0

−𝑏𝑣

1

2

1

2

1

0

2

1

2

2

2

2

−𝑏𝑣 2

2

2

2

2

𝑐

1

−𝑏𝑣 1

2

1

2

1

0

Integral 𝐽 1

2

can be calculated and given by

1

2

7

1

2

2

3

[ 𝐽

1

2

]

2

3

[ 2 𝑋

7

1

2

]

2

2

2

Prof. Avadhut Surekha Prakash Manage

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.

  1. The viscous drag which opposes the Brownian motion.

  2. 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 Δ( Δ is root mean square

distance). Let A be the cross section area of the cylinder. Let n 1

and n 2 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 = 𝑛 1

Prof. Avadhut Surekha Prakash Manage

The number of Brownian particles crossing at P side

1

2

1

The number of Brownian particles in the cylinder on the Q

side = 𝑛

2

The number of Brownian particles crossing at Q side

1

2

2

So the total number of particles crossing the middle section R

from left to right =

1

2

1

2

By the definition of diffusion co-efficient, the number of

particles crossing the middle section R from left to right

𝑑𝑛

𝑑𝑥

Where D= diffusion coefficient, -

𝑑𝑛

𝑑𝑥

concentration grdient and

t is the time taken for diffusion.

From equation ( 1 )&( 2 )

1

2

Where (𝑛

1

2

𝑑𝑛

𝑑𝑥

2

2