Calculating Carbon Fixation: Ideal Gas Law for Forests | 25oC & 1 atm

[yang09]

Homework Statement

Each square meter of an actively growing forsest fixes on average each year 1 kg of carbon from the atmosphere into plant matter through photosynthesis. The atmosphere is approximately 20% O₂ and 80% N₂, but contains 0.046% CO₂ by weight.

a.) What volume of air at 25^oC and 1 atm is needed to provide this 1 kg of carbon?
b.) How much carbon is present in the entire atmosphere lying above each square meter of the Earth's surface? (Hint: 1 atm is equivalent to 1.0332 X 10⁴ kg/m²
c.) At the current rate of utilization, how long would it take to use all the CO₂ in the entire atmosphere directly above a forest

Homework Equations

Ideal Gas Law: PV = nRT
R = 0.0821

The Attempt at a Solution

I know for Part a, I am going to use the ideal gas law and "n" will be the moles of carbon, but do not know how to incorporate the percent by weight of Carbon dioxide(0.046%) to find the moles of carbon. Any help or advice is appreciated

 

[Borek]

1 kg of carbon - how many kg of carbon dioxide? Let's call it x.

If x is 0.046% of some mass - what is 100%?

Now you have to convert this mass of air to volume. IMHO you can't, as question is ambiguous. You aren't told if it is 20/80 by mass, or by volume. Make your pick and continue.

 

[yang09]

Im not really following your line of reasoning. Can you please explain further. From what I understand from your post is that our unknown is going to be the kg of Carbon Dioxide. Is that correct? Where do we go from there? What does IMHO stand for as well? Thanks

 

[Borek]

IMHO = In My humble Opinion.

No, x is not an unknown, it just a mass that you should calculate, from a simple stoichiometry. Sorry if the choice of the symbol was confusing.

 

[yang09]

So x is the kg of Carbon dioxide. But we are not given a balanced equation to convert the 1kg of Carbon to kg of Carbon Dioxide. How would i go about and find the mass of Carbon Dioxide? I know that the 0.046% is the percent weight of Carbon Dioxide in the atmosphere so if I were to use the percent weight of Carbon Dioxide, I could only use this formula: 0.046% = [mass of Carbon Dioxide / Total mass in atmosphere( Oxygen + nitrogen + dioxide)] X 100.
Is it possible to incorporate the 1 kg of Carbon to the percent weight of Carbon Dioxide?

 

[Borek]

How much carbon is needed to produce 1 kg of carbon dioxide? Think simple combustion. How many moles of dioxide can be produced from 1 mole of carbon? How many moles of carbon present in one mole of carbon dioxide?

Or even better - think just about percent composition of the compound.

You know formula, you know molar mass, you have everything needed.

 

[yang09]

So from the Photosynthesis formula, I know that 1 mol of Carbon(glucose) produces 6 moles of carbon dioxide. Here is my calculation: (1000g Carbon) X (1mol Carbon/12 g Carbon) X (6 mol CO2/1mol C) = 500 moles of CO2 (seems kind of big). My "n" that I used in the ideal gas law was the moles of CO2. Is that right? With 500 moles of CO2, I got a volume of 12232.9 L of air. This can't be right...

 

[Borek]

I know that 1 mol of Carbon(glucose) produces 6 moles of carbon dioxide.

1 mole of glucose is not 1 mole of carbon.

Apparently you have problems with moles/formulas. What does formula CO₂ tell about the compound composition? If there is a mole of CO₂, how many molecules of CO₂ does it contain? How many atoms of C? How many moles of C?

 

[yang09]

I assumed that the moles from glucose was the same as the moles of carbon because in the photosythesis process, CO2 is converted to carbon chains, which are then converted to glucose. I guess my thought process behind that was wrong then. To go from 1 mol of CO2 to molecules or atoms, I know that you just multiply by avogadros number. But to go from 1 mol of CO2 to 1 mol of C, would you do the same thing? I don't believe we should.

 

[Borek]

Answer precisely - one by one - my questions from the previous post.

 

[yang09]

The compound composition of CO2 are 2 atoms of Oxygen and 1 atom of Carbon. I am basing this off of the formula
1 mole of CO2 = 6.022 X 10^23 molecules of CO2
Since you have 1 atom of Carbon, you would also have 1 mole of carbon
Am I right?

 

[Borek]

Since you have 1 atom of Carbon, you would also have 1 mole of carbon
Am I right?

Yes, that's correct.

 

[yang09]

So going back to the question, I know that one mole of carbon dioxide produces one mole of carbon. To clarify things up, in the ideal gas law, is the "n" we are solving for the moles of carbon dioxide since the question wants us to find the volume of air needed to support the 1 kg of Carbon. If it is, then I would use stoichiometry to find the moles of CO2 starting from the 1kg of Carbon. Is this reasoning correct?

 

[Borek]

If it is, then I would use stoichiometry to find the moles of CO2 starting from the 1kg of Carbon. Is this reasoning correct?

Yes, that's what I suggested in my first post in this thread.

 

[yang09]

Thanks Borek for the help and explaining the things to me. It really cleared things up. I got the answer now for part A

For Part B, I have a feeling that we will be using the ideal gas law again, except our pressure will be 1.0332 X 10^4 kg/ m² instead of 1 atm and we will be solving for "n", the moles of Carbon. To find the amount of Carbon in the atmosphere, I assumed that the 0.046% weight of CO2 in the atmosphere weighed 0.046 g and that the total mass of all gases in the atmosphere was 100 g. Is that correct? From here, I solved for the mass of of Carbon dioxide and through stoichiometry, solved for the amount of carbon in the atmosphere in grams. Is my work correct?

 

[yang09]

Forget what I said in the previous post about my work for Part B. I did not use the Ideal Gas law. What I did was that I used the percent weight of CO2 in the atmosphere and I set up the equation: 0.046 = (mass of CO2/mass of all gases in the atmosphere) X 100%.
I assumed that the masses of all gases in the atmosphere was 100g, so that would have made the mass of CO2 to be 0.046g. From here, I used stoichiometry to solve for the amount of Carbon in the atmosphere in units of gram. Does this sound right?

 

[Borek]

Why 100 g? You are given mass, almost directly.

 

[yang09]

Thanks Borek. I got an answer to it