Latest Articles

Pressure of Gas

Pressure of gas

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_1x² / l
this is the force exerted on wall A due to this molecule.
Force on wall A due to all other molecules
F = Σmv_1x²/L (3)
As all directions are equivalent
Σv_1x²=Σv_1y²=Σv_1z²
Σv_1x²= 1/3Σ((v_1x)² + (v_1y)² +( v_1z)² )
= 1/3 Σv₁²
Thus F = (m/3L) Σv₁²
N is total no. of molecules in the container so
F = (mN/3L) (Σ(v₁)²/N)
Pressure is force per unit area so
     P = F/L²
      =(M/3L³)(Σ(v₁)²/N)
where ,M is the total mass of the gas and if ρ is the density of gas then
     P=ρΣ(v₁)²/3N
since Σ(v₁)²/N is the average of squared speeds and is written as v_mq² known as mean square speed
Thus, v_rms=√(Σ(v₁)²/N) is known as root mean squared speed rms-speed and v_mq² = (v_rms)²
Pressure thus becomes
     P = (1/3)ρv_mq²                     (4)
or     PV = (1/3) Nmv_mq²                     (5)
from equation (4) rms speed is given as
     v_rms = √(3P/ρ)
               = √(3PV/M)                    (6)