4D.5


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Renee Grange 1I
Posts: 56
Joined: Fri Aug 30, 2019 12:16 am
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4D.5

Postby Renee Grange 1I » Tue Feb 11, 2020 12:15 am

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

Postby Abigail Sanders 1E » Tue Feb 11, 2020 12:34 am

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.


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