At what temperarture, the kinetic energy of a gas molecule is half of the value at `27^(@)C.` ?

Explanation

Given, `T= 27^(@)C =27+273=300K`
Kinetic energy of a gas molecule `E=(3)/(2)kT`
Where, k is Boltazmann's constant.
`therefore" " EpropT`
`Rightarrow" " (E_(1))/(E_(2))=(T_(1))/(T_(2))`
`Rightarrow" " (E)/(E//(2))=(300)/(T_(2))`
`When" " T_(2)=150K`
`then" " T_(2)=150-273=-123^(@)C`