Answer
$\mu=-\frac{2 a v}{c_p(2 a-R T v)}=0 \longrightarrow v=0 $
This gas does not have an inversion line.
Work Step by Step
The equation of state of this gas can be expressed as $$
T=\frac{v}{R}\left(P+\frac{a}{v^2}\right) \longrightarrow\left(\frac{\partial T}{\partial v}\right)_P=\frac{v}{R}\left(-\frac{2 a}{v^3}\right)+\frac{1}{R}\left(P+\frac{a}{v^2}\right)=-\frac{2 a v}{R v^2}+\frac{T}{v}=\frac{R T v-2 a}{R v^2}
$$ Substituting into the Joule-Thomson coefficient relation, $$
\mu=-\frac{1}{c_p}\left(v-T\left(\frac{\partial v}{\partial T}\right)_P\right)=-\frac{1}{c_p}\left(v-\frac{R T v^2}{R T v-2 a}\right)=-\frac{2 a v}{c_p(2 a-R T v)}
$$ The temperature at $\mu=0$ is the inversion temperature, $$
\mu=-\frac{2 a v}{c_p(2 a-R T v)}=0 \longrightarrow v=0
$$ Thus the line of $v=0$ is the inversion line. Since it is not physically possible to have $v=0$, this gas does not have an inversion line.
