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Te Jung Yang 4K
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Joined: Fri Sep 28, 2018 12:18 am


Postby Te Jung Yang 4K » Sun Feb 10, 2019 9:39 pm

I don't understand why the equation is change in U is equal to q + w. Why is it not change in U is equal to the change in q + the change in w?

Kyle Golden Dis 2G
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Joined: Fri Sep 28, 2018 12:17 am

Re: DeltaU=q+w

Postby Kyle Golden Dis 2G » Sun Feb 10, 2019 9:41 pm

q and w inherently denote change because it is a transfer of energy.

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Re: DeltaU=q+w

Postby 305113590 » Sun Feb 10, 2019 9:46 pm

Both q and w are considered to be change already. Since heat is flowing between the system and surroundings, there is already a change in heat being transferred/absorbed/released. Same applies for work. Work is a broad term, but it's the merely the idea that something is capable of doing stuff. When we look at the equation of -P∆V and -nRTln(V2/V1), there is an area under the curve that is indicative of work with infinitesimal changes. Both q and w already are going through changes, so it isn't necessary to write a delta.

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Re: DeltaU=q+w

Postby taryn_baldus2E » Sun Feb 10, 2019 9:48 pm

The change in q and w can be assumed whereas internal energy is a state function.

Joon Chang 2F
Posts: 59
Joined: Fri Sep 28, 2018 12:25 am

Re: DeltaU=q+w

Postby Joon Chang 2F » Sun Feb 10, 2019 9:49 pm

q and w are both path functions and both represent transfers of energy between systems and their surroundings.

Jchellis 1I
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Joined: Fri Sep 28, 2018 12:15 am

Re: DeltaU=q+w

Postby Jchellis 1I » Sun Feb 10, 2019 10:47 pm

Yes but q is energy lost as heat it does no work.

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