Ten moles of `(O_2)` gas are kept at temperature `T`. At some higher temperature `2T`, fourty percent of molecular oxygen breaks into atomic oxygen. Find change in internal energy of the gas.

Answer

A, B, D

Explanation

Initial energy Using the relation, `U = (nf)/2 RT`
we have, `n = 10, f = 5` for diatomic `O_(2)` gas
`U_i = (10 xx 5)/2(RT) = 25 RT`
Final energy fourty percent means 4 moles `O_(2)` breaks into `O`. So, it will become `8 moles` of monoatomic gas (O). remaining 6 moles are of diatomic gas `(O_(2)`. But now the new tempreature is (2 T).
`:. Uf =(8 xx 3)/2(R) (2T) +(6 +5)/2(R)(2T) =54 RT`
So, change in internal energy, `DeltaU = Uf -U_i = 29RT`.