Moving boundary work

I'm not real clear on the question, but W depends on the change in V. As Mr Miller points out p = dW/dV, and p is likely to be a function of the volume.

Moving boundary work can be equated from $Delta W = p dV$.

$Delta W$ literally means "change in work". But there is no "change in work" in the same sense that there is a change in pressure, or a change in temperature, or a change in volume, all of which are thermodynamic properties of a system. Work is not a thermodynamic property. Work is an amount of energy transfer between a system and its surroundings. And the amount of work done between two equilibrium states depends on the process between the two states.

Therefore, the amount of boundary work done (the work expanding or contracting the boundaries of the system) between two equilibrium states is $$W_{1-2}=int_{V1}^{V2} p(V)dV$$

In order to evaluate the integral you need to know how pressure varies with volume, that is you need to know the function $p(V)$.

Hope this helps.

To find the issue of doing that, let us write down the explicit functional dependencies:

$$ dW = - PdV$$

Now, the problem is to integrate both sides, we need the pressure as some pure function of volume. Depending on the process the relation between pressure and volume is different.

For example, consider an adiabatic change where $ q=0$, here we can't use the ideal gas law to write pressure as a function of volume because the ideal gas law makes the pressure a function of both volume and temperature. Hence, we use other adiabatic law.

So, the moral of the story is that unless you specify the type of process, the kind of dependency pressure has on volume is not ascertained and hence we can't find a function $W$ which is purely dependent on volume such that its differential is pressure without specifying the path.