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Topic: molar percentage oxygen at high temperature  (Read 196 times)

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Offline joincyberspace

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molar percentage oxygen at high temperature
« on: July 26, 2020, 06:17:21 AM »
This is my first post, I read quite a bit of available answers in this forum (and other sources) but I cant find my mistakes nor the answer to the problem. I am not experienced in chemistry, thus I put this under "undergraduate" and hope that someone of you can give me some hints on how this can be solved. Thank you for your time!

I - The main problem/question:

A combustion calculator gives me the quantity of oxygen in a volume (say 1m3) at various temperatures in, "mol frac", starting at 20.49 [mol frac] at normal condition (T=20°C and p0=101300Pa), decreasing to e.g 5.00 [mol frac] at 500°C.

I have difficulties to understand the unit [mol frac] especially in the conext of the change of the density. The user manual didnt help; the only hint I found was  that "assumes that the molecular weight of the gaseous mixture throughout the domain is approximately that of air, 29 g/mol." which doesnt help me at all.

Thus, I started to calculate "something" to see some pattern but created more questions...

II - The problems I met on the way:

Using the following equation (Eq. 1) should help
M = m / n (Eq. 1)

At normal conditions (101300 Pa, 20°C) the air has a density of about 1.2 kg/m3.

Knowing that oxygen as molecule (O2) has a molar mass M=2x16u (u as the atomic mass unit) and
that the share at normal conditions is 23.135 [mass-%] I can derive
the amount of substance n [mol] by using Eq. (1) as following:

n(O2)= (1200g *23.135%)  / 32 g/mol =8.68 mol oxygen per m3 air
the mass m would be
8.68 mol x 32g/mol =277.8g oxygen per m3 air

Thus, in one m3 air, there are 8.68 mol oxygen equivalent to 277.8g oxygen (a check delivered: 289/1200=23.15% of 1.2kg)
Q1: Is this correct, isn't it?

However:
The Gay Lussac's Law of definite proportions says, 1 mol of a gas requires a volume of 22.4 litres.
1m3 has 1000 litres, thus 1000/22.4 =44.64 ["mol-spaces"]
Q2: Is there a better wording for "mol-spaces"?

I will use the vol-% of oxygen, given to 20.942%:
and that 1 mol oxygen has m=32g:
44.64 x 20.942% gives 9.37 ["mol-spaces"] for oxygen
32g/mol x  =299.9 [g] oxgen per m3 air

Q3: Where is this inconsistency coming from; or to be honest: what is wrong in my calculations?

When it comes to the original question, the mol-fraction at a certain temperature I am lost as the density of air changes.
At 500°C I calculate the density of air with the changing air pressure using Eq. (2):

ρ_gas,L = p_0/[287.1∙ (T+273.15)]      (2)

Using Eq. (2) for 500°C, this results in a density of 0.49kg/m3 while at 20°C it would be 1.20 kg/m3.

Thus, for "no change of the oxygen concentration", I would expect still less mol per m3 for the higher temperatures. I expect a reduction to 0.49/1.20=0.40 for all moles as the "mol-density" would decrease

From the upper section I learned that at normal condition, there were 8.68 mol oxygen in one m3, thus, at the higher temperature, that would mean 8.68 mol x 0.40 = 3.54mol - which would be still 21% (as the other fractions would also decrease)
Q4: Is this reduction factor determination reasonable?

Phew, that became quite a long question, I hope it is not too confusing.
Thank you so much for reading, considering and eventually answering to my "stupid" questions.
Jo
« Last Edit: July 26, 2020, 06:28:07 AM by joincyberspace »

Offline MNIO

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Re: molar percentage oxygen at high temperature
« Reply #1 on: July 27, 2020, 03:22:37 PM »
so first of all..
  mole fraction of chemical "A" = moles of "A" / total moles present

examples
  (1) 0.21 moles O2 molecules + 0.79 moles N2 molecules
           total moles = 0.21 + 0.79 = 1.00 moles
           mole fraction O2 = 0.21 / 1.00 = 0.21
  (2) 0.42 moles O2 + 1.58 moles N2
           mole fraction O2 = 0.42 / (0.42 + 1.58) = 0.42 / 2.00 = 0.21
  (3) 0.60 mole O2 + 2.37 mol N2 + 0.03 mol CO2
           mole fraction CO2 = 0.03 / (0.60 + 2.37 + 0.03) = 0.03 / 3.00 = 0.01
           mole fraction O2 = 0.60 / 3.00 = 0.20
           mole fraction N2 = 2.37 / 3.00 = 0.79
           note that 0.01 + 0.20 + 0.79 = 1

*********
Second.  Mole fraction doesn't depend on density (for the most part).  The less dense the gas, the less moles of ALL the gases present.  The ratios stay the same!  So it doesn't make sense to say "mole fraction O2 = 20.49 @ RTP and 5.00 @ 500°C"... unless a reaction is taking place consuming some of the oxygen.  Where did you get that information from?

**********
Third. In part II you're confounding mole fraction with mass fraction.  They are two different critters. let's keep then straight and well defined. 
  mole fraction A = mole A / total moles present
  mass fraction A = mass A / total mass present

if we say the mole fraction of O2 in air is 0.21 and N2 in air is 0.79, then assuming 1.00 mole air
  then mass O2 = 0.21 mol * (32.00g / mol) = 6.72g
         mass N2 = 0.79 mol * (28.02g / mol) = 22.14g
         mass fraction O2 = 6.72g / (6.72g + 22.14g) = 0.23
again
  mole fraction O2 = 0.21
  mass fraction O2 = 0.23

likewise since mole and mass fractions sum to 1
  mole fraction N2 = 1 - 0.21 = 0.79
  mass fraction N2 = 1 - 0.23 = 0.77

********
your Q1
  1 m3 air * (1.2kg air / m3 air) * (0.23kg O2 / kg air) = 0.276kg O2 = 276g O2
that is correct. 
then
  276g O2 * (1 mol / 32.00g) = 8.63 mol O2
close enough... both are correct.

BUT#1.. I don't know where you came up with "Gay-Lussac's law of definite proportions says 1 mol gas = 22.41L"  That isn't correct. 

From the ideal gas law
  PV = nRT
  V/n = RT/P
  V/n = (0.08206 Latm/ molK) * 273.15K / 1.00 atm = 22.41 L/mol
meaning
  the MOLAR VOLUME (V/n) of an IDEAL GAS = 22.41 L/mol @ T = 273.15K and P = 1.00 atm
we call that T = 273.15 and P = 1.00 atm "standard temperature and pressure".. aka.. STP

again
  @ STP, molar volume of an ideal gas = 22.41 L/mol.

BUT#2.. in your example, you have 2 issues
  (1) you don't have STP... you have T = 293.15K not 273.15K
  (2) you're just looking at the O2  you need to consider ALL moles of gas present

from the ideal gas law at T = 293.15K
  V/n = (0.08206 Latm/molK) * 293.15K / 1.00 atm = 24.06 L/mol

moles of gas present
  moles O2 = 276g * (1 mol / 32.00g) = 8.63
  moles N2 = 1m3 air * (1.2kg / m3) * (0.77kg N2/kg air) * (1 mol / 0.028kg) = 33.0
then
  TOTAL moles gas present = 8.63 + 33.0 = 41.63
  volume / mol = (1m3 / 41.63) * (1000L / 1m3) = 24.02 L/mol
  exactly as it should be

*********
Q2
  is there a better name than "mol-spaces"

you have defined mole spaces as
  1000L / (22.42 L/mol) = 44.64 "mol-spaces"

that value is "gas volume / molar volume @ STP" which tells you how many moles of an ideal gas would be present at STP in a volume of 1000L.  Nothing more.

then you're taking
  (mole of all gas @ STP in 1000L) * (mole fraction O2) = 9.37 mol O2
  9.37 mol * 32.00g/mol = 299.9g
that would be ok but..
  (1) we don't have STP
  (2) there's a better way

if you calc'd this correctly you would have
  1000 / 24.02 = 41.63 "mol-spaces"
  41.63 mol spaces * 0.21 = 8.74 mol O2
  8.74 mol O2 * 32.0 g/mol = 280g O2   close enough.

The better way would be this
  1 m3 air      1000L air     1 mol air    0.21 mol O2    32.0g O2
 ------------ x ---------- x ------------ x -------------- x ---------- = 279.9 g O2
         1          1 m3 air    24.02 L air       1 mol air       mol O2

the math is called "dimensional analysis".. actually it's "factor label method" aka "unit factor method" etc.  It's the basis for math in chemistry.  You should spend some time learning it.

***********
Q3  where is the inconsistency coming from?
  (1) you incorrectly used 22.41 L/mol (didn't account for T = 20°C, nor entire gas)
  (2) you're creating more work than you need to.  learn dimensional analysis
       it will save you much trouble!
  (3) You have incorrect data.  the mole fraction of O2 in air @ 500°C ≠ 5.00
        I have no idea where that came from, but you need to revisit it

btw... your equation for density is correct.  How you came up with mole fraction O2 = 5.00 is the mystery.

**********
Q4   is the reduction factor from 21% to 5% O2 in air going from 20°C to 500°C reasonable

     NO!  it's not correct

   

« Last Edit: July 27, 2020, 06:42:53 PM by MNIO »

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