1. Moisture Variables & The Equation of State for Moist Air
SO 254 – Spring 2017
LCDR Matt Burich

2. Ideal Gas Law  Eqn. of State for Dry Air
This we previously derived at the beginning of the course:
𝑝𝑉=𝑛 𝑅 ∗ 𝑇 ⇒ 𝑝=𝜌𝑅𝑇
Where 𝑅=287 J K −1 kg −1 is the gas constant for dry air
Thus, in reality we have:
𝑝 𝑑 = 𝜌 𝑑 𝑅 𝑑 𝑇
𝑑= “dry”
In the real atmosphere, this is an oversimplification since there is almost always water vapor present to some extent.
The water vapor itself obeys the ideal gas law and exerts a pressure (vapor pressure) which we give the symbol 𝑒
Thus:
𝑒= 𝜌 𝑣 𝑅 𝑣 𝑇
𝑣= “vapor”
𝑅 𝑑 =287 J K −1 kg −1 , so what is 𝑅 𝑣 ?
Recall how we calculated 𝑅 𝑑 …

3. Ideal Gas Law  Eqn. of State for Dry Air
𝑅→ 𝑅 ∗
universal gas constant
𝑅 𝑑 = 𝑅 ∗ 𝑀 𝑑
~78% N 2 ~21% O 2
~1% Ar
𝑅 ∗ = J K −1 mol −1
N 2 : ~14 g mol − =21.84
O 2 : ~16g mol − =6.72
Ar: ~40 g mol − =0.4
28.96 g mol −1
𝑅 𝑑 = 𝑅 ∗ 𝑀 𝑑 = J K −1 mol − g mol − g 1 kg =287 J K −1 k g −1
𝑅 𝑣 = 𝑅 ∗ 𝑀 𝑤 = J K −1 mol −1 18 g mol − g 1 kg =461.9 J K −1 k g −1
What is 𝑀 𝑤 ?
Molecular weight of water
H 2 : ~1 g mol −1 2 =2
O: ~16g mol −1 1 =16
18 g mol −1

4. So we have: 𝑝 𝑑 = 𝜌 𝑑 𝑅 𝑑 𝑇 and 𝑒= 𝜌 𝑣 𝑅 𝑣 𝑇
Where 𝑅 𝑑 =287 J K −1 kg −1 and 𝑅 𝑣 =461.9 J K −1 kg −1
From Dalton’s Law of Partial Pressures, total air pressure is simply the sum of the constituent gasses:
𝑝= 𝑝 𝑑 +𝑒
A problem is, concentration of water vapor is highly variable in the atmosphere (e.g., Annapolis in July vs. Phoenix in July). The gas constant for air of varying moisture content would thus be continuously changing
constant = changing
Therefore, to construct an equation of state for moist air with varying amounts of water vapor, we’ll proceed in such a way as to retain the dry air gas constant in the expression.
A volume of moist air contains some mass of dry air plus some mass of water vapor so that its density can be expressed as:
𝜌= 𝑚 𝑑 + 𝑚 𝑣 𝑉 = 𝜌 𝑑 + 𝜌 𝑣

5. 𝑝=𝜌 𝑅 𝑑 𝑇 𝑣 So we now have: 𝑝 𝑑 = 𝜌 𝑑 𝑅 𝑑 𝑇 𝑒= 𝜌 𝑣 𝑅 𝑣 𝑇 𝑝= 𝑝 𝑑 +𝑒
𝑝 𝑑 = 𝜌 𝑑 𝑅 𝑑 𝑇
𝑒= 𝜌 𝑣 𝑅 𝑣 𝑇
𝑝= 𝑝 𝑑 +𝑒
𝜌= 𝜌 𝑑 + 𝜌 𝑣
And we can write:
𝜌= 𝑝 𝑑 𝑅 𝑑 𝑇 + 𝑒 𝑅 𝑣 𝑇 ⇒ 𝑝−𝑒 𝑅 𝑑 𝑇 + 𝑒 𝑅 𝑣 𝑇 ⇒ 𝑝 𝑅 𝑑 𝑇 − 𝑒 𝑅 𝑑 𝑇 + 𝑒 𝑅 𝑣 𝑇
We factor out the measurable/known terms:
This is a constant
𝜌= 𝑝 𝑅 𝑑 𝑇 1− 𝑒 𝑝 + 𝑒 𝑅 𝑑 𝑝 𝑅 𝑣 ⇒ 𝑝 𝑅 𝑑 𝑇 1− 𝑒 𝑝 1− 𝑅 𝑑 𝑅 𝑣
=0.622≡𝜀
𝜌= 𝑝 𝑅 𝑑 𝑇 1− 𝑒 𝑝 1−𝜀
Solving for 𝑝:
𝑝= 𝜌 𝑅 𝑑 𝑇 1− 𝑒 𝑝 1−𝜀
And we define the virtual temperature 𝑇 𝑣 as:
𝑇 𝑣 = 𝑇 1− 𝑒 𝑝 1−𝜀
so:
𝑝=𝜌 𝑅 𝑑 𝑇 𝑣

6. Eqn. of State for Moist Air
𝑝= total pressure
𝜌= total density
𝑝=𝜌 𝑅 𝑑 𝑇 𝑣
where
𝑇 𝑣 = 𝑇 1− 𝑒 𝑝 1−𝜀
𝑅 𝑑 =287 J K −1 kg −1
Virtual temperature 𝑇 𝑣 is the temperature that dry air would have to have in order to have the same density as moist air at the same pressure
In effect, the denominator of 𝑇 𝑣 is a “correction factor” to air temperature to account for the presence of water vapor
Example
Compute the virtual temperature of the air for a typical summer afternoon in Annapolis where 𝑇=90℉ (32℃) and 𝑝=1010 mb if the density of water vapor alone at 90℉ is 1.78× 10 −2 kg/ m 3

7. What is a big takeaway from this??
Example
𝑇 𝑣 = 𝑇 1− 𝑒 𝑝 1−𝜀
Compute the virtual temperature of the air for a typical summer afternoon in Annapolis where 𝑇=90℉ (32℃) and 𝑝=1010 mb if the density of water vapor alone at 90℉ is 1.78× 10 −2 kg/ m 3
𝑒= 𝜌 𝑣 𝑅 𝑣 𝑇
𝑒= 𝜌 𝑣 𝑅 𝑣 𝑇 = 1.78× 10 −2 kg m J K −1 kg − K =2508 Pa
𝑇 𝑣 = 𝑇 1− 𝑒 𝑝 1−𝜀 = 305 K 1− hPa 1010 hPa (1−0.622) =307.9 K (94.8℉)
This states that to make dry air the same density as moist air at 90℉, the temperature of the dry air would need to be 94.8℉
What is a big takeaway from this??
Taking dry air from 90℉ to 94.8℉ does what to its density?
Lowers it (thus air with higher water vapor content is less dense than drier air at the same temperature/pressure!)

8. Recall in the example of East Coast cyclogenesis:
Air with higher water vapor content (humid air) is less dense than drier air at the same temperature and pressure
This is also evident through examination of the molecular weights of dry air (28.96 g mol −1 ) vs. water (18 g mol −1 )
For a moist air parcel, in effect, lighter water molecules have displaced a portion of heavier air molecules making the parcel lighter than it would be if it were all dry air!
Recall in the example of East Coast cyclogenesis:
air
H 2 O

9. A Few Other Moisture Variables
Saturation vapor pressure ( 𝑒 𝑠 )
Consider a closed box of initially dry air at temperature 𝑇 with a layer of water at the bottom
Depending on temperature, water molecules will begin to evaporate into the air
Once in vapor form, some of these molecules will then begin to condense back into the liquid water at the bottom
Once the rate of evaporation equals the rate of condensation, the air is said to be saturated with respect to water
The vapor pressure that exists at that point is 𝑒 𝑠 =
What is the effect of changing temperature?
Higher temp  more evaporation  higher 𝑒 𝑠

10. A Few Other Moisture Variables
Water vapor mixing ratio (𝑤)…or simply, mixing ratio: a ratio of the mass of water vapor 𝑚 𝑣 in a volume of air to the mass of dry air 𝑚 𝑑 in the volume:
𝑤≡ 𝑚 𝑣 𝑚 𝑑
Normally quoted in units of grams of water vapor per kilogram of dry air (though in calculations, water vapor mass must be converted to kg for the units to cancel)
If no phase changes of water are occurring in the volume (e.g., condensation, evaporation, etc.), or if they’re cancelling out, then 𝑤 is constant and is said to be conserved
In terms of mixing ratio, the calculation for virtual temperature becomes:
𝑇 𝑣 ≅𝑇(1+0.61𝑤)
Thus if no water vapor were present, 𝑤=0 and 𝑇 𝑣 =𝑇

11. A Few Other Moisture Variables
Similarly, saturation mixing ratio ( 𝑤 𝑠 ): a ratio of the mass of water vapor 𝑚 𝑣𝑠 in a volume of air that is saturated with respect to water to the mass of dry air 𝑚 𝑑 in the volume:
𝑤 𝑠 ≡ 𝑚 𝑣𝑠 𝑚 𝑑
Per the eqn of state
For a unit volume:
𝑤 𝑠 = 𝜌 𝑣𝑠 𝜌 𝑑 = 𝑒 𝑠 𝑅 𝑣 𝑇 (𝑝− 𝑒 𝑠 ) 𝑅 𝑑 𝑇 = 𝑒 𝑠 𝑅 𝑑 𝑇 (𝑝− 𝑒 𝑠 ) 𝑅 𝑣 𝑇
Since 𝑅 𝑑 𝑅 𝑣 =0.622 and, in Earth’s atmosphere, 𝑝≫ 𝑒 𝑠
𝑤 𝑠 = 𝑒 𝑠 𝑝
This states that for a given temperature, saturation mixing ratio is inversely proportional to pressure (makes sense because as 𝑝↓, 𝑚 𝑑 ↓ while 𝑚 𝑣𝑠 remains unchanged)

12. A Few Other Moisture Variables
Dew point ( 𝑇 𝑑 ): Qualitatively, the dew point is the temperature to which air must be cooled at constant pressure in order to become saturated with respect to water…that is, relative humidity (𝑅𝐻) becomes 100 percent
Once air is slightly over saturated with respect to water (supersaturated), the rate of condensation will exceed the rate of evaporation and liquid (visible) moisture will begin to appear (i.e., cloud droplets)
Mathematically, relative humidity is equal to 100 percent when the mixing ratio and saturation mixing ratio become equivalent
𝑅𝐻=100 𝑤 𝑤 𝑠