## Homework Question 8.47

$\Delta U=q+w$

Lily Guo 1D
Posts: 64
Joined: Fri Sep 29, 2017 7:03 am

### Homework Question 8.47

For 8.47, why does the solutions manual use the equation (delta)H = U + P(delta)V instead of just U = q+ w? I'm also confused as to why work would be positive instead of negative because I thought that expansion work done ON a system is negative.

Felicia Fong 2G
Posts: 31
Joined: Sat Jul 22, 2017 3:00 am

### Re: Homework Question 8.47

I think expansion work done on a system is positive. Work done by the system would be negative.

Sarah_Stay_1D
Posts: 57
Joined: Sat Jul 22, 2017 3:00 am
Been upvoted: 1 time

### Re: Homework Question 8.47

Lily Guo 1D wrote:For 8.47, why does the solutions manual use the equation (delta)H = U + P(delta)V instead of just U = q+ w? I'm also confused as to why work would be positive instead of negative because I thought that expansion work done ON a system is negative.

You are in a sense actually using ΔU= q + w. First you can rearrange the equation ΔH = ΔU + PΔV to be ΔU = ΔH - PΔV. Sense we know that at a constant pressure ΔH=q, we can substitute q for ΔH. We also know that w = -PΔV , so we can substitute w for -PΔV. So, know you have the familiar equation ΔU = q + w. Then you just substitute the values given in the problem.

Angela 1K
Posts: 80
Joined: Fri Sep 29, 2017 7:05 am

### Re: Homework Question 8.47

But that ultimately give you a negative work, whereas work done ON a system should be positive???

Lily Guo 1D
Posts: 64
Joined: Fri Sep 29, 2017 7:03 am

### Re: Homework Question 8.47

Angela 2I wrote:But that ultimately give you a negative work, whereas work done ON a system should be positive???

I just realized my mistake. Work done ON a system is positive, but EXPANSION work means that work is done on the surroundings/BY the system because the system is expanding. Therefore, work would be negative. That's my bad for not catching that, sorry!