Adiabatic free expansion of real (Van der Waal's model) gas below/at/above inversion temperature For an adiabatic free expansion, $W = 0$ and $Q=0$. Therefore, by the first law of thermodynamics, $\Delta U = Q -W = 0$.
For a Van der Waals model of a real gas, $\Delta U = n C_v \Delta T - a\,n^2 \left(\frac{1}{V_2}-\frac{1}{V_1} \right)$. This means that $n C_v \Delta T - a\,n^2 \left(\frac{1}{V_2} -\frac{1}{V_1} \right) = 0$, or $n C_v \Delta T =  a\,n^2 \left(\frac{1}{V_2} -\frac{1}{V_1} \right)$. Now, since the gas is expanding, $V_2 > V_1$. So, $a \, n^2 \left(\frac{1}{V_2}-\frac{1}{V_1} \right) < 0$. Therefore, $n C_v \Delta T<0$. This means that the temperature of the gas decreases in an adiabatic free expansion.
Will the temperature always decrease? I mean, what if the gas is above its inversion temperature or at it? Does inversion temperature play no role in adiabatic free expansion?
 A: Your calculation indeed shows that the temperature of the van der Waals gas always decreases in adiabatic free expansion. It is only when the gas flows between two regions of different pressures that the sign of the variation of temperature depends on the temperature. Enthalpy, rather than energy, is conserved in this case. Using the expression of pressure for the van der Waals gas
   $$P={Nk_BT\over V-Nb}-{N^2a\over V^2}$$
enthalpy reads
   $$\eqalign{
   H=U+PV&\simeq {3\over 2}Nk_BT-{N^2a\over V}
   +\left({Nk_BT\over V-Nb}-{N^2a\over V^2}\right)V\cr
   &={3\over 2}Nk_BT-{2N^2a\over V}
   +Nk_BT\left(1-{Nb\over V}\right)^{-1}\cr
   &\simeq {3\over 2}Nk_BT-{2N^2a\over V}
   +Nk_BT\left(1+{Nb\over V}\right)^{-1}\cr
   &={5\over 2}Nk_BT-{2N^2a\over V}+{N^2bk_BT\over V}\cr
   &={5\over 2}Nk_BT+{N^2\over V}\big(bk_BT-2a\big)\cr
   }$$
which can be rewritten as
   $$H=C_pT+{N^2bk_B\over V}(T-T_i)$$
where the inversion temperature is
   $$T_i={2a\over bk_B}$$
For $T>T_i$, the expansion leads to an increase of the temperature.
A: this is more a comment on your answer: in any free expansion the pressures in each side are not kept constant: the pressure decreases in one side and increases in the other side.  The correct statement is: "It is only when the gas flows between two regions of different pressures (each kept constant) that the total entalpy $H_1 +H_2$ is conserved. For a real gas, it follows that the sign of $\Delta T$ depends on T".
