The temperature at which mean kinetic energy of an ideal gas molecule will be doubled that at `27^(@)C` is

Options

A. `54^(@)C`
B. `13.5^(@)C`
C. `600^(@)K`
D. `150^(@)K`

Answer

C

Explanation

`K.E_(2) prop T_(1) and KE_(2) prop T_(2)`
`(KE_(2))/(KE_(1))=(T_(2))/(T_(1))`
`therefore (2KE_(1))/(KE_(1))xxT_(1)=T_(2)`
`2xx300=T_(2)`
`therefore T_(2)=600K`