## DeltaU=q+w

$\Delta U=q+w$

Te Jung Yang 4K
Posts: 63
Joined: Fri Sep 28, 2018 12:18 am

### DeltaU=q+w

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
Posts: 67
Joined: Fri Sep 28, 2018 12:17 am

### Re: DeltaU=q+w

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

305113590
Posts: 66
Joined: Fri Sep 28, 2018 12:28 am

### Re: DeltaU=q+w

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.

taryn_baldus2E
Posts: 62
Joined: Fri Sep 28, 2018 12:24 am

### Re: DeltaU=q+w

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

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

Jchellis 1I
Posts: 61
Joined: Fri Sep 28, 2018 12:15 am

### Re: DeltaU=q+w

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