The total mass of water vapor in the air in the apartment can be calculated using the following equation:
Mass (kg) = Volume (m3) x Density (kg/m3)
Density of water vapor at 25°C and 56% relative humidity is 0.0075 kg/m3.
Therefore, the total mass of water vapor in the air in the apartment is:
Mass (kg) = 20 m x 10 m x 7 m x 0.0075 kg/m3 = 105 kg
An apartment has the dimensions 20 m by 10 m by 7 m. The temperature is 25°C, and the relative humidity is 56 percent. What is the total mass (in kg) of water vapor in the air in the apartment?
2 answers
It is way lower:
To find the mass of water at a given temperature and relative humidity, you have to know the mass of water vapor in a cubic meter of air when the air is saturated at the temperature.
First calculate the volume of the room in m3. Multiply the 3 dimensions together. Volume = 15*8*5 = 600 m3.
I found data for the vapor density of saturated water vapor at 20 C and 30 C which is 0.017 and 0.03 kg/m3.
Linearly interpolate to get the value at 25 C to get 0.0065 kg/m3. This is an approximation as the curve is not linear.
The total mass of water in the room at saturation is 600 m3 x 0.0065 kg/m3 = 3.9 kg. As the relative humidity is 57%, the actual mass of water vapor is 0.57 x 3.9 kg = 2 kg approximately.
So yeah, about 2kg.
To find the mass of water at a given temperature and relative humidity, you have to know the mass of water vapor in a cubic meter of air when the air is saturated at the temperature.
First calculate the volume of the room in m3. Multiply the 3 dimensions together. Volume = 15*8*5 = 600 m3.
I found data for the vapor density of saturated water vapor at 20 C and 30 C which is 0.017 and 0.03 kg/m3.
Linearly interpolate to get the value at 25 C to get 0.0065 kg/m3. This is an approximation as the curve is not linear.
The total mass of water in the room at saturation is 600 m3 x 0.0065 kg/m3 = 3.9 kg. As the relative humidity is 57%, the actual mass of water vapor is 0.57 x 3.9 kg = 2 kg approximately.
So yeah, about 2kg.