**Postulates:**

**1. **A gas is composed of a very large number of tiny particles called molecules.

2. The molecules of a gas are perfectly elastic spheres.

3. The molecules of a gas are identical in all respects.

4. They move in all directions with all possible velocities.

5. During their random motion, they collide with each other and with the walls of the container in which they are kept.

6. Between collisions, the molecules move in a straight line with a constant velocities.

7. The time of collision is very small compared with the time of transit.

8. The average distance travelled by a molecule between successive collision is called the **mean free path.**

9. The force of attraction between individual molecule negligible..

10. The volume of the molecule is negligible compared to the volume of the gas.

This is animated picture of the molecules of the gas in motion. The picture above shows the molecule in constant motion. The random motion makes the molecules collide with each other and with the walls of the container. The collision is perfectly elastic. The molecules are perfectly spherical in shape. They are elastic in nature.

Now let us find the pressure exerted by the gas.

1. Let us have a cubical vessel whose side = 1 m in length.

2. Let us fill the cubical vessel with the gas.

3. Let there be 'n' number of molecules in the vessel.

4. These molecules hit the wall in their random motion and exert a force on the walls.

5. The force experienced by the walls per unit area gives the pressure of the gas.

6. Let a single molecule move parallel to the x axis with a velocity ' v '. It strikes the wall and rebounds with a velocity '-v-

7. Momentum of the molecule on the wall before impact = mv

8.Momentum of the molecule after the impact = -mv

9. Change in momentum = mv - (-mv) = 2mv

10. Time taken to make one to and fro motion ( 2 metres) =` (displacement)/(velocity)` = `2/v`

11. Rate of change of momentum =change in momentum / time =` (2mv)/(2/v) ` = `mv^2`

12. Newton's second law states that rate of change of momentum gives force.

Hence force on the wall due to single molecule = mv^{2}

13. The cubical container has directions x,y and z .

Hence average molecules = n/3

Hence force on the wall due to n/3 molecules =`( nmv^2)/3`

14. Force per unit area gives pressure.

The area of the face of the cube = 1 square meter

Therefore Pressure on the wall due to these molecules is `P = nmv^2/ 3`

15. Let us represent the average velocities of the molecules by C

Hence C can replace v

Thus we have P = `(n mC^2)/3`

16. nm is the mass of the gas per unit volume. So nm gives the density ρ of the gas.

Hence `P = (1 rho C^2 )/3` **Therefore `P = (1rhoC^2)/3` **** **

**Relation between pressure and kinetic energy:**

**1. **Pressure =` (n mC^2)/3 `

2. But nm + M ( the total mass of the gas per unit volume)

3. Hence P =`(1 MC^2)/3`

4. Multiply and divide by 2 .

we get `P = 2/3 xx (1 MC^2)/2`

5. Since `(1 MC^2)/2 ` = Kinetic energy we get `P = (2 KE)/3 `

Hence the pressue of a gas is equal to two-thirds of the mean kinetic energy per unit volume of the gas.

**Temperature :**

1. `P = 1/3 nmC^2`

2. Multiplying both sides by V we get `PV = 1/3 mn VC^2`

3. nV = N = Avagadro number

4. Hence` PV = 1/3 NmC^2`

5. But PV = RT ( R is a gas constant)6. Therefore `1/3 NmC^2 = RT`

7. Multiply both sides by 3/2 we get `(3/2) xx( 1/3) xx(NmC^2) = RT xx 3`

This gives us `1/2 NmC^2 = (3RT)/2`

8. Let us rearrange as `1/2 m C^2 = 3(RT) /N` where `R/N` is a constant K (Boltzmann's consan)

9. Therefore mean kinetic energy of the molecule of the gas ` (3 KT)/2`

**Hence kinetic energy of a molecule is directly proportional to the Kelvin temperature.**

**Q:1 How much energy is possessed by a helium atom moving at 30.0 m/s?**

**Sol:**

Formula `KE = 1/2(mv^2)`

Atomic mass of Helium atom is `4.0026`

Hence

KE = `0.5 xx 4.0026 xx 30^2`

=` 0.5 xx4.0026 xx 900`

= `1801.17` m/s

= `1.80 xx10^ 3` ^{ }m/s

Answer : **`1.80 xx 10^3` m/s**