Answer
$\left(\frac{\partial T}{\partial P}\right)_v=\frac{v-b}{R}$
Work Step by Step
The van der Waals equation of state can be expressed as $$
T=\frac{1}{R}\left(P+\frac{a}{v^2}\right)(v-b)
$$ Taking the derivative of $T$ with respect to $P$ holding $v$ constant,
$$
\left(\frac{\partial T}{\partial P}\right)_v=\frac{1}{R}(1+0)(v-b)=\frac{v-b}{R}
$$ which is the slope of the $v=$ constant lines on a $T-P$ diagram.
