## U clarifications

$\Delta U=q+w$

Kirsty Star 2H
Posts: 48
Joined: Fri Sep 28, 2018 12:24 am

### U clarifications

How does the equation ΔU = q + w change when ΔP is 0 and ΔV is 0 respectively?

monikac4k
Posts: 56
Joined: Fri Sep 28, 2018 12:25 am

### Re: U clarifications

Hey Kirsty!
When the change in pressure is 0, both q and w should be considered when calculating the change in internal energy of the system.
However, when the change in volume is 0, we can only focus on q. Remember that the calculation for work is dependent upon the change in volume times the external pressure. If delta v is 0, then the calculation for work should come out to zero as well.

BenJohnson1H
Posts: 68
Joined: Fri Sep 28, 2018 12:17 am

### Re: U clarifications

So essentially, in this case, deltaU=deltaV

2c_britneyly
Posts: 62
Joined: Fri Sep 28, 2018 12:16 am

### Re: U clarifications

When ΔV=0, ΔU=q+w=q-PΔV. Therefore when we substitute ΔV=0, the term -PΔV turns into zero, and we're left with ΔU=q.
When ΔP=0, then ΔU=qp+w. Heat at a constant pressure is enthalpy, thus the equation will be ΔU=ΔH-PΔV