Why does adiabatic expansion occur in the Carnot cycle?

Question

Several people have asked this question before and I haven't found any of the explanations even remotely satisfying.

To begin, Let's imagine our system consists of an ideal gas enclosed in a cylinder with non-conducting lateral walls, a frictionless, non-conducting piston on one end and a conducting base.

Now, I can rationalize the first step of the Carnot Cycle - heat energy added to the system is converted into mechanical work performed on the surroundings. Nothing terribly interesting going on here.

However, things get a little hairy for me in the second step. Once the system is no longer in thermal contact with the hot reservoir, there's no input of energy into the system. Despite this, the system still performs mechanical work on its surroundings. The fact that we can achieve mechanical work in the absence of energy input seems to suggest that any such system consisting of an ideal gas above 0K would spontaneously perform mechanical work on its surroundings, presumably until its temperature gets arbitrarily close to 0K. How could this be the case?

Answer 1

Why Does Adiabatic Expansion Occur in the Carnot Cycle?

The short answer is the external pressure is purposely reduced allowing the adiabatic expansion which in turn lowers the temperature of the gas to that of the cold temperature reservoir. This permits the reversible isothermal compression to take place to reject heat to the cold reservoir as required by the second law. The final reversible adiabatic compression completes the cycle. Now, to some of your specific questions:

Once the system is no longer in thermal contact with the hot reservoir, there's no input of energy into the system. Despite this, the system still performs mechanical work on its surroundings.

The mechanical work occurs because the external pressure is gradually reduced. The expansion work comes at the expense of the internal energy of the gas which, in turn for an ideal gas, lowers the temperature. Note that the positive work done by the adiabatic expansion exactly equals the negative work done by the adiabatic compression, for a net adiabatic work of zero.

The fact that we can achieve mechanical work in the absence of energy input seems to suggest that any such system consisting of an ideal gas above 0K would spontaneously perform mechanical work on its surroundings,

No it does not "spontaneously" perform work on its surroundings. As indicated above the pressure of surroundings has to be purposely reduced for the expansion to occur.

Hope this helps.

Answer 2

The need for adiabatic steps is revealed by reviewing the necessities of an optimally efficient heat engine cycle:

Your requirement: Take a very large warm body at temperature $T_H$, and obtain useful work from it for a very long time. Also, don't create any entropy in the process, so that the highest possible efficiency is maintained.

"Impossible. (1) When I remove thermal energy from the warm body, entropy flows with it. (2) Work carries no entropy. (3) Entropy can't be destroyed. Thus, I'm left with a buildup of entropy."

OK, you can have a single additional large cool body at temperature $T_C$.

"OK, I'll plan to dump entropy there through heat transfer $Q$. For reversible heating, the entropy transfer is $ΔS = Q/T$, so I'll pull out a lot of thermal energy from the warm body (high $Q$, high $T$), and sacrifice a little energy to the cool body (low $Q$, low $T$) to dump the exact same amount of entropy. I'll output the difference as work."

I'm impressed! This is the fundamental concept of all heat engines—and the reason why one can't turn heat entirely into work. But you're still talking conceptually; what exactly is it that you're going to do?

"I guess I'll take a cool fluid, put it next to the warm body to heat it up and pressurize it, and let it expand."

You'll encounter a temperature gradient—i.e., a spatial variation in temperature—when you put the cool fluid next to the warm body. Energy flowing down a gradient always produces entropy. Your mechanism isn't as efficient as it could be.

"I'll very slowly compress the cool fluid while insulated (i.e., adiabatically) to bring it to the temperature of the warm body. Then I'll just let the fluid expand at that temperature and do work. No temperature gradients, ever."

You don't have the space to expand that much. The warm body is essentially infinite, and you need to produce work indefinitely.

"I'll work in cycles, expanding and compressing, to reset the system completely at the end of each cycle. I'll apply the compression at the lower temperature as part of the reset. Oh, and to address your objection about about moving suddenly between the two reservoirs, I'll wrap up the isothermal expansion at the higher temperature with some adiabatic expansion to cool the fluid down so that there's no entropy generation when I put it next to the cool body. The work I collect during this adiabatic expansion will exactly pay for the work I expended to achieve the adiabatic compression."

Summarize clearly the sequence you've developed.

"Starting at an arbitrary point: adiabatic compression from $\boldsymbol{T_C}$ to $\boldsymbol{T_H}$ (this eliminates any temperature gradient), isothermal expansion at $T_H$ (this provides output work), adiabatic expansion from $\boldsymbol{T_H}$ to $\boldsymbol{T_C}$ (also to eliminate any temperature gradient), and isothermal compression at $T_C$ (this dumps entropy)."

You've solved the required problem.

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The fact that we can achieve mechanical work in the absence of energy input seems to suggest that any such system consisting of an ideal gas above 0K would spontaneously perform mechanical work on its surroundings, presumably until its temperature gets arbitrarily close to 0K. How could this be the case?

Yes, a gas will spontaneously expand to fill its available volume; this maximizes entropy because there are more positional possibilities when the entire space is explored. Furthermore, it will push a frictionless piston until its pressure matches the surrounding pressure; this also maximizes entropy. (The entropy reduction from the reduced volume of the surroundings is outweighed by the entropy increase from the gas expansion, until the pressures equalize.) Yes, the gas will tend to cool as this occurs. One doesn't have to heat a system to extract energy from it! Any gradient (e.g., in pressure) is a potential energy source.

Answer 3

Would you be surprised that if you compress a spring from its relaxed state then it could perform work on something else? You would not, so why would you be surprised if a compressed gas can perform mechanical work? Instead you should be surprised by your own statement that you "can rationalize the first step of the Carnot Cycle - heat energy added to the system is converted into mechanical work performed on the surroundings." That is the really puzzling thing because what does the word "converted" mean in the context of heat energy absorbed?

Anyhow, starting from Kelvin's axiom, namely, that in a cycle consisting of two stages (1) an isothermal step and (2) and adiabatic step the work done on the environment is either zero (reversible) or negative (irreversible), which really nobody denies, you must do something else than just these two steps to get positive work done. The Carnot cycle offers such process during which the externally absorbed entropy from the higher temperature source when it is transported reversibly to the lower temperature entropy sink performs positive work in an exactly analogous way as the falling water from a higher gravitational potential to a lower one can perform an equivalent amount of work by moving an amount of electric charge from a lower electric potential to a higher one.