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.
  • v1x , v1y, v1z 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 v1x and rebounds with the same velocity in the backward direction as the collisions are perfectly elastic.

    Pressure of gas
  • If m is the mass of molecule, the change in Momentum during collision is
         mv1x - (-mv1x) = 2 mv1x          (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/v1x
  • Number of impacts of this molecule with A in unit time is
         n = I/Δt = v1x/ 2l               (2)
    Rate of change of momentum is
         ΔF = ΔP/Δt
    from (1) and (2)
         ΔF = mv1x2 / l
    this is the force exerted on wall A due to this molecule.
  • Force on wall A due to all other molecules
         F = Σmv1x2/L           (3)
  • As all directions are equivalent
    Σv1x2= 1/3Σ((v1x)2 + (v1y)2 +( v1z)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

    where ,M is the total mass of the gas and if ρ is the density of gas then
    since Σ(v1)2/N is the average of squared speeds and is written as vmq2 known as mean square speed
    Thus, vrms=√(Σ(v1)2/N) is known as root mean squared speed rms-speed and vmq2 = (vrms)2

  • Pressure thus becomes
         P = (1/3)ρvmq2                     (4)
    or     PV = (1/3) Nmvmq2                     (5)
    from equation (4) rms speed is given as
         vrms = √(3P/ρ)
                   = √(3PV/M)

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