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

Show that pressure exerted by the gas is two - thirds of average kinet

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