The TRUTH? η = 1 – (Qc / Qh) = 1 – (Tc / Th)

[Tom Booth]

It's the action of the expanding gas "consuming" energy that I am interested in. Is internal energy really reduced when the gas expands, other than by conduction losses to the surrounding colder surfaces of the engine. And if so, how exactly?

The internal energy of the entire expanding volume isn't reduced, but the internal energy per unit volume is, because the total internal energy of the gas is spread out more, into a bigger space.
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This statement (bold) is, or would be, a violation of conservation of energy.

Your statement is basically that energy can GO OUT from the working fluid as WORK but all the energy is still in the working fluid but just spread out more.

That would double the total amount of energy.

Say 1000 joules gone out as "work" but the same 1000 joules still in the working fluid but just spread out in a greater volume.

So the original 1000 joules has split and turned into 2000, 1000 going out as "work" and 1000 remaining but just spread out in a larger volume.

[Tom Booth]

It's the action of the expanding gas "consuming" energy that I am interested in. Is internal energy really reduced when the gas expands, other than by conduction losses to the surrounding colder surfaces of the engine. And if so, how exactly?

The internal energy of the entire expanding volume isn't reduced, but the internal energy per unit volume is, because the total internal energy of the gas is spread out more, into a bigger space.

By the same token, the temperature of individual molecules of air doesn't change, but the bulk temperature of the gas does reduce, because the thermometer, or finger end, isn't getting hit so often because the molecules are more spread out or rarified in the expanded volume; the pressure falls.
...
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Same problem with this later statement regarding the temperature of individual molecules.

Unless, the expansion is "free expansion into a vacuum".

But we are discussing engines. Gas expanding and doing external work.

An "individual gas molecules" strikes the piston, the piston moves. Kinetic energy was transfered. The individual gas molecule looses energy, slows down, is now "colder", has, as an individual molecules LESS energy, not as you suggest above, that the temperature (energy) of the individual molecules stays the same.

That violates conservation of energy.

What you are claiming is the individual molecule transfered energy to the piston but still retains the same amount of energy, doubling the energy.

That would clearly be a violation of conservation of energy.

Unless of course you're talking about isothermal expansion, but that only means heat is added to compensate.

When a particle transfers heat to the piston and loose energy it then perhaps contacts the hot plate, gaining the energy back, but that is not energy simply "spreading out" into a larger volume.

In an adiabatic expansion, (doing work on a piston) not only is the energy "spread out" reducing the temperature, but the individual molecules cool down, move slower, loose energy as well.

[Tom Booth]

In any example, we would use the minimum amount of gas needed to do the work at hand.

By degrades, I mean into a lower temperature state. It's the action of the expanding gas "consuming" energy that I am interested in. Is internal energy really reduced when the gas expands, other than by conduction losses to the surrounding colder surfaces of the engine. And if so, how exactly?

The isolated room would of course contain high and low energy states, like a small version of the world. Any possible mechanism could be considered to perform work, using only the energy internal to the room. I realize this leads to many complex calcs, but if boiled down to the end, does the room reach 0k?

Your question clearly (IMO, see highlighted), relates to expansion with work output, or expansion of a gas doing work on a piston in an engine not "free expansion" of the gas into a vacuum.

[Stroller]

Yeah. We already covered the additional input energy required to perform work on a piston earlier in the conversation. In the response quoted, I was only considering the free expansion of the gas, so we can isolate one thermodynamic issue at a time. It would have been clearer if I'd specified that for people joining the conversation later.

[VincentG]

Tom that was a good response and about sums up what I would ask.

By "free expansion", I believe Stroller and I had settled on the baseline of the gas expanding against one atmosphere. So it's still doing "work".

I think, for now, thinking in terms of an isothermal expansion is less complex than adiabatic. The key is ignoring any reduction of internal energy from any kind of direct thermal transfer of the gas to a colder body.

[Tom Booth]

Tom that was a good response and about sums up what I would ask.

By "free expansion", I believe Stroller and I had settled on the baseline of the gas expanding against one atmosphere. So it's still doing "work".

I think, for now, thinking in terms of an isothermal expansion is less complex than adiabatic. The key is ignoring any reduction of internal energy from any kind of direct thermal transfer of the gas to a colder body.

Well, "adiabatic" IS without any kind of thermal transfer in or out of the gas. To me that is a less complex scenario than isothermal which for every joule of work done by the gas an equivalent amount of heat is added to replace it.

I guess either way has its complexities though. Like, in an adiabatic expansion, what's causing the expansion? Momentum of the piston? The flywheel? The internal energy of the gas? Both?

[Tom Booth]

Yeah. We already covered the additional input energy required to perform work on a piston earlier in the conversation. In the response quoted, I was only considering the free expansion of the gas, so we can isolate one thermodynamic issue at a time. It would have been clearer if I'd specified that for people joining the conversation later.

So we are in agreement then?

[VincentG]

Well, "adiabatic" IS without any kind of thermal transfer in or out of the gas. To me that is a less complex scenario than isothermal which for every joule of work done by the gas an equivalent amount of heat is added to replace it.

Very true, but my point being that an adiabatic expansion is reducing in temperature as it expands. This, I think, leads to further complications.

[Tom Booth]

Well, "adiabatic" IS without any kind of thermal transfer in or out of the gas. To me that is a less complex scenario than isothermal which for every joule of work done by the gas an equivalent amount of heat is added to replace it.

Very true, but my point being that an adiabatic expansion is reducing in temperature as it expands. This, I think, leads to further complications.

Well, OK.

But just out of curiosity. What complications?

Your question is, (I think) about heat being "destroyed" i.e. converted.

A measurable cooling of the gas via work output seems like an example of that. Cooling above and beyond simple dispersal into a larger volume.

Isothermal expansion where there is no clear influence on temperature is not as clear cut, or maybe you could explain how it could be.

What is hard to understand is the mechanism by which a gas can expand and loose 100% of its energy to work driving a piston and then remain expanded.

How does it do that?

My answer is IT DOESN'T!!!!

After an adiabatic expansion the piston MUST return. It has to.

Visualizing how that's possible is challenging, but I think it is easier if it is kept in mind that temperature is AVERAGE kinetic energy.

The very very very hot gases do the majority of the work driving the piston, but they loose that energy just as quickly. The gas is heated, expands and does work but the net effect is the gas looses energy and contracts immediately afterwards.

Matt, fool and others don't think that is possible. The second law, the Carnot limit don't take it into account.

The Carnot theorem/limit is a violation of conservation of energy. It says the gas takes in heat energy, expands and does work but retains the heat energy at the same time which then needs to be removed.

How does it need to be removed when it already went out as WORK 100% ???

The gas cannot remain expanded if 100% of the energy to do that expanding was transfered out as work. That would be 200%

100% used for work output and the same 100% retained to keep the gas expanded.

[Stroller]

So we are in agreement then?

Earlier in the conversation with Vincent I showed that doubling the volume of 100cc of working fluid in a 1cm^2 area cylinder against atmosphere would require 36.5J of energy passed into the working fluid.

Then I showed that lifting a 102g piston as well as pushing against atmosphere would require an additional 1J, which ended up in the piston mass as gravitational potential energy.

Then I explained that because the piston mass is gaining GPE as it rises, the extra 1J doesn't increase the temperature of the working fluid. Nor does the work done raising the piston cool it down.

Please check my working and see if there's anything wrong with the reasoning or math.

[Tom Booth]

So we are in agreement then?

Earlier in the conversation with Vincent I showed that doubling the volume of 100cc of working fluid in a 1cm^2 area cylinder against atmosphere would require 36.5J of energy passed into the working fluid.

Then I showed that lifting a 102g piston as well as pushing against atmosphere would require an additional 1J, which ended up in the piston mass as gravitational potential energy.

Then I explained that because the piston mass is gaining GPE as it rises, the extra 1J doesn't increase the temperature of the working fluid. Nor does the work done raising the piston cool it down.

Please check my working and see if there's anything wrong with the reasoning or math.

So your talking about an isothermal expansion. Correct?

What is the final pressure? (Internal, of the working fluid, at the end of the expansion).

A similar problem here:

https://physicstasks.eu/2179/work,-pres ... -expansion

Suggests doubling the volume (in an isothermal expansion) halves the pressure.

P2 = P1/2

The math however is quite frankly over my head.

[Stroller]

Please go back and read the conversation.
With the input of 36.5J, the temperature increases from 300K to 600K and doubles the volume from 100cc to 200cc.
The final pressure balances atmospheric, plus a negligible fraction for holding up the weight of the piston, which I didn't calculate.

[Tom Booth]

Please go back and read the conversation.
With the input of 36.5J, the temperature increases from 300K to 600K and doubles the volume from 100cc to 200cc.
The final pressure balances atmospheric, plus a negligible fraction for holding up the weight of the piston, which I didn't calculate.

So not isothermal then?

If the temperature has increased (doubled from 300k to 600k).

600k at the end of the expansion?

[Stroller]

Sorry, my mistake, I just went back and read what I calculated.

In the more complex case of joules needed at constant volume, a reasonable engineering approximation would be to multiply the Cv value of air by the mass of air and the change in Temperature.

0.718 x 0.12g x 300 = 25.85J

This is 9.65J less than the free expansion case. So to answer your other question of what T the air will be once we allow it to expand freely after it reached 600K at constant volume, we just need to work out how much 100cc of freely expanding air will heat up if we put 25.85J of heat energy into it.

Q 25.85J / (Cp value of air is 1.01 x mass of air 0.12g) = Delta T = 213.28K
We add this result to the initial temperature of 300K and we get 513.28K

Pressure remains constant, but now I need to think whether this accounts for the change in volume too.

[Tom Booth]

Please go back and read the conversation.
With the input of 36.5J, the temperature increases from 300K to 600K and doubles the volume from 100cc to 200cc.
The final pressure balances atmospheric, plus a negligible fraction for holding up the weight of the piston, which I didn't calculate.

Are you sure that is correct?

The temperature doubles AND the volume doubles??

I would think it would be one of the other but not both.