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Ideal gases and partial volume 
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#1
Apr2913, 07:32 PM

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What does it mean by 21% oxygen and 79% nitrogen by volume?
Because won't the oxygen and nitrogen have the same volume which is the volume of the whole container? n_{T}RT/V_{T}=n_{O2}RT/V_{T}+n_{N2}RT/V_{T} so why would we say oxygen is 21% by volume since the volume of the oxygen and nitrogen is the same which is the volume of the container? I managed to find a link that kinda says that "by volume" represents the (w/v) ratio rather than the (v/v) ratio. Here it is at the bottom (http://forum.onlineconversion.com/sh...ad.php?t=11470). So if it was w/v the w would represent the weight of the gas in g while if the v represent the volume of the entire container? But if the "by volume" is w/v how can I use that ratio to get my mole fraction? Because if I had 10000m3 at 1 atm and 298K with 21% oxygen and 79% nitrogen by volume (w/v), then the mass of oxygen is 2.1x10^{9}g and the mass of nitrogen would be 7.9x10^{9}g. So now if I used the formula mass/mr equals moles, I'd get a different total number if I used 10000000dm3/24dm3. So I'm quite confused about what this "by volume" means and partial volume. Hope you guys can enlighten me here. Thanks :) 


#2
Apr2913, 09:40 PM

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PF Gold
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It is confusing. In practice, the term volume fraction really means mole fraction, and the term volume percent really means mole percent. Why they use the term volume fraction has never made sense to me.
Chet 


#3
Apr3013, 01:01 AM

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AM 


#4
Apr3013, 01:48 AM

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Ideal gases and partial volume
Thanks again :) 


#5
Apr3013, 06:49 AM

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Chet 


#6
Apr3013, 07:10 AM

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#7
Apr3013, 11:28 AM

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The volume occupied by a gas depends on the number of molecules of the gas, as well as the external force/area (pressure) on the gas molecules and the average translational kinetic energy (temperature) of the gas molecules. It does not depend directly on the weight or mass of the molecules. AM 


#8
Apr3013, 11:34 AM

P: 645

But actually I was reading the Amagat's Law for the additive volumes, I don't really understand how VT=V1+V2...+Vn. If I had a mixture of oxygen and nitrogen won't both Vo2 and Vn2 be the same since they are in the same container? So how can there be a partial volume here? 


#9
Apr3013, 01:42 PM

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You can think of each gas a occupying only part of the volume  in proportion to n. You can determine this from the equation of state. For an ideal gas V = nRT/P so if n is the no. of moles of gas, 21% of which are oxygen and 79% of which are nitrogen, then V_{ox} = .21nRT/P and V_{nitr} = .79nRT/P. Or you can think of the different gases providing partial pressures which combine to the total pressure. The total pressure is determined by the force/area caused by the change in momentum per unit time of all the gas molecules due to collisions of gas molecules with the walls of the container. If 21% of the molecules are O_{2} and 79% are N_{2}, then the partial pressures are in the same proportion. AM 


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Apr3013, 10:52 PM

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#11
May113, 12:21 AM

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#12
May113, 01:50 AM

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#13
May113, 10:45 AM

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Hi sgstudent,
You are obsessing and getting bogged down on this. Please don't do this to yourself. Your time is very valuable, and you need to move on to much more important learning. Andrew Mason did a wonderful job of trying to provide motivation for how the term volume fraction for gases came about, but in the great scheme of things, it is not important whether you are comfortable with every last detail of this. All you really need to remember is that those percentages by volume mean mole percentages. The terms volume mixing ratio and volume fraction are synonymous with mole fraction. Chet 


#14
May213, 02:57 AM

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#15
May213, 07:05 AM

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#16
May213, 07:46 AM

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Sorry about that I meant the Amagat's law part of this link http://en.wikipedia.org/wiki/Partial_pressure



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