**Relation between water vapour and air pressure**

We know, if there is water vapour or if air is moist then its density also changes. Density of moist or vapour containing water vapour is less than that of dry air. That means, as the water vapour increases in air, its density also decreases.

In atmosphere there always exists some water vapour. Through evaporation process enough water are evaporated from canals, ponds, water reservoir, rivers, oceans etc and mixes with atmosphere. From cloud, rain, fog, dew etc natural phenomena it is proved that there is enough water vapour in air.

*Amount of water vapour in atmosphere depends on the sources of water, latitude, location of the place above the sea level etc.*

Suppose, air of mass in at pressure P_{1} and temperature T_{1}K, has volume of V_{1} and density ofρ_{1}. and at pressure P_{2} and temperature T_{2}K if the volume of that air is V_{2} and density is ρ_{2}, then

ρ_{1} = m/V_{1 } or, V_{1} = m/ρ_{1}

and, ρ_{2} = m/2 or, V_{2} = m/ρ_{2}

Now, inserting the values of V_{1} and V_{2} in the relation.

P_{1}V_{1}/T_{1} = P_{2}V_{2}/T_{2}

we get,

P_{1}m/ρ_{1}T_{1} = P_{2}m/ρ_{2}T_{2} = Constant

or, P_{1}/ρ_{1}T_{1} = P_{2}/ρ_{2}T_{2} = Constant

or, ρ_{1}T_{1}/ P_{1} = ρ_{2}T_{2}/P_{2 }= Constant …. ….. (1)

then, ρT/P = Constant

This equation gives the relation between pressure and density of water vapour at temperature T. If temperature remains constant, i.e., if, T_{1} = T_{2} then from equation (1)

we get, ρ_{1}/P_{1} = ρ_{2}/P_{2} = constant.

or, ρ/P = constant

so, ρ ∞ P