Given that U = (3/2)PV does this mean that ΔU = Δ(3/2)PV for an ideal gas? Hence when finding the change in internal energy using a P-V diagram, can we simply apply this equation instead of using ΔU = Q+W?
Yes, if you can know ##\Delta(PV)## you can know the change in internal energy. For this formula, you have to know both the initial and final pressure and volume. A more often seen version of this concept is using the other side of the ideal gas law, so that with constant N: $$\Delta U = \frac{3}{2}Nk_T \Delta T$$ Thus, for a closed system, only a change in temperature will lead to a change in energy. Specifically, isothermal processes on ideal gases have the condition ##\Delta U =0##. But ##\Delta U=Q+W## is the first law of Thermodynamics. It can come in very useful when you're finding either ##\Delta U## or ##Q## or ##W##. Often you will have to use this law in some form or another in many problems.