Asked by Catelyn
For a certain reaction at constant pressure, ∆H = -15kJ, and 22 kJ of expansion work is done on the system. What is ∆U for this process?
So I know
∆= ∆U + Work flow
If work is being done on the system, shouldn't work flow be positive? So I assumed it was:
-15kJ = ∆U + 22 kJ
-37kJ = ∆U
but the answer is 7kJ.
What am I doing that's wrong? Thanks.
So I know
∆= ∆U + Work flow
If work is being done on the system, shouldn't work flow be positive? So I assumed it was:
-15kJ = ∆U + 22 kJ
-37kJ = ∆U
but the answer is 7kJ.
What am I doing that's wrong? Thanks.
Answers
Answered by
Devron
Expansion is lost, so 22kJ should be -22kj. Plugging in your values, you will get 7kj.
-15kJ=∆U + (-22kJ)
-15kj+22kj=∆U
7kj=∆U
-15kJ=∆U + (-22kJ)
-15kj+22kj=∆U
7kj=∆U
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