The Latent Heat of vaporization of a given mass of liquid is, say, L
If we boil said mass of liquid, the vapor phase has a larger volume. Let us assume this increase in volume is ΔV.
Now by the first law of thermodynamics, Q=ΔU+W.
Rearranging, ΔU=Q-W Work(W) is PΔV, where P is the atmospheric pressure.
So is ΔU=L-PΔV?
Or is the latent heat of vaporization(L) = ΔU itself?
The specific latent heat, that is divided by the mass, is (by use of your notation):
$$l=\frac{L}{m}$$ where $m$ is the mass of the substance we are examining.
One way to determine the specific latent heat is that it is the difference of enthalpies of the phase transition.
So:
$$l= h''-h'$$
where $h''$ is the enthalpy of the substance when it turned all to vapor (so no more liquid state is present) and $h'$ is when there liquid just starts to boil, so it is all in liquid phase, i.e. before (actually, exactly at the moment) the phase transition starts.
The change of specific internal energy during the phase change can be provided as:
$$\Delta u= h_{2}-h_{1}-p(v_{2}-p_{1})$$
where terms indexed with $1$ represent the states at the beginning of the process and $2$ represents the end of the process. So the process can start even in a "cold" liquid and end in an "overheated" vapor.
So you can see that your equation:
$$\Delta U = L - p\Delta V$$ could be interpreted as correct, but you have to be more precise when writing equations as to exactly specify the start and the end of the process.
