# 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

Sponsored Links

The density of air varies as the temperature and moisture content in the air varies. When the temperature increases the higher molecular motion results in an expansion of the volume and thus decreasing the density.

The density of a gas, either it is dry air, water vapor or a mixture of dry air and water vapor - moist or humid air, can be calculated on basis of 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 influences density. Water vapor is a relatively light gas compared to diatomic Oxygen and diatomic Nitrogen - the dominant components in air.

When water vapor content increases in the moist air the amount of Oxygen and Nitrogen decreases per unit volume and the density decreases because the mass is decreasing.

Based on the 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 will reduce the density of the moist air - **Dry air is more dense than moist air.**

Sponsored Links