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Posted: Tue Feb 11, 2020 12:15 am
by Renee Grange 1I
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?

Re: 4D.5

Posted: Tue Feb 11, 2020 12:34 am
by Abigail Sanders 1E
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.