Show that pressure exerted by the gas is two - thirds of average kinetic energy per unit volume of the gas molecules.

Explanation

The intemal cnergy of the gas is given by
`U=(3)/(2)NkT`
The above equation can also be written as
`U =(3)/(2)PV`
Since PV =NKT
`P=(2)/(3) (U)/(V) =(2)/(3)u` ........(1)
From the equation (1), we can state that the pressure of the gas is equal to two thirds of internal energy per unit volume or internal energy density `(u=(U)/(V))` .
Writing pressure in terms of mean kinetic energy density using equation
`P = (1)/(3)nmbar(V^(2)) =(1)/(3)rhobar(V^(2))` ......(2)
where `rho = nm = "mass density"` (Note n is number density)
Multiply and divide R.H.S of equation (2) by 2, we get
`P=(2)/(3)((rho)/(2)bar(V^(2)))`
`P=(2)/(3)bar(KE)` ..(3)
From the equation (3), pressure is equal to `(2)/(3)` of mean kinetic energy per unit volume.