Answer
$\frac{{\partial P}}{{\partial T}} = \frac{{nR}}{{V - nb}}$
$\frac{{\partial V}}{{\partial P}} = \frac{{nb{V^3} - {V^4}}}{{P{V^3} + 2ab{n^3} - a{n^2}V}}$
Work Step by Step
We have the equation $\left( {P + \frac{{a{n^2}}}{{{V^2}}}} \right)\left( {V - nb} \right) = nRT$.
Write $F\left( {T,P,V} \right) \equiv \left( {P + \frac{{a{n^2}}}{{{V^2}}}} \right)\left( {V - nb} \right) - nRT$.
The partial derivatives are
$\frac{{\partial F}}{{\partial T}} = - nR$, ${\ \ }$ $\frac{{\partial F}}{{\partial P}} = V - nb$
$\frac{{\partial F}}{{\partial V}} = - \frac{{2a{n^2}}}{{{V^3}}}\left( {V - nb} \right) + P + \frac{{a{n^2}}}{{{V^2}}}$
$ = P + \frac{{2ab{n^3}}}{{{V^3}}} - \frac{{a{n^2}}}{{{V^2}}}$
1. Calculate $\frac{{\partial P}}{{\partial T}}$
Suppose that $P$ is defined implicitly as a function of $P$ and $V$ by the equation
$F\left( {T,P,V} \right) = \left( {P + \frac{{a{n^2}}}{{{V^2}}}} \right)\left( {V - nb} \right) - nRT = 0$
Using the Chain Rule we differentiate $F$ with respect to $T$:
$\frac{{\partial F}}{{\partial T}}\frac{{\partial T}}{{\partial T}} + \frac{{\partial F}}{{\partial P}}\frac{{\partial P}}{{\partial T}} + \frac{{\partial F}}{{\partial V}}\frac{{\partial V}}{{\partial T}} = 0$
Since $\frac{{\partial T}}{{\partial T}} = 1$, $\frac{{\partial V}}{{\partial T}} = 0$, so
$\frac{{\partial F}}{{\partial T}} + \frac{{\partial F}}{{\partial P}}\frac{{\partial P}}{{\partial T}} = 0$
$\frac{{\partial P}}{{\partial T}} = - \frac{{\partial F}}{{\partial T}}/\frac{{\partial F}}{{\partial P}}$
$\frac{{\partial P}}{{\partial T}} = \frac{{nR}}{{V - nb}}$
2. Calculate $\frac{{\partial V}}{{\partial P}}$
Suppose that $V$ is defined implicitly as a function of $T$ and $P$ by the equation
$F\left( {T,P,V} \right) = \left( {P + \frac{{a{n^2}}}{{{V^2}}}} \right)\left( {V - nb} \right) - nRT = 0$
Using the Chain Rule we differentiate $F$ with respect to $P$:
$\frac{{\partial F}}{{\partial T}}\frac{{\partial T}}{{\partial P}} + \frac{{\partial F}}{{\partial P}}\frac{{\partial P}}{{\partial P}} + \frac{{\partial F}}{{\partial V}}\frac{{\partial V}}{{\partial P}} = 0$
Since $\frac{{\partial P}}{{\partial P}} = 1$, $\frac{{\partial T}}{{\partial P}} = 0$, so
$\frac{{\partial F}}{{\partial P}} + \frac{{\partial F}}{{\partial V}}\frac{{\partial V}}{{\partial P}} = 0$
$\frac{{\partial V}}{{\partial P}} = - \frac{{\partial F}}{{\partial P}}/\frac{{\partial F}}{{\partial V}}$
$\frac{{\partial V}}{{\partial P}} = - \frac{{V - nb}}{{P + \frac{{2ab{n^3}}}{{{V^3}}} - \frac{{a{n^2}}}{{{V^2}}}}} = \frac{{nb{V^3} - {V^4}}}{{P{V^3} + 2ab{n^3} - a{n^2}V}}$
