A concert hall of volume 2000 m3

contains air at a temperature of 25 ◦C and relative humidity 80 %. Given
that the density of water vapour in saturated air at 25 ◦C is 22.8 g/m3 and that the temperature and
pressure of the concert hall remain constant what mass of water vapour must be removed from this air to
reduce the relative humidity to 50 %?

To get the humidity change: 0.8-0.5=0.3

Humidity change times density of water vapour gives the density of water vapour.
0.3*22.8= 6.84
This times volume gives the mass of water vapour to be removed.
2000*6.84=13.7

To find the mass of water vapor that needs to be removed from the air in order to reduce the relative humidity, we can follow these steps:

Step 1: Calculate the initial mass of water vapor in the air.
- First, we need to calculate the saturation vapor density at 25°C, which is the maximum amount of water vapor that can be present in the air at this temperature. The given value is 22.8 g/m³.
- The relative humidity of 80% means that the air is holding 80% of the maximum amount of water vapor it can hold at 25°C.
- Therefore, the initial mass of water vapor can be calculated as (80% of saturation vapor density).

Initial mass of water vapor = 0.8 * 22.8 g/m³

Step 2: Calculate the final mass of water vapor required to achieve a relative humidity of 50%.
- We need to find the new relative humidity which is 50%.
- Therefore, the final mass of water vapor can be calculated as (50% of saturation vapor density).

Final mass of water vapor = 0.5 * 22.8 g/m³

Step 3: Find the difference between the initial and final masses of water vapor.
- The mass of water vapor that needs to be removed can be calculated by subtracting the final mass from the initial mass.

Mass of water vapor to be removed = Initial mass - Final mass

Now, you can plug in the values and calculate the required mass of water vapor to be removed from the air in the concert hall.