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

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#### 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....