Can $V\text dP$ be thought of as an increase in internal energy? Because, physically, if the volume is constant and the pressure increases, won't it mean that there is an increase in that internal energy?
If not, how else should we perceive it?
 A: It’s called an isochoric (constant volume) heat transfer process. For an ideal gas the ratio of pressure to temperature is constant and the change in internal energy is $C_{V}\Delta T$.
Hope this helps 
A: If the volume does not change, then there is no work being done on or by the system.  Any change in internal energy would be associated with the process which caused the change in pressure: heat or gas being added or removed.
A: Let's illustrate this for an ideal gas.  (In general, we should be considering thermodynamic potentials.)  Using the ideal gas law $pV=nRT$, we can put this in differential form by noting that
$$d(nRT) = d(pV)\Longrightarrow nR dT = Vdp+pdV,$$
where we have assumed that the system is closed ($n=$ constant). Now, during a constant volume process, $pdV=0$, and so we get
$$nRdT=Vdp.$$
Finally, for an ideal gas, the internal energy is given by
$$U=\frac{f}{2}nRT,$$
where $f$ is the number of degrees of freedom (per particle) of the gas, in which case
$$dU = \frac{f}{2}nRdT.$$
Putting these together, we get
$$Vdp=\frac{2}{f}dU,$$
and so while $Vdp$ itself is not a change in energy of the system, it is proportional to the change in internal energy during a constant-volume process. Finally, since the volume is constant, no work is done on the system, and so the total change in energy is due only to heating.
A: For a single phase fluid of constant composition, the change in molar internal energy can be determined for any arbitrary variation of T vs P by integrating the equation $$dU=C_vdT-\left[P-T\left(\frac{\partial P}{\partial T}\right)_V\right]\left(\frac{\partial V}{\partial P}\right)_TdP$$This integral will depend only on the two end states, and not on the chosen variation in T vs P used to evaluate the change.  You can see that the second term is definitely not VdP.  The integrand in the second term depends on the equation of state for the fluid.  For an ideal gas, the integrand is zero for all variations of T vs P.
