homework and exercises - Adiabatic expansion in van der Waals gas

Given a Van der Waals gas with state equation:

$$\left( P+\frac{N^2 a}{V^2}\right)\left( V-Nb \right)=NkT,$$ show that the equation of an adiabatic process is: $$\left( V-Nb\right)T^{C_V}=\text{constant}.$$

I began by setting $đQ=0$ in

$$\mathrm dU=đQ+đW,$$ one then gets $$0=\mathrm dU+P~\mathrm dV.$$

Now given $U=\frac{3}{2}NkT-\frac{N^2 a}{V},$ I plugged it's derivatives into

$$\mathrm dU=\left( \frac{\partial U}{\partial T}\right)_V~\mathrm dT+\left( \frac{\partial U}{\partial V} \right)_T~\mathrm dV,$$ from which I obtained $$0=C_V~\mathrm dt+\left(P+\frac{N^2 a}{V^2} \right)_T~\mathrm dV=C_V~\mathrm dT + \frac{NkT}{V-Nb}~\mathrm dV,$$ using $V~ đW$'s equation.

Dividing by $T$ and integrating gives

$$C=\log{T^{C_V}}+\log{(V-Nb)^{Nk}},$$ which is equivalent to $$C'=(V-Nb)^{Nk}T^{C_V},$$ for $C$ and $C'$ constants.

Now the expression so obtained seems very similar to what I was looking for, but I can't seem to get rid of the $Nk$ exponent. Anyone got a different approach to this problem, or a way to get the desired formula?

Answer

The correct answer is $(V-Nb)T^{C_V/Nk}=\text{const}$, the problem statement is just wrong.