Answer
a) $\frac{dV}{dP}=\frac{nb-V}{P-\frac{n^2a}{V^2}+\frac{2n^3ab}{V^3}}$
b) $\frac{dV}{dP}=-4.0405$ $L$atm$^{-1}$
Work Step by Step
a) $(P+\frac{n^2a}{V^2})(V-nb)=nRT\\ PV-nbP+\frac{n^2a}{V}-\frac{n^3ab}{v^2}=nRT\\ P\frac{dV}{dP}+V-nb-\frac{n^2a}{V^2}\frac{dV}{dP}+\frac{2n^3ab}{V^3}\frac{dV}{dP}=0\\ P\frac{dV}{dP}-\frac{n^2a}{V^2}\frac{dV}{dP}+\frac{2n^3ab}{V^3}\frac{dV}{dP}=nb-V\\ \frac{dV}{dP}(P-\frac{n^2a}{V^2}+\frac{2n^3ab}{V^3})=nb-V\\ \frac{dV}{dP}=\frac{nb-V}{P-\frac{n^2a}{V^2}+\frac{2n^3ab}{V^3}}$
b) $\frac{dV}{dP}=\frac{(1)(0.04267)-10}{2.5-\frac{(1)^2(3.592)}{(10)^2}+\frac{2(1)^3(3.592)(0.04267)}{(10)^3}}\\ \frac{dV}{dP}=-4.0405$
