Baking Soda and Vinegar?

A place for general potato gun questions and discussions.
SpudBlaster15
First Sergeant 3
First Sergeant 3
Seychelles
Posts: 2400
Joined: Wed Oct 18, 2006 11:12 pm
Has thanked: 2 times
Been thanked: 10 times

Fri Jun 10, 2011 3:27 pm

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.
User avatar
saefroch
Staff Sergeant 2
Staff Sergeant 2
Posts: 1679
Joined: Mon Jun 08, 2009 8:47 am
Location: U.S.A.- See Map

Fri Jun 10, 2011 3:30 pm

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.
tbillion
Private
Private
Posts: 7
Joined: Fri Jun 10, 2011 10:23 am

Fri Jun 10, 2011 3:39 pm

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.
tbillion
Private
Private
Posts: 7
Joined: Fri Jun 10, 2011 10:23 am

Fri Jun 10, 2011 3:42 pm

also found nothing when i searched tech's diagram.. care to shed more light on that?
User avatar
Technician1002
Captain
Captain
Posts: 5189
Joined: Sat Apr 04, 2009 11:10 am

Fri Jun 10, 2011 5:40 pm

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.
User avatar
inonickname
First Sergeant 4
First Sergeant 4
Posts: 2606
Joined: Sun Dec 07, 2008 3:27 am

Fri Jun 10, 2011 8:39 pm

SpudBlaster15 wrote:
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.
Ino is pretty terrible at chemistry ( :wink: ), 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. :cry:
You bastard :P the 1.36 came from using 20 instead of 19.7. Or I guess he might have a gauge that reads to 0.1 PSI? :roll:
PimpAssasinG wrote:no im strong but you are a fat gay mother sucker that gets raped by black man for fun
tbillion
Private
Private
Posts: 7
Joined: Fri Jun 10, 2011 10:23 am

Fri Jun 17, 2011 9:47 pm

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/
Post Reply