Kinetic Theory of Gases - RCUB
Transcript of Kinetic Theory of Gases - RCUB
Kinetic Theory of Gases
Prof. Avadhut Surekha Prakash Manage D.M.S. Mandalβs 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 1860-1880βs. 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.
Kinetic Theory of Gases
Prof. Avadhut Surekha Prakash Manage D.M.S. Mandalβs Bhaurao kakatkar College, Belgaum
Maxwellβs law of distribution of velocities
This law is to find the number of molecules which have a
velocity within small interval (ie. π to π + ππ).
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 π’ to π’ + ππ’ is π(π’)ππ’
similarly the probability of molecules having velocity component
π£ to π£ + ππ£ is π(π£)ππ£ and the probability of molecules having
velocity component w to π€ + ππ€ is π(π€)ππ€. Thus the number of
molecules whose velocity lies between π’ to +ππ’ , π£ to π£ + ππ£ and
w to π€ + ππ€ is given by
ππ = π π(π’)ππ’ π(π£)ππ£ π(π€)ππ€
ππ = π π(π’) π(π£) π(π€) ππ’ ππ£ ππ€ β¦ β¦ . . (1)
If velocity components π’, π£ and π€
along the three axis. The space formed
is called as the velocity space. A
molecule having velocity components
π’, π£ , π€ be represented by a point π. Let
the small volume ππ’ ππ£ ππ€ around π
Kinetic Theory of Gases
Prof. Avadhut Surekha Prakash Manage D.M.S. Mandalβs Bhaurao kakatkar College, Belgaum
contains ππ number of molecules. Thus molecules have a resultant
velocity c is given by
π2 = π’2 + π£2 + π€2
Differentiating on both side, we get
0 = 2π’ ππ’ + 2π£ ππ£ + 2π€ ππ€ β΅ π = ππππ π‘πππ‘
β΄ π’ ππ’ + π£ ππ£ + π€ ππ€ = 0 β¦ β¦ β¦ β¦ β¦ (2)
The probability density π(π’) π(π£) π(π€) should depend only
on the magnitude but not the direction of π.
β΄ π(π’) π(π£) π(π€) = π(π2) (π€βπππ π(π2) ππ ππ’πππ‘πππ ππ π2)
Differentiating on both side, we get
π[π(π’) π(π£) π(π€)] = 0 β΅ π2 = ππππ π‘πππ‘
πβ²(π’)ππ’π(π£)π(π€) + πβ²(π£) ππ£π(π’)π(π€) + πβ²(π€)ππ€π(π£)π(π£) = 0
Dividing above equation by π(π’) π(π£) π(π€), we get
πβ²(π’)
π(π’)ππ’ +
πβ²(π£)
π(π£)ππ£ +
πβ²(π€)
π(π€)ππ€ = 0 . . . . . . . . . . (3)
Multiplying the equation (2) by an arbitrary constant (Ξ²)
and adding to equation (3), we get
[πβ²(π’)
π(π’)+ π½π’] ππ’ + [
πβ²(π£)
π(π£)+ π½π£] ππ£ + [
πβ²(π€)
π(π€)+ π½π€] ππ€ = 0
As ππ’, ππ£ & ππ€ can not be zero independently, so we get
πβ²(π’)
π(π’)+ π½π’ = 0
πβ²(π£)
π(π£)+ π½π£ = 0
πβ²(π€)
π(π€)+ π½π€ = 0
πβ²(π’)
π(π’)= βπ½π’
Kinetic Theory of Gases
Prof. Avadhut Surekha Prakash Manage D.M.S. Mandalβs Bhaurao kakatkar College, Belgaum
Integrating wrt u, we get
log π(π’) = β1
2 π½π’2 + log π
(πβπππ πππ π ππ πππ‘πππππ‘πππ ππππ π‘πππ‘ )
Taking antilog on both side, we get
π(π’) = ππβ12
π½π’2
= ππβππ’2 (Where, π = β
1
2 π½ = ππππ π‘πππ‘)
Similarly, π(π£) = ππβππ£2 & π(π€) = ππβππ€2
Substituting π(π’), π(π£) & π(π€) in eqn (1), we get
ππ = π (ππβππ’2) (ππβππ£2
) (ππβππ€2) ππ’ ππ£ ππ€
ππ = ππ3 πβπ(π’2+π£2+π€2)ππ’ ππ£ ππ€
β΅ π2 = π’2 + π£2 + π€2
ππ = ππ3 πβππ2ππ’ ππ£ ππ€ β¦ β¦ β¦ (4)
In this equation constant π and π is found to be
π = βπ
2πππ and π =
π
2ππ , where π is the mass of the gas particle,
π is Boltzmann constant and π is temperature of gas
Then equation (4) becomes
ππ = π (π
2πππ)
32
πβ ππ2
2ππ ππ’ ππ£ ππ€ β¦ β¦ (5)
This is one form of Maxwellβs law of distribution of
velocities.
Kinetic Theory of Gases
Prof. Avadhut Surekha Prakash Manage D.M.S. Mandalβs Bhaurao kakatkar College, Belgaum
The number of molecules lies
within box of volume ππ’ ππ£ ππ€ whose
velocity lies within π to π + ππ is the
same as number of molecules within
the spherical shell of inner radius π and
outer radius π + ππ. Hence volume
ππ’ ππ£ ππ€ can be replaced volume of the
shell 4ππ2ππ. Therefore equation (5)
becomes
ππ = π (π
2πππ)
32
πβ ππ2
2ππ 4ππ2ππ
ππ = 4ππ (π
2πππ)
32
πβ ππ2
2ππ π2ππ
The above equation is called as Maxwellβs equation of
distribution of velocity which gives the value of number
molecules ππ having velocity π to π + ππ.
Mean or Average velocity (πͺππ)
The average speed is the sum of all the velocities ranging from 0 to β divided by total number of molecules N.
β΄ πΆππ£ =β« ππ π π
β
0
π
Substituting the value of ππ in in above equation
πΆππ£ =
β« (4ππ (π
2πππ)
32
πβ ππ2
2ππ π2ππ) π πβ
0
π
Kinetic Theory of Gases
Prof. Avadhut Surekha Prakash Manage D.M.S. Mandalβs Bhaurao kakatkar College, Belgaum
πΆππ£ = 4π (π
2πππ)
32
β« πβ ππ2
2ππ π3ππ
β
0
πΆππ£ = 4π (π
π)
32
β« πβππ2 π3ππ
β
0
π
β΅ β« πβππ2 π3ππ
β
0
=1
2π2
( Where π =π
2ππ )
πΆππ£ = 4π (π
π)
32 1
2π2
πΆππ£ = β4
ππ
β΄ πΆππ£ = β4
π (π
2ππ)
β΄ πΆππ£ = β8ππ
ππ
Where π is the mass of gas molecule. If π is molecular
weight which can be given by π = πππ΄. Then π =π
ππ΄ Therefore
the above equation becomes.
πΆππ£ = β8ππ
π (πππ΄
)
Kinetic Theory of Gases
Prof. Avadhut Surekha Prakash Manage D.M.S. Mandalβs Bhaurao kakatkar College, Belgaum
β΄ πΆππ£ = β8πππ΄π
ππ
β΄ πΆππ£ = β8π π
ππ β΅ πππ΄ = π
Root Mean Square Velocity (πͺπππ)
The rms speed is the square root of the sum of all the squared velocities ranging from 0 to β divided by total number of molecules N.
β΄ πΆπππ = ββ« ππ π π2β
0
π
Substituting the value of ππ in in above equation
πΆπππ =ββ« (4ππ (
π2πππ
)
32
πβ ππ2
2ππ π2ππ) π π2β
0
π
πΆπππ = β4π (π
2πππ)
32
β« πβ ππ2
2ππ π4ππ
β
0
= β4π (π
π)
32
β« πβππ2 π4ππ
β
0
( Where π =π
2ππ )
Kinetic Theory of Gases
Prof. Avadhut Surekha Prakash Manage D.M.S. Mandalβs Bhaurao kakatkar College, Belgaum
= β4π (π
π)
32
π 3
8(
π
π5)
12
= β3
2π
= β3
2 (π
2ππ)
πΆπππ = β3ππ
π
Substituting π =π
ππ΄ in above equation. We get
πΆππ£ = β3ππ
(πππ΄
)
β΄ πΆππ£ = β3πππ΄π
π
β΄ πΆππ£ = β3π π
π β΅ πππ΄ = π
Most Probable Velocity (πͺππ)
The most probable velocity is the velocity possessed by maximum number of molecules. For this condition is,
Kinetic Theory of Gases
Prof. Avadhut Surekha Prakash Manage D.M.S. Mandalβs Bhaurao kakatkar College, Belgaum
π[ππ]
ππ= 0
β΄π
ππ(4ππ (
π
2πππ)
32
πβ ππ2
2ππ π2ππ) = 0
4ππ (π
2πππ)
32 π
ππ( πβ
ππ2
2ππ π2ππ) = 0
4ππ (π
2πππ)
32
( πβ ππ2
2ππ [2π] + [β 2ππ
2ππ] πβ
ππ2
2ππ π2) = 0
4ππ (π
2πππ)
32
πβ ππ2
2ππ ( 2π β [ ππ
ππ] π2) = 0
2π β [ ππ
ππ] π2 = 0
β΄ ππ3
ππ= 2π
β΄ π2 =2ππ
π
β΄ π = β2ππ
π
Thus most probable velocity,
πΆππ = β2ππ
π
Substituting π =π
ππ΄ in above equation. We get
Kinetic Theory of Gases
Prof. Avadhut Surekha Prakash Manage D.M.S. Mandalβs Bhaurao kakatkar College, Belgaum
πΆππ£ = β2ππ
(πππ΄
)
β΄ πΆππ£ = β2πππ΄π
π
β΄ πΆππ£ = β2π π
π β΅ πππ΄ = π
Relation between πͺππ , πͺππ & πͺπππ
πΆππ βΆ πΆππ£ βΆ πΆπππ = β2π π
π: β
8π π
ππβΆ β
3π π
π
= β2: β8
πβΆ β3
= 1 βΆ 2
βπβΆ β
3
2
= 1: 1.128 βΆ 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.
Kinetic Theory of Gases
Prof. Avadhut Surekha Prakash Manage D.M.S. Mandalβs 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 π2πΏ.
No. of collisions suffered = No. molecules in the cylinder
by molecule A of volume Ο π2πΏ
= No. of molecules per unit volume
X Volume of cylinder
= π π π π2πΏ
= π π π2πΏ
Now mean free path of molecule is given by Ξ»,
Ξ» =πππ‘ππ πππ π‘ππππ π‘πππ£πππππ ππ π‘πππ π‘
ππ. ππ ππππππ ππππ π π’πππππ
=πΏ
π π π2πΏ
Kinetic Theory of Gases
Prof. Avadhut Surekha Prakash Manage D.M.S. Mandalβs Bhaurao kakatkar College, Belgaum
πππ π, π =π£ π‘
π π π2οΏ½Μ οΏ½ π‘=
π£
π π π2οΏ½Μ οΏ½β¦ . . (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 π£1 and π£2 and ΞΈ
be the angle between them .Then
relative π£π is given by,
π£π = βπ£12 + π£2
2 β 2π£1π£2 cos π β¦ . (1)
If πππ is the number of molecules per unit volume in
between ΞΈ to π + ππ with a given direction the average relative
velocity π£οΏ½Μ οΏ½ is given by
π£οΏ½Μ οΏ½ =β« π£ππππ
π β¦ β¦ β¦ . . (2)
From eqn (1), we get
π£οΏ½Μ οΏ½ =β«(βπ£1
2 + π£22 β 2π£1π£2 cos π)πππ
π
Kinetic Theory of Gases
Prof. Avadhut Surekha Prakash Manage D.M.S. Mandalβs Bhaurao kakatkar College, Belgaum
To find the expression for
πππ , 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 + ππ then number of molecules
lying in this region is πππ .
π πΌ ππππ ππ π πβπππ
πππ πΌ ππππ ππ π§πππ πππ‘π€πππ π to π + ππ
β΄πππ
π=
ππππ ππ π§πππ πππ‘π€πππ π to π + ππ
ππππ ππ π πβπππ
=2ππ sin π πππ
4ππ2
=1
2sin π ππ
πππ =1
2π sin π ππ β¦ β¦ β¦ β¦ . . (3)
Substituting πππ in equation (3), we get
π£οΏ½Μ οΏ½ =β« βπ£1
2 + π£22 β 2π£1π£2 cos π (
12 π sin π ππ)
π
π£οΏ½Μ οΏ½ =1
2π
β« βπ£12 + π£2
2 β 2π£1π£2 cos π sin π ππ
π
Kinetic Theory of Gases
Prof. Avadhut Surekha Prakash Manage D.M.S. Mandalβs Bhaurao kakatkar College, Belgaum
π£οΏ½Μ οΏ½ =1
2β« βπ£1
2 + π£22 β 2π£1π£2 cos π sin π ππ
π
0
π£οΏ½Μ οΏ½ =1
2(
1
3π£1π£2[(π£1
2 + π£22 β 2π£1π£2 cos π)
32]
0
π
)
π£οΏ½Μ οΏ½ =1
6π£1π£2[(π£1
2 + π£22 + 2π£1π£2)
32 β (π£1
2 + π£22 β 2π£1π£2)
32]
π£οΏ½Μ οΏ½ =1
6π£1π£2
[(π£1 + π£2)3 β (π£1 β π£2)3]
π£οΏ½Μ οΏ½ =1
6π£1π£2
[(π£13 + π£2
3 + 3π£12π£2 + 3π£1π£2
2)
β (π£13 β π£2
3 β 3π£12π£2 + 3π£1π£2
2)]
π£οΏ½Μ οΏ½ =1
6π£1π£2
(2π£23 + 6π£1
2π£2)
π£οΏ½Μ οΏ½ =1
3π£1
(π£22 + 3π£1
2) β¦ β¦ . (4)
Since close gas molecules have almost same velocity.
Hence we can assume that π£2 β π£1 β π£
β΄ π£οΏ½Μ οΏ½ =π£2 + 3π£2
3π£
β΄ π£οΏ½Μ οΏ½ =4
3π£
Substituting π£οΏ½Μ οΏ½ in mean free path, π. We get
π =π£
π π π2 (43 π£)
Kinetic Theory of Gases
Prof. Avadhut Surekha Prakash Manage D.M.S. Mandalβs Bhaurao kakatkar College, Belgaum
π =3
4π π π2
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 π£2, we get an
intermediate average relative velocity οΏ½Μ οΏ½1
.
οΏ½Μ οΏ½1 =β« οΏ½Μ οΏ½ππ2
β
0
π
Where ππ2 is the number of molecules whose velocity ranging
from π£2 to π£2 + ππ£2
β΄ οΏ½Μ οΏ½1 =β« οΏ½Μ οΏ½ 4ππ (
ππ)
32
πβππ£22 π£2
2ππ£2β
0
π
.
β΄ οΏ½Μ οΏ½1 =4ππ (
ππ)
32
β« οΏ½Μ οΏ½ πβππ£22 π£2
2ππ£2β
0
π
β΄ οΏ½Μ οΏ½1 = 4π (π
π)
32
β« οΏ½Μ οΏ½ πβππ£22 π£2
2ππ£2
β
0
Kinetic Theory of Gases
Prof. Avadhut Surekha Prakash Manage D.M.S. Mandalβs Bhaurao kakatkar College, Belgaum
Since οΏ½Μ οΏ½ takes different values in for π£1 > π£2 (π£2
2+3π£12
3π£1)and π£2 >
π£1 (π£1
2+3π£22
3π£2)
β΄ οΏ½Μ οΏ½1 = 4π (π
π)
32
{β« (π£2
2 + 3π£12
3π£1) πβππ£2
2 π£2
2ππ£2
π1
0
+ β« (π£1
2 + 3π£22
3π£2) πβππ£2
2 π£2
2ππ£2
β
π1
}
Introducing the probabilities of occurrence of π£1 , we get the final
average οΏ½Μ οΏ½2 for relative velocity
οΏ½Μ οΏ½2 = 4π (π
π)
32
β« οΏ½Μ οΏ½1 πβππ£12 π£1
2ππ£1
β
0
β΄ οΏ½Μ οΏ½2 = 4π (π
π)
32
β« [4π (π
π)
32
{β« (π£2
2 + 3π£12
3π£1) πβππ£2
2 π£2
2ππ£2
π1
0
β
0
+ β« (π£1
2 + 3π£22
3π£2) πβππ£2
2 π£2
2ππ£2
β
π1
}] πβππ£12 π£1
2ππ£1
β΄ οΏ½Μ οΏ½2 =16π3
πβ« [{β« (
π£22 + 3π£1
2
3π£1) πβππ£2
2 π£2
2ππ£2
π1
0
β
0
+ β« (π£1
2 + 3π£22
3π£2) πβππ£2
2 π£2
2ππ£2
β
π1
}] πβππ£12 π£1
2ππ£1
β΄ οΏ½Μ οΏ½2 =16π3
π[ π½1 + π½2]
Kinetic Theory of Gases
Prof. Avadhut Surekha Prakash Manage D.M.S. Mandalβs Bhaurao kakatkar College, Belgaum
Where
π½1 = β« {β« (π£2
2 + 3π£12
3π£1) πβππ£2
2 π£2
2ππ£2
π1
0
} πβππ£12 π£1
2ππ£1
β
0
&
π½2 = β« { β« (π£1
2 + 3π£22
3π£2) πβππ£2
2 π£2
2ππ£2
β
π1
} πβππ£12 π£1
2ππ£1
β
0
Integral π½1& π½2 can be calculated and given by
π π½1 = π½2 =
1
8β2(
π
π7)
12
β΄ οΏ½Μ οΏ½2 =16π3
π[ π½1 + π½2]
β΄ οΏ½Μ οΏ½2 =16π3
π[ 2 π
1
8β2(
π
π7)
12
]
β΄ οΏ½Μ οΏ½2 =4
β2ππ = β
8
ππ
β΄ οΏ½Μ οΏ½2 = β8
π (π
2ππ)
β΄ οΏ½Μ οΏ½2 = β2π 8ππ
ππ
Kinetic Theory of Gases
Prof. Avadhut Surekha Prakash Manage D.M.S. Mandalβs Bhaurao kakatkar College, Belgaum
β΄ οΏ½Μ οΏ½2 = β2β 8ππ
ππ
β΄ οΏ½Μ οΏ½2 = β2β 8ππ
ππ
β΄ οΏ½Μ οΏ½2 = β2π£
Where π£ is the average velocity of molecule having mass π
Therefore the mean free path, π becomes
π =π£
π π π2(β2π£)
π =1
β2π π π2
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.
Kinetic Theory of Gases
Prof. Avadhut Surekha Prakash Manage D.M.S. Mandalβs 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.
Einsteinβs 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 = π1β π΄
Kinetic Theory of Gases
Prof. Avadhut Surekha Prakash Manage D.M.S. Mandalβs Bhaurao kakatkar College, Belgaum
The number of Brownian particles crossing at P side
=1
2π1β π΄
The number of Brownian particles in the cylinder on the Q
side = π2β π΄
The number of Brownian particles crossing at Q side
=1
2π2β π΄
So the total number of particles crossing the middle section R
from left to right =1
2(π1 β π2) β π΄ . . (1)
π By the definition of diffusion co-efficient, the number of
particles crossing the middle section R from left to right
=βπ·ππ
ππ₯π΄ π‘ β¦ β¦ (2)
Where D= diffusion coefficient, -ππ
ππ₯ concentration grdient and
t is the time taken for diffusion.
From equation (1)&(2)
1
2(π1 β π2) β π΄ = βπ·
ππ
ππ₯π΄ π‘
Where (π1 β π2) = βππ
ππ₯ π₯
β΄ β1
2
ππ
ππ₯ π₯2 π΄ = βπ·
ππ
ππ₯π΄ π‘
β΄ π· =π₯2
2π‘β¦ . (3)
Kinetic Theory of Gases
Prof. Avadhut Surekha Prakash Manage D.M.S. Mandalβs Bhaurao kakatkar College, Belgaum
π Let P1 and P2 be the osmotic pressure acting on the surfaces P
and Q respectively.
Then π1 = π1ππ & π2 = π2ππ
β΄ ππππ π π’ππ ππππππππππ = π1 β π2 = (π1 β π2)ππ
The force exerted due to this pressure difference
= (π1 β π2)πππ΄
This force acts on all the particles in the cylinder PQ in which
the number of particles, where π =π1+π2
2
β΄ Force experience by each particle, πΉ = (π1 β π2)πππ΄
nΞA
β΄ πΉ = (π1 β π2)ππ
nΞ
β΅(π1 β π2)
Ξ= β
ππ
ππ₯
β΄ πΉ = βππ
n
ππ
ππ₯β¦ . . (4)
π Simultaneously particle also experience viscous drag (stokes
law) , which is given by
πΉ = 6ππππ£ β¦ . . (5)
Where Ξ· is the coefficient of viscosity
π is the radius of particles
π£ is the velocity of the particles
Kinetic Theory of Gases
Prof. Avadhut Surekha Prakash Manage D.M.S. Mandalβs Bhaurao kakatkar College, Belgaum
From equation (4) & (5)
πΉ = βππ
n
ππ
ππ₯= 6ππππ£
ππ£
ππππ₯
= βππ
6πππ
Where ππ£ is number of particles moving from left to right
throgh unit area in unit time and hence ππ£
βππ
ππ₯
gives diffusion
coefficient D
β΄ π· =ππ
6πππβ¦ . . (6)
From equation (3) & (6)
π₯2
2π‘=
ππ
6πππ
β΅ π =π
ππ΄
β΄π₯2
2π‘=
π π
6ππππ
β΄ π₯2 =π π
π
π‘
3πππ
The above equation is Einsteinβs equation which gives the
value of square of rms displacement of Brownian particles.