Correct Answer - We know, `(1)/(2) mv^(bar2) = (3)/(2) kT`
or `(1)/(2) mv^(bar2) prop T`
Therefore, higher the temperature of the gas, more will be the average kinetic energy possessed by the gas molecules, thus, the temperature of the gas given the measure of the average kinetic energy of the gas molecules.
If `T = 0`, then `(1)/(2) mv^(bar2) = 0`
Therefore, absoulte zero is that temperature at which the molecules motion stops.