# How Much Water In Gallons Per Cubic Foot Of Air At 74 Degrees F And 40% Relative Humidity?

Relative humidity specifies the partial pressure of water relative to the saturation pressure of water at a given temperature. The saturation pressure of water at 74 °F can be found here. It is estimated to be 2857.14 Pa.
Using the ideal gas law, one can compute the density of water in grams per cubic meter.
P*V = n*R*T, where n is the number of moles. We can solve for n/V to get
n/V = P/(R*T)
Multiplying this by the number of grams per mole (18.01528), and dividing by the density of water in grams per cubic meter, we get the volume fraction that water occupies at the temperature of interest. In compatible units, we have R = 8.314472 J/°K/mol, T = 296.483 °K, water density at 74 °F is about 997451 g/m^3, so our volume fraction at saturation is
(2857.14*18.01528)/(8.314472*296.483*997451) ≈ 2.093*10^-5
At 40% relative humidity, we have 40% that value, about 8.37349*10^-6.
There are 1728/231 gallons per cubic foot, so you will have
1728/231*8.37349*10^-6 gallons ≈ 62.64*10^-6 gallons of water per cubic foot
It will take about 15,965 cubic feet of air at that temperature and humidity to hold 1 gallon of water.
I estimate the errors in the various formulas to be less than 1% in total. The vapor pressure of water is calculated from an approximate formula, and the density of water is calculated using linear interpolation of this chart. The ideal gas law is said to hold for water vapor over some reasonable temperature and pressure range.
Apparently, air pressure has nothing to do with it, except that it must be high enough to prevent boiling.
thanked the writer. 