Open Systems

[MarkE]

Once again Mark you are missinterpreting my post. I made no assumptions at all.
The higher the pressure,the more compressed the bubbles of gas are forming on each plate. Each plate will continue to produce gas until such time as the pressure reaches the point as to when the gas becomes liquid-and we are no where near that point.

The erroneous assumption that you made and you continue to make is that the gas production rate in moles/Joule is constant independent of the external pressure.  If you don't believe me then construct an experiment. 

So i leave you to think about these two posabilities,and ponder as too which would be true.
1-the output of gas production remains the same,as the compressed gas= ambiant pressure amount.
2-The lower gas production due to rising pressure=a higher heat production by the cell.
One or a combination of the two are true-->but energy IS conserved.
Incorrect,as the energy has already been accounted for in the production of heat,and energy stored within the gas-->as you have accepted as being true already. Storeing the gas at pressure dose not change the ratio of energy used(input) to the energy converted into heat and stored in the hydroxy gas.

You have led yourself down a garden path.  The gas generation rate in molecules / Joule, IE moles / Joule depends is limited by the net available energy B E C A U S E as we have both agreed:  ENERGY IS CONSERVED.  You have agreed that a portion of the input energy goes into the phase change from liquid to gas.  Now it is down to how much energy.  The amount of energy required depends on the pressure and temperature.  Increase the pressure as you have and you indentically increase the work required to evolve a given number of gas molecules.  The energy stored in the compressed gas volume identically comes from the battery energy source, and idnetically takes away from the enrgy available electrolyze additional water molecules. 

If your idea were correct, then we could destroy energy by pumping down your cylinder with a vacuum pump.  Then we would have to add external work to move your piston while according to your idea evolving the same number of gas molecules per Joule of input battery energy as at 1 ATM or ~7 ATMs as you propose.  Have you ever travelled to the mountains? 

Did you ever notice that water boils faster the higher the altitude, IE the lower the ambient pressure?  The same principles are at work here.  You are "boiling" the water by electrolyzing it.  At a lower ambient pressure it takes less work to do the "boiling" part that than it does at high ambient pressure. 

.Please feel free to show us where energy is destroyed,and not conserved.

You are unfortunately the one doing the funny accounting.

As i stated,producing and storing the gas at pressure dose NOT change the energy output of the system. If less gas is produced when under pressure,then more heat is produced as a result-->energy is conserved.

Less gas is produced.  An identically reduced amount of heat is evolved during the subsequent combustion.  The adiabatic heating increases which further decreases the gas generation rate in absolute molecules / Joule.  Again, if you do not believe me, do the math or do an experiment.

The pneumatic cylinder makes no impact on the energy stored in the gas,as the amount of gas remains the same regardless of wether the cylinder is there or not. The rest of the energy was converted into heat during the electrolisis process.

This is again wrong.  You have drawn your boundaries incorrectly resulting in incorrect accounting and conclusions.  This is first year college chemistry.

I am afraid you are wrong this time Mark-but please feel free to prove me incorrect.

[profitis]

Quote frm markE:'  Increase the pressure as you have and you indentically increase the work required to get a given amount of gas molecules'

Unquote

True.this is found by substituting pressure(p) for concentration (c) in the nernst equasion.HOWEVER,you forgot something very important: conversion of liquid to gas cools the liquid down and increases dramaticaly its entropy.this increase in entropy more (much more) compensates for the losses incurred on the nernst pressure joules.you gain,if you do it right

[profitis]

Put simply: h2 + o = h2o + energy - gas to liquid phase change energy.. and h2o + energy = h2 + o + liquid to gas phase change energy.phase change energy are direct as usable heat to and from environment at no cost.E(gas electrode)= RT/nf ln P.increase P and you decrease T at the same time from phase change liquid to gas.

[tinman]

       

At a lower ambient pressure it takes less work to do the "boiling" part that than it does at high ambient pressure.  You are unfortunately the one doing the funny accounting.Less gas is produced.  An identically reduced amount of heat is evolved during the subsequent combustion.  The adiabatic heating increases which further decreases the gas generation rate in absolute molecules / Joule.  Again, if you do not believe me, do the math or do an experiment.This is again wrong.  You have drawn your boundaries incorrectly resulting in incorrect accounting and conclusions.  This is first year college chemistry.

The erroneous assumption that you made and you continue to make is that the gas production rate in moles/Joule is constant independent of the external pressure.

I made no such assumption-->once again this is yet another faulse claim you make. The loss in gas production per joule due to pressure increase is made up for in the way of heat output energy by the system.

You have agreed that a portion of the input energy goes into the phase change from liquid to gas.  Now it is down to how much energy.

The energy that went into the phase change from liquid to gas will be the same as the energy returned when that phase change is reversed from gas to liquid-->unless you have found a way to destroy energy?.

The amount of energy required depends on the pressure and temperature.  Increase the pressure as you have and you indentically increase the work required to evolve a given number of gas molecules.

That is correct,and that increase in work results in a higher heat output,and a lower number of gas molecules-->once again,energy is conserved.

Did you ever notice that water boils faster the higher the altitude, IE the lower the ambient pressure?  The same principles are at work here.  You are "boiling" the water by electrolyzing it.

Now i think you are taking the trip up the garden path. We are not boiling the water by electrolyzing it.. We are dissociating the water into H2 and O2. I have no idea as to how you think this is steam-->which is the result of boiling water.

The energy stored in the compressed gas volume identically comes from the battery energy source, and idnetically takes away from the enrgy available electrolyze additional water molecules.

That is right,the energy stored within the gas is the exact amount that was drawn from the battery to create it-->the rest of the energy supplied by the battery during this process is transformed into heat.. So please explain how reducing the pressure of that stored gas by enlarging the storage device changes any of the energy accounting after the gas has been produced.

If your idea were correct, then we could destroy energy by pumping down your cylinder with a vacuum pump.

Once again,another incorrect statement. To pump down the cylinder requires work to be done,and thus energy is required to do so. You are now transforming energy,not destrying it.

Then we would have to add external work to move your piston while according to your idea evolving the same number of gas molecules per Joule of input battery energy as at 1 ATM or ~7 ATMs as you propose.

And again-another faulse accusation. I never said the same number of gas molecules are produced per joule of input energy as the storage pressure go's up. I clearly stated that more energy from the battery would be transformed to heat,and less to gas production as the pressure builds within the storage medium-->energy is conserved.\

Have you ever travelled to the mountains?

We dont have mountains here-->are you not happy with YOUR garden path?

How you ever came about the idea that by increasing the size of the storage vessel ,and inturn decreasing the gas pressure uses or depletes the stored energy of the gas-- >is way beyond the garden path Mark.
Remove the pneumatic ram from the equasion,so as we just have our storage tank. We produce the gas until such time as that pressure in the tank reaches 120psi gage pressure. Energy has been conserved-->the amount of energy depleted from the battery is the exact amount of heat energy disipated by the circuit as a whole,and stored within the gas-->we have NOT destroyed any energy.
Now that all the energy has been accounted for,your saying that !some how! we are going to loose some of the stored energy by increasing the tank size,and decreasing the pressure of the stored gas. You MarkE will be the first to destroy energy.

You say that this is basic math,and yet you have screw'd it up-->is google not working right for you ATM?.

[MarkE]

  At a lower ambient pressure it takes less work to do the "boiling" part that than it does at high ambient pressure.  You are unfortunately the one doing the funny accounting.Less gas is produced.  An identically reduced amount of heat is evolved during the subsequent combustion.  The adiabatic heating increases which further decreases the gas generation rate in absolute molecules / Joule.  Again, if you do not believe me, do the math or do an experiment.This is again wrong.  You have drawn your boundaries incorrectly resulting in incorrect accounting and conclusions.  This is first year college chemistry.
I made no such assumption-->once again this is yet another faulse claim you make. The loss in gas production per joule due to pressure increase is made up for in the way of heat output energy by the system.
The energy that went into the phase change from liquid to gas will be the same as the energy returned when that phase change is reversed from gas to liquid-->unless you have found a way to destroy energy?.

Of course not, but your invalid claims are tantamount to being able to both create and destroy energy:  Simply change the ambient pressure.  And that is of course absurd.

That is correct,and that increase in work results in a higher heat output,and a lower number of gas molecules-->once again,energy is conserved.

Energy is completely conserved all the way through.  There is no surplus for free work from operating the piston, just as no deficit could be created by drawing a vacuum.

Now i think you are taking the trip up the garden path. We are not boiling the water by electrolyzing it.. We are dissociating the water into H2 and O2. I have no idea as to how you think this is steam-->which is the result of boiling water.

"Boiling" note the quotes refers to the liquid to gas phase change.  That change requires less energy into a lower pressure atmosphere than into a higher pressure atmosphere.  That difference leaves more or less energy to break chemical bonds electrolyzing the water.  Higher pressure atmosphere => lower number of molecules of H₂ and O₂ evolved.  If you don't believe me then build your apparatus and attempt to wow the world.

That is right,the energy stored within the gas is the exact amount that was drawn from the battery to create it-->the rest of the energy supplied by the battery during this process is transformed into heat..

There you go with the screwy accounting again.  The enrgy from the battery goes into:  chemical bond splitting, and the phase change.  That phase change results in stored energy in latent heat of vaporization and the pressure * volume product.  When you remove some of that work from the system by way of the moving piston, it has left the building, Elvis is on stage no more and the amount of energy that you can recover from what remains is now diminished.  The accounts remain balanced, and your scheme has failed to produce the excess energy you claim.

So please explain how reducing the pressure of that stored gas by enlarging the storage device changes any of the energy accounting after the gas has been produced.

It is not ex facto.  It is intrinsic to the phase change.

Once again,another incorrect statement. To pump down the cylinder requires work to be done,and thus energy is required to do so. You are now transforming energy,not destrying it.

You are simply and utterly wrong.  Take any volume of gas, and reduce the pressure, IE remove gas, and the remaining energy in the volume has been reduced:  Ideal Gas Law, Charles Law, and Boyle's Law all apply.

And again-another faulse accusation. I never said the same number of gas molecules are produced per joule of input energy as the storage pressure go's up. I clearly stated that more energy from the battery would be transformed to heat,and less to gas production as the pressure builds within the storage medium-->energy is conserved.\

Yes energy is conserved and no surplus becomes available because of the extra pressure and no extra can be extracted by operating the cylinder.  Only by applying funny accounting do you challenge those facts.

We dont have mountains here-->are you not happy with YOUR garden path?

I am not joining you on your stroll down yours.

How you ever came about the idea that by increasing the size of the storage vessel ,and inturn decreasing the gas pressure uses or depletes the stored energy of the gas-- >is way beyond the garden path Mark.

Actually do the math or do the experiment and maybe you will see past your false assumptions.

Remove the pneumatic ram from the equasion,so as we just have our storage tank. We produce the gas until such time as that pressure in the tank reaches 120psi gage pressure. Energy has been conserved-->the amount of energy depleted from the battery is the exact amount of heat energy disipated by the circuit as a whole,and stored within the gas-->we have NOT destroyed any energy.

Now you're thinking.  And for those exact reasons your proposed scheme fails.

Now that all the energy has been accounted for,your saying that !some how! we are going to loose some of the stored energy by increasing the tank size,and decreasing the pressure of the stored gas. You MarkE will be the first to destroy energy.

Oops, now you've jumped right on your garden path again.

You say that this is basic math,and yet you have screw'd it up-->is google not working right for you ATM?.

Obviously your hand calculator is not working for you.  Neither is basic logic.