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Say we have the graph of a reversible cyclic process, in which an isochoric, isobaric, and general process take place.

enter image description here

How would we determine in which path - AB, BC or CA, the gas would attain it's maximum or minimum temperature?

It would have to be of the form y = mx + c ie. P as a function of V ie. it has to pass the Vertical Line Test to make sure it's a function. So, in this case, it cannot be the isochoric process AB.

This is actually a graph from a very popular problem, so I'm aware the gas reaches it's maximum temperature during the BC process.

I'm aware of how to calculate the maximum or minimum temperature once we have determined the process or path at which it occurs thanks to this Stack exchange post. Writing the equation of the line, turning it into a quadratic equation, and determining the maxima or minima isn't too difficult.

But what if we had a Carnot Cycle as such:

enter image description here

We're given the pressure and volume at every point. In this case, how would we determine the path at which the gas attains it's maximum temperature? And how would we determine that temperature in this case (as each process is a curve)?

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Assuming all the above processes are being performed reversibly over an ideal gas, the gas follows the ideal gas state equation $$pV=nRT$$ in every moment.

As we are interested in temperature in the respective path segments, we can rewrite the equation as $$p = \frac{nRT}{V}.$$

Then we can, really or mentally, draw hyperboles as isoterms $p=\mathrm{f}(V)_T$.

For a given $T$, $pV = \text{const}$ and $pV$ is proportional to $T$.

For better visualization of the gas isotherms:

enter image description here


For every point of the process paths on the diagram, the absolute temperatute $T$ is proportional to the area of the rectangular with diagonal from the origin to the given point.

Therefore, either from the drawn hyperbole set, either from the rectangle area, we can see that:

  • The lowest temperature is at the point A.
  • The highest temperature is near the middle of the path segment BC.
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  • $\begingroup$ Comments have been moved to chat; please do not continue the discussion here. Before posting a comment below this one, please review the purposes of comments. Comments that do not request clarification or suggest improvements usually belong as an answer, on Chemistry Meta, or in Chemistry Chat. Comments continuing discussion may be removed. $\endgroup$
    – Buck Thorn
    Dec 6, 2023 at 8:13

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