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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 $VdP $ term will occur as $$\\d H = dU + d (PV) = d U + PdV + VdP\\\\$$

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 $$\mathrm dH=\mathrm dU+\mathrm d(PV)=\mathrm dU+P\,\mathrm dV+V\,\mathrm dP$$

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 $$\\d H = dU + d (PV) = d U + PdV + VdP\\\\$$

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 $VdP $ term will occur as $$\\d H = dU + d (PV) = d U + PdV + VdP\\\\$$

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 $$\\\Delta H = \Delta U + \Delta (PV) = \Delta U + P\Delta V + V\Delta P\\\\$$

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 $$\\d H = dU + d (PV) = d U + PdV + VdP\\\\$$