At T(K), the ratio of kinetic energies of 4g of `H_(2)` (g) and 8g of `O_(2)` (g) is

Options

A. `1 : 4`
B. `4 : 1`
C. `2 : 1`
D. `8 : 1`

Answer

d

Explanation

`KE = 4nRT implies (KE_(1))/(KE_(2))=n_(1)/n_(2)`
No. of moles of `H_(2) (n_(1))=4/2=2`
No. of moles of `O_(2) (n_(2))=8/32=1/4`
`therefore (KEH_(2))/(KEO_(2))=2/(1//4)=8/1`
`KEH_(2):KEO_(2)=8:1`