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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!
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