Change in internal energy of an ideal gas

[awesomeness]

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?

 

[Matterwave]

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.