Energy change of isentropic expansion of ideal gas

I'm trying to get a better grasp on dealing with different processes thermodynamic, and this one is giving me trouble.


The differential of energy is given by
$$dU= \delta Q + \delta W~.$$ I'm interested in the ideal gas for a closed system, so $\delta W = -pdV$. For an isentropic process, there is no heat exchange, so $\delta Q =0$. This gives us the result that
$$dU = -pdV ~.$$
Further, for the reversible process, the external pressure against which the system expands is the same as the pressure of the gas, which for an ideal gas is $$p=\frac{NkT}{V}$$
Substituting this into the differential of the energy, we can integrate both sides to find
$$\Delta U = -Nk\int\_{V\_0}^{V\_f}\frac{T(V)dV}{V}$$
In this case, if I'm not mistaken the temperature is not constant (this is an isentropic, not isothermal process), so how does one calculate the energy change of this compression? Is there a relation between $V$ and $T$ I am forgetting?


