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

[Stroller]

Alternatively, we can calculate the Work W that has to go into the gas to keep it at a constant temperature T as it expands from V1 to V2

W = RTln(V2/V1)

Gas constant R is 0.082 L atm K−1 mol−1

[MikeB]

Charles law states V1/T1 = V2/T2
So if we have a 1cm^2 tube with 100cc of air under the piston, then we need to double its volume to lift our piston by 1m. If we start with 300K air then 100/300 = 200/T2 = 600K
3.65 Joules are needed to raise 100cc of air from 300K to 600K
We used 1J raising the 102g piston 1m. That piston now has 1J extra gravitational potential energy.
Is the gas still at 600K having done the work?

"Is the gas still at 600K having done the work?"
In this example, yes it is, by definition - surely?
However you seem to be implying that you added 3.65 joules of energy, but I would postulate that you had to add (3.65 + 1) joules.

[VincentG]

Stroller, thanks for your time here. I'll let you sort out MikeB's excellent comment before I chime in. I'd say he's barking up the right tree.

[Stroller]

Charles law states V1/T1 = V2/T2
So if we have a 1cm^2 tube with 100cc of air under the piston, then we need to double its volume to lift our piston by 1m. If we start with 300K air then 100/300 = 200/T2 = 600K
3.65 Joules are needed to raise 100cc of air from 300K to 600K
We used 1J raising the 102g piston 1m. That piston now has 1J extra gravitational potential energy.
Is the gas still at 600K having done the work?

"Is the gas still at 600K having done the work?"
In this example, yes it is, by definition - surely?
However you seem to be implying that you added 3.65 joules of energy, but I would postulate that you had to add (3.65 + 1) joules.

Thanks Mike, I phrased my post badly. 3.65J would double the volume of freely expanding air. But the air has to expand against the force of gravity acting on the 102g piston mass, which is where we use the extra 1J to lift it 1m. So, as you said, that's an extra Joule for 4.65 total.

[VincentG]

Thanks Mike, I phrased my post badly. 3.65J would double the volume of freely expanding air. But the air has to expand against the force of gravity acting on the 102g piston mass, which is where we use the extra 1J to lift it 1m. So, as you said, that's an extra Joule for 4.65 total.

Putting the weight aside for a moment...
So 3.65j to raise 100cc from 300k to 600k(at 100cc still I assume)? But then when allowed to freely expand to 200cc, isn't the temperature lower than 600k? Or is 3.65j the energy needed to expand 100cc at 300k to 200cc at 600k?

Just want to clear this part up before we move on.

[Stroller]

Or is 3.65j the energy needed to expand 100cc at 300k to 200cc at 600k?

Yes. This is what Charles law tells us:

"When the pressure on a sample of a dry gas is held constant, the Kelvin temperature and the volume will be in direct proportion."

Hence:

V1/T1 = V2/T2

The equation shows that, as absolute (Kelvin) temperature increases, the volume of the gas also increases in proportion.

[Tom Booth]

.... pressure drops during an isothermal expansion.

The pressure and temperature won't drop if the right amount of heat/energy is being put into the gas as it expands.
...

What determines the "right amount"?

Aside from the opinion of "Stroller".

A "perfect" Carnot expansion would be isothermal followed by adiabatic. Pressure dropping all the way and temperature and pressure dropping during the final leg.

[VincentG]

Stroller can answer other questions as he sees fit, but I would like to focus on the very basics for a moment. I'll chime back in after I get home today.

[Stroller]

The pressure and temperature won't drop if the right amount of heat/energy is being put into the gas as it expands.
...

What determines the "right amount"?
Aside from the opinion of "Stroller".

"Opinion"

We can calculate the Work W (in the form of heat energy) that has to go into the gas to keep it at a constant temperature T as it expands from V1 to V2

W = RTln(V2/V1)

Where R is the gas constant, T is the temperature, ln is the natural logarithm.

As discussed above, any extra work done by the gas, shoving pistons up against the inertia of flywheels for example, will require more energy input. There is no free lunch.

[Fool]

Take a bicycle pump, extend it, and block the outlet orifice with your finger.
Push the pump handle in and feel the warmth on your finger where it covers the outlet orifice as the air is compressed. That's kinetic energy being converted to heat.
No let go of the pump handle. It is forced outwards by the warm compressed gas and the gas cools at it expands. That's heat being converted back to kinetic energy.

Just a nit pick. You have described a two-adiabatic-process cycle. Compression followed by expansion. For that kind of a cycle heat in and heat out are both zero, adiabatic, isentropic.

The work in also equals work out, summing to zero.

What happened to work and temperature can be attributed to internal energy. Internal energy is stored energy, and not heat.

An isothermal expansion converts internal energy to work, with heat returning the internal energy. Again heat is not converted to work. The textbooks an processers all speak as if you do, so it's very difficult to separate the fact that heat is not hot, or internal energy. Temperature is just one variable in internal energy.

Work = P•∆V

Thermal internal energy U:
U=M•Cv•T

Heat requires two masses at two different temperatures and a thermal conduction connection.

I was taught wrong in school. It took a lot to break that bad thought process. Probably won't find this spelled out in any book, but the concepts are there.

[Fool]

How is Stroller's "opinion" any different from other posters here. Answer, science. He is using science.

[Stroller]

heat is not hot, or internal energy. Temperature is just one variable in internal energy.

Work = P•∆V

Thermal internal energy U:
U=M•Cv•T

Heat requires two masses at two different temperatures and a thermal conduction connection.

I was taught wrong in school. It took a lot to break that bad thought process. Probably won't find this spelled out in any book, but the concepts are there.

I'd just add that the two masses that heat flows between in my example are the air mass rising to a higher temperature as it is pressurised by the pump action, and the part of Vincent's finger covering the outlet orifice of the pump. Pressure and delta V do the work of pushing the pump handle outwards as you correctly stated.

[Tom Booth]

The pressure and temperature won't drop if the right amount of heat/energy is being put into the gas as it expands.
...

What determines the "right amount"?
Aside from the opinion of "Stroller".

"Opinion"

We can calculate the Work W (in the form of heat energy) that has to go into the gas to keep it at a constant temperature T as it expands from V1 to V2

W = RTln(V2/V1)

Where R is the gas constant, T is the temperature, ln is the natural logarithm.

As discussed above, any extra work done by the gas, shoving pistons up against the inertia of flywheels for example, will require more energy input. There is no free lunch.

I was thinking "right amount" in terms of having a running engine. Heat Addition for high temperature and pressure all the way to BDC would probably stall the engine.

A Stirling engine anyway. You can get away with that in an IC engine maybe. Just blow all the excess heat and pressure out the exhaust. Not very efficient though.

[Tom Booth]

Take a bicycle pump, extend it, and block the outlet orifice with your finger.
Push the pump handle in and feel the warmth on your finger where it covers the outlet orifice as the air is compressed. That's kinetic energy being converted to heat.
No let go of the pump handle. It is forced outwards by the warm compressed gas and the gas cools at it expands. That's heat being converted back to kinetic energy.

Just a nit pick. You have described a two-adiabatic-process cycle. Compression followed by expansion. For that kind of a cycle heat in and heat out are both zero, adiabatic, isentropic.

The work in also equals work out, summing to zero.

What happened to work and temperature can be attributed to internal energy. Internal energy is stored energy, and not heat.

An isothermal expansion converts internal energy to work, with heat returning the internal energy. Again heat is not converted to work. The textbooks an processers all speak as if you do, so it's very difficult to separate the fact that heat is not hot, or internal energy. Temperature is just one variable in internal energy.

For heat exchange between solids and liquids generally.

For gases: "internal energy is a function of temperature only".

Work = P•∆V

Thermal internal energy U:
U=M•Cv•T

Heat requires two masses at two different temperatures and a thermal conduction connection.

I was taught wrong in school. It took a lot to break that bad thought process. Probably won't find this spelled out in any book, but the concepts are there.

[VincentG]

Yes. This is what Charles law tells us:

"When the pressure on a sample of a dry gas is held constant, the Kelvin temperature and the volume will be in direct proportion."

Hence:

V1/T1 = V2/T2

The equation shows that, as absolute (Kelvin) temperature increases, the volume of the gas also increases in proportion.

Ok, it may just be me, but I don't have this completely straight.

The joules needed to raise 100cc from 300k to 600k at constant volume = ?

Then, the temperature of the gas after free adiabatic expansion(with no additional heat input) from 100cc at 600k to 200cc is = ?

The joules needed to heat 100cc at 300k through a free expansion to 200cc at 600k = ?

Now of course, "free expansion" is still fighting the 14.7psi of the atmosphere. But that is the baseline for this correct?

Thanks for your patience. These are things I think need to be established before discussing any further dynamic scenarios.