## Work

$w=-P\Delta V$
and
$w=-\int_{V_{1}}^{V_{2}}PdV=-nRTln\frac{V_{2}}{V_{1}}$

Kate_Santoso_4F
Posts: 72
Joined: Fri Sep 28, 2018 12:29 am

### Work

If a problem indicates constant volume in one way or another, do we always assume w=0?

Seohyun Park 1L
Posts: 73
Joined: Fri Sep 28, 2018 12:29 am

### Re: Work

I'm assuming you do assume w=0 because the only work we consider in reactions is usually pressure-volume expansion and in constant volume, delta v= 0.

Sarah Fatkin 4I
Posts: 71
Joined: Fri Sep 28, 2018 12:27 am

### Re: Work

Yep! For example, if a system has constant pressure, we know that delta(V) = 0, so the work is zero.

Max Hayama 4K
Posts: 63
Joined: Fri Sep 28, 2018 12:16 am

### Re: Work

Because the system is not doing any work against the surroundings to change the volume (or vise versa), it is safe to assume that under constant volume that no work is being done by or on the system.

Danny Elias Dis 1E
Posts: 60
Joined: Fri Sep 28, 2018 12:19 am

### Re: Work

Is this situation also the case where deltaU = q?

Posts: 60
Joined: Fri Sep 28, 2018 12:26 am

### Re: Work

Danny Elias Dis 1E wrote:Is this situation also the case where deltaU = q?

Yes, because ΔU = q+w, and if w = 0, then ΔU = q+0 = q.