A room has relative humidity of 20% at 20 degrees Celsius with dimensions 3m \times 3m \times 2.7m. Determine the amount of water vapour in grams and liters that must be added to the air of the room in order to raise the relative humidity to 60%?

1 Answer
Jan 26, 2018

"mass of water to be added" = 168.2 g

"Liters of water to be added" = 168.2xx10^-3" L"

Explanation:

The information that I am using to work this problem is obtained from this reference on Relative Humidity .

Compute the volume of the room:

3" m"xx 3" m"xx2.7" m" = 24.3" m"^3

Using the equation:

R.H. = "actual vapor density"/"saturated vapor density"xx100

and, at 20^@" C", the "saturated vapor density" = 17.3" g/m"^3 we may obtain the mass density of water vapor in the air at R.H. = 20%

"actual vapor density" = (20%)/100(17.3" g/m"^3)

"actual vapor density" = 3.46" g/m"^3

Please observe that the equation for relative humidity is linear, therefore, multiplying the relative humidity by 3 will multiply the actual vapor density by 3, which is the same as adding twice the original amount:

"change in actual vapor density" = 2(3.46" g/m"^3)

"change in actual vapor density" = 6.92" g/m"^3

We can compute the number of grams to be added by multiplying the above by the volume of the room:

"mass of water to be added" = (6.92" g/m"^3)(24.3" m"^3)

"mass of water to be added" = 168.2 g

To convert grams of water to liters of water we multiply by 1xx10^-3" L/g"

"Liters of water to be added" = 168.2xx10^-3" L"