Is work a state function for an isobaric process? For an isobaric process, work is $P(V_2-V_1)$. This quantity seems to be depending only on initial and final volume. Is work a state function for an isobaric process?
 A: Welcome to this community!
A very short first answer is: no, work is not a function of state in an isobaric process. For several reasons:
1 – Suppose the state is given by the pair $(p,V)$. If the work between states $(p_0,V_0)$ and $(p_1, V_1)$, with $p_0=p_1$, were a function of state, it should be expressible in terms of $(p_0, V_0)$ alone, or of $(p_1,V_1)$ alone (that's why we say "function of state" and not "function of states"). But that's clearly impossible, since its expression requires $V_0$ and $V_1$.
This fact can be seen immediately. Let's take 1 mol of a monatomic ideal gas. Someone tells us its state: $T=\mathrm{300\ K}, V=\mathrm{1\ m^3}$. Now let's ask: what are its pressure and internal energy? The answers are "$p=\mathrm{2494.35\ Pa}$" and "$U=\mathrm{3741.52\ J}$". We can answer because pressure and internal energy are functions of state, the formulae being $p(T,V)=nRT/V$ and $U(T,V)=\frac{3}{2}nRT$. Now let's ask: what is the work done in an isobaric (or some other) process? Can we answer? No, because knowing the present state isn't enough.
In some specific cases, like in the isobaric process you mention, the amount of work done is path-independent, which means that we only need the initial and final states to calculate it. So we can say "for such processes, the work done is a function of the initial and final states". But that's different from being a "function of state": for the latter you only need one state – the one you're at.

2 – The expression "function of state" has an important physical meaning: that the quantity is a property of the state. This is true of pressure and internal energy, they're properties of each single state. Work, as is clear from its definition, is never a property of a state. If someone asks us "what's the pressure of the state $(T=\mathrm{300\ K}, V=\mathrm{1\ m^3})$?", we understand the question and can give an answer. Imagine someone asking you "what's the work of the state $(T=\mathrm{300\ K}, V=\mathrm{1\ m^3})$?"; wouldn't you reply "how do you mean?". See the passage from Zemansky & Dittman below.

3 – A quantity is called a function of state if its value is instantaneously determined by the state for all kinds of processes available to the system. So the fact that $p\,\mathrm{d}V$ can be rewritten as $\mathrm{d}f$ for some processes doesn't make $f$ a function of state.

4 – If we put enough constraints on our system, reducing its state space to a one-dimensional (simply connected) space, then every form $A\,\mathrm{d}B$ can always be rewritten as $\mathrm{d}C$, because on a one-dimensional simply connected manifold every form is exact. So if we start with a two-dimensional state space such as $(T,V)$ and we reduce its dimensionality and available processes by a constraint such as $p=\text{const}$ or $T=\text{const}$, then not only the infinitesimal work, but any other infinitesimal (more precisely, differential form) becomes an exact differential.
If we consider isobaric (or adiabatic) processes for a system with state $(T,V,X)$, where $X$ is some other independent state variable (for example chemical composition or magnetization), then the infinitesimal work cannot be written as an exact differential.

In view of the last point let me also emphasize that it's best to keep in mind that thermostatics and thermodynamics are not only about simple fluids and ideal gases. It's best not to generalize results that are only valid for specific substances in specific situations. This is especially important if in the future you'll work with, or do research about, say, atmospheric or oceanic thermomechanics, or piezoelectric systems, or plasmas – just to give examples that are not quirky or remote subjects; and they're quite cool! :)
In general the work done or received by a thermodynamic system has other contributions besides changes in volume. Systems capable of electrodynamic interactions are an example. In such cases, the work would depend on the path even in an isobaric process. And in many thermodynamic systems work can also be done by deformation without changes in volume (rubber is an everyday example).
A good text about such matters for example is:

*

*Astarita: Thermodynamics: An Advanced Textbook for Chemical Engineers (Springer 1990).

Let me quote from § 1.3, Work and kinetic energy, of this book:

It is important to realize that, in analogy with the total rate of heating
$Q_\text{t}$, the total net rate of work $W_\text{t}$ is not a function of state. The analogy is closer than may appear at first sight. The reason why $W_\text{t}$ cannot be a function of state is that body forces contribute to it, and body forces are active owing to the influence of bodies other than the one considered (for instance, gravity is due to the pull of the Earth).

Let me also quote from

*

*Zemansky, Dittman: Heat and Thermodynamics: An Intermediate Textbook (McGraw-Hill 1997)

§ 3.5:

There is no function of the thermodynamic coordinates representing the work in a body. The physical interpretation of thermodynamic work is that the performance of work is an external activity or process that leads to a change in the state of the body, namely, to increase or decrease the energy. The phrase "work in a body" has no meaning. Notice that work is not a quantity that itself moves, but is a process that moves another quantity, namely, energy of some type.

I have also checked Callen's Thermodynamics, Hill's An Introduction to Statistical Thermodynamics, Buchdahl's The Concepts of Classical Thermodynamics, Müller's Thermodynamics, and can't find any statements saying that "work becomes a function of state" in isobaric or adiabatic processes. Some of these authors emphasize the fact that for some processes the work becomes "easier to calculate" because it depends only on the initial and final states – but none of these authors go so far as saying that work is therefore a function of state for those processes.
The point is that the implication "function of state${}\Rightarrow{}$exact differential" is true, but the reverse "exact differential${}\Rightarrow{}$function of state" is generally false because it involves an integration constant or universal reference state. In the case of work there is no universal reference state, since work is always related to two specific states – the initial and final one – which are different from situation to situation. This again connects to Zemansky & Dittman's statement above.
A: Good question.
Once you have specified a path, the thermodynamic differentials become exact differentials. The reason they are 'inexact' to begin with is that different paths end up in different changes in the state variables.

Note
: All statements made for reveriblse processes
A: It is convenient and instructive to think of thermodynamics in terms of differential geometry.
The set of equilibrium states is a manifold $\mathcal{M}$. The dimensionality of the manifold depends on the physical context; for simple fluids, it is usually between 1 and 3. The manifold of equilibrium states can be parametrized by many sets of parameters, e.g. $\{U, V, N\}$, $\{T, p, N\}$, $\{S, V,\mu\}$, etc.
In this context, the functions of state are the functions that map equilibrium states to numbers $f: \mathcal{M}\ni m \mapsto f(m)\in \mathbb{R}$*. When doing physics, we tend to simplify the notation and write functions of parameters instead of functions of points, e.g. $U(m) = U(S(m),V(m),N(m)) \rightarrow U(S,V,N)$.
Differentials, e.g. $\Delta W = -p dV$**, are vectors co-tangent to $\mathcal{M}$ - they describe paths on the manifold.
The exact differentials on the manifold of equilibrium states are the exterior derivatives $df$ of the functions of state $f$.
To the point: is work a function of state? In principle, no. In this case, you could make a point for it but it is not a useful way of thinking about it. The argument could look like this:
Let us consider a reversible process *** which takes us along some curve $\gamma$ on the manifold $\mathcal{M}$, e.g. the isobaric process takes as along the curve parametrized by volume with pressure and number of moles fixed $\gamma_{p_0, N_0}: V \mapsto \{V,p_0, N_0\}$. The curve is a 1-dimensional manifold so every differential form is exact****, including the form of work $\Delta W$. Therefore, the work $W_\gamma = \int_\gamma \Delta W$ does not depend on the path*****. In this case, work can be seen as a function $W: \gamma \times \gamma \rightarrow \mathbb R$, i.e. work is a function that maps a pair of points: initial and final onto real numbers. By fixing initial point $\gamma(t_0) = m_0$, you can make work a function: $W(m_0, \cdot): \gamma\rightarrow \mathbb{R}$ and make it a function of state in this sense. Then you could write $\Delta W = d W(m_0,\cdot)$.
So it is possible to make work a function of state in some cases. I want to emphasize that it is mostly useless and can cause some confusion, so I would refrain from doing it. The term function of state should stay reserved for the functions that really contain no information besides what state they describe, like $U(S, V, N)$ or $\mu(T, P, N)$.
*In principle, you could consider functions of state that map to something different than reals, but that's not the point.
**I will write $\Delta W$ because I can't write $d$-bar in this editor.
***Irreversible processes do not specify a curve on the manifold of equilibrium states.
****Every differential restricted to the space of vectors cotangent to $\gamma$.
*****Also, there is exactly one path to get you to the desired point.
