AOS2 Discussion Activity #1
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Subject
Chemistry
Date
Feb 20, 2024
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Uploaded by MegaGiraffeMaster41
AOS 2: Air Pollution Discussion Activity #1 worksheet
Scan (or complete electronically) this assignment and upload to the bruinlearn course site as soon as
possible after the end of the discussion section. Remember to put everyone’s name on this sheet (your
name and ID goes at the top) and rate the level of participation. Feel free to discuss the questions among
yourselves or with the TA.
1) Calculate the volume of the air inside the
garage in
cm
3
. The area of the garage floor
covers a rectangle of 8 m by 8 m and its height
is 3 m.
Convert units:
How many cubic centimeters
(cm
3
) are in one cubic meter (m
3
)?
One cubic meter is equivalent to one million
(1,000,000) cubic centimeters
The volume of air in the garage is therefore
192,000,000 cm^3
Get the volume of the garage in cm
3
. [compute the volume in m
3
first, then convert to cm
3
with the
conversion factor from above]
(In this and subsequent questions, show major steps of the calculations)
1m = 100cm
1 m^3 = 1,000,000 cm^3 (100 x 100 x 100)
8m x 8m x 3m = 192m^3
192 x 1,000,000cm^3 =
192,000,000cm^3
2) The mixing ratio of carbon monoxide (CO) in the garage is 10.0 ppm.
(a) What is the CO concentration in molecules/cm
3
?
Remember that 1 cm
3
of air contains 2.50×10
19
molecules (thus, the total concentration of all air molecules is 2.50×10
19
molecules/cm
3
), and that 1 ppm
means “one part per million” (or 1 molecule of something per million of all molecules).
1
AOS 2 Discussion Activity #1 worksheet
C = (# of molecules) / volume
2.5 x 10^19 x 192,000,000 = 4.8 x10^27 = total # of molecules
4.8 x 10^27 x (10/1,000,000)
# of CO2 molecules = 4.8 x10^22
(4.8 x 10^22) / 192,000,000 = concentration
Concentration
=
2.5 x 10^14 cm^3 molecules/cm^3
(b) What is the total number CO molecules inside the entire garage
(within the volume found in
problem #1)
?
The total number of CO molecules inside the entire garage is 4.8 x 10^22 cm^3
3) Assume that the 10.0 ppm mixing ratio of carbon monoxide is steady, the residence time of CO is 8.0
hours, and the CO is being emitted by leakage from a natural gas space heater in the garage.
What is the
source rate of CO from this heater?
Since the mixing ratio of CO and hence its concentration is steady, we can use the steady-state box model
concentration formula, where
q
= concentration (molec/cm
3
),
S
= source rate (molec/hr),
τ
= residence
time (hr), and
V
= volume (cm
3
). Then, just solve for source rate
S
:
q
⋅
⇒
=
⋅
=
S
V
τ
q V
S
τ
[Note that (
q
×
V
) is the same calculation as was done in problem 2b, so just substitute that result here]
Concentration (q) = 2.5 x 10^14 cm^3
Residence time (
τ
) = 8 hrs
Volume (v) = 192,000,000 cm^3
Source rate (s) = ?
S = (q x v) /
τ
S = 6 x10^21 molecules C02/hr
4)
Explain in your own words
how adding a new sink process affects the pollutant residence time and
the steady-state concentration of air pollutant molecules inside a box.
With residence time being the average time for a substance to spend in a reservoir before being removed,
adding a new sink process will lead to pollutants leaving the box at a faster rate and therefore a lower
residence time. Because of this, the time at which steady-state concentration is achieved is far faster than
2
AOS 2 Discussion Activity #1 worksheet
it was before with the rate of removal being higher. Basically, the steady-state concentration of air
pollutant molecules inside a box will be lower than before.
5) Suppose we open the garage door, which reduces the CO residence time to 2 hours. By how much can
the CO source rate increase without exceeding a steady-state CO mixing ratio of 10 ppm? (Multiplicative
factor is okay; ignore the non-uniform distribution of CO between the space heater and the open garage
door.)
The source rate can increase by up to 4 times without exceeding a steady-state CO mixing ratio of 10
ppm, as that counteracts the fact that the residence time decreased by 4 times (from 8 to 2).
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I
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