Question

On a hot summer day in Washington, D.C., the temperature is 104°F, and the relative humidity is 78%. How much water vapor does each cubic meter of air contain?

______kg

Answer 1

temperature, T = 104 F = 40 C = 313K

The Saturation Vapour Pressure of air at 30 degree celsius is:
es=Ae^(−B/T)
Where A≈2.53×1011 Pa
and B=5420K. T is the temperature in kelvin.

es≈2.53×10(^11) * e−5420/313
es≈7633.96 Pa

Now the relative humidity is given by :
RH = {100*e)/ es
where , e denotes the actual vapour pressure .
e = {RH*es}/100 = {78*7633.96}/100   = 5954.488 Pa

THe specific humidiry is approx given by :
SH = 0.622 *(e/P) , where P = atmospheric pressure = 10^5 Pa
putting the values and calculating SH :
SH = 0.622*(5954.488/10^5) = 0.037 kg of moisture per kg of air
Now, the air density would be :
rho = P/RT
R = 287Jkg^-1 K^-1 is the gas constant , T = air temperature
rho = 10^5 / 287*313 = 1.113kg/m^3
hence the water vapour is
W = rho x SH = 1.113 * 0.037 = 0.0412 kg of moisture per m^3 of air

or approx 4.12 grams of water vapour per cubic meter of given air