Context: I'm asking about classical thermodynamics, that is "ideal gas", closed system, reversible processes etc.
Why is the $VdP$ term omitted in calculation of work during isothermal process in a closed system? Why only the PdV term considered? Clearly, during isothermal process not only V changes but also the pressure P drops/grows too, so $VdP$ is not zero as would have been during isobaric process.
The terse reason is that it was never there in the first place. The total differential for the internal energy is defined to be, \begin{gather*} dU = \delta w + \delta q \end{gather*} where $\delta q$ is the inexact differential of heat and $\delta w$ is the inexact differential for work. Often, the only meaningful work being done is $P-V$ work, which, following on the line integral definition of linear work, \begin{gather*} w = \int_\gamma \vec{F} \cdot d\vec{r} = \int_\gamma P dV \end{gather*} The differential for $P-V$ work is thus in the form, \begin{gather*} \delta w_{rev} = - P dV \end{gather*} where the subscript refers to the fact that this only applies for reversible processes. For a non-reversible process we would just use the external pressure to the system $P_{ex}$ rather than the constantly equilibrated pressure of the system.
As an aside, based on a Clausius analysis of entropy, we can use the differential, \begin{gather*} \delta q_{rev} = T dS \end{gather*} Then, combining the two relations, we get the total differential for the internal energy for reversible processes, \begin{gather*} dU = TdS - PdV \end{gather*} This is called the fundamental thermodynamic relation because it reveals that the internal energy is a natural function of $S$ and $V$. Because the internal energy is a state function, this will also hold for arbitrary processes. All we lose is the ability to identify each term with either heat or work specifically. Using a Legendre transform to define the enthalpy, \begin{gather*} H = U + PV \\ \implies dH = dU + VdP + P dV \end{gather*} actually does have a $VdP$ term, though this is not due to work but the product rule of calculus.
The thermodynamic work is closed system/reversible process is $$dW = -P dV$$ There is no $V dP$ term in this formula, the why is answered in any undergraduate textbook (if your question is where this formula comes from, then you should revise your post to indicate that). The result has nothing to do with ideal gas, it is the general expression for reversible work in closed system.
There is no $VdP$ work because mass isn't moved into and out of a closed system. The work in a closed system is boundary work, the work required to expand or contract the boundaries of the system, i.e., change its volume.
I know that VdP is a "flow work" of some sorts... but I wondered why it is never interpreted as the "changing pressure" part of the isothermal work... I'm stuck in somekind of a dumb idea about isothermal work and I can't get out :
You need to go back to the basic definition of work, which is force times the displacement of material in contact with the force. For the closed system the force is the pressure and the displacement is the change in volume.
So for a closed system a change in volume must always be associated with the change in pressure in order for work to be done. Without a displacement of volume (and the mass contained there in), a change in pressure does no work. For example, if you heat a rigid vessel containing a gas its pressure will increase but since the volume doesn't change no work is done. There is no displacement of mass. With regard to the open system $\Delta(PV)$ is the work required to move mass into and out of system.
Finally, with regard to a reversible isothermal process for an ideal gas there is a change in volume along with the change in pressure where PV is a constant, o work is done.
Hope this helps.
It depends on what you mean by "work". When people say $\delta W = p_0dV$ is the work, where $p_0$ is the pressure of the environment and $dV$ is the increase of volume of the system and decrease of volume of the environment, then depending on its sign they mean the work done on or absorbed by the environment (or by the system). In a reversible process $p=p_0$ so $\delta W_{rev}=pdV$. But there is another way of looking at the balance of work.
Let us start with simplest case. The energy equation $dU=TdS-pdV+\mu dN$ holds for the simplest body. Assume that it can exchange $S,V,N$ at potential levels $T_0,p_0,\mu_0$ representing thermal, pressure and chemical reservoirs having combined energy change $dU_0=T_0dS_0-p_0dV_0+\mu_0 dN_0$, then the total energy variation is $dU+dU_0=0$ with $dV+dV_0=0$ and $dN+dN_0=0$ but $dS+dS_0=\sigma \ge 0$. Here $\sigma >0$ is the irreversibly produced entropy in the system at temperature $T$; if and only if the process is reversible $\sigma =0.$
From energy conservation, therefore $$0=dU+dU_0\\=T_0dS_0 +T(\sigma - dS_0)-(p_0-p)dV_0+(\mu_0-\mu)dN_0$$ and after rearrangement $$-T\sigma =(T_0-T)dS_0-(p_0-p)dV_0+(\mu_0-\mu)dN_0 \tag{1}$$ You could write this Eq.1 as $$-T\sigma =\Delta T dS_0-\Delta p dV_0+\Delta \mu dN_0 \tag{2}$$ where $\Delta T,\Delta p, \Delta \mu$ denote the energetic potential drop to which the transported work quantities $dS_0,dV_0,dN_0$ are subjected, respectively. Each term is a separate work term:
If and only the process is reversible, ie, $\sigma = 0$, these work terms are in perfect balance with each other.
Now in the special case that the process is adiabatic $dS_0=0$ and closed $dN_0=0$ then $-T\sigma = -(p_0-p)dV_0$. Since the process cannot now be balanced with another work term it must, per force, be irreversible and all of the $\Delta p dV_0$ work is dissipated and turns to heat. On the other hand if it is still closed but diathermal then we have $-T\sigma =\Delta T dS_0-\Delta p dV_0$ and this can now be reversibly so that the thermal and spatial works be balanced, $\Delta T dS_0=\Delta p dV_0$, and that would be a Carnot engine operating at temperatures $T_0$ and $T$ while absorbing thermal energy (heat) $T_0S_0$ and rejecting absorbing thermal energy (heat) $TS_0$, resp.
So the punchline: the $\Delta p dV_0$ work term represents all the pressure-volume work that is done during the energetic exchange between the system and its environment.
From the comments, it looks like you’re still wondering why $V\,dP$ doesn’t appear in the definition of work or in the First Law when it appears in the differential form of the enthalpy $H\equiv U+PV$ (or $dH=dU+d(PV)=dH+P\,dV+V\,dP$).
First, what is thermodynamic work anyway? It’s the raising of many particle energies in concert. (In contrast, heating is the broadening of the distribution of particle energies.) This requires a so-called generalized force—which is sometimes just literally a mechanical force—and a so-called generalized displacement. Not just a position, a displacement. So right away we have an asymmetry: We're talking about the value of one parameter and a difference in another parameter. The generalized force $F$ drives the change at any moment, and the generalized small displacement $dx$ describes the extent of particle energy elevation. The infinitesimal work is then in the form of $F\,dx$, integrated to give path-dependent work $\int F\,dx$.
For instance, pushing up on a body locked in place does nothing to raise its gravitational potential energy; a table doesn’t do work on the objects sitting on it. It’s the displacement that embodies the energy transfer process.
(Other examples are a mechanical force and displacement, a pressure and a volume shift, a surface tension and an area change, a stress and a volumetric strain, an electric field and extent of polarization, a magnetic field and extent of magnetization, and a chemical potential and amount of shifted matter. These pairs are called conjugate thermodynamic variables; their product gives units of energy.)
This summarizes why $P\,dV$ looks the way it does when defining expansion–compression work. Why, then, is enthalpy defined using the complete term $\boldsymbol{PV}$, with $\boldsymbol{d(PV)}$ appearing in the differential form? Doesn’t this give $\boldsymbol{P\,dV+V\,dP}$? Yes, but $PV$ here is a unique term. It conceptually describes the work done on the atmosphere—at surrounding pressure $P$—to make room for the entire system at volume $V$. (In other words, it's the result of evaluating the expansion work in the special case of $\int_0^V P\,dV$ when the surrounding pressure $P$ is constant.) This lets us account for the fact that Nature prefers denser systems when the surrounding pressure is higher. Again, this is a special case; systems don’t generally wink in and out of existence. When do they effectively wink in and out of existence? When a chemical reaction trades some compounds for others—and we use the enthalpy to characterize the so-called heat of reaction. When matter is pushed in and out of a control volume—and we use the so-called shaft work $V\,dP$ to analyze the energy required/collected when the inlet and outlet pressures differ.
Note carefully the difference between these processes and the process of simple expansion and compression, in which the work done by the system is again simply $P\,dV$ (where $P$ is the external pressure).
Now, isn't this just splitting hairs regarding "making room for the system"? Can't we just model expansion as removing the entire system and inserting it again at the larger volume? Yes, and the enthalpy change would be $\Delta H=-\int P\,dV+\Delta (PV)$. Note that we've accounted for the loss in internal energy of the system from doing the expansion work. If the surrounding pressure remains constant, this reduces to $\Delta H=0$.
To summarize: It would be a mistake to conflate $P\,dV$, or infinitesimal expansion work, with one term of the differential enthalpy, which looks similar but has an additional $V\,dP$. Work is not equivalent to enthalpy. Terms such as "shaft work" and "flow work" arose in part as a convenient shorthand to refer to enthalpy changes, but they aren't a part of the generalized framework of thermodynamic work discussed above—and it's this thermodynamic work that alters the internal energy, as described by the First Law $\Delta U=Q-W$ (for work done by the system).
