Consider a cubical vessel with perfectly elastic walls containing large number of molecules say N let l be the dimension of each side of the cubical vessel.

v_{1x} , v_{1y}, v_{1z} be the x, y, and z component of a molecule with velocity v.

Consider the motion of molecule in the direction perpendicular to the face of cubical vessel.

Molecule strikes the face A with a velocity v_{1x} and rebounds with the same velocity in the backward direction as the collisions are perfectly elastic.

If m is the mass of molecule, the change in Momentum during collision is
mv_{1x} - (-mv_{1x}) = 2 mv_{1x} (1)

The distance travelled parallel to x-axis an is between A to A´ and when molecule rebounds from A´ and travel towards A is 2L

Time taken by molecule to go to face A´ and then comeback to A is
Δt = 2l/v_{1x}

Number of impacts of this molecule with A in unit time is
n = I/Δt = v_{1x}/ 2l (2)
Rate of change of momentum is
ΔF = ΔP/Δt
=nΔP
from (1) and (2)
ΔF = mv_{1x}^{2} / l
this is the force exerted on wall A due to this molecule.

Force on wall A due to all other molecules
F = Σmv_{1x}^{2}/L (3)

As all directions are equivalent
Σv_{1x}^{2}=Σv_{1y}^{2}=Σv_{1z}^{2}
Σv_{1x}^{2}= 1/3Σ((v_{1x})^{2} + (v_{1y})^{2} +( v_{1z})^{2} )
= 1/3 Σv_{1}^{2}
Thus F = (m/3L) Σv_{1}^{2}

N is total no. of molecules in the container so
F = (mN/3L) (Σ(v_{1})^{2}/N)

Pressure is force per unit area so
P = F/L^{2}
=(M/3L^{3})(Σ(v_{1})^{2}/N)
where ,M is the total mass of the gas and if ρ is the density of gas then
P=ρΣ(v_{1})^{2}/3N
since Σ(v_{1})^{2}/N is the average of squared speeds and is written as v_{mq}^{2} known as mean square speed
Thus, v_{rms}=√(Σ(v_{1})^{2}/N) is known as root mean squared speed rms-speed and v_{mq}^{2} = (v_{rms})^{2}

Pressure thus becomes
P = (1/3)ρv_{mq}^{2} (4)
or PV = (1/3) Nmv_{mq}^{2} (5)
from equation (4) rms speed is given as
v_{rms} = √(3P/ρ)
= √(3PV/M) (6)