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?

  • $\begingroup$ Your notation is inconsistent in (2). $Q_1$ and $Q_2$ are not initial and final points in the state space. They are not state variables. Also, the integrals go through cycles, so the cannot be specified by initial and final points at all. $\endgroup$ – joigus Feb 22 at 22:08
  • $\begingroup$ @joigus you're right, that is part of what I'm asking. This is an excerpt of the proof, I did not choose the notation. $\endgroup$ – Angelo Di Bella Feb 22 at 22:15
  • $\begingroup$ if you go around the cycle twice (and come back to the same starting state), is Q the same as going around once? $\endgroup$ – Chet Miller Feb 22 at 22:17

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.