**Postulates of Kinetic Theory of Gases**

- All gases consist of molecules and the molecules of a gas are all alike but differ with the molecules of different gases.
- The volume of molecules is negligible compared to the volume occupied by the gas.
- The molecules are in random motion and the velocity in all directions are ranging from zero to a maximum.
- During the motion, the molecules collide with each other and to walls of the containing vessel. The collision of molecules is perfectly elastic.
- The molecules behave like a perfectly elastic sphere.
- The duration of a collision between two molecules is negligible compared to the time between two successive collisions on the wall.
- The attractive or repulsive force between the molecules is negligible, and no force is exerted by molecules except during collision.

**Calculation of pressure**

Pressure is exerted by the gas molecules due to the continuous bombardment between the molecules and the walls of the containing vessel. Consider a cubical vessel of side L containing N number of molecules in gas inside it. Let the mass of each molecule be m

_{1}, m_{2}, m_{3},…………., m_{n}, and are moving with velocities c_{1}, c_{2}, c_{3},………., c_{n} respectively. Let the components of velocity along x, y, and z-axis be u, v, and w respectively as shown in the figure.Then we know

c

_{1}= u_{1}^{2}+ v_{1}^{2}+ w_{1}^{2}c

_{2}= u_{2}^{2}+ v_{2}^{2}+ w_{2}^{2}: : : :

c

_{n}= u_{n}^{2}+ v_{n}^{2}+ w_{n}^{2}The first molecule moves with velocity u

_{1}along x-axis and collides to the face of the cube with momentum mu_{1}. Since the collision is perfectly elastic, the molecule rebounds with the same velocity u_{1}in the reverse direction with momentum -mu_{1}. Then the change in momentum of the molecule along x-axis due to the collision is = mu_{1}– (-mu_{1}) = 2mu_{1}.The time between the two collisions = 2l/u

_{1}The rate of change in momentum of the molecule along x-axis = Change in momentum / Time taken

= 2mu

_{1}/(2l/u_{1})= mu

_{1}^{2}/ lThis is the force exerted by the molecule along x-axis. In the same way, the force exerted by the other molecules along x-axis are mu

_{2}^{2}/ l, mu_{3}^{2}/ l, mu_{N}^{2}/ l.The total force exerted by n molecules along x-axis is,

F

_{x}= mu_{1}^{2}/ l + mu_{2}^{2}/ l + mu_{3}^{2}/ l + mu_{N}^{2}/ lNow, the pressure exerted by N molecules of the gas along x-axis is

P

_{x}= F_{x}/A = F_{x}/ l^{2}= 1/l^{2}m/l [u_{1}^{2}+ u_{2}^{2}+ u_{3}^{2}+ …. +u_{N}^{2}]= m/l

^{3}[u_{1}^{2}+ u_{2}^{2}+ u_{3}^{2}+ …. +u_{N}^{2}]Similarly, the pressure exerted by the gas molecules along y and z axis are

P

_{y}= m/l^{3}[ v_{1}^{2}+ v_{2}^{2}+ v_{3}^{2 }+ ………. + v_{n}^{2}]P

_{z}= m/l^{3}[ w_{1}^{2}+ w_{2}^{2}+ w_{3}^{2}+ ………. + w_{n}^{2}]Since from large number of molecules, pressure is same in all direction, So,

P

_{x}= p_{y}= p_{z}Then the total pressure exerted by gas in the vessel

P = (1/3) (p

_{x}+ p_{y}+ p_{z})P = (1/3) [m/l

^{3}(u_{1}^{2}+ u_{2}^{2}+ u_{3}^{2}+ …. +u_{N}^{2}) + m/ l^{3}(v_{1}^{2}+ v_{2}^{2}+ v_{3}^{2}+ ………. + v_{n}^{2}) + m/ l^{3}(w_{1}^{2}+ w_{2}^{2}+ w_{3}^{2}+ ………. + w_{n}^{2})]P = (m/3l

^{3}) [(u_{1}^{2}+ v_{1}^{2}+ w_{1}^{2}) + (u_{2}^{2}+ v_{2}^{2}+ w_{2}^{2}) +…………………………. + (u_{n}^{2}+ v_{n}^{2}+ w_{n}^{2})]P = (m/3l

^{3}) [ c_{1}^{2}+ c_{2}^{2}+ c_{3}^{3}+ ……… + c_{n}^{2}]
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