Open Systems

[tinman]

The change in internal energy depends on the degrees of freedom of the molecules and the change in volume.  How fast you change the volume determines the power.If you use a small tank as your reservoir raising the pressure in a larger vessel then you will lose a greater percentage of energy with each operation than if you use a large tank as the high pressure reservoir.  Remember that when you calculate energy you need to use absolute temperature and pressure values.

What im trying to ask is-if we take 2 vessels with a volume of say 5 ltr's each,and fill one of those vessels to a pressure of say 40psi,and we then open the valve so as gas flows into the empty vessel,then in an ideal situation where there is no losses,we would have 20psi in each vessel. So we have halved the pressure but doubled the volume. Dose this then mean that we still have the same amount of energy in those two vessels combined as we did at the start with the 40psi in just the one vessel?. If so,then to obtain a higher energy level than when we started,i would just have to raise the end pressure in the combined vessels by say 1psi to 21psi each. Would this then be proof enough that by making the gas do work can indeed raise the energy level of that gas..?.

[MarkE]

If you have two vessels each of some volume V and you pressurize one to 1 ATM and the other to 1 ATM + 40 psi with N₂ and O₂ then the math works as such:

E_START = K₁*V * (40+15+15) = 70*K₁*V

If the energy were distributed across both vessels evenly, then each would have 35*K₁*V which is the 20psi you predict.

But since:  P*V^1.4 = K  this does not happen and you will have less than 20psi in each vessel.

To more directly answer your question:  Consider that if equalizing pressure between two vessels left us with more energy than we had at the start.   Then in order to operate pneumatic machinery we could just keep coupling larger and larger storage tanks to each other and then use the energy stored at reduced pressure and very high volume to do our work and recompress a new seed quantity of gas.  That of course does not work.  We'll see what happens in your experiment.

[tinman]

If you have two vessels each of some volume V and you pressurize one to 1 ATM and the other to 1 ATM + 40 psi with N₂ and O₂ then the math works as such:

E_START = K₁*V * (40+15+15) = 70*K₁*V

If the energy were distributed across both vessels evenly, then each would have 35*K₁*V which is the 20psi you predict.

But since:  P*V^1.4 = K  this does not happen and you will have less than 20psi in each vessel.

To more directly answer your question:  Consider that if equalizing pressure between two vessels left us with more energy than we had at the start.   Then in order to operate pneumatic machinery we could just keep coupling larger and larger storage tanks to each other and then use the energy stored at reduced pressure and very high volume to do our work and recompress a new seed quantity of gas.

Exactly

Now,whats with the and you pressurize one to 1 ATM
There is no !pressurizing to 1 ATM,as it is already at 1 ATM. As 1 ATM in the tank is equal to the pressure of the outer enviroment,then there is a 0 potential difference,and no work can be done.  Our gauges will read 0,both differential nad gauge pressure. So why say !!pressurize to 1 ATM? ,when you dont have any pressure at 1 ATM. If we have a vessel pressurized at 1 ATM,dose that vessel have any stored energy in it?-->if so,then our whole atmostphere must be 1 big storage of the same energy/

[MarkE]

Exactly

Now,whats with the and you pressurize one to 1 ATM
There is no !pressurizing to 1 ATM,as it is already at 1 ATM. As 1 ATM in the tank is equal to the pressure of the outer enviroment,then there is a 0 potential difference,and no work can be done.

The internal energy in the gas is based on absolute pressure.  You will reduce errors by always going back to absolute pressure and temperature.

Our gauges will read 0,both differential nad gauge pressure. So why say !!pressurize to 1 ATM? ,when you dont have any pressure at 1 ATM. If we have a vessel pressurized at 1 ATM,dose that vessel have any stored energy in it?-->if so,then our whole atmostphere must be 1 big storage of the same energy/

Yes the atmosphere stores a tremendous amount of energy.

[tinman]

Ok,i have just messured the two tank's,and i was out by double lol.
The small tank is exactly 10 ltr's,and the big tank is exactly 20 ltr's.

There is a reason i will be using the small tank for the pressurized tank,and that will become apparent later on in the testing. The two tanks will be joined via a ball valve that will be opened to let the pressurized gas drain into the large tank. Each tank will have it's own temperature gauge,but only the small tank will need a pressure gauge-as the pressure at the end of each test will be the same in both tanks.

So a quick bit of calculating for you Mark,if you could please.
Our small tank(10LTR) will have a pressure of 40psi gauge pressure at the start of the test,and the large tank will have 1 ATM. The temperature will be close to 25*C-avaerage ambiant temperature here at the moment-->will give exact temperature on day of test. But going on these measurement's,can you calculate what we should have in the way of temperature in each tank,and the pressure in both tanks?.