- 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)

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