Solution to Set 2

Problem 1

Part a

$(P+{{aN^{2}} \over v^{2}})(v-Nb)=NkT$

say $V={v \over N}$

$Pv+{aN^{2}v \over v^{2}}-PNb-{aN^{2}Nb \over v^{2}}-NkT=0$

by multiplying both sides by $v^{2}$ we get

${Pv^{3}}+aN^{2}V-PNbv^{2}-aN^{3}b-NkTv^{2}=0$

by dividing both sides by $PN^{2}$ we get

${v^{3} \over N^{3}}+{av \over PN}-{bv^{2} \over N^{2}}-{ab \over P}-{kTv^{2} \over PN^{2}}=0$

so

$V^{3}+V{a \over P}-V^{2}b-{ab \over P}-V^{2}{kT \over P}=0$

and combining terms we get

$V^{3}-V^{2}(b+{kT \over P})+V{a \over P}-{ab \over P}=0$

part b

$P={nRT \over (v-nb)^{2}}-{an^{2} \over v^{2}}$

Taking the derivative we get

${dP \over dv}={-nRT \over v-nb}+{2an^{2} \over v^{3}}=0$

Multiply the derivative by ${-2 \over v-nb}$ to get

${2nRT \over (v-nb)^{3}}-{4an^{2} \over v^{3}(v-nb)}=0$

Taking the derivative again we get

${d^{2}P \over d^{2}v}={2nRT \over (v-nb)^{3}}-{6an^{2} \over v^{4}}=0$

By setting the derivatives equal to each other we get

${2nRT \over (v-nb)^{3}}-{4an^{2} \over v^{3}(v-nb)}={2nRT \over (v-nb)^{3}}-{6an^{2} \over v^{4}}$

Which reduces to

$6(v-nb)=4v$

$6v-6nb=4v$

$6nb=2v$

$3nb=v_{c}$

Now we can say

${-nRT \over v_{c}-nb}+{2an^{2} \over v_{c}^{3}}=0$

Plugging $v_{c}$ in we get

${-nRT \over (3nb-bn)^{2}}+{2an^{2} \over (3nb)^{3}}=0$

Now we solve for T

${-nRT \over (2bn)^{2}}+{2an^{2} \over 27n^{3}b^{3}}=0$

${-RT \over 4b^{2}n}+{2a \over 27nb^{3}}=0$

${2a \over 27nb^{3}}={RT \over 4b^{2}n}$

${8a \over 27b}=RT$

${8a \over 27bR}=T_{c}$

Now $T_{c}$ and $v_{c}$ can be plugged in to find $P_{c}$

say

$P={nR({8a \over 27bR}) \over 3nb-nb}-{an^{2} \over (3nb)^{2}}$

$P={8na \over 27b(2nb}-{a \over 9b^{2}}$

$P={4a \over 27b^{2}}-{a \over 9b^{2}}$

$P={4a \over 27b^{2}}-{3a \over 27b^{2}}$

$P={a \over 27b^{2}}$

$P_{c}={a \over 27b^{2}}$