One mole of an ideal diatomic gas undergoes a transition from A to B along a path AB as shown in (figure). The change in internal energy of the gas during the transition is `(gamma=3//5)`

Options

A. `- 20 kJ`
B. `20 J`
C. `- 12 kJ`
D. `20 kJ`

Explanation

`T_(A) = (P_(A) V_(A))/(nR) , T_(B) = (P_(B) V_(B))/(nR)`
`(Delta U)_(A rarr B) = nC_(V) Delta T = n. (5R)/(2) (T_(B) - T_(A))`
`= (5 nR)/(2) ((P_(B) V_(B) - P_(A) V_(A))/(nR))`
`= (5)/(2) (P_(B) V_(B) - P_(A) V_(A))`
`= (5)/(2) (2 xx 6 - 5 xx 4) xx 10^(3) = - 20 kJ`