)
A PVPV vs. PP graph simply shows that the relationship P_1V_1 = P_2V_2 ~~ "const"P1V1=P2V2≈const holds. If it were really the case that (0,0)(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 VV on PP; if it were true that P = 0P=0 implies PV = 0PV=0, then changes in volume would not affect the pressure under the conditions of Boyle's law (constant TT and nn), which directly contradicts Boyle's law, or 0*oo = 00⋅∞=0, which is mathematically incorrect.
I also graphed the data above myself and replicated the graph, so that at truly constant nn and TT, we have the true values of VV for n = "0.0025 mols"n=0.0025 mols and T = "298.15 K"T=298.15 K:
![]()
Really, how you should view PVPV vs. PP is that as P uarrP↑, V darrV↓ and vice versa. No matter how PP changes, PVPV should not.
Thus, as P->0P→0, V->ooV→∞ such that PV = "const"PV=const, i.e. P_1V_1 = P_2V_2P1V1=P2V2.
So, if we had:
P_1 = "2 atm"P1=2 atm
V_1 = "5 L"V1=5 L
P_2 = "0 atm"P2=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.
