Question

As you can see from Figure5.20,5.20,the critical point is the unique point on the original van der Walls isotherms (before the Maxwell construction) where both the first and second derivatives ofPPwith respect toVV(at fixedTT) are zero. Use this fact to show that

V_c=3Nb, P_c =$\frac{1}{27}\frac{a}{b^2}$ and kT_c=$\frac{8}{27}\frac{a}{b}$

Short answer

V_c=3Nb, P_c=$\frac{1}{27}\frac{a}{b^2}$and kT_c=$\frac{8}{27}\frac{a}{b}$

Step-by-step solution

van der Waal's equation

$P=\frac{NkT}{(V-Nb)}-\frac{a{N}^2}{V^2}$ (1)

Partial differentiation of the equation w.r.t V

we get

$\frac{\delta P}{\delta V}=-\frac{NkT}{{(V-Nb)}^2}+\frac{2a{N}^2}{V^3}$ (2)

Again differentiating we get

$\frac{{\delta }^{2}P}{{\delta }^{2}V}=\frac{NkT}{{(V-Nb)}^3}-\frac{6a{N}^2}{V^4}$ (3)

At critical point

$$\begin{gathered} \frac{\delta P}{\delta V}=0 \\ \frac{{\delta }^{2}P}{{\delta }^{2}V}=0 \end{gathered}$$

$$\begin{gathered} \frac{Nk{T}_c}{{(V_c-Nb)}^2}=\frac{2a{N}^2}{{V_c}^3}and \\ \frac{Nk{T}_c}{{(V_c-Nb)}^3}=\frac{6a{N}^2}{{V_c}^4} \end{gathered}$$

Step 3:  Finding Vc, Tc, Pc.

On equating the above equations we get

$V_c=3Nb$ (4)

Substituting the equation (4) in equation (2)

$$\begin{gathered} \frac{Nk{T}_c}{{(3Nb-Nb)}^2}=\frac{2a{N}^2}{{\left(3Nb\right)}^3} \\ \Rightarrow k{T}_c=\frac{3}{27}\frac{a}{b} \end{gathered}$$

Substituting the above values in equation (1)

$$\begin{gathered} P_c=\frac{Nk({\displaystyle \raisebox{1ex}{$3a$}\!\left/ \!\raisebox{-1ex}{$27b)$}\right.}}{(3Nb-Nb)}-\frac{a{N}^2}{{\left(3Nb\right)}^2} \\ \Rightarrow P_c=\frac{1}{27}\frac{a}{b^2} \end{gathered}$$