Within the first integral, the volume of an ideal gas is subtracted, and it must be compensated by the second integral. Now the argument in the first integral describes the deviation of the volume of a real fluid from ideal gas behavior.
The second rearrangement concerns the lower integration boundary of the first integral in Equation (10). Instead of Pr, the lower boundary is changed to 0, which has to be compensated again, whereas the second integral of Equation (10) can be solved immediately
(11)
For pure fluids we have
(12)
and we get a molar quantity (molar quantities are designated here by small letters) which we can either designate as 
or 
:
(13)
