If the average kinetic energy of gas molecule at `27^(@)C` is `6.21xx10^(-21)J`, then the average kinetic energy at `227^(@)C` will be

Options

A. `52.2xx10^(-21)J`
B. `5.22xx10^(-21)J`
C. `10.35xx10^(-21)J`
D. `11.35xx10^(-21)J`

Answer

C

Explanation

`(KE_(1))/(KE_(1))=(T_(2))/(T_(1))=(500)/(300)=(5)/(3)`
`K.E_(2)=(5)/(3)xx6.21xx10^(-21)`
`=10.35xx10^(-21)J`