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. <br/> Suppose, you push the container by $dx$ amount, So, the gas is compressed and work is done on the gas. <br/>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**. <br/>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.<br/>**But if you consider enthalpy change for a system, there $V\Delta P $ term will occur as** $$\\\Delta H = \Delta U + \Delta (PV) = \Delta U + P\Delta V + V\Delta P\\$$