The pressue of a fluid is a linear function of volume (P=a+bV)and the internal energy of the fluide is
`U=34+ 3PV`(S.I.units). Find a,b,w, `DeltaE` and q for change is state from `( 100 Pa, 3m^(3))` to
`(400 Pa, 6m^(3))`

Explanation

`W = - underset(V_(1))overset(V_(2))intPdV" "(100 = a + 3b, 400= a +6b, a= - 200 & b = 100)`
`W = - underset(V_(1))overset(V_(2))int(a + bV)dV" "- {[aV]+[(bV^(2))/(2)]}_(V_(1))^(V_(2))`
` =- [-600+13.5 xx 100] =- 750`
`DeltaU = 6300`