Question

A closed, rigid tank having a volume of 3 mº contains moist air in equilibrium with liquid water at 80°C. The respective masses present initially are 10.4 kg of dry air, 0.88 kg of water vapor, and 0.17 kg of liquid water. If the tank contents are heated to 160°C, determine (a) the final pressure, in bar. (b) the heat transfer, in kJ.

Answer 1

To calculate the mass of water that has passed into the air, we need to determine the final amount of water vapor in the air when the room becomes saturated. The initial amount of water vapor is known to be 0.88 kg. We can use the ideal gas law to calculate the final mass of water vapor and subtract the initial mass to find the mass of water that has passed into the air.

To calculate the mass of water that has passed into the air, we need to determine the final amount of water vapor in the air when the room becomes saturated. The initial amount of water vapor is known to be 0.88 kg. The equilibrium vapor pressure of water at 19°C is 0.02168 atm. From this information, we can use the ideal gas law to determine the final mass of water vapor:

Moles of water vapor = (Pressure * Volume) / (Gas constant * Temperature) = (0.02168 atm * 180 m³) / (0.0821 atm·L/mol·K * 19°C + 273.15)

Using the moles of water vapor, we can then calculate the mass of water vapor:

Mass of water vapor = Moles of water vapor * Molar mass of water = Moles of water vapor * 18.015 g/mol

Finally, we subtract the initial mass of water vapor from the final mass to find the mass of water that has passed into the air.