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physics/thermodynamics question
Not a trick question -
What's the ratio of air mass between a fixed volume of air at STP and an identical fixed volume of air pressurized to 2 bar gauge, stabilized to the same ambient temperature? I figure there's 3 times as much air by mass in the pressurized container based on the ideal gas law. What's the proper way to go about this? Thanks, Sixto 87 300D |
#2
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(p1 v1)/t1=(p2 v2)/t2
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Yeah, that's what I was thinking too
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So wouldn't it be twice as much mass in the pressurized container? Assuming 1 bar is standard pressure. As I remember PV = nRT.
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2004 VW Jetta TDI (manual) Past MB's: '96 E300D, '83 240D, '82 300D, '87 300D, '87 420SEL |
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Those are variations of the ideal gas law. Can ideal gas laws apply to typical air at STP and STP + 2 bar?
Volume doesn't change, temperature doesn't chage (after equalizing) and the ideal gas constant doesn't change (if it even applies) so we're down to P1/n1 = P2/n2. P1 = 1 bar absolute, P2 = 3 bar absolute. If P2/P1 = 3, n2/n1 = 3. That's how I came up with 3:1 air mass ratio. But I don't know if all the stuff before getting to the P/n ratios is valid for a non ideal gas. Nor if a mole is directly proportional to mass in this situation. Sixto 87 300D |
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Quote:
The 'n' is the sum of the number of moles [6.02 * 10E24 molecules] for each different kind of molecule in the gas mixture [O2 + N2 + Ar + H2O + CO2 +....] |
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In your OP, you said 2 bar, not 3 bar.
I think air is reasonably close to being an ideal gas. Are you trying to get an exact answer or approximate? Mole is indeed directly proportional to mass.
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2004 VW Jetta TDI (manual) Past MB's: '96 E300D, '83 240D, '82 300D, '87 300D, '87 420SEL |
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Oh wait, never mind, now I see you meant 1 bar + 2 bar. So yes, there would be 3 times as much mass in the pressurized container.
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2004 VW Jetta TDI (manual) Past MB's: '96 E300D, '83 240D, '82 300D, '87 300D, '87 420SEL |
#9
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Quote:
In your original statement you were comparing STP and at 2 bar. Is that not 2X, not STP +2?
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1982 300SD " Wotan" ..On the road as of Jan 8, 2007 with Historic Tags |
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My assumption is 2 bar on a gauge is 3 bar absolute.
The reason I ask is the prevalence of "nitrogen filled" tires. How much of the air in the tire isn't nitrogen by virtue of the tire mounting process? My guess is the tire is full of regular air even if they use nitrogen to seat the beads. Then inflating to a typical 30 psi means the tire contains 2 parts by weight pure nitrogen for each part of regular air. So what they claim is a nitrogen fill is at best 89% nitrogen (66% pure nitrogen + 70% nitrogen in the third by mass that's regular air). Is there a better way to calculate this? Sixto 87 300D |
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Quote:
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The green caps mean "another sucker rolls by."
When they say 89% nitrogen, I assume that they mean that they have a supply of 89% nitrogen, rather than the 80% nitrogen supply that I have in my garage. There are "nitrogen generators" that are used. These tire shops are not buying cylinders of pure N2. |
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There is way tocalculate it because of mixing at the rim. All they really care about is getting rid of water vapor, the amount of N2 is not really what they seek.
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Quote:
http://en.wikipedia.org/wiki/Boyle's_law Danny
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Quote:
B |
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