How does one produces a thermodynamic cycle from a microscopic description? When I attempt to do so, I find that the priors of my equation of state are dependant upon the Lagrange multipliers and thus the two cannot be varied independently from one another. The flexibility of my cycles appears limited compared to what seems to be implied in the introductory literature on the subject.
For instance, let $Q$ be the set $\{0,1\}$ and let $E$ be the function $E:Q\to\mathbb{R}$, implemented (for simplicity) as $E(q)=q$ and let $V$ be the function $V:Q\to\mathbb{R}$ implemented as $V(q)=q$. Assume the statistical prior $\overline{E}$ and $\overline{V}$. Then, the probability distribution that maximizes the entropy is the Gibbs ensemble. In this case:
$$ Z=\sum_{q\in Q}e^{-\beta ( E(q)-p V(q))}=e^{- 0}+e^{-\beta-\beta p}=1+e^{-\beta-\beta p} $$
where $\beta$ and $p$ are the Lagrange multipliers.
In this simple example, the priors $\overline{E}$ and $\overline{V}$ are:
$$ \overline{E}=-\frac{1}{Z}\frac{\partial Z}{\partial \beta} = \frac{-1}{1+e^{-\beta-\beta p}} \left( -e^{-\beta-\beta p} \right)=\frac{e^{-\beta-\beta p}}{1+e^{-\beta-\beta p}}\\ \overline{V}=-\frac{1}{Z}\frac{\partial Z}{\partial p} = \frac{-1}{1+e^{-\beta-\beta p}} \left( -e^{-\beta-\beta p} \right)=\frac{e^{-\beta-\beta p}}{1+e^{-\beta-\beta p}} $$
The entropy, defined as $S=k_B\left( \ln[Z]+\beta (\overline{E} + p \overline{V} ) \right)$ is
$$ S=k_B\ln[1+e^{-\beta-\beta p}]+k_B\beta \left( \frac{e^{-\beta-\beta p}}{1+e^{-\beta-\beta p}} \right)+k_B p\beta \left( \frac{e^{-\beta-\beta p}}{1+e^{-\beta-\beta p}} \right) $$
I would now like to draw various thermodynamic cycles over $p$, $\overline{V}$, $\beta$, $\overline{E}$ and $S$.
As an example, lets say I attempt to draw this cycle illustrated in the following image. It represents a very common example of a cycle involving $p$ and $\overline{V}$.
Let's look at the arrow going from 4 to 1. You will notice that $\overline{V}$ does not change over this line but $p$ does. However, since $\overline{V}$ does depend upon $p$ in the equation we just derived, how can that be? One thing that could happen is to use $\beta$ to cancel out the effect increasing $p$ has on $\overline{V}$? For instance, if we say that the temperature is increased to maintain the relation $\beta-\beta p$ constant as $p$ is varied, then $\overline{V}$ would remain constant. Is this the interpretation --- the second Lagrange-conjugate pair, in this case $\beta$ and $\overline{E}$, is used to offset the change that would normally occur in $\overline{V}$ as $p$ is increased?
Is it then correct to say that no cycle is possible over a thermodynamic system comprised only of $\beta$ and $\overline{E}$. For instance, the system described by:
$$ Z=\sum_{q \in Q}e^{-\beta E(q)} $$
cannot admit a cycle over $\beta$ and $\overline{E}$ because there isn't a second conjugate-Lagrange pair to offset the changes on $\overline{E}$ as $\beta$ is varied?