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Teledyne API - T100 UV Fluorescence SO2 Analyzer
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5.4.4.1. CONVERTING MICROGRAMS PER CUBIC METER TO PARTS PER MILLION
The conversion between micrograms per cubic meter and parts per million is based on
standard conditions (0
standard conditions (0
o
C and 101.325 kPa) where one mole of an ideal gas occupies
22.414 L. Thus, converting the mass of the pollutant M
p
in grams to its equivalent
volume V
p
in liters at standard temperature and pressure (STP) takes the following
equation:
V
p
= [(M
p
)/(MW)] x 22.414 L
.
mol
-1
Where MW is the molecular weight of the pollutant in units of grams per mole. For
readings made at temperatures and pressures other than standard conditions, the standard
volume, 22.414 L
readings made at temperatures and pressures other than standard conditions, the standard
volume, 22.414 L
.
mol
-1
, must be corrected. The ideal gas law to make the correction can
be used:
(22.414 L
.
mol
-1
) x [(T
2
)/(273 K)] x [(101.325 kPa)/(P
2
)]
Where T
2
and P
2
are the absolute temperature (in Kelvin) and absolute pressure (in
kilopascals) at which the readings were made. Because parts per million is a volume
ratio, it can be written as:
ratio, it can be written as:
ppm = (V
p
)/(V
a
+ V
p
)
where V
a
is the volume of the air in cubic meters at the temperature and pressure at
which the measurement was taken. Then combine equations to yield:
ppm = {[(M
p
)/(MW)] x (22.414 L
.
mol
-1
) x [(T
2
)/(273 K)] x [(101.325 kPa)/(P
2
)]}
-------------------------------------------------------------------------------------------
(V
a
) x (1000 L
.
m
-3
)
where M
p
is the mass of the pollutant of interest in micrograms. The factors converting
micrograms to grams and liters to millions of liters cancel one another. Unless
otherwise stated, it is assumed that V
otherwise stated, it is assumed that V
a
= 1.00 m
3
Example:
A 1-m
3
sample of air was found to contain 80 µg
.
m
-3
of SO
2
. The temperature and
pressure were 25.0
o
C and 103.193 kPa when the air sample was taken. What was the
SO
2
concentration in parts per million?
Solution:
First, determine the MW of SO
2
:
MW of SO
2
= 32.06 + 2(15.9994) = 64.06 g
.
mol
-1
Next, convert the temperature from Celsius to Kelvin:
25
o
C + 273 K = 298 K
Using the equation derived above, Concentration is:
{[(80 µg)/(64.06 g
.
mol
-1
)] x (22.414 L
.
mol
-1
) x [(298 K)/(273 K)] x [(101.325 kPa)/(103.193 kPa)]} / [(1 m
3
) x (1000 L
.
m
3
)]
= 0.030 ppm of SO
2
06807C DCN6650