( P + a N 2 v 2 ) ( v − N b ) = N k T {\displaystyle (P+{{aN^{2}} \over v^{2}})(v-Nb)=NkT}
say V = v N {\displaystyle V={v \over N}}
P v + a N 2 v v 2 − P N b − a N 2 N b v 2 − N k T = 0 {\displaystyle Pv+{aN^{2}v \over v^{2}}-PNb-{aN^{2}Nb \over v^{2}}-NkT=0}
by multiplying both sides by v 2 {\displaystyle v^{2}} we get
P v 3 + a N 2 V − P N b v 2 − a N 3 b − N k T v 2 = 0 {\displaystyle {Pv^{3}}+aN^{2}V-PNbv^{2}-aN^{3}b-NkTv^{2}=0}
by dividing both sides by P N 2 {\displaystyle PN^{2}} we get
v 3 N 3 + a v P N − b v 2 N 2 − a b P − k T v 2 P N 2 = 0 {\displaystyle {v^{3} \over N^{3}}+{av \over PN}-{bv^{2} \over N^{2}}-{ab \over P}-{kTv^{2} \over PN^{2}}=0}
so
V 3 + V a P − V 2 b − a b P − V 2 k T P = 0 {\displaystyle V^{3}+V{a \over P}-V^{2}b-{ab \over P}-V^{2}{kT \over P}=0}
and combining terms we get
V 3 − V 2 ( b + k T P ) + V a P − a b P = 0 {\displaystyle V^{3}-V^{2}(b+{kT \over P})+V{a \over P}-{ab \over P}=0}
we get
we get
