Exact Heat Differential in Reversible Processes

Question

From Clausius' theorem for a reversible process $C:$

$$\oint_C\frac{\delta Q_\text{rev}}{T}=0,\tag{1}$$

doesn't this imply that the differential $\delta Q_\text{rev}$ is exact? Or does $T$ serve the purpose of some integrating factor. I'm asking this because I ran into a description of entropy where, considering a closed, reversible process consisting of two sub processes $C_1$ and $C_2$ (each starting and ending at the same points on the $PV$-diagram):

$$\oint_{C_1}\frac{\delta Q_\text{rev}}{T}+\oint_{C_2}\frac{\delta Q_\text{rev}}{T}=\oint_{Q_1}^{Q_2}\frac{\delta Q_\text{rev}}{T}+\oint_{Q_2}^{Q_1}\frac{\delta Q_\text{rev}}{T}=0,\tag{2}$$

where $Q_1$ and $Q_2$ are respectively the heats corresponding to the initial and final points of the process. Wouldn't this be possible if and only if $Q$ is a state function in this case?

Answer 1

I hope this clarifies somewhat the situation. $$ \oint_{C}\frac{\delta Q_{\textrm{rev}}}{T}=0,\:\forall C $$ means that, $$ \delta Q_{\textrm{rev}}=A\left(X,Y\right)dX+B\left(X,Y\right)dY $$ for some state variables $X$ and $Y$. With, $$ \frac{\partial A}{\partial Y}-\frac{\partial B}{\partial X}\neq0 $$ (not an exact differential, thus) because, $$ A\neq\frac{\partial S}{\partial X} $$ $$ B\neq\frac{\partial S}{\partial Y} $$ for any state function $S\left(X,Y\right)$ These $X$, $Y$ could be $P$ and $V$, for example. Now, with $T$ in the picture, you can assert, $$ \frac{A\left(X,Y\right)}{T\left(X,Y\right)}=\frac{\partial}{\partial X}S\left(X,Y\right) $$ $$ \frac{B\left(X,Y\right)}{T\left(X,Y\right)}=\frac{\partial}{\partial Y}S\left(X,Y\right) $$ $T^{-1}\left(X,Y\right)$ is a so-called integrating factor; and $S\left(X,Y\right)$ (entropy) is a function of state that results from making the differential exact.