The PV product is a measure of the gas PE, which is also measured by nRT, none of which are mass specific until we talk about particular types of molecules.

Well first dear tinman I have not talked about mass in this discussion.  Why you bring it up is a secret of your own.

What you keep missing is basic thermodynamics that has been established for almost 200 years.  If you are just going to proverbially jam your fingers in your ears and shout "na, na, na" there is no discussion.

The whole thread is about the stored energy within the HHO gas-->the mass of gas. You are the one that took it on yourself to deviate from the intended(and well defined) position of this thread.
I have shown and explaind how and where the energy went to increase the pressure of the mass of gas. The energy required to increase the gas mass pressure is accounted for in the form of heat disipated by the system as a whole.
As i explained above-and have proven-you have applied laws based around an ideal gas,and also not applicable to my setup. You continually take threads and side track them,and provide faulse and missleading information-->you have done this time and time again.

Now,time for you to back up your claim that these ideal gas laws apply to my system-AS IT IS
Show me the tests carried out by the guru's that developed these 200 year old law's useing HHO gas as there test subject.
Show me a decrease in heat energy within the stored gas when the piston is opperated--you cannot
And the reason you cant is because you know(well you should know) that there is no lose in heat energy within the gas when the piston is opperated-->how is your thermodynamics doing now in this situation.
Next i want you to go and do some reserch on how the ideal gas laws were calibrated-->what type of vessel was used in these experiments. Then come back and tell us all-->i'll give you a little hint-->it was of a fixed volume, Now,is my gas containment vessel of a fixed size when the piston starts to opperate?

About time you put your money where your mouth is Mark,and start showing some truth to your claims.

And a comment from profitis you refuse to challenge because you know he is right.
Quote: Mark E and libre will continue to ignore the most important aspect of the system @tinman.the phase change liquid>gas.the gain is here and here alone

Now,spend 5 minutes and go look up profitis's claim,and let us know what you find.


It's time you stopped useing DC power messurements to try and calculate AC power.

 



Now, as the gas in the pump cools and heat is transferred to the environment it eventual becomes the same temperature as the environment.  The product of PV before compression is the same as after and the internal energy of the gas is the same as before. BUT, at that point all the heat created by the application of work has left the system.

Now, having done this simple experiment go back and analyse your previous statements and realise you are completely misguided about even the simplest thermodynamic processes.

BUT.. why don't you perform the following experiment. It will show you that your assumptions and theories are completely erroneous.

Take a bike pump and block of the exit with your thumb. Compress the air rapidly... what happens ? Two things should be obvious. You expend energy (work) to move the piston and 2, the air inside the piston heats up.

There is probably little point in attempting to educate you in what conventional science has known for > 200 years and what repeated experimental evidence shows.

ow, hold the piston in the pumped position till the pump cools down.

Done rapidly enough the first action is approximately equal to and adiabatic process, and over time the heat LEAVES the cylinder (It must do, as it is hotter than the environment).

This experiment alone should be enough to give the lie to your conjecture that the internal energy of a gas is the product of PV and does not change during a thermodynamic process.

Why did the gas heat up?  Since internal energy of the gas is proportional to the molecular quantity * temperature and temperature increased then the internal energy of the gas must have increased. Clearly, the work done was converted to heat in the system.

There is probably little point in attempting to educate you in what conventional science has known for > 200 years and what repeated experimental evidence shows.

BUT.. why don't you perform the following experiment. It will show you that your assumptions and theories are completely erroneous.

Take a bike pump and block of the exit with your thumb. Compress the air rapidly... what happens ? Two things should be obvious. You expend energy (work) to move the piston and 2, the air inside the piston heats up. Now, hold the piston in the pumped position till the pump cools down.

Done rapidly enough the first action is approximately equal to and adiabatic process, and over time the heat LEAVES the cylinder (It must do, as it is hotter than the environment).

This experiment alone should be enough to give the lie to your conjecture that the internal energy of a gas is the product of PV and does not change during a thermodynamic process.

Why did the gas heat up?  Since internal energy of the gas is proportional to the molecular quantity * temperature and temperature increased then the internal energy of the gas must have increased. Clearly, the work done was converted to heat in the system.

Now, as the gas in the pump cools and heat is transferred to the environment it eventual becomes the same temperature as the environment.  The product of PV before compression is the same as after and the internal energy of the gas is the same as before. BUT, at that point all the heat created by the application of work has left the system.

Now, having done this simple experiment go back and analyse your previous statements and realise you are completely misguided about even the simplest thermodynamic processes.

I am not the one stuck here-->lets explain where you are going wrong Mark,MH,and the sheeple that follow.

It is undeniable that when some one here(on this forum) believes they have a device that can deliver more energy than it consume's,the guru'sZ(including my self) demand accurate test measures be made before anything else. Meassurement error is normally the call here.
But when it comes time for the guru's to try and prove some one wrong,near enough seems to be good enough. We will look at each of the below law's,and show why they are not applicable to my setup,and have been missused to explain away a free source of energy.

Boyle's law-->For a fixed mass of gas at constant temperature, the volume is inversely proportional to the pressure.That means that, for example, if you double the pressure, you will halve the volume. If you increase the pressure 10 times, the volume will decrease 10 times.
You can express this mathematically as
pV = constant
Is this consistent with pV = nRT ?
You have a fixed mass of gas, so n (the number of moles) is constant.

Now why isnt the above applicable in my system? The answer being that the volume increases when the cylinder is opperated-the pressure decreases when the cylinder is opperated,and thus the temperature decreases when the cylinder is opperated.<--Do not mix up a temperature decrease with a decrease in heat energy within my systemAs the temperature dose not remain constant,boyle's law dose not apply. There is an interesting thing that go's along with boyle's law--Quote:If you want to increase the pressure of a fixed mass of gas without changing the temperature, the only way you can do it is to squeeze it into a smaller volume. That causes the molecules to hit the walls more often, and so the pressure increases. The same applies for decreasing the pressure without changing the temperature of a fixed mass of gas,you just increase the volume to decrease the pressure.

Charles law
For a fixed mass of gas at constant pressure, the volume is directly proportional to the kelvin temperature.

Now who knows why Charles law dose not apply with my setup?-->the answer is the pressure of the fixed mass of gas drops when the piston is opperated-->it dose NOT remain at a constant pressure.

Quote: Basic thermodynamic processes are defined such that one of the gas properties (P, V, T, or S) is constant throughout the process.
Which is constant within my system when the piston is opperated?.

So you see,Mark has applied the ideal gas law,that is based around an ideal gas(Quote:a hypothetical gas whose molecules occupy negligible space and have no interactions, and which consequently obeys the gas laws exactly).
Deviations-the ideal gas law neglects both molecular size and intermolecular attractions, it is most accurate for monatomic gases at high temperatures and low pressures.
So who here can show me that H2O1 is an ideal gas --Where H is a fuel and O is an oxidizer for that fuel-->couldnt find a gas further from an ideal gas if we tried.

What is funny to see ,is the sheep in the flock follow Mark blindly without even bothering to do a little reserch them self-->including you MH,the one that insist that i read books to gain knowledge,and yet you failed to do it your self here and now.
@MH,either you didnt read the thread from the start,or you have just blindly followed the deviated and incorrect path that Mark has set.

Fact's
Regardless of any change in temperature.pressure.volume-,the mass of gas remains constant,and the energy stored within that gas remains constant.

When the volume is increased(the piston is opperated)the pressure decreases,and the temperature decreases-->but the heat energy remains the same throughout the system--the heat just dosnt disappear as this would be destroying energy-->the heat energy remains constant regardless of the temperature drop.