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### Internal Energy

Posted: Wed Jan 24, 2018 10:51 pm
If delta P = 0 (constant pressure), why is delta U = q + w? I thought that w = -PdeltaV. If P is 0, there would be no work then right?

### Re: Internal Energy

Posted: Wed Jan 24, 2018 10:52 pm
William Lan 2l wrote:If delta P = 0 (constant pressure), why is delta U = q + w? I thought that w = -PdeltaV. If P is 0, there would be no work then right?

In w = -P * delta V, you would plug in the constant pressure for P, not the change in P which would be delta P.

### Re: Internal Energy

Posted: Thu Jan 25, 2018 12:10 am
no change in pressure means ∆P=0, not P=0. the equation for work uses P not ∆P and it is unlikely P=0

### Re: Internal Energy

Posted: Thu Jan 25, 2018 9:43 pm
A change in pressure(delta P) does not necessarily mean that pressure is 0 it just means that the pressure is constant. The equation for work uses the actual pressure not the change in the pressure.

### Re: Internal Energy

Posted: Thu Jan 25, 2018 9:52 pm
If it was delta P instead of P in the equation, then pressure would equal 0 when it is constant.

### Re: Internal Energy

Posted: Thu Jan 25, 2018 10:42 pm
William Lan 2l wrote:If delta P = 0 (constant pressure), why is delta U = q + w? I thought that w = -PdeltaV. If P is 0, there would be no work then right?

That is a variation of the original equation. It all depends on what the environment/surrounding/situation is and whether the reaction is reversible or not.