8.47

[Jessica Jones 2B]

I am very confused as to why the solutions manual has it as -15kJ + 22 kJ when I would expect it to be -15kJ - 22 kJ?
wouldn't it be (delta U)= (delta H)-(P delta V)?

[Deap Bhandal L1 S1J]

Actually delta u equals q + w, so delta u = delta h +pv. I think you are confusing it with the equation delta h = delta u -pv which is a derivation of the first one.

[Kaylin Krahn 1I]

The work is positive because the work was done onto the system instead of by the system.

[Brigitte Phung 1F]

I think that because w = -P∆V for expansion problems, the 22 kJ in expansion work becomes equal to w = -P∆V = -(22 kJ). Therefore, when plugging it back into the equation, ∆U = ∆H - P∆V = -15 kJ - (-22 kJ) = - 15 kJ + 22 kJ = +7 kJ. Hope this helps!