# Postulates of Kinetic Theory of Gases | Calculation of Pressure | Physics Grade XI

### Kinetic Theory of Gas

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#### Viva Voice Questions # Postulates of Kinetic Theory of Gases | Calculation of Pressure

Postulates of Kinetic Theory of Gases

1. All gases consist of molecules and the molecules of a gas are all alike but differ with the molecules of different gases.
2. The volume of molecules is negligible compared to the volume occupied by the gas.
3. The molecules are in random motion and the velocity in all directions are ranging from zero to a maximum.
4. During the motion, the molecules collide with each other and to walls of the containing vessel. The collision of molecules is perfectly elastic.
5. The molecules behave like a perfectly elastic sphere.
6. The duration of a collision between two molecules is negligible compared to the time between two successive collisions on the wall.
7. 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 m1, m2, m3,…………., mn , and are moving with velocities c1, c2, c3,………., cn­ 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
c1 = u12 + v12 + w12
c2 = u22 + v22 + w22
:        :        :         :
cn = un2 + vn2 + wn2

The first molecule moves with velocity u1 along x-axis and collides to the face of the cube with momentum mu1. Since the collision is perfectly elastic, the molecule rebounds with the same velocity u1 in the reverse direction with momentum -mu1. Then the change in momentum of the molecule along x-axis due to the collision is = mu1 – (-mu1) = 2mu1.
The time between the two collisions = 2l/u1

The rate of change in momentum of the molecule along x-axis = Change in momentum / Time taken
= 2mu1/(2l/u1
= mu12/ l
This 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 mu22/ l, mu32/ l, muN2/ l.

The total force exerted by n molecules along x-axis is,
Fx = mu12/ l + mu22/ l + mu32/ l + muN2/ l

Now, the pressure exerted by N molecules of the gas along x-axis is
Px = Fx/A = Fx/ l2 = 1/l2 m/l [u12 + u22 + u32 + …. +uN2]
= m/l3 [u12 + u22 + u32 + …. +uN2]

Similarly, the pressure exerted by the gas molecules along y and z axis are
Py = m/l3 [ v12 + v22 + v3+ ………. + vn2]
Pz = m/l3 [ w12 + w22+ w32+ ………. + wn2]

Since from large number of molecules, pressure is same in all direction, So,
Px = py = pz

Then the total pressure exerted by gas in the vessel
P = (1/3) (px + py + pz)
P = (1/3) [m/l3 (u12 + u22 + u32 + …. +uN2) + m/ l3 (v12 + v22+ v32+ ………. + vn2) + m/ l3 (w12 + w22+ w32+ ………. + wn2)]
P = (m/3l3) [(u12 + v12 + w12) + (u22 + v22 + w22) +…………………………. + (un2 + vn2 + wn2)]
P = (m/3l3) [ c12 + c22 + c33 + ……… + cn2] ## Popular Subjects

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