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%%--------------------------------------------------------------------------------------   \section{Work from mixing gases}   \label{sec:workout}     What is the entropy increase associated with mixing dissimilar gases, and how do we extract usable work from this?     Once two distinct gases, $A$ and $B$, mix, there is an increase in entropy and it will take some work to separate them again. We can also extract work from their mixing (maximum if quasistatic).     Questions:   \begin{enumerate}   \item What is entropy change?   \item Hence what is the maximum work extractable from the mixing process?   \item How does it vary with dissimilarity of particles?   \item How could one extract this work?   \end{enumerate}     Answers:   \begin{enumerate}   \item For {\it distinguishable} particles, mixing two volume-$\frac{V}{2}$ boxes of $\frac{N}{2}$ ideal-gas particles each incurs an entropy increase of $\Delta S = N\ln2$.     \item The maximum work extractable from this process is $W_{\rm max} = NT\ln2$. (This is also the minimum work required to recover the original state.)     \item The reason I ask is this: perhaps it is the case that increasingly dissimilar particles have more different attributes which can be harnessed to extract more work. However, there is clearly a maximum amount of work that can be extracted from any mixing process -- we cannot gain infinite energy! This leads me to conclude that $NT\ln2$ is the maximum work out (for two containers of equal volume, with no initial pressure or temperature difference).     \item \label{pnt:mixwork} Essentially the way to extract work from mixing particles A and B is to have it so that the diaphragm is permeable to particles A and not to B. Then the partial pressure of A will push it over to B's side, reducing the total pressure in side A. Then we can do $p\,dV$ work.   \begin{itemize}   \item {\it How much work?} Say we have two containers, 1 and 2, which have initial volumes $V_1^{\rm i}$ and $V_2^{\rm i}$ respectively. Box 1 contains $N^{\rm A}$ particles of type A and box 2 contains $N^{\rm B}$ particles of type B. Everything is in contact with a thermal reservoir at fixed temperature, and we ensure all processes happen sufficiently slowly that equilibrium is maintined at that temperature. For now, assume that the particles obey the ideal gas equation of state.\\   The protocol is as follows: the partition is permeable to particles of type A. They diffuse into box 2 until the partial pressures equalise. An impermeable but movable partition is then inserted, and the total pressures in the two boxes equalise by expanding the volume of box 2. This last stage is where work is done.\\   After some calculation, I find that $W=-\frac{V_1^{\rm i}}{V}N^{\rm A}T\ln\frac{N^{\rm A}}{N}$.     \item For the simpler case where the boxes are initially the same size with equal particle numbers, $W=NT\ln2$ theoretically. But here, we get only $W=\frac{1}{4}NT\ln2$ for the same conditions. What's going on?\\   Remember that at the end of this process, the gases are not totally mixed. Thus there is still some work left to extract. But does it add up when we take this into account? {\bf CHECK}.   \end{itemize}       %Here's an idea {\it Update: bad idea: need to carry out mixing process isothermally and reversibly}. The total number of particles $N$ is fixed, but not necessarily the final volume. If we set $V^{\rm final} < V^{\rm init}_1 + V^{\rm init}_2$, then the system will end in a pressurised state; this pressure can do work. \begin{itemize} \item It would cost work to diminish the total volume -- how much? \item Ignoring this, I calculated (I think) the entropy change for mixing to a volume $\eta V < V$. I found $\Delta S \approx N\ln(2\eta)$, which means the theoretical work is $W\approx NT\ln(2\eta)$. \item But then, allowing the ideal gas to expand from volume $\eta V$ to volume $V$ only gave a work of $W=NT\ln\eta$. Check calculations. \end{itemize}     \end{enumerate}     {\bf To do:}   \begin{itemize}   \item Extension beyond ideal gas? -- Hard, and perhaps not much to learn from the exercise.   \end{itemize}