What is kinetic energy of 1 gm of `O_(2)` at `47^(@)C` ?

Options

A. `2.17 xx 10^(2) J`
B. `2.24 xx 10^(2) J`
C. `1.24 xx 10^(2) J`
D. None of these

Answer

C

Explanation

`T = 47 + 273 = 320 K`
`K.E. = (3)/(2) RT` (For 1 mole)
`= (3)/(2) xx (8.314 JK^(-1) mol^(-1)) xx (320 K)`
`= 3990.7 J mol^(-1)`
Molar mass of `O_(2) = 32 g mol^(-1)`
`:.` K.E. (per gram of `O_(2)`) `= (3990.7)/(3) Jg^(-1)`
`= 124.7 J g^(-1)`
`= 1.247 xx 10^(2) Jg^(-1)`