Suppose an air conditioner is used to cool a room that is 7.44 m x 11.70 m x 2.19 m when the outside temperature is 28 °C. At this temperature, the vapor pressure of water is 28.4 torr. The partial pressure of water in the air is 85.5% of the vapor pressure of water.

How much water, in grams, is removed from the air each time the air from the room is cycled through the air conditioner?

mass of H₂O:

The mass of water removed from the air by an air conditioner can be calculated using the ideal gas law and vapor pressure. This involves determining the saturation point where the partial pressure equals the known vapor pressure at a specific temperature and then converting the moles of vapor to mass.

Explanation:

To calculate the mass of water removed from the air when cycling through an air conditioner, we use the ideal gas law and vapor pressure concepts. Given that the vapor pressure of water at 19°C is 0.02168 atm, we can determine the mass of water that would saturate the air in a closed room of 180m³.

Initially, there is no water vapor in the room. When the room is saturated with water vapor, the partial pressure of water vapor will be equal to the vapor pressure at 19°C. We use the ideal gas law (PV = nRT) to find the number of moles (n) of water vapor present in the room when saturated, where P stands for pressure (0.02168 atm), V for volume (180m³), n for moles of water, R for the ideal gas constant (0.0821 L·atm/K·mol), and T for temperature in Kelvin. The volume should be converted from cubic meters to liters. After finding the moles of water vapor, we then convert it to mass by using the molar mass of water (18.01528 g/mol).