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
Douglas K.
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"#

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