# Amount Of Hydrogen

To address a lot of questions about hydrogen safety or leaks in the cell, let's talk about the amount of hydrogen we are filling in the cell.

The glass tube dimensions are:

- Outer diameter: 40.0 mm

- Inner diameter: 35.0 mm

- Length: 300.0 mm

So, the internal volume of the cell is: π⋅r^{2}⋅L = 3.14159 * 0.175^{2} * 3.0 = 0.2886 liters

Then the mica boards are: 3.2 mm x 11.5 mm x 300.0 mm

So the volume of each mica part is: 0.032 x 0.115 x 3.0 = 0.01104 liters

The total volume of the 3 parts of mica is : 3 x 0.01104 = 0.03312 liters

Finally the stainless steel rod in the center of the cell is: ø6.0 mm x 300.0 mm

The volume of the rod is: π⋅r^{2}⋅L = 3.14159 * 0.03^{2} * 3.0 = 0.0085 liters

Thus the available volume inside the cell is: 0.2886 - 0.03312 - 0.0085 = 0.2470 liter = 2.47⋅10^{-4} m^{3}

The ideal gas law will give us the number of mole of hydrogen in the cell:

PV = nRT => n = PV / RT

When we load the cell with pure hydrogen at ambient temperature (295 °K) at 3.5 bars we have:

n = 3.5⋅10^{5} Pa * 2.47⋅10^{-4} m^{3} / (8.3144621 J·K^{−1}mol^{-1} * 295 °K) = 0.0352 mol

The atomic mass of dihydrogen is: 2.0159 gram / mol

So at maximum, we are filling the cell with: 0.0352 * 2.0159 = **71 milligrams of H2**

## Comments

V=diameter^2*pi/4

I apologize for being a little questionable on this one it's just I had always used this formula and thought it would be appropriate for this one here. I would appreciate being told if I am wrong about this because I am also working with reactor volumes.

Like I say I am just a mechanic so if you guys do things different in science cool but am I way wrong here on this one?

- Does daylight have access to the room, I guess not?

- How quickly was vacuuming/press ure reduction done, in comparison to Celani? If pressure is reduced too fast, maybe hydrogen outgasses violently, perhaps it kills the wire.

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