Question
Asked 28th Feb, 2015

How do I calculate the gas concentration in PPM?

I have asked two times the following question:
I want to calculate the gas concentration, in ppm, in the closed chamber of volume 1025 cm3. I have a cylinder of mixture of NH3 and air with the 10 and 90% ratio respectively. we are flowing this gas mixture to the chamber at flow rate of 100 sccm. Now how can i calculate the NH3 concentration in ppm inside the closed chamber ?
But my seniors and teachers answer it in a different manner I don’t know to which answer I believe, I am bit confused… the answers I got in response of my questions are as:
Hi Kaleem,
the concentration in ppm does not depend on flow rate or volume as long as the chamber has been evacuated before you introduce the mixture of NH3/air. The ppm measure simply says how many µg/g (weight) or molecules per million molecules (molar ratio) of NH3 there are in the mix.
Assuming that your 1:10 mix is calculated in weight, the concentration is 100,000 µg/g or ~170000 molecules/1,000,000 molecules of mix. The latter, higher molecular ratio arises from the fact that the molar mass of NH3 is lower than the average for air, i.e. per unit weight there are more molecules of NH3 than of the gas mix that makes up air.
If your gas mix of 1:10 is in molecules/molecules (partial pressures), the concentration of NH3 is ~59,000 µg/g or 100,000 molecules/1,000,000 molecules of mix.
The 2nd way of calculations of the question (We have a closed chamber having volume 1000 cm3. 1 sccm of pure oxygen was then incorporated in chamber from flow controller. Air is already present in chamber in which oxygen is 20.95%. How we can calculate the concentration of oxygen in ppm in that chamber?):
a) If we suppose that ppm are cm^3/m^3, we can calculate oxygen concentration as:
V(O2)/V(total) = (209.5+x)cm^3/10^-3m^3.
b) If ppm = mg/l:
m(O2)/V(total) = (209.5+x)*32/(Vm*1).
Here x= is volume of oxygen added, x(cm^3)=1*t ((cm^3/min)*min), t(min) is time during which oxygen was added, 32 is molar mass of O2, Vm is molar volume, 1 = 1litre, volume of your chamber.
The 3rd way of calculations :
xA = mA / mtotal
Parts Per Million (ppm) = xA*106
i.e mole fraction of component (XA *106 )
There are other answers even I know that they are not right….… Now tell me the right way which is most accurate to calculate the gas concentration in PPM......Thanx

Most recent answer

4th Aug, 2021
Chidinma Peace Okonkwo
Nnamdi Azikiwe University, Awka
@mumtaz danish , u are absolutely right
1 Recommendation

Popular answers (1)

1st Mar, 2015
Arinze Okoye
University of Nigeria
The conversion equations depend on the temperature at which the conversion is wanted (usually about 20 to 25 °C). At an ambient sea level atmospheric pressure of 1 atm (101.325 kPa or 1.01325 bar), the general equation is:
\mathrm{ppmv} = \mathrm{mg}/\mathrm{m}^3\cdot \frac{(0.08205\cdot T)}{M}
and for the reverse conversion:
\mathrm{mg}/\mathrm{m}^3 = \mathrm{ppmv}\cdot \frac{M}{(0.08205\cdot T)}
where:
mg/m3 = milligrams of pollutant per cubic meter of air at sea level atmospheric pressure and T
ppmv = air pollutant concentration, in parts per million by volume
T = ambient temperature in K = 273.15 + °C
0.08205 = Universal gas constant in atm·m3/(kmol·K)
M = molecular mass (or molecular weight) of the air pollutant
Notes:
1 atm = absolute pressure of 101.325 kPa or 1.01325 bar
mol = gram mole and kmol = 1000 gram moles
Pollution regulations in the United States typically reference their pollutant limits to an ambient temperature of 20 to 25 °C as noted above. In most other nations, the reference ambient temperature for pollutant limits may be 0 °C or other values.
Although ppmv and mg/m3 have been used for the examples in all of the following sections, concentrations such as ppbv (i.e., parts per billion by volume), volume percent, mole percent and many others may also be used for gaseous pollutants.
Particulate matter (PM) in the atmospheric air or in any other gas cannot be expressed in terms of ppmv, ppbv, volume percent or mole percent. PM is most usually (but not always) expressed as mg/m3 of air or other gas at a specified temperature and pressure.
For gases, volume percent = mole percent
1 volume percent = 10,000 ppmv (i.e., parts per million by volume) with a million being defined as 106.
Care must be taken with the concentrations expressed as ppbv to differentiate between the British billion which is 1012 and the USA billion which is 109 (also referred to as the long scale and short scale billion, respectively).
5 Recommendations

All Answers (6)

1st Mar, 2015
Arinze Okoye
University of Nigeria
The conversion equations depend on the temperature at which the conversion is wanted (usually about 20 to 25 °C). At an ambient sea level atmospheric pressure of 1 atm (101.325 kPa or 1.01325 bar), the general equation is:
\mathrm{ppmv} = \mathrm{mg}/\mathrm{m}^3\cdot \frac{(0.08205\cdot T)}{M}
and for the reverse conversion:
\mathrm{mg}/\mathrm{m}^3 = \mathrm{ppmv}\cdot \frac{M}{(0.08205\cdot T)}
where:
mg/m3 = milligrams of pollutant per cubic meter of air at sea level atmospheric pressure and T
ppmv = air pollutant concentration, in parts per million by volume
T = ambient temperature in K = 273.15 + °C
0.08205 = Universal gas constant in atm·m3/(kmol·K)
M = molecular mass (or molecular weight) of the air pollutant
Notes:
1 atm = absolute pressure of 101.325 kPa or 1.01325 bar
mol = gram mole and kmol = 1000 gram moles
Pollution regulations in the United States typically reference their pollutant limits to an ambient temperature of 20 to 25 °C as noted above. In most other nations, the reference ambient temperature for pollutant limits may be 0 °C or other values.
Although ppmv and mg/m3 have been used for the examples in all of the following sections, concentrations such as ppbv (i.e., parts per billion by volume), volume percent, mole percent and many others may also be used for gaseous pollutants.
Particulate matter (PM) in the atmospheric air or in any other gas cannot be expressed in terms of ppmv, ppbv, volume percent or mole percent. PM is most usually (but not always) expressed as mg/m3 of air or other gas at a specified temperature and pressure.
For gases, volume percent = mole percent
1 volume percent = 10,000 ppmv (i.e., parts per million by volume) with a million being defined as 106.
Care must be taken with the concentrations expressed as ppbv to differentiate between the British billion which is 1012 and the USA billion which is 109 (also referred to as the long scale and short scale billion, respectively).
5 Recommendations
11th Mar, 2015
Mumtaz Danish
University of Karachi
Dear Kaleem,
I have performed some simple calculations for you. Hope this will be beneficial to you.  
5 Recommendations
25th May, 2015
Meixia He
Southeast University (China)
Dear Professor Mumtaz Danish, I have a question that whether you have a defined referrance about the defenition of PPM? Thank you!
1 Recommendation
7th Mar, 2019
Sabry Abdallah
Tanta University
Parts per Million by Volume (or mole) in Air
In air pollution literature ppm applied to a gas, always means parts per million by volume or by mole. These are identical for an ideal gas, and practically identical for most gases of air pollution interest at 1 atm. Another way of expressing this value is ppmv. [1]
One part per million (by volume) is equal to a volume of a given gas mixed in a million volumes of air:
1ppm = 1 gas volume /106 air volume
A micro liter volume of gas in one liter of air would therefore be equal to 1 ppm:
1ppm = 1µL gas/ 1 L air
Today's more and more there is an interest to express gas concentrations in metric units, i.e. µg/m3. Although expressing gaseous concentrations in µg/m3 units, has the advantage of metric expression, it has the disadvantage of being greatly influenced by changes in temperature and pressure. Additionally, because of difference in molecular weight, comparisons of concentrations of different gases are difficult. [2]
To convert ppmv to a metric expression like µg/m3, the density of the concerning gas is needed. The density of gas can be calculated by the Law of Avogadro's, which says: equal volumes of gases, at the same temperature and pressure, contain the same number of molecules. This law implies that 1 mole of gas at STP a volume of 22.71108 liters (dm3) enfolds, also mentioned as the molar volume of ideal gas. Standard Temperature and Pressure (STP) is defined as a condition of 100.00 kPa (1 bar) and 273.15 K (0°C), which is a standard of IUPAC. [3] The amount of moles of the concerning gas can be calculated with the molecular weight.
1ppm = Vm/M = 1µg gas/ 1L air
Where:
Vm = standard molar volume of ideal gas (at 1 bar and 273.15 K) [3] [22.71108 L/mol]
M = molecular weight of gas [g/mol]
For converting ppm by mole, the same equation can be used. This can be made clear by the following notation:
1ppm = 1µmole gas / 1 mole air = Vm/M * 1 1µg gas /1L air
By checking the dimensions of the most right part of the equation, there will be found a dimensionless value, like the concentration in ppm is.
To calculate the concentration in metric dimensions, with other temperature and pressure conditions the Ideal Gas Law comes in handy. The volume (V) divided by the number of molecules (n) represents the molar volume (Vn) of the gas with a temperature (T) and pressure (P).
Vn= V/n=R T/P
Where:
Vn = specific molar volume of ideal gas (at pressure P and temperature T) [L/mol]
V = volume of the gas [m3]
n = amount of molecules [mol]
R = universal gas law constant [3] [8.314510 J K-1 mol-1] or [m3 Pa K-1 mol-1]
T = temperature [K]
P = pressure [Pa]
With this equation it comes clear that the percentage notation by ppm is much more useful, because the independency of the temperature and pressure.
Parts per Million by Weight in Water
The concentration in ppm of gas in water is meanly meant by weight. To express this concentration with metric units the density of water is needed.
The density of pure water has to be by definition 1000.0000 kg/m3 at a temperature of 3.98°C and standard atmospheric pressure, till 1969. Till then this was mean definition for the kilogram. Today's the kilo is defined as being equal to the mass of the international prototype of the kilogram [4]. Water with a high purity (VSMOW) at a temperature of 4°C (IPTS-68) and standard atmospheric pressure has a density of 999.9750 kg/m3. [5]
The density of water is effected by the temperature, pressure and impurities, i.e. dissolved gasses or the salinity of the water. Even the concerning concentration of gas dissolved in the water is affecting the density of the solution. By nature there's a chance that water contains a certain concentration of Deuterium which influences the density of the water. This concentration is also called the isotopic composition [6].
Accurate calculations on these conversions are only possible when the density of the water is measured. In practice the density of water is therefore set to 1.0 ·103 kg/m3. When calculating the conversion with this value you gets:
1ppm=1mg gas/1Kg water = 1/?w * 1mg gas /1L water
Where:
?w = density of water [1.0 ·103 kg/m3]
Reference
[1] Never, N. , Air Pollution Control Engineering. McGraw-HILL, Singapore 1995.
[2] Godish, T. , Air Quality. Lewis Publishers, Michigan 1991.
[3] Cohen, E.R. and Taylor, B.N., J. Res. Nat. Bur. Stand. 92 (1987) 85-95. (International Union of Pure and Applied Chemistry (IUPAC))
[4] n/a, Kilogram. International prototype of the kilogram, www.bipm.org/en/scientific/mass/prototype.html.
[5] Marsh, K.N., Ed., Recommended Reference Materials for the Realization of Physicochemical Properties. Blackwell Scientific Publications, Oxford.
1 Recommendation
17th Mar, 2020
Md Tawabur Rahman
Khulna University of Engineering and Technology
i use two different gases. one is target gas whose ppm is known and another is 99% N2. i calculate the conc. of desired gas by following formula.
desired con. of target gas in ppm =(flow rate of target gas/(flow rate of target gas + flow rate of N2))*known ppm of target gas
hope it will help!
4th Aug, 2021
Chidinma Peace Okonkwo
Nnamdi Azikiwe University, Awka
@mumtaz danish , u are absolutely right
1 Recommendation

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