Question

Assuming the expression for the pressure P exerted by an ideal gas, prove that the kinetic energy per unit volume of the gas is `3/2` P.

Answer 1

According to the kinetic theory of gases, the pressure P exerted by the gas is

p = `1/3 ρv_(rms)^2`

= `1/3 M/V v_(rms)^2`

where `v_(rms)` is the rms speed of the gas molecules; M, V and ρ are the mass, volume and density of the gas, respectively.

∴ `PV = 1/3 Mv_(rms)^2`

= `2/3 (1/2 Mv_(rms)^2)`

Here, `1/2Mv_(rms)^2` is the total kinetic energy (KE) of all molecules of the gas. 

∴ PV = `2/3 (KE)`

∴ KE = `3/2 PV`

∴ `(KE)/V = 3/2 p`

Thus, the kinetic energy per unit volume of an ideal gas is `3/2` P.