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Topic: Why don't we consider VΔP when we define Q?  (Read 2685 times)

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Offline zillai

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Why don't we consider VΔP when we define Q?
« on: March 27, 2018, 07:28:01 AM »
We only define Q = ΔU + Wexp(expansion work = -PΔV).
If heat can cause ΔU and work, why work is defined only as expansion work in the first place where there are other forms of works, such as isochoric work (=VΔP)?

Let's say, there is a container with ideal gas T, P, V and V is constant. I give it heat and T, P changes to 2T, 2P. In this constant volume process, expansion work, Wexp is 0. That's why Qv = ΔU. In the mean time(in this constant volume process), ΔH = Qv + VΔP and VΔP is work done in constant volume process. My question is that why can't we just say this heat Q = ΔU + VΔP? Do we just ignore the non-expansion work? or is it included in ΔU?

One more thing. I googled it all day and someone wrote VΔP is work in flow process. I can't imagine how matter can flow in the constant volume container, and even though it can flow, what is the relationship between VΔP? They didn't explain it why so....
Enlighten me please and thank you in advance. 

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