I don't think Pdv **on its own** represents any meaningful quantity in free expansion, as you rightly pointed out that, work done in free expansion is zero, since there is no external pressure to work against. Thus pressure of the gas remains same*(see edit), even though the volume changes. Now, consider the set-up, where you have a adiabatic container, having a removable partition, with one side containing ideal gas(for simplicity) and the other side being vacuum. If we now, remove the partition, and allow the gas to fill the container and come into equilibrium, then at this stage entropy has increased. This increase in entropy is given by, 
Integral of Pdv/T, between the limits Vi & Vf. Here Temperature T, will also remain constant, since the container is adiabatic and W=0.[See the link for mathematical form][1]

The increase in entropy takes place because of the **finite pressure difference** that exists between the two partitions, which is an **irreversibility** and hence leads to entropy change. 
  [1]: https://goo.gl/photos/cWeM7d3FNDWBGvPDA


* Edit: The pressure doesn't remain constant in free expansion, it changes. Consider the same  set up i have described above, since temperature is constant and amount of substance remains unchanged, therefore according to boyle's law, as volume increases pressure decreases.