# $\oint{A}=0\implies$ A is a State function?

If $$A$$ is a thermodynamic variable (ex:Pressure, volume, entropy). then If $$\oint{A}=0$$, then does it imply that $$A$$ has to be a state function? I'm trying to prove that Entropy is a state function. In a reversible process, $$\oint{\frac{dQ}{T}}=0$$ (taking it for granted). From this can I deduce that entropy has to be a state function?

## 1 Answer

If $$\oint \mathrm dA = 0$$, then $$A$$ needs to be a state function which implies it is path independent.

## Why? In the above image, if we go from $$X$$ to $$Y$$ by path 1 and return from $$Y$$ to $$X$$ by path 2, then $$\oint \mathrm dA = 0$$ since we started from $$X$$ and returned back to $$X$$. Therefore,

$$\oint \mathrm dA =\left( \int _X ^Y \mathrm dA \right)_{\text{path 1}} + \left( \int _Y ^X \mathrm dA \right)_{\text{path 2}}=0$$

So,

\begin{align} \left( \int _X ^Y \mathrm dA \right)_{\text{path 1}}&=-\left( \int _Y ^X \mathrm dA \right)_{\text{path 2}}\\ \left( \int _X ^Y \mathrm dA \right)_{\text{path 1}}&=\left( \int _X ^Y \mathrm dA \right)_{\text{path 2}} \end{align}

Thus no matter what path you take, $$\int \mathrm d A$$ will always be the same. The above procedure can be repeated with any two arbitrary paths and the value of $$\int \mathrm d A$$ will always come out to be the same.