How Do You Determine The Number Of Moles In A Volume Of Gas?

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  • #1
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Summary:
gas volume measure
Sadly, I don't even know what I don't know. Could somebody generate a ballpark figure of how many moles of hydrogen you might have in a sphere 2 meters across, 75F and roughly 16 psi?

Even with all that information I'm clueless. (I'll now resume twig-fishing for termites...)
 

Answers and Replies

  • #2
ChemAir
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What have you calculated so far?

You need to get clarification on the pressure, 16 psi (absolute or gauge). Calculate the volume of your sphere and convert the temperature to absolute and you should be good.

Keep your units straight.

n=(PV)/(RT)
 
  • #3
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I seem to be dropping a stitch somewhere as my answer doesn't make any sense.
Volume 35.5 meters times 16psi divided by 297K doesn't seem to provide a large enough value...
 
  • #4
russ_watters
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I seem to be dropping a stitch somewhere as my answer doesn't make any sense.
Volume 35.5 meters times 16psi divided by 297K doesn't seem to provide a large enough value...
Please show your actual math, because those numbers have inconsistent units and don't match the equation...

Your volume is wrong too.
 
  • #5
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My actual math was me asking Google what is the volume of a two meter sphere.

I multiplied that by the pressure units, then divided by the radius times the temp.

I was going to ask about the pressure unit to use, but didn't. Pascals maybe?
 
  • #6
russ_watters
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My actual math was me asking Google what is the volume of a two meter sphere.

I multiplied that by the pressure units, then divided by the radius times the temp.

I was going to ask about the pressure unit to use, but didn't. Pascals maybe?
[sigh] Please put more effort into this. This is a teaching forum and we don't just hand out answers, but if you just put a little bit of effort into it, you'll find what you are trying to do is easy. @ChemAir gave you the equation you need. Notice any variables that you didn't include in your math....?

You can use google, but you need to actually look at what it is giving you and make sure it understood your question correctly. I just did the same google and got the same wrong answer. But it was easy to see why the answer was wrong/how it misinterpreted the question.
 
  • #7
ChemAir
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Different units of the gas constant (R, in PV=nRT):

8.314 m3Pa/mol K
0.08314 liter bar/mol K
0.08206 liter atm/mol K
62.36 liter mmHg/mol K
0.7302 ft3 atm/lbmol Rankine
10.73 ft3 psia/lbmol Rankine

K=Kelvin
 
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  • #8
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Not trying to be trying.

If I may explain, I'm 61 years old and recovering from my second stroke.

I was asked to try and do more logic puzzles of a mathematical nature and I came up with this exercise after hearing about a large balloon of hydrogen being used as a makeshift firework. I was trying to be as neutral as I could pursuant to the dangerous activities rules of this forum, of which that most certainly is, and had found out the joules per mole of detonating hydrogen, but not the amount of moles in that volume of gas.

I'm unfortunately familiar with large energy releases and could relate if the number was anywhere near 2.5 megajoules, the typical energy released by an exploding main wheel on a F-14 Tomcat. At 500 feet it feels like a punch in the chest, at 40 feet that much energy will flatten 50 men and radiate them out like trees at Tunguska.

Been there, done that. Gives you pulmonary barotrauma if your looking at the source with your mouth hanging open.
 
  • #9
BillTre
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hearing about a large balloon of hydrogen being used as a makeshift firework.

I had a science teacher in Jr. High who filled a balloon with hydrogen (from a water dialysis experiment) tie a string on it, let go up to the ceiling and lite the string. Boom!
Reversing the process.
He was a lot of fun, but would probably get in trouble today.
 
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  • #10
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High school chemistry teacher was Mr. O'Conner. Was in the last damage control team to leave the USS Hornet before she went down.

He'd bubble the hydrogen side through a soap and glycerin solution and then call for volunteers to scoop up the suds in their hand and hold it to a Bunsen burner flame. Aww good times.

I always found the brisance notable.

Two friends and I were lab assistances and once we cleaned up the lab after the "allotropes of sulfur" lesson. And its role as an oxidizer. So after heating it into various goos we got to mix powdered sulfur with ground rubber in small batches and ignite them. So when we cleaned up my friends and I had a pile of left over material from all the work stations about four inches high and we asked Mr. O'Conner if we could light it.

Yeah, set the ceiling tiles on fire. Mr. O'Conner's boss, who taught in the mornings, didn't notice for three days.

Seems like it directed more heat upwards than when we were permitted to ignite the left over iron oxide and powered aluminum experiments earlier. (Wasn't nearly as much either.)
 
  • #11
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This did not work out.

I'm going to have to call this a fail on my part.
 
  • #12
jim mcnamara
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Let's work on the V part. I want you to see how to get that value and then do it. Fill in the red parts with the correct number.

Your sphere is 2 meters in diameter. You need radius (Radius is one half of diameter). Why radius?
Volume of a sphere is ##4/3 * \pi * R^3 ##
We will call radius R. (you need to change that to a number). We'll use 1.33 as four thirds. You see why?

So ##V = 1.33 * \pi * R * R * R## (R cubed is ##R*R*R ##)
What answer do you get for V ?

If you get this far, we will take some steps further. We help with homework here a lot, so we expect student effort. You aren't a student and it seems to me your background may not be what you need to meet all of other folks expectations. So I'm taking the long way. Hopefully we got a value for V, which has units meters3 Units are specifications to tell the reader what the number represents -- like meters, degrees, minutes, seconds, etc, and special units like Hertz, or Watts, or Amperes.)
 
  • #13
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Just to let you know, I found this a few minutes ago and am working on this.

Thank you Mr. McNamara.
 
  • #14
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So if the radius is one meter and the cube of one is still one, then wouldn't the answer be 1.33 divided by pi times one or .422?

That looks very wrong to me.
 
  • #15
hutchphd
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It is ##(4/3)\pi R^3##. So yes it would be wrong. Times pi! (not divided by pi )
 
  • #16
jim mcnamara
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So, since R ==1, then 1 * 1 * 1 = 1. So.

V = 1.33 * 3.14 * 1. Right?

Break out the calculator.... and give us your V.
Helpful hint:
Windows has a builtin calculator, kind of a "windows-ized" hand calculator on a screen. Type 'calculator' in the search box at the very bottom of the screen.

Approximately 4.18 cubic meters Note - added "cubic meters" which is the unit.
 
  • #17
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So I got 4.17

(As an aside, I got off the phone with my team earlier this evening and my issues are most likely pharmacological relative to a new antiseizure med regime and I'm going to taper off and quit one. This should also help with the exuberance of speech that's been creeping into my posts.)
 
  • #19
Borek
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So I got 4.17

Listing units won't hurt (while we can guess from the context you mean m3 it never hurts to include them in the answer).

I got 4.19 m3 so we are in the same ballpark.
 
  • #20
256bits
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So I got 4.17
So now you have the cubic metres of volume. But I get 4.19 cubic metres also.

And if you remember your chemistry, at STP ( standard temperature and pressure - 1 atm , 273K or 0 degrees celcius ) , one mole of a gas occupies 22. 4 litres, that is true for any gas.
Also 1 m3 is the same volume as 1000 litres.

4.19 m3 = 4.19 m3 x 1000 l / m3 = 4190 litres.

and,
4190 litres = 4190 litres x 1 mole / 22.4 l = 187.1 moles of gas <-- at 0C temperature and 1 atm pressure
as a roundabout guestimate for your conditions.

You have to make adjustments for your temperature and pressure from STP to obtain a more exact answer.
Using P1/(T1n1) = P2/(T1n2) , solve for n2.
 

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