A #PV# vs. #P# graph simply shows that the relationship #P_1V_1 = P_2V_2 ~~ "const"# holds. If it were really the case that #(0,0)# is a point on the graph, then it would not matter what the volume is, as long as it is finite.

*But Boyle's law states that volume must affect pressure if temperature and mols of gas are held constant.*

In other words, you would then be ignoring the dependence of #V# on #P#; if it were true that #P = 0# implies #PV = 0#, then changes in volume would *not* affect the pressure under the conditions of Boyle's law (constant #T# and #n#), which directly contradicts Boyle's law, or #0*oo = 0#, which is mathematically incorrect.

I also graphed the data above myself and replicated the graph, so that at truly constant #n# and #T#, we have the true values of #V# for #n = "0.0025 mols"# and #T = "298.15 K"#:

Really, how you should view #PV# vs. #P# is that as #P uarr#, #V darr# and vice versa. No matter how #P# changes, #PV# should not.

Thus, as #P->0#, #V->oo# such that #PV = "const"#, i.e. #P_1V_1 = P_2V_2#.

So, if we had:

#P_1 = "2 atm"#

#V_1 = "5 L"#

#P_2 = "0 atm"#

Then

#V_2 = lim_(P_2->0) (P_1/P_2)V_1#

but since #P_2 = (nRT)/(V_2)#, we have that #V_2 -> oo# when #n# and #T# are constant (as usual in Boyle's law). On the other hand, given a *nonzero* #P_2#, say, #P_2 = "0.05 atm"#, we at least have a finite volume, which clearly increases as #P_2# decreases:

#V_2 = ("2 atm")/("0.05 atm")("5 L")#

#=# #"200 L"#

If you repeat with smaller and smaller #P_2#, you can see the pattern:

For #P_2 = "0.005 atm"#, #V_2 = (("2 atm")/("0.005 atm"))("5 L")#

#=# #"2000 L"#

For #P_2 = "0.0005 atm"#, #V_2 = (("2 atm")/("0.0005 atm"))("5 L")#

#=# #"20000 L"#

etc.