# $p$-$V$ diagram and internal energy

I recently learned that if the initial state is equal to the final state in the $$p$$-$$V$$ diagram then the internal energy is equal to zero so that means $$Q=W$$. My question is how can you know that only from the $$p$$-$$V$$ diagram without knowing the amount of Q? ( I'm taking generally not only in the case of ideal gasses)

## 1 Answer

Because U is a state funcion of P and V. This means that U isn't influenced by how a system arrived to a particular state (to a particular P and V) but just by the P and V that the system has at the moment. Therefor, if you draw a closed curve on the Clapryron plane, you will mean that (after some changes that U doesn't care about) you will be back with the same P and V, so the same U.

Also, note that you can say that U is constant, but you can't determine its value.

• But isn't it possible to rise the temperature of lets say solid with out changing its pressure and volume? In this case work will be zero and initial and final state will be equal but we obviously have heat transfer... Sorry if my question seems stupid but my book only show the relation between p, v and t for ideal gasses – Naifqar Feb 11 at 9:57
• To my understand, it's not possible. Even if you consider real gasses, the famous PV=nRT becomes the Van der Waals formula, but even in this formula changing T would result in a change in P and/or V. – Mauro Giliberti Feb 11 at 10:13