Why does adiabatic expansion occur in the carnot process?

(Spoiler: Why adiabatic expansion happens in Carnot cycle doesn't really answer the question for me.)

In the Carno cycle, the open system is first brought into contact with the warm reservoir, which then expands isothermally.

In the second step, the system is separated from the reservoir and is now thermally insulated, resulting in another small adiabatic expansion.

I ask myself, why is a little internal energy converted into volume work after the separation? Isn't the equilibrium already reached after the isothermal expansion? (obviously not, but why?).

• These are quasistatic reversible processes - the system is always in equilibrium. Commented Jun 25, 2022 at 19:56
• It's not possible to answer until you suggest what alternative procedure you have in mind. Of course some other procedure would not be called a Carnot cycle, but in any case it's not clear at the moment what your question really is. Commented Jun 26, 2022 at 12:58

At the end of the reversible isothermal expansion the external pressure is deliberately further slowly reduced, but now according to $$Pv^{\gamma}$$ = constant, instead of $$Pv$$=constant, to allow the expansion to continue adiabatically and reversibly.

But why does the external pressure decrease further? Wouldn't that mean that some other work would have to be applied to lower the pressure of the environment?

No. In my answer to associated with the link you initially introduced, I described a process by which one grain of sand at a time is horizontally removed from the top of a frictionless piston surface in order to infinitesimally reduce the external pressure. The removal theoretically requires no work.

But how can I imagine that it requires no energy? Wouldn't it cost me mv^2/2 to move it, even if it's frictionless?

No. The idea is to do positive work to get it moving then negative work to bring it to rest on the adjacent platform. That makes the total change in kinetic energy zero. From the work energy theorem the net work done on an object equals its change in KE. Ergo, the net work done on each grain of sand is zero.

Hope this helps.

• But why does the external pressure decrease further? Wouldn't that mean that some other work would have to be applied to lower the pressure of the environment?
– iwab
Commented Jun 25, 2022 at 21:23
• @iwab I will update my answer in response to your follow up question. But the answer is no Commented Jun 25, 2022 at 21:26
• thank you! But how can I imagine that it requires no energy? Wouldn't it cost me mv^2/2 to move it, even if it's frictionless?
– iwab
Commented Jun 26, 2022 at 21:26
• @iwab Se my answer in my update. Commented Jun 26, 2022 at 22:02

Because both heat transfer steps must be isothermal to be reversible and generate no entropy. The Carnot cycle works by exploiting the temperature difference between two reservoirs. It uses reversible adiabatic compression and expansion to adjust the working fluid's temperature between the heat transfer steps. Thus, there is no entropy generation at any step in the cycle.

You are in control of the piston, the piston does not move spontaneously, the external pressure is set up (controlled) by you at every time so that the process is quasistatic. So the system is at equilibrium at all times, but you keep changing the external pressure to make it expand (adiabatically in this stage). The external pressure is not kept fixed, so the gas has no other option than to expand and do work.

• I understand, but in my mind this is still a bit in conflict with the meaning of the cycle, shouldn't the system do work on the environment, that's the whole idea behind the carnot process, how much work is theoretically possible with given heat. If I now change the pressure myself, or pull/push on the piston, my system has not done any work of its own. That confuses me a little bit right now :/
– iwab
Commented Jun 25, 2022 at 22:25
• That is incorrect. The gas makes a work $dW=PdV$, where P is the gas's pressure. That means that if the external pressure is almost the same (just slightly less to allow it to move slow so P is well defined), then the gas will do work into the external system.
– user338734
Commented Jun 25, 2022 at 22:42
• That makes sense, yes. I just think about the net work. The small pressure reduction of the environment must happen through some other work (e.g. a valve that must be opened, or another piston that is pulled out), in total the work that the gas does minus the one needed for pressure reduction = 0. Maybe i'm looking at this from a little wrong position?
– iwab
Commented Jun 25, 2022 at 22:48
• and then I think that the work needed to make the circumstances to allow adiabatic expansion could be used directly.
– iwab
Commented Jun 25, 2022 at 22:53
• You do not need any work to change the external pressure. You can have a piston with weights that move horizontally, so you can adjust the piston pressure at will with doing as little work as you want, if you do it slow enough, and friction is minimal. What I mean this this is that the details do not matter, you only need to know that you can change the pressure without spending any energy, and the result is a transfer on energy PdV from the gas to the environment (regardless of if you spend energy changing the pressure, which you don't need to spend under ideal conditions)
– user338734
Commented Jun 25, 2022 at 22:56

In the second step, the system is separated from the reservoir and is now thermally insulated, resulting in another small adiabatic expansion. I ask myself, why is a little internal energy converted into volume work after the separation?

The need for adiabatic steps is revealed by reviewing the necessities of creating an 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.

And there you go: four sequences of adiabatic and isothermal compression and expansion. There's actually no other way to reversibly obtain work from heat using only two reservoirs.

I'm back with a side point: You've been talking about expanding and compressing the fluid. These steps are driven by pressure differences, or gradients, which tend to generate entropy (just like the heat transfer down a temperature gradient that you addressed above). In addition, any mechanism that applies or collects work involves some degree of friction, which also tends to generate entropy.

"I'll perform every mechanical step very slowly, easing up on the externally applied pressure to allow expansion and gently applying it again to obtain compression. If I adjust the pressure using a series of weights, for example, I'll slide them into place slowly over low-friction surfaces to avoid dissipating heat and generating entropy. In real life, I'll never get rid of gradients—after all, things don't move without a slight force difference, and heat doesn't flow without a slight temperature difference—but with skilled engineering I can reduce them to be arbitrarily close to zero.

"You are aware that the power output, meaning the work done per unit time, will correspondingly approach zero, right? Since the Carnot cycle proceeds infinitely slowly to avoid generating entropy, it provides maximum efficiency and maximum work but zero power (work collected per unit time). In a real engine, I would sacrifice efficiency (not that I have any choice—all real macroscale processes generate entropy and dissipate heat) to push processes to run conveniently fast, wasting some heat but providing useful power."

That is acceptable.

• Thank you for the detailed answers, but still I don't understand this passage: "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." Why does this process happen at all then? So why does the gas continue to expand by itself after insulation to cool down? Couldn't it have reached the same volume already with the isothermal expansion?
– iwab
Commented Jun 26, 2022 at 3:29
• I wonder, wouldn't there have to be some kind of pressure change in the environment to allow the gas to expand/compress adiabatically in the first place? Wouldn't that need some other work, so to say the energy to set up the right conditions to allow adiabatic expansion/compression?
– iwab
Commented Jun 26, 2022 at 3:32
• The adiabatic step is needed to avoid a temperature difference between the system and the hot and cold reservoirs. Heat transfer over a temperature difference would generate entropy, which we can't afford in this perfectly efficient (Carnot) heat engine cycle. Commented Jun 26, 2022 at 7:02
• And yes, you must constantly adjust the applied pressure to nearly match the system pressure to maintain quasistatic conditions and minimize (in the ideal case, eliminate) entropy generation. An example is adding or removing weights from a piston pressing on the system. Commented Jun 26, 2022 at 20:08
• This is the way, ok keep the surrounding pressure lower than your system's pressure, you will get the work from the internal energy of the system. But doesnt it require some energy to lower the pressure somehow? Wouldn't adding or removing weights from a piston require energy?
– iwab
Commented Jun 26, 2022 at 20:23