Percent Change in Concentration

[henry3369]

Homework Statement

The world burns approximately 3.8 x 10¹² kg of fossil fuel per year.

1. Use the combustion of octane as the representative reaction and determine the mass of carbon dioxide (the most significant greenhouse gas) formed per year.

2. The current concentration of carbon dioxide in the atmosphere is approximately 368 ppm (by volume). By what percentage does the concentration increase each year due to fossil fuel combustion? Approximate the average properties of the entire atmosphere by assuming that the atmosphere extends from sea level to 15 km and that it has an average pressure of 381 torr and average temperature of 275 K. Assume Earth is a perfect sphere with a radius of 6371 km.

Homework Equations

Volume of a sphere = (4/3)π(r³)
Percent change = (new - old)/old * 100

The Attempt at a Solution

I already calculated #1.
1. 1.2 x 10^6 g

2. V_atm = V_{earth+atmosphere} - V_earth
V_atm = (4/3)π(6386³ - 6371³) = 7.66899 x 10⁹ km³ = 7.66899 x 10¹⁸ m³

368 ppm = 368 g/m³

Using mass obtained from #1, and finding concentration of the increase in CO2:
1.2 x 10¹⁶g/7.66899 x 10¹⁸ m³ = 0.001564743 g/m³

New concentration = 368 + 0.001564743
Percent change = (new-old)/old * 100 = 0.00042519%

So the answer is 0.42%, and it seems that the way to achieve this is to have the V_atm = 7.66899 x 10¹⁵ m³ rather than 7.66899 x 10¹⁸ m³. Is there an error in my volume calculation that I am not catching?

 

[siddharth23]

Homework Statement

The world burns approximately 3.8 x 10¹² kg of fossil fuel per year.

1. Use the combustion of octane as the representative reaction and determine the mass of carbon dioxide (the most significant greenhouse gas) formed per year.

2. The current concentration of carbon dioxide in the atmosphere is approximately 368 ppm (by volume). By what percentage does the concentration increase each year due to fossil fuel combustion? Approximate the average properties of the entire atmosphere by assuming that the atmosphere extends from sea level to 15 km and that it has an average pressure of 381 torr and average temperature of 275 K. Assume Earth is a perfect sphere with a radius of 6371 km.

Homework Equations

Volume of a sphere = (4/3)π(r³)
Percent change = (new - old)/old * 100

The Attempt at a Solution

I already calculated #1.
1. 1.2 x 10^6 g

2. V_atm = V_{earth+atmosphere} - V_earth
V_atm = (4/3)π(6386³ - 6371³) = 7.66899 x 10⁹ km³ = 7.66899 x 10¹⁸ m³

368 ppm = 368 g/m³

Using mass obtained from #1, and finding concentration of the increase in CO2:
1.2 x 10¹⁶g/7.66899 x 10¹⁸ m³ = 0.001564743 g/m³

New concentration = 368 + 0.001564743
Percent change = (new-old)/old * 100 = 0.00042519%

So the answer is 0.42%, and it seems that the way to achieve this is to have the V_atm = 7.66899 x 10¹⁵ m³ rather than 7.66899 x 10¹⁸ m³. Is there an error in my volume calculation that I am not catching?

I didn't go through the entire calculation. But it says 368 ppm by volume, for which you've taken 368 g/m³. But 'g' is a unit of mass. Check that part.

 

[henry3369]

I didn't go through the entire calculation. But it says 368 ppm by volume, for which you've taken 368 g/m³. But 'g' is a unit of mass. Check that part.

I looked up ppm to g/m^3 and a website told me it was a 1:1 conversion. Can you please tell me what I'm supposed to convert ppm to? I'm familiar with the term.

 

[henry3369]

I didn't go through the entire calculation. But it says 368 ppm by volume, for which you've taken 368 g/m³. But 'g' is a unit of mass. Check that part.

According to: http://chemwiki.ucdavis.edu/Physica...sics/Units_Of_Concentration#Parts_per_Million

ppm = 1mg/L which, can be converted to 1g/m³

 

[siddharth23]

I looked up ppm to g/m^3 and a website told me it was a 1:1 conversion. Can you please tell me what I'm supposed to convert ppm to? I'm familiar with the term.

Well I'd say that 368 ppm by volume will mean 368 m³ in 1000000 m³. Now that you know the volumes, use the density to find out the amount of CO₂ by weight in air. (density = mass / volume)

 

[henry3369]

Well I'd say that 368 ppm by volume will mean 368 m³ in 1000000 m³. Now that you know the volumes, use the density to find out the amount of CO₂ by weight in air. (density = mass / volume)

d = m/V = 44.0087/368 = 0.1196 g/m^3

new concentration = 0.1196 + 000156 = 0.12116

(new-old)/old * 100 = 1.3%

This is incorrect though. What am I doing incorrectly?

 

[Borek]

d = m/V = 44.0087/368 = 0.1196 g/m^3

Why do you assume 1 mole of carbon dioxide occupies 368 cubic meters? Just because there are 368 m³ per 1000000 m³? Would you assume volume of 0.368 m³ if you were told 386 ppm means 0.368 m³ in 1000 m³?