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 KJ`
C. `-12 KJ`
D. `20 KJ`

Answer

A

Explanation

Here, `P_(1)= 5Kpa, P_(2)= 2Kpa`
`V_(1)= 4m^(3), V_(2)= 6m^(3), r= 3//5`
As change is internal energy
`DeltaU= n C_(v)DeltaT= (PV)/(nR)`
`DetlaT=T_(2)-T_(1)=(P_(2)V_(2)-P_(1)V_(1))/(nR), C_(v)= R/(gamma-1)`
`DeltaU=(nR)/(gamma-1)((P_(2)V_(2)-P_(1)V_(1))/(nR))=(P_(2)V_(2)-P_(1)V_(1))/(gamma-1)`
`=(5xx4-2xx6)/(3/5-1)=(200-12)/(-2/5)= -8/(2/5)=-20KJ`