**Hello dear students, we know that the gas possess very negligible force of attraction between its molecules, due to this every gas obeys the following assumptions.**

- The gas molecules are always in state of random motion. i.e. gas molecules are moving in all possible directions with all possible velocities.
- Due to their random motion, gas molecules always collide with each other and with wall of container.
- Between any two successive collisions molecules travels in a straight line with constant velocity.
- The distance covered by molecules during the collision is called free path.
- As the collision is perfectly elastic kinetic energy of molecules before and after collision is same.

All above discussion implies that gas molecules exert pressure on the wall of vessel.

**Let’s derive an expression for the pressure exerted by gas molecules……………!**

Consider a perfect gas is filled in cubical vessel of side ‘*l’ as shown below,*

By assumptions of kinetic theory of gases, the gas molecules are constantly moving in all possible direction with all possible velocities. Therefore they possess momentum thus exerts pressure on wall of container.

N= number of molecules of gas.

m=mass of each molecule of gas.

M=Nm= Total mass of gas.

A= *l ^{2}=* Area of face of cube.

V=* l ^{3}*= Volume of cube.

ρ=M/V= Density of gas.

Imagine that the gas molecules are moving in XYZ co-ordinate axes with different velocities C_{1,} C_{2…}C_{N}. Let u_{1},v_{1},w_{1} are the components of velocity C_{1} along X, Y and Z axes respectively.

Consider a gas molecule of mass m is moving along X axis with velocity u_{1}.Then its initial momentum is given as,

initial momentum=mu_{1 }

The gas molecule will collide on wall of container and moves in opposite direction with velocity – u_{1}, therefore the final momentum of gas molecule is,

final momentum=-mu_{1}

The Change in momentum of gas molecule per unit collision

= Final momentum- Initial momentum.

= -mu_{1}– mu_{1}

= -2 mu_{1}

∴ The Change in momentum of gas molecule per unit collision = |-2 mu_{1} |

∆P = 2 mu_{1} ……….(1)

The distance travelled by gas molecule during this collision is ‘2l’ with velocity u_{1}.

∴ Time taken by molecule during this collision,

t = 2l/u_1 ……………(2)

Then according to Newton’s second law, force exerted by this molecule is given as,

∴f_{1} = change in momentum)/time

This is an expression for the pressure exerted by the gas when contained in a vessel.

**Some important points to note…..!**

- Pressure of gas increases with increase in density of gas.
- Pressure of gas is directly proportional to square of rms speed of gas molecules.

**Let’s solve some numerical on the same article…..!**

Ex:1) Find the rms speed of gas molecules if the pressure exerted by the gas of mass 0.5 kg filled in a cubical vessel of volume 2.5 cm^{2} is 2.5×10^{6} Pa.

Solution:

Here, M=0.5 kg, V=2.5 cm^{2}=2.5×10^{-4} m^{2}, P=2×10^{6} Pa.

The formula for the pressure