4A3 7th Edition

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Jasmin Argueta 1K
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Joined: Fri Sep 28, 2018 12:16 am

4A3 7th Edition

Postby Jasmin Argueta 1K » Wed Feb 06, 2019 5:26 pm

In regards to this problem, part c asks for the internal energy of the system. The work done is 28J and the internal energy is also 28J. I know internal energy is change in U=q+w, although since there is no heat being added would this mean q is zero? Making it an adiabatic system? This is the only way I could get the answer to be 28J but the solutions Manuel doesn't mention anything about this being an adiabatic system

David Effio 1H
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Joined: Wed Feb 14, 2018 3:01 am

Re: 4A3 7th Edition

Postby David Effio 1H » Wed Feb 06, 2019 7:02 pm

I would assume this is correct. Since the problem does not really mention any temperature change so trying to calculate for change in temperature using the heat capacity would get you 0.

Miriam Sheetz 2B
Posts: 64
Joined: Fri Sep 28, 2018 12:25 am

Re: 4A3 7th Edition

Postby Miriam Sheetz 2B » Sat Feb 09, 2019 2:45 pm

I agree that you can assume that q=0 because if no heat is added/removed from the system then delta U is equal to work.

Emily Huang 1E
Posts: 30
Joined: Fri Sep 28, 2018 12:20 am

Re: 4A3 7th Edition

Postby Emily Huang 1E » Sat Feb 09, 2019 2:58 pm

q would equal zero so there is no heat input or output. You can assume the system is adiabatic even though it doesn't explicitly state that

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