Why don't we consider VΔP when we define Q?

Question

We only define $Q = \Delta U + W_\text{exp}$ (expansion work = $-P\Delta V$). If heat can cause $\Delta U$ 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 ($=V\Delta P$)?

Let's say, there is a container with ideal gas $T, P, V$ and $V$ can't be changed by this container. I give it heat therefore $T, P$ changes to $2T$, $2P$. In this constant volume process, expansion work, Wexp is 0. That's why $Q_v = \Delta U$. In the mean time (in this constant volume process), $\Delta H = Q_v + V\Delta P$ and $V\Delta P$ is work done in constant volume process. My question is that why can't we just say this heat $Q = \Delta + V\Delta P$? Do we just ignore the non-expansion work? or is it included in $\Delta U$? or because the definition of $Q$ is $\Delta U+P\Delta V$ originally?

One more thing. I googled it all day and someone wrote that $V\Delta P$ 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 $V\Delta P$? They didn't explain it why so my question get bigger and bigger.

Answer 1

Let's start by the fundamental ideas. Assume, you have a container with volume $V$ with a piston on it of cross sectional area $A$ and container is filled with gas.
Suppose, you push the container by $\mathrm dx$ amount. So, the gas is compressed and work is done on the gas.
By the definition of work classically, $$\mathrm dW=F\,\mathrm dx$$ Now if we use the fact that $P=\frac FA$, and $\mathrm dV=A\,\mathrm dx$, we will have $$\mathrm dW=\frac FAA\,\mathrm dx= P\,\mathrm dV$$ So, the classical definition of work only tells us the that work done on the gas should be $$\int_{V_1}^{V_2}P\,\mathrm dV=P\,\Delta V$$ if $P$ is constant.
So, the definition tells us there is no $V\,\Delta P$ term in the expresion for work. And, therfore, in the first law of thermodynamics, i.e. $\Delta Q = \Delta U - W_\text{on the system}$, also, that expression doesn't occur from work, and it is not also included in $\Delta U$ either.
But if you consider enthalpy change for a system, there $V\,\mathrm dP$ term will occur as $$\mathrm dH=\mathrm dU+\mathrm d(PV)=\mathrm dU+P\,\mathrm dV+V\,\mathrm dP$$

Answer 2

As an answer to your comment, it is not that we ignore the P$\Delta V$ term. Let us consider the first law statement, $$\mathrm{d}U = \mathrm{d}Q+ \mathrm{d}W$$ And we have, $\mathrm{d}W = P\mathrm{d}V$. Notice that I use differentials instead of delta terms.

Considering any process, Intgerating the expression and setting appropriate limits, we actually don’t get $\Delta W = P\Delta V$. This is because the pressure is not constant and cannot be taken out of the integral. Hence the $V \Delta P$ is technically not ignored.(This was also explained in Soumik’s answer). Only for an isobaric process $V\Delta P$ term can be ignored. For all other processes involving an ideal gas, if we work with $PV = nRT$, differentiating we get, $$P\Delta V + V\Delta P = nR\Delta T$$ And incase of an isochoric process, we get the following relation, $$V\Delta P = nR\Delta T$$