Short Cut Formulas For Gas Densities At Various Standard Conditions

Α. Let us try to make density calculations handy, at least for ideal gas conditions as above, remembering that quantity of 1 kmol of gas
(α) has mass equal to gas molecular weight M
(β) has volume equal to molar volume Vm
As we remember from chemistry, Vm = 22.4 lt/gmol = 22.4 m3/kgmol at 0 oC and 1 Atm a (*). A more precise value is 22.414 m3/kgmol, though difference can be insignificant in practical calculations.
To calculate gas density at some standard condition (or even non standard), we have to find molar volume at this condition by ideal gas law, then divide (α)/(β). Molar gas constant R need not be used in the calculation.
Β. Examples of gas density at several standard conditions P, T.
1. P= 1 Atm a, T=15 oC = 288.15 oK
Vm=22.414*(288.15/273.15)=23.64 m3/kgmol, ρ=Μ/23.64 kg/m3
2. P=1.0 Bar a, T=25 oC=298.15 oK
Vm=22.414*(1.01325/1.0)*(298.15/273.15)=24.79 m3/kgmol, ρ=M/24.79 kg/m3
3. P= 14.696 psi a, T=60 oF = 519.67 oR, density is required in lb/ft3.
We can first convert Vm = 22.414 m3/kgmol to ft3/lbmol, taking into account 1 ft = 0.3048 m and 1 lb = 0.4536 kg; so Vm = 22.414/0.3048^3 ft3/(1/0.4536)lbmol = 359.04 ft3/lbmol (at 1 Atm a =14.696 psi a and 0 oC = 491.67 oR).
Thus for considered standard condition Vm=359.04*(14.696/14.696)*(519.67/491.67) = 379.5 m3/lbmol, ρ=Μ/379.5 lb/ft3.
If conversions looks difficult in this specific case due to English units, we can find density in metric units, then convert it:
T=60 oF = 15.56 oC=288.71 oK, P = 14.696 psi a = 1 Atm a
Vm=22.414*(288.71/273.15)=23.691 m3/kgmol, ρ=Μ/23.691 kg/m3 = (M/0.4536)/(23.691/.3048^3) lb/ft3 = (M*0.3048^3)/23.691*0.4536) lb/ft3 = M/379.5 lb/ft3.
Γ. Concerning ground level concentrations estimated in dispersion models, conversion of ppm (v) to mg/m3 of a substance in air at some standard condition is similarly facilitated by consideration of air molar volume. This is indicated in http://www.cheresour...r-mgm3-to-ppm-v '> http://www.cheresour...r-mgm3-to-ppm-v . Air volume Vm contains 1 kgmole of air, and 1*ppm(v)*1E-6*M kg of the substance. Finding Vm at "standard" condition, mg/m3 is then specified.
Δ. With simple thinking and some skill in unit conversions (rather necessary for a Chemical Engineer) we can do a lot using a calculator, even out of the office.

Note (*): Chemistry courses in USA schools seem to give Vm in metric units and at 0 oC, 1 Atm a, see Schaum's Colledge Chemistry, Chapter 8, Molar Volume.

Kkala,

Should I thank you for repeating what I have written in my blog entry. Well, I don't think so.

Please read kkala's contribution again. Spirit is different, "try to use elementary knowledge with minimum memorizing". This is another option to shortcut formulas. "Standard" conditions of two examples were chosen same, for the reader to see that results are also same. For this purpose M was retained as letter (molecular weight), instead of using an arithmetic value.

We follow 1 atm & 0 Deg C (273 Deg K) as the standard condition and refer is as Normal Condition for Gases. ( NM3...) Is it okay?

1. W. Badger and J Banchero in "Introduction to Chemical Engineering" (McGrawHill, 1955) write "While standard conditions for scientific work are always 760 mm Hg and 0 oC, much engineering work is based on 30 in Hg and 60 oF", Chapter 1 (Introduction) page 6.
2. However standard conditions do not seem to be unique today, needing definition (whenever referred) to avoid misunderstandings. E.g. standard conditions can be 1 atm a and 60 oF or any other reported in previous posts, including thread http://www.cheresou...mgm3-to-ppm-v/ .
3. 1 atma and 0 oC are usually called "Normal" conditions in several countries (Europe, etc). In a broad sense these make an additional set of "standard" conditions. However 200 standard m3 of gas does not usually mean 200 Normal m3 of gas. 200 Nm3 refers to 1 atma and 0 oC, while 200 std m3 refers to some other conditions among the "standard" ones.