Derive an xpression for pressure of gas in a container
Let us consider a cubic enclosure of side L with n number of gas molecules inside of mass m inside it.
According to kinetic theory, the molecules are constantly moving, collisions are elastic and pressure experienced by the walls of the enclosure is due to the impact of these molecules with the walls.
Let a molecule travel with a velocity c with its components along x axis be u, y axis be v and z axis be w and it collides with the wall of the enclosure.
Momentum before collision = mu
and after collsion it becomes –mu.
Thus change in momentum in time ∆t be mu-(-mu) = 2mu
Let Fx = Average Impulsive force experienced by the wall due striking of n molecules in the x direction.
Thus total change in momentum in time ∆t in x diection will be Fx∆t ---1.
Now per collision a molecule has to travel distance 2L from one end to the other, let the time taken for a molecule to collide with the wall once = t
Thus we have t = 2L/v
Now number of collision per 1 second that is collision frequency will be = 1/t= u/2L
Again number of collisions by 1 molecule hitting the wall in time ∆t will be = (u/2L)∆t
Again number of collisions n molecules hitting the wall will be n(u/2L)∆t
Total change in momentum of the molecules travelling along x axis will be = 2mu× n(u/2L)∆t
= (mnu2/L)∆t ---2
By 1 and 2.
Fx ∆t = (mnu2/L)∆t
=> Fx = (mnu2/L) ----3.
Similarly force experienced by the wall because of molecules moving in y and zdirection will be
Fy ∆t = (mnv2/L)∆t
=> Fy = (mnv2/L) --- 4
and
Fz ∆t = (mnv2/L)∆t
=> Fz = (mnv2/L) ---5
Again we have from components of velocity
c2 = u2 + v2 + w2
Since the particles are not acted with any forces we assume they move in all directions with same velocity
u = v = w
=> c2 = 3u2
=> u2 = v2 = w2 = = 1/3 c2
Hence,
Fx =1/3(mnc2/L)
Now A = L2 for a cube
Fx/A = Px = 1/3mnc2/L3
Again Fx/A = Px and L3 =V (Volume)
Px = 1/3mnc2 /V
But as gas has uniform pressure in all direction thus Px = Py = Pz = P
=> P = 1/3mnc2 /V
=> PV = 1/3mnc2