Kinetic Theory

hii sir, I wants to know about the kinetic theory please tell me in detail about the The kinetic theory ?

[Shaleen]

The kinetic theory describes a gas as a large number of submicroscopic particles (atoms or molecules), all of which are in constant, random motion. The rapidly moving particles constantly collide with each other and with the walls of the container.
Assumptions
The theory for ideal gases makes the following assumptions:
The gas consists of very small particles known as molecules.
These particles have the same mass.
The number of molecules is so large that statistical treatment can be applied.
These molecules are in constant, random, and rapid motion.
The rapidly moving particles constantly collide among themselves and with the walls of the container. All these collisions are perfectly elastic. This means, the molecules are considered to be perfectly spherical in shape, and elastic in nature.
Except during collisions, the interactions among molecules are negligible. (That is, they exert no forces on one another.
This implies:
1. Relativistic effects are negligible.
2. Quantum-mechanical effects are negligible. This means that the inter-particle distance is much larger than the thermal de Broglie wavelength and the molecules are treated as classical objects.
Properties
Pressure and kinetic energy
Pressure is explained by the kinetic theory as arising from the force exerted by molecules or atoms when they hit the walls of a container. Consider a gas of N molecules, each of mass m, enclosed in a cuboid of volume V=L3. When a gas molecule collides with the wall of the container perpendicular to the x axis and bounces off in the opposite direction with the same speed (an elastic collision), the change in momentum is given by:
\Delta p = p_{i,x} - p_{f,x} = p_{i,x} - (-p_{i,x}) = 2 p_{i,x} = 2 m v_x\,
where p is the momentum, i and f indicate initial and final momentum (before and after collision), x indicates only the x-direction is being considered and v is the speed of the particle (which is the same before and after the collision).
If one considers the average speed of the particle in the container is \bar{v}_x, one can then consider the average change in momentum due to the collision of n particles:
\Delta \bar{p}_{tot}=n \times 2m\bar{v}_x
The force on the wall is given by:
\bar{F}=\frac{\Delta \bar{p}_{tot}}{\Delta t}

or what is the same: the average change in momentum per second.
The question is then to calculate how many particles n will collide with the wall during a one second interval (on average).
Since the particle travels v_x metres per second, if the particle is at less than v_x metres away from the wall, it will hit it before a second has elapsed. Hence, any particle in a volume \bar{v}_x \times L^2 (where L^2 is the area of the wall) will hit the end of the container. However for this calculation only one of the ends of the container is of interest (not the two walls perpendicular to x).
Half the particles will be travelling in the positive x direction and the other half will be travelling in the negative x direction such that only half the particles in the volume defined will hit the wall of interest ( a factor of a half is thus required). Now, to know how many particles are in that volume we need to multiply by the density :
n=\frac{1}{2} \times \bar{v}_x \times L^2 \times \frac{N}{L^3}
n= \frac{\bar{v}_x N}{2L}
The total force on the wall is then
F = \frac{\bar{v}_xN}{2L} \times 2m\bar{v}_x
F = \frac{N m \overline{v}_x^2}{L}
Since the motion of the particles is random and there is no bias applied in any direction, the average speed in each direction is identical:
\bar{v}^2_x=\bar{v}^2_y=\bar{v}_z^3
The total speed v is given by:
\bar{v}^2=\bar{v}^2_x+\bar{v}^2_y+\bar{v}_z^3
\bar{v}^2=3\bar{v}^2_x
Therefore:
\bar{v}^2_x=\frac{\bar{v}^2}{3}
and the force can be written as:

F = \frac{Nm\overline{v^2}}{3L}.
This force is exerted on an area L2. Therefore the pressure of the gas is

P = \frac{F}{L^2} = \frac{Nm\overline{v^2}}{3V}

where V=L3 is the volume of the box.

In terms of the kinetic energy K.E.:
P =\frac{2}{3} \times \frac{N}{V} \times{K.E.}