A room is filled with saturated moist air at 25°C and a total pressure of 100 kPa. If the mass of dry air in the room is 100 kg, the mass of water vapor is: a. 0.52 kg
b. 1.97 kg
c. 2.96 kg
d. 2.04 kg
e. 3.17 kg

the mass of water vapor is=d. 2.04 kg

Explanation:

The humidity ratio by mass is given as

x = mw / ma                                  (1)

where

mw = mass of water vapor

ma = mass of dry air

And also humidity ratio by pressure by  Ideal Gas Law is given as

x = 0.62198 pw / (pa - pw)                            (2)

where

pw = partial pressure of water vapor in moist air

pa = atmospheric pressure of moist air

Equating both equation we have that

Humidity ratio =  0.62198 pw / (pa - pw)= mw / ma      (3)

From the from the water table at 25°C,

water vapor partial pressure = 23.8 torr

 1 atm =760 torr.

1 atm = 101.325 kPa,

So, 760 torr =101.325 kPa,

Therefore 23.8 torr =3.173kPa

Humidity ratio =  0.62198 pw / (pa - pw)= mw / ma  

0.62198 x 3.173 / (100-3.173)= mw / 100

1.9736 /96.827=mw/100

0.0204=mw/100

mw=2.04kg