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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
- Any gas consist large number of molecules. These molecules
are identical, perfectly elastic and hard sphere.
- Gas molecules do not have preferred direction of motion, their
motion is completely random.
Gas molecules travels in straight line.
The time interval of collision between any two gas molecules
is very small.
- The collision between gas molecules and the walls of
container is perfectly elastic. It means kinetic energy and
momentum in such collision is conserved.
- The motion of gas molecules is governed by Newton's laws of
motion.
- 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
(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
1
2
3 𝑣 1
)and 𝑣
2
1
𝑣
1
2
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
- The unbalanced force due to collision of unbalanced
molecules in the opposite direction.
The viscous drag which opposes the Brownian motion.
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
