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Increase of entropy

Here we imagine a cycle in which we go from point 1 to point 2 by an irreversible process and back by a reversible process. (See here for an explanation of the picture.)

\begin{figure}\begin{center}\mbox{\epsfig{file=rev_irrev.eps,width=7truecm,angle=0}}
\end{center}\end{figure}

The entropy change for a reversible process is given by

\begin{displaymath}
S_2-S_1=\int_1^2 {{}\raise0.44ex\hbox{\bf\symbol{'040}}\llap{d}Q^{\rm rev}\over T }
\end{displaymath}

What can we say about an irreversible process between the same endpoints? Clearly the entropy change is the same (that's what we mean by saying entropy is a function of state.) But if we consider a cycle involving an irreversible process from 1 to 2 and a reversible process to return to 1, we have a cycle for which Clausius's theorem holds:

\begin{eqnarray*}
\oint {{}\raise0.44ex\hbox{\bf\symbol{'040}}\llap{d}Q\over T }...
...mbol{'040}}\llap{d}Q^{\rm irrev}\over T }\!\!\!&<&\!\!\!S_2-S_1.
\end{eqnarray*}



Hence in general,

\begin{displaymath}
{}\raise0.44ex\hbox{\bf\symbol{'040}}\llap{d}Q\le T{\rm d}S,
\end{displaymath}

and so for an isolated system, where $ {}\raise0.44ex\hbox{\bf\symbol{'040}}\llap{d}Q=0$,
$\mbox{\large\colorbox{yellow}{\rule[-3mm]{0mm}{10mm} \
$\displaystyle {\rm d}S\ge 0.
$  }}$


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Previous: Proof of Clausius's theorem
Judith McGovern 2004-03-17