Enviro Chem.- simple ideal gas law calculation.

[luna02525]

Homework Statement

1. Ozone has a maximum concentration of 4.7 x 1012 molecules/ cm3 at an altitude of 20 km. The total pressure at this altitude is 100 torr. Using the ideal gas law, what is the mole fraction of O3 at that altitude? Now express the mole fraction in ppb.

Homework Equations

PV=nRT (ideal gas law)

X/(X_a + X_b...) mole fraction

temperature @ 20 km = -50 Celsius = 223 K

100 torr = .132 atm

The Attempt at a Solution

So far I have taken concentration of ozone and put it into mol/L.

4.7 x 10^12 molecules/cm^3 = 7.81 x 10^-9 mol/L

I'm not sure how to find a mole fraction without additional information. All the info for ideal gas law is also all there so I'm confused. I forgot how to take mol/L into just moles I guess. I've looked online for more information and reading up on the general chemistry but am unable to find any information related to this problem. Also notable is the fact that I do not have a general chemistry textbook (my brother sold it for money) which would ideally give me a refresher in PV= nRT and finding the solution to this problem.

So, any help would be appreciated.

 

[chemisttree]

Now, take the concentration of ozone and convert it into a pressure using the gas law. Compare that pressure with the total pressure at that altitude. What would such a comparison represent?

 

[luna02525]

Now, take the concentration of ozone and convert it into a pressure using the gas law. Compare that pressure with the total pressure at that altitude. What would such a comparison represent?

ok so I took the concentration of ozone and plugged it into

PV=nRT V= 1 L

P(1 L) = (7.81 x 10^-9 mol)(.082 Latm/kmol)(223 K)
P = 1.43 x 10^-7 atm

so, now I know the pressure of ozone at 20 km

in order to find the mole fraction I would place the pressure of ozone at 20 km over the total air pressure at 20 km?

so,

(1.43 x 10^-7 atm)/(.132 atm) = 1.08 x 10^-6 ozone per total air

1) Is this correct?
2) To find ppb now I just multiply the ozone mole fraction by 1/(10^-9)?

 

[chemisttree]

That's how I would do it.

 

[luna02525]

Thanks a lot for your help!

 

[chemisttree]

Alternatively, you could leave the ozone concentration in terms of moles/L and then, using the gas law, calculate the concentration of gas using the gas law. That ratio would give you the ozone concentration as well.

You might want to calculate it both ways to convince yourself of your answer.