textbook 8.93 c


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Jessica Wakefield 1H
Posts: 58
Joined: Thu Jul 27, 2017 3:00 am

textbook 8.93 c

Postby Jessica Wakefield 1H » Sat Mar 17, 2018 1:36 pm

For part c of 8.93 you have to calculate the change in internal energy of the system using your answers from part a and b.
The value of w you find in part a is +3.72 kj/mol and the value of dH you find in part b is -3267.5 kj/mol and for c you are supposed to get -3263.8 kj/mol but I was wondering why you would do -3267.5+3.72 instead of -3.72 because work is being done by the system against the atmosphere so I would have thought w would be subtracted.

Leah Thomas 2E
Posts: 51
Joined: Fri Sep 29, 2017 7:06 am

Re: textbook 8.93 c

Postby Leah Thomas 2E » Sat Mar 17, 2018 2:01 pm

When you are calculating work, the negative sign is already taken care of when you do -PΔV=-Δnrt so, therefore, all you have to do is ΔU=q+w and don't have to look at the sign.

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