# Density of Dry Air, Water Vapor and Moist Humid Air

## Calculate the density of dry air, water vapor or the mixture of air and water vapor - moist or humid air

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The density of humid air varies with temperature and moisture content in the air. When the temperature increases the higher molecular motion results in an expansion of the volume and decrease of density.

The density of a gas, dry air, water vapor or a mixture of dry air and water vapor like moist or humid air, can be calculated with the Ideal Gas Law.

### Density of Dry Air

The density of dry air can be expressed as:

ρ_{a}= 0.0035 p_{a}/ T(1)

where

ρ_{a}= density dry air (kg/m^{3})

p_{a}= partial pressure of air (Pa, N/m^{2})

T= absolute dry bulb temperature (K)

### Density of Water Vapor

The density of water vapor can be expressed as:

ρ_{w}= 0.0022 p_{w}/ T(2)

where

p_{w}= partial pressure water vapor (Pa, N/m^{2})

ρ_{w}= density water vapor (kg/m^{3})

T= absolute dry bulb temperature (K)

### Density of Moist Air - an Air Vapor Mixture

The amount of water vapor in air will influence the density. Water vapor is a relatively light gas compared to diatomic Oxygen and diatomic Nitrogen - the dominant components in air.

When the vapor content increase in moist air the amount of Oxygen and Nitrogen are decreased per unit volume and the density of the mix will also decrease since the mass is decreasing.

Based on specific volume of moist air density can be expressed as:

ρ = 1 / v

= (p / R_{a}T) (1 + x) / (1 + x R_{w}/ R_{a})(3)

where

v= specific volume of moist air per mass unit of dry air and water vapor (m^{3}/kg)

R_{a}= 286.9 - the individual gas constant air (J/kg K)

R_{w}= 461.5 - the individual gas constant water vapor (J/kg K)

x= specific humidity or humidity ratio (kg/kg)

p= pressure in the humid air (Pa)

Density of dry air can be expressed as:

ρ_{da}= p / R_{a}T(4)

where

ρ_{da}= density dry air (kg/m^{3})

Combining (4) and (3):

ρ = ρ_{da}(1 + x) / (1 + x R_{w}/ R_{a})(5)

The gas constant ratio between water vapor and air is

R_{w}/ R_{a }=(461.5 J/kg K) / (286.9 J/kg K)

= 1.609

Inserting the ratio in (5):

ρ = ρ_{da}(1 + x) / (1 +1.609x)(6)

**Note!** As we can see from (6) increased moisture content reduces the density of the moist air - **dry air is more dense than moist air.**

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