We only define Q = ΔU + Wexp$Q = \Delta U + W_\text{exp}$ (expansion work = -PΔV$-P\Delta V$). If heat can cause ΔU$\Delta U$ 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$=V\Delta P$)?
Let's say, there is a container with ideal gas T, P, V$T, P, V$ and V$V$ can't be changed by this container. I give it heat therefore T, P$T, P$ changes to 2T$2T$, 2P$2P$. In this constant volume process, expansion work, Wexp is 0. That's why Qv = ΔU$Q_v = \Delta U$. In the mean time (in this constant volume process), ΔH = Qv + VΔP$\Delta H = Q_v + V\Delta P$ and VΔP$V\Delta P$ is work done in constant volume process. My question is that why can't we just say this heat Q = ΔU + VΔP$Q = \Delta + V\Delta P$? Do we just ignore the non-expansion work? or is it included in ΔU$\Delta U$? or because the definition of Q$Q$ is ΔU+PΔV$\Delta U+P\Delta V$ originally?
One more thing. I googled it all day and someone wrote that VΔP$V\Delta P$ 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$V\Delta P$? They didn't explain it why so my question get bigger and bigger.... Enlighten me please and thank you in advance.
