Baking Soda and Vinegar?
-
- First Sergeant 3
- Posts: 2400
- Joined: Wed Oct 18, 2006 11:12 pm
- Has thanked: 2 times
- Been thanked: 10 times
Lorem ipsum dolor sit amet, consectetur adipiscing elit. Cras nec placerat erat. Vivamus dapibus egestas nunc, at eleifend neque. Suspendisse potenti. Sed dictum lacus eu nisl pretium vehicula. Ut faucibus hendrerit nisi. Integer ultricies orci eu ultrices malesuada. Fusce id mauris risus. Suspendisse finibus ligula et nisl rutrum efficitur. Vestibulum posuere erat pellentesque ornare venenatis. Integer commodo fermentum tortor in pharetra. Proin scelerisque consectetur posuere. Vestibulum molestie augue ac nibh feugiat scelerisque. Sed aliquet a nunc in mattis.
Last edited by SpudBlaster15 on Thu Jul 15, 2021 12:17 am, edited 3 times in total.
Atmospheric pressure in psi is 14.7
Almost no appreciable amount of CO<sub>2</sub> will be dissolved in the water... and the use of a dessicant would just take up even more volume in the chamber... that's before you consider the fact that vinegar is only 5% acetic acid, so you have a whole heck of a lot of water to absorb.
If using sulfuric acid and a metal (as it is now called, tech's diagram is fairly ancient), the gas produced will be hydrogen, which really doesn't appreciably react with water.
Concentrated acids are very dangerous, I highly recommend against using them.
Almost no appreciable amount of CO<sub>2</sub> will be dissolved in the water... and the use of a dessicant would just take up even more volume in the chamber... that's before you consider the fact that vinegar is only 5% acetic acid, so you have a whole heck of a lot of water to absorb.
If using sulfuric acid and a metal (as it is now called, tech's diagram is fairly ancient), the gas produced will be hydrogen, which really doesn't appreciably react with water.
Concentrated acids are very dangerous, I highly recommend against using them.
5 years old but still full of useful information that is new to me!, thanks for the thorough explanation!
I have been using high concentrated acids for years, for different purposes. for my purpose now, if I use a high concentrated acid, and only need to create a minimal amount of gas to inflate something so that it will not collapse upon itself when outside pressure is applied then I am interested. In this solution I need to be super mindful of the weight of the reactants due to the fact that the more they weigh will reduce my over all buoyancy.
I have been using high concentrated acids for years, for different purposes. for my purpose now, if I use a high concentrated acid, and only need to create a minimal amount of gas to inflate something so that it will not collapse upon itself when outside pressure is applied then I am interested. In this solution I need to be super mindful of the weight of the reactants due to the fact that the more they weigh will reduce my over all buoyancy.
- Technician1002
- Captain
- Posts: 5189
- Joined: Sat Apr 04, 2009 11:10 am
I just googled the image of the extinguisher. Most of the time the manufacture did not provide any refill instructions as you had to use a fire extinguisher service company to maintain certification. I don't know the acid strength or quantity of bicarb in the water. Sorry I don't have any more details on those antiques.
- inonickname
- First Sergeant 4
- Posts: 2606
- Joined: Sun Dec 07, 2008 3:27 am
You bastardSpudBlaster15 wrote:Ino is pretty terrible at chemistry (tbillion wrote:1.36 is the number of atmospheres in the jug? what is an atmosphere 4.7 is atmospheric pressure? increasing the volume to place in the jug by 5 i get because i want 5 psi. where did the 1.36 and the 4.7 come from is where i am lost at this point. gases is new science to me.), so I'll explain this a bit more thoroughly.
One atmosphere is equivalent to 14.7PSI absolute pressure. You are looking to raise the internal pressure in the jug to 5PSI gauge pressure, which is the differential between the interior of the container and the atmosphere. This can also be expressed as 19.7PSI absolute pressure.
Dividing the absolute pressure after the addition by the initial (atmospheric) pressure will give you the ratio of the final pressure to the starting pressure, which is actually 1.34, not 1.36 as was stated above.
Finally, multiplying this ratio (1.34) by the starting volume (5L) and subtracting the initial volume gives you the final atmospheric pressure gas volume equivalent which must be added to the vessel in order to raise the pressure by 0.34atm (5PSI).
V<sub>addition</sub> = (1.34*5L) - 5L
V<sub>addition</sub> = 1.7L
Using Boyle's law is much simpler, but the above approach is more intuitive.
EDIT: Also, way to kick up a 5 year old thread, for a second I thought BLB had actually returned to the forum.


PimpAssasinG wrote:no im strong but you are a fat gay mother sucker that gets raped by black man for fun
well thanks to all that helped when i first read this forum I didn't even know that a gas law was what I was looking for, thank you for putting up with my questions I really appreciate that!
If you would like to see how i have applied this information you can check my temporary website @ http://travisgillespie.netau.net/
If you would like to see how i have applied this information you can check my temporary website @ http://travisgillespie.netau.net/