## HW 8.47

Alyssa McLeod 3E
Posts: 13
Joined: Wed Sep 21, 2016 2:59 pm

### HW 8.47

For a certain reaction at constant pressure, $\Delta U=-15kJ$, and 35kJ of expansion work is done by the system. What is $\Delta U$ for this reaction?

I got the correct answer by using the formula $\Delta U=\Delta H+w$ but when I checked my answer via an old post here on chemistry community, someone made the point that work should be negative because $w=-P\Delta V$ and the volume change must be positive because expansion occurs. But plugging a negative work into the equation $\Delta U=\Delta H+w$ yields the wrong answer, so what am I missing? (Note: I don't have the solutions manual, so my answer (7kJ) could just be wrong?)

Rachel Kipp 1I
Posts: 16
Joined: Wed Sep 21, 2016 2:55 pm

### Re: HW 8.47

$\Delta H=\Delta U+P\Delta V$ so $\Delta U = \Delta H - P\Delta V$ and $w = -P\Delta V$ which $= +22kJ$, since the work is done ON the system. We also know $\Delta H = -15kJ$ from the givens in question 47

Plug in this info and you will get, $-15kJ + 22kJ = \Delta U = +7kJ$

nicoleclarke_lec1M
Posts: 25
Joined: Sat Jul 09, 2016 3:00 am

### Re: HW 8.47

My book stated that the work done by the system was 22 kJ, so I simply used the formula Δu=q+w, and made the work negative since the work is done BY the system, and therefore the system is losing energy. Also, since the equation is at constant pressure, q is really q sub p, and i used Δu=15+(-22) to get that Δu=7 kJ.

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