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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 $dx$ amount, So, the gas is compressed and work is done on the gas.
By the definition of work classically, $$\\dW = F.dx\\\\$$ Now if we use the fact that $\\P= \frac{F}{A}\\\\$ , and $\\dV = Adx, \\\\$ we will have ,$$\\dW= \frac{F}{A}.Adx = PdV\\\\$$ So, the classical definition of work only tells us the that work done on the gas should be $$\\\int_{V_1}^{V_2} PdV = P\Delta V \\\\$$ 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. $\Delta Q = \Delta U - W_{on the system}$, 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\Delta P $ term will occur as $$\\d H = dU + d (PV) = d U + PdV + VdP\\\\$$