## 4D.5

$\Delta U=q+w$

Renee Grange 1I
Posts: 56
Joined: Fri Aug 30, 2019 12:16 am
Been upvoted: 1 time

### 4D.5

In this question it states 22 KJ of expansion work is done on the system by compressing it into a smaller volume.
They set +22 kJ = -P delta V
Then state -22 kJ = P delta V
Why is this step necessary? How can work be negative is the system is being compressed?

Abigail Sanders 1E
Posts: 112
Joined: Wed Sep 11, 2019 12:16 am

### Re: 4D.5

This is because in the question, they are asking for the change in internal energy which is equivalent to the equation ∆U=∆H-P∆V. W=-P∆V so, this equation can be written as ∆U=∆H+w. They are simply setting -22kJ equivalent to P∆V to substitute it into the equation ∆U=∆H-P∆V. You are correct that the work on the system is positive. However w=-P∆V and the answer key is simply finding P∆V.