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Air conditioners dry the air while they cool the air. Suppose an air conditioner is used to cool a room that is [tex]7.90 \, \text{m} \times 11.52 \, \text{m} \times 2.25 \, \text{m}[/tex] when the outside temperature is [tex]37 \, ^\circ\text{C}[/tex]. At this temperature, the vapor pressure of water is [tex]47.1 \, \text{torr}[/tex]. The partial pressure of water in the air is [tex]81.5\%[/tex] 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?

Answer :

The resulting mass is the amount of water removed from the air each time the air in the room cycles through the air conditioner.

To calculate how much water is removed from the air by the air conditioner, we first need to determine the amount of water vapor initially present in the air in the room. The question specifies that the partial pressure of water in the air is 81.5% of the vapor pressure of water at 37 °C, which is 47.1 torr.

The room's volume is given as 7.90 m x 11.52 m x 2.25 m. First, we calculate the room's volume:

  • Volume = length × width × height
  • Volume = 7.90 m × 11.52 m × 2.25 m
  • Volume = 205.416 m³

Next, we calculate the partial pressure of the water vapor in the room:

  • Partial pressure = 81.5% of vapor pressure at 37 °C
  • Partial pressure = 0.815 × 47.1 torr
  • Partial pressure = 38.3915 torr

Now we need to convert the pressure from torr to atmospheres, as the ideal gas law uses pressure in atmospheres:

  • 1 atm = 760 torr
  • Partial pressure in atm = 38.3915 torr / 760 torr/atm
  • Partial pressure in atm = 0.050515 atm

Using the ideal gas law (PV = nRT), we can calculate the number of moles of water vapor (n) in the room. We will use R = 0.0821 L·atm/mol·K and convert the temperature from Celsius to Kelvin by adding 273 to the Celsius temperature:

  • T(K) = 37 + 273 = 310 K
  • V(L) = 205,416 m³ × 1000 L/m³ = 205,416,000 L
  • n = PV / RT
  • n = (0.050515 atm × 205,416,000 L) / (0.0821 L·atm/mol·K × 310 K)
  • n = moles of water vapor

Finally, we can calculate the mass of water vapor (m) using the molar mass of water (18.015 g/mol):

  • m = n × molar mass of water
  • m = moles of water vapor × 18.015 g/mol
  • m = mass of water in grams

The resulting mass is the amount of water removed from the air each time the air in the room cycles through the air conditioner.

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