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Topic: C02 pressure spikes  (Read 23272 times)

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C02 pressure spikes
« on: September 18, 2004, 06:42:54 PM »
Hello! I actually want to ask a question for a little tid bit of research I want to do. I am a writer for a paintball magazine, and I wish to do an article on the diference between Compressed air and C02. I have taken only a college introductory level of chemistry and an honors level high school chemistry course. I was wondering if anyone could help me find the best way to visually show the pressure, and velocity spikes that would occur due to the C02 being compressed and over several diferent temperatures.
  I have some information(like pressure at a certain temperature, size of the container and all that), and I was going to use the ideal gas constant equation, but I realized that that wouldn't show the inefficiency I wish to show.
  Or if anyone else has any ideas I'd be glad to listen to them. Also, if I do use the help I get from here, I will be sure to mention my thanks in the article itself :) .
  I also know many things already, so don't feel like you have to give me other things, I'm just wondering about showing the velocity spikes graphically(just so noone gets confused or anything like that). Thankyou!

                 Raven  8)

Offline billnotgatez

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Re:C02 pressure spikes
« Reply #1 on: September 19, 2004, 11:20:59 AM »
Raven - When you publish your article please post where we can read it here. It seems very interesting.

 It seems to me (and I could be wrong) that for the pressures and temperatures you will be operating at - you can use the universal gas equation / ideal gas law and not need Van der Waals equation. You might want to consider discussing the triple point of Carbon Dioxide (it sublimes at standard temperature and pressure). Also remember that air is the combination of the partial pressures of all its constituents. Since I am only interested in atmospheric Carbon Dioxide, I can not be of much more help unless I study it more. You might try
and search Ideal Gas Laws.

By the way what do you mean by velocity spikes?

This is something I ran across that I use
 when dealing with atmospheric stuff

Notes to the administrators – when you moved this item from where it was originally posted I think that a more appropriate place could have been selected. For instance, you might have chosen physical chemistry or general chemistry. Then someone with expertise would see this question.


« Last Edit: September 19, 2004, 11:28:05 AM by billnotgatez »

Offline Donaldson Tan

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Re:C02 pressure spikes
« Reply #2 on: September 19, 2004, 12:28:39 PM »
I move the topic here.

Isn't this a physics problem?
"Say you're in a [chemical] plant and there's a snake on the floor. What are you going to do? Call a consultant? Get a meeting together to talk about which color is the snake? Employees should do one thing: walk over there and you step on the friggin� snake." - Jean-Pierre Garnier, CEO of Glaxosmithkline, June 2006


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Re:C02 pressure spikes
« Reply #3 on: September 19, 2004, 01:54:47 PM »
I thought it was an analysis thing. I was using the ideal gas law, I graphed it, but I havn't tried the Compressed air in the equation yet. Well, heres the problem with C02 being used as a compressed gas.

  The problem is that it goes into its liquid form too readily. When you try to pull it out at a regular pressure, you pull out a lot of liquid too, which causes velocity spikes, and if your using regulators, you freeze them(which is not good at all). This is why C02 requires a few things, like anti-siphon tubes, and expansion chambers(a chamber system used between the ASA and the marker to allow the C02 expand) and most times you still get liquid C02, which can cause problems on a paintball marker. Compressed air however, is almost completely a gas when confined(its three point is spread out, so it takes a LOT of pressure to get it as a liquid), which means you can pull out of it an exact pressure, without liquid in there to throw off the pressure(since liquid is a diferent weight/size than gas).

  Also, with changes in temperature during the day, C02 can be thrown around its stages quite frequently, this means that the settings get thrown off a lot as well, and that you then get oddjumps in pressure. I was just hoping that I could somehow show this more graphically... not that I need to, but if I can, I want to try... and because I actually like doing this for some wierd wierd reason(next year I hope to go to college for this type of thing). Any ideas? I only did C02 using the ideal gas law(I had pressure and volume at a certain temperature, so I used PV=nRT and found n, which I could then change the temperature and leave Pressure as a variable), and the increase seemed quite steady, but I havn't done CA yet.

  I did C02 over 10 degrees(from 78 to 89 degrees fahrenheight, which needed converting to Kelvin). I'll try CA in a little. And I have to go take my car somewhere, but I'll check some of my books for van der waals, I remebmer his principles and forces, but the equation is blowing my mind. Thanks again!

                 Raven 8)


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Re:C02 pressure spikes
« Reply #4 on: September 19, 2004, 03:21:41 PM »
I believe it is the wrong track trying to explain this through the ideal gas equation (although ideal graph can be used as a reference). If air and CO2 were ideal, they would behave exactly the same, which obviously isn't the case. Van der Waals equation is more appropriate.

This article is pretty good in applying Van der Waals to different gases in another apllication, illustrating differences. The examples illustrate pressure/vol but pressure/temp can also be drawn. It also is handy in providing the a, b constants for different gases, including air and CO2 (a is much larger for CO2 than the other gases).
« Last Edit: September 19, 2004, 04:08:01 PM by Demotivator »


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Re:C02 pressure spikes
« Reply #5 on: September 19, 2004, 04:25:57 PM »
I thought so... hmm... but my problem is that van der walls' would be rather complicated for the mixture I am dealing with which is 78 percent nitrogen and 21 percent oxygen. It works well for the C02, but I don't know if I could pull it off with the mixture... any other ideas? Should I even try to represent it graphically?

                Raven  8)


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Re:C02 pressure spikes
« Reply #6 on: September 19, 2004, 06:36:41 PM »
But your 78 percent nitrogen and 21 percent oxygen is just  air. The a and b constants have already been calculated for the mixture as the composite, air,  by the article (in the second table).
          a                     b
Air   1.372516   0.037209


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Re:C02 pressure spikes
« Reply #7 on: September 20, 2004, 12:43:58 AM »
Hmm... I have C02 at a certain pressure, temperature, and size of the tank... I determined how many moles there were using PV=nRT, but I tried to use that and it didn't work in vander waal's equation( at a higher temperature I got 200 psi... when at the lower tmperature it was 850 psi). So, I thought, I was dumb and hopeful to try to use that same mole value, and I was going to use van der waals and keep n as a variable, but now its a really big equation. Anyone know an easier way to find this, or should I just try to push through it to find the mole value?

                 Raven  8)
P.S. I'd try harder right now, and probably find the answer, but its getting rather late, so I gotta call it quits for tonight, but tomorrow I'll start back in on it after school and check on this topic during school to see if I can get anywhere.


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Re:C02 pressure spikes
« Reply #8 on: September 20, 2004, 08:17:57 AM »
If I use .08206 the pressure is in atm isn't it?

                 Raven  8)


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Re:C02 pressure spikes
« Reply #9 on: September 20, 2004, 11:00:52 AM »
Yes, if R = .08206 the pressure is in atm.

Ok, tutorial time  :rolleyes2:
As far as the problem you're having, you must be calculating incorrectly. I don't know your value for n from PV=nRT  but I'll assume it's 2 moles.
Then in van der waal's equation:
(P + an2 / V2)( V - nb ) = nRT
for CO2, a = 3.59   b= .0427
assume n =2, V = 0.5 liter
for T = 300 K:

(P + 3.59(22)/0.52)(0.5 -2x .0427) = 2(.082)(300)
(P + 57.44)(.4146) = 49.2
P + 57.44 = 49.2/.4146 = 118.7
P = 118.7 - 57.44= 61.3 atm

Now try T= 200 K:
(P + 3.59(22)/0.52)(0.5 -2x .0427) = 2(.082)(200)
(P + 57.44)(.4146) = 32.8
P + 57.44 = 32.8/.4146 = 79.1
P = 79.1 - 57.44= 21.7 atm

So the pressure goes down as the temp goes down as expected.

If you want to keep n variable and solve for it in Van der waal equation then it is true that it looks complicated and there is no straightforward arithmetic way to solve for n. But it can be solved by the technique of iteration, meaning simple guess and refine. You make an initial guess at n and see how far the equation is off from being true, then make a better guess, and so on until the equation is close enough to true.

For example,
n + 3 = 7
guess n = 2 , then 5= 7
guess n = 3 , 5 = 6 (that's better)
guess n = 3.5,  6.5 = 7 now it's close
guess n = 4,  7 = 7 that's the answer


  • Guest
Re:C02 pressure spikes
« Reply #10 on: September 20, 2004, 11:14:40 AM »
my equation works out this way:
n= 1.41 (derived from the PV=nRT equation)
V= .59(20 fl. ounces)
R= .08206
T= 304.26
a= 3.59
b= .0427
p=16.05 atm
16.05 atm*760= forgot the number torr= 236 psi

this was for 88 degree's fahrenheight, and at 70 degrees the psi was 850. I then tried to leave n as a variable, since I knew that at 70 degrees(from actual date not my own, given to me by my editor) a 20 oz. C02 bottle has a pressure of 850 degrees. However, like you said, I couldn't find an arithmetic value for n. I bet my calculations are off, I keep doing the calculations in between class, but I only get diferences of .001 in my n value. I don't know why I keep screwing up, but when I get home I will try again fresh and new(I'll use the PV=nRT equation to find the n, then I'll go and use it in van der waals equation and hope that works out). Any thoughts on what I did wrong? I bet I did something very simple and stupid,lol, god I hate mondays,lol. Thankyou for your patience.

                          Raven  8)


  • Guest
Re:C02 pressure spikes
« Reply #11 on: September 20, 2004, 01:27:38 PM »
The equation is
[P + (an*n/V*V)](V-nb) etc
[(P + an*n)/V*V](V-nb) etc

(P+7.14)/(.3481)=66.91 is wrong
should be
P + (7.14/.3481) = 66.91
P + 20.51 = 66.91
P = 46.4 atm

Also, the n=1.41 you derived from pv=nrt would be off because CO2 is not very ideal.
A quick calculation shows me that n is closer to 1.7 from van der waal.

I found this graph which dramatically shows how CO2 (because of its compressibility) is different from ideal gas or nitrogen, etc

« Last Edit: September 20, 2004, 02:40:13 PM by Demotivator »


  • Guest
Re:C02 pressure spikes
« Reply #12 on: September 20, 2004, 11:51:34 PM »
all times

I end up with a final pressure of 52.22 atm which equals 767.42 psi. Oh crap, I just substitued 850 psi, and got a temperature of 125 degrees fahrenheight... somethings not right.

           Raven  8)


  • Guest
Re:C02 pressure spikes
« Reply #14 on: September 21, 2004, 07:57:37 AM »
The problem is with copyright now, which is why I was trying to do it myself, and because I actually enjoy doing this stuff, but I geuss I can somehow work this out.

              Raven  8)

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