U clarifications


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

U clarifications

Postby Kirsty Star 2H » Tue Feb 05, 2019 4:58 pm

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

Postby monikac4k » Tue Feb 05, 2019 5:02 pm

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

Postby BenJohnson1H » Tue Feb 05, 2019 10:13 pm

So essentially, in this case, deltaU=deltaV

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

Re: U clarifications

Postby 2c_britneyly » Wed Feb 06, 2019 12:27 am

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


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