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 change in mass of water vapor in the air as we increase the humidity from 20% to 85%.
Here's the step-by-step process to calculate it:
Step 1: Calculate the initial and final mass of water vapor:
The initial mass of water vapor in the air can be calculated using the formula:
Initial mass = initial relative humidity * saturation vapor pressure * volume / (specific gas constant * temperature)
Given:
Initial relative humidity = 20% = 0.20
Saturation vapor pressure at 20°C = 2.338 kPa (source: steamtablesonline.com)
Volume = 30 m³
Specific gas constant (R) = 8.314 J/(mol·K) (source: engineeringtoolbox.com)
Temperature = 20°C = 20 + 273.15 K = 293.15 K
Calculating the initial mass of water vapor:
Initial mass = 0.20 * 2.338 kPa * 30 m³ / (8.314 J/(mol·K) * 293.15 K)
Step 2: Calculate the final mass of water vapor:
The final mass of water vapor can be calculated using the same formula as above, but with the final relative humidity:
Final relative humidity = 85% = 0.85
Calculating the final mass of water vapor:
Final mass = 0.85 * 2.338 kPa * 30 m³ / (8.314 J/(mol·K) * 293.15 K)
Step 3: Find the change in mass of water vapor:
Change in mass = Final mass - Initial mass
Step 4: Convert the change in mass from moles to kilograms:
The molecular weight of water (H₂O) is 18 g/mol.
So, change in mass (in kilograms) = change in mass (in moles) * molecular weight (in kg/mol)
Now you have the amount of water required to increase the humidity from 20% to 85% in kilograms.