The temperature in a room is kept at a constant 20 °C but the humidity is only 20%. If the room has a volume of 30m³, and the room’s door and windows closed, how many kilograms of water must be put into the vaporizer to increase the humidity from 20% to 85% ( the molecular weight of water is 18)
To find out how many kilograms of water must be put into the vaporizer, we need to calculate the amount of water vapor currently present in the room and then determine the additional amount required to reach the desired humidity level.
First, let's calculate the initial amount of water vapor in the room. We'll assume that the volume is completely filled with air.
1. Calculate the initial mass of air in the room:
Air has a density of approximately 1.2 kg/m³.
Mass of air = Density x Volume
Mass of air = 1.2 kg/m³ x 30 m³ = 36 kg
2. Calculate the initial moisture content in the air:
The humidity is given as a percentage. We can convert it to a decimal form:
Initial humidity = 20% = 0.20
Initial moisture content = Initial humidity x Mass of air
Initial moisture content = 0.20 x 36 kg = 7.2 kg
Next, we need to calculate the additional moisture content required to reach the desired humidity level.
1. Calculate the additional moisture content:
Desired humidity = 85% = 0.85
Additional moisture content = Desired humidity x Mass of air - Initial moisture content
Additional moisture content = 0.85 x 36 kg - 7.2 kg = 26.4 kg
Finally, we need to convert the additional moisture content to kilograms of water.
1. Calculate the mass of water:
The molecular weight of water is given as 18 g/mol, which is equivalent to 0.018 kg/mol.
Mass of water = Additional moisture content / Molecular weight of water
Mass of water = 26.4 kg / 0.018 kg/mol ≈ 1466.67 mol
However, we need to take into account that water vapor is diatomic, so we divide the number of moles by 2:
Mass of water = 1466.67 mol / 2 ≈ 733.33 kg
Therefore, approximately 733.33 kilograms of water must be put into the vaporizer to increase the humidity from 20% to 85% in the room.