Question

The average kinetic energy of gas molecule at `27^(@)C` is `6.21xx10^(-21)` J. Its average kinetic energy at `127^(@)C` will be
A. `12.2xx10^(-21)J`
B. `8.28xx10^(-21)J`
C. `10.35xx10^(-21)J`
D. `11.35xx10^(-21)J`

Answer 1

Correct Answer - B
as, KE `prop`T `rArr E_(127)/E_(27)`=((127+273))/((27+273))`
`E_(127) =6.21xx120^(-21)xx(400)/(300)`
`=.28xx10^(-21) J`