# Can two quasi-static adiabates intersect? [closed]

I saw a question in my physics book that I've been trying to understand lately.

The question is simple: given two quasi-static adiabatic curves, could you find a point of their intersection? Also, what would be the efficiency and entropy of such process?

My thoughts: I think that if one of the curves is reversible and the other isn't, it is possible to construct such an example, but I am not able to find neither the entropy nor the efficiency of such a process.

## closed as off-topic by Norbert Schuch, GiorgioP, ZeroTheHero, heather, FGSUZApr 19 at 11:06

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• The question about intersection is clear, but you don't seem to have specified the 'process' (or your textbook hasn't). – Philip Wood Apr 13 at 15:43
• This question is the same as the one addressed at this link. – Jeffrey J Weimer Apr 14 at 13:07
• See the addendum to my answer in response to your request for an example. Hope this helps. – Bob D Apr 14 at 22:17

Two adiabatic curves can intersect, but one of them has to be irreversible. Therefore, one of the quasi-static processes has to be irreversible. Essentially, that means one of them has to involve friction. All reversible processes are quasi-static, but not all quasi-static processes are reversible. A process can be carried out very slowly (quasi-statically) but still involve friction. Friction makes it irreversible.

My thoughts: I think that if one of the curves is reversible and the other isn't, it is possible to construct such an example, but I am not able to find neither the entropy nor the efficiency of such a process.

A colleague of mine (@Chet Miller) and I compared a quasi-static reversible adiabatic process to a quasi-static process with piston friction, meaning an irreversible quasi-static process. Chet did the analysis of the friction case (the hard part). The diagram below shows the two processes starting at the same initial conditions (the point where the two curves intersect) to a specific final volume. See the curves below.

As you can see, for the process with friction the ideal gas had a higher final temperature than the process without friction. This was because only part of the work done by the gas was productive (expanding against a weight and atmosphere) whereas the rest of the work done by the gas had to overcome friction, which generated heat that was returned to the gas. Since for any adiabatic process:

$$\Delta U=-W$$

And since for any ideal gas, any process

$$\Delta U=C_{V}(T_{i}-T_{f})$$

A higher final temperature means less productive work done for the irreversible process, and therefore less efficiency.

Hope this helps.