## delta U=q+w

$\Delta U=q+w$

Ajith Raja 2L
Posts: 24
Joined: Wed Nov 18, 2015 3:00 am

### delta U=q+w

So I was looking at the reversible expansion area in the course reader in Ch.9. There is a part where as the piston pushes out, heat comes in to replace the energy so delta U is 0 since there is no change in temp. But couldnt you also say that q=0 since q=CdeltaT would make q 0? Or does the mcat equation not apply to this area? Thank you.

Shailyn_Moore_3C
Posts: 21
Joined: Wed Sep 21, 2016 2:59 pm

### Re: delta U=q+w

Well to my knowledge it seems that the delta U equals 0 since if heat is going into the system q= some positive number but if equal amount of work is being done by the system (through expansion) than w= some negative number. So it kind of a situation where you 5+(-5)= 0. The heat and work cancel each other out