How to calculate amount of water in m^3 Air?

you could look at a psychrometric chart like this one

http://www.truetex.com/psychrometric_chart.gif

Or you could do this...

from ln(P1/P2) = (dHvap/R) x (1/T2 - 1/T1)

where

P1 = unknown vapor pressure of water at T1

T1 = the temperature you are "given"

P2 = a reference pressure.. like 760mmHg @ 373.15K

.. . ..ie.. the normal bp of H2O

T2 = 373.15K

dHvap = heat of vaporization of water = -40680 J/mol

R = 8.314 J/molK

so.. you plug in the data for your given Temp and calculate P1.. and that is MAXIMUM vapor pressure of water at that temperature, then you can multiply relative humidity by that number (for example, if your at 23°C, that P1 will be about 0.0277atm. then 60% rh would have vapor pressure = 0.6 x .0277 atm = 0.0166atm

then..

... PV = nRT

... n = mass / mw

so that

... mass = mw x PV/(RT)

... .. ... .. = (18.02 g/mol) x (0.0166atm) x (1m³ x 1000L/1m³) / (0.08206 Latm/molK x 296K)

... .. ..... .= 12.3g

and of course

.. mass air = (29 g/mol) x (1atm - 0.0166atm) x (1m³ x 1000L/1m³) / (0.08206 Latm/molK x 296K)

... .. ..... ... = 1174g

so that mass H2O / mass dry air = 0.010lb H2O / lb dry air

and as you can see from that chart the actual value is 0.0092 and I'm only off by 10% which is probably due to the fact that I'm using the clausius clapeyron equation and ideal gas law to estimate vapor pressure and PVnT relationship.

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that help?