Answer:
To calculate the fraction of water that is condensed from the air, we need to compare the amount of water vapor in the incoming air to the amount of water vapor in the outgoing air.
First, let's calculate the amount of water vapor in the incoming air:
- The air contains 10.0 mole% water, which means that 10.0% of the air is water vapor.
- Since we know the total pressure of the incoming air is 1.0 atm, we can calculate the partial pressure of the water vapor using Dalton's law of partial pressures.
- The partial pressure of the water vapor is equal to 10.0% of 1.0 atm, which is 0.1 atm.
Next, let's calculate the amount of water vapor in the outgoing air:
- The temperature of the outgoing air is 15.6°C, and the pressure is 3.00 atm.
- Using the vapor pressure tables, we can find that the vapor pressure of water at 15.6°C is approximately 0.126 atm.
- The partial pressure of the water vapor in the outgoing air is 0.126 atm.
To calculate the fraction of water condensed, we can subtract the partial pressure of the water vapor in the outgoing air from the partial pressure of the water vapor in the incoming air:
0.1 atm - 0.126 atm = -0.026 atm.
Since we have a negative value, it means that no water is condensed from the air in this process.
Moving on to the relative humidity of the outlet air at 95.0°C:
- Relative humidity is the ratio of the partial pressure of water vapor in the air to the vapor pressure of water at that temperature, expressed as a percentage.
- The vapor pressure of water at 95.0°C is approximately 0.961 atm.
- The partial pressure of water vapor in the outlet air is equal to the vapor pressure of water at 95.0°C, which is 0.961 atm.
To calculate the relative humidity, we can use the formula:
Relative humidity = (Partial pressure of water vapor / Vapor pressure of water) x 100%
= (0.961 atm / 0.961 atm) x 100%
= 100%
Therefore, the relative humidity of the outlet air at 95.0°C is 100%.
Lastly, let's calculate the volume ratio of the outlet air at 95.0°C to the feed air at 85.0°C:
- To calculate the volume ratio, we need to consider the ideal gas law: PV = nRT.
- Since the process is isobaric, the pressure remains constant at 1.0 atm.
- The temperature of the outlet air is 95.0°C, which is equal to 368.15 K, and the temperature of the feed air is 85.0°C, which is equal to 358.15 K.
- The volume ratio can be calculated using the formula:
Volume ratio = (V_outlet air / V_feed air) = (n_outlet air x T_feed air) / (n_feed air x T_outlet air)
Here, we need to know the number of moles of air at both temperatures. However, the question only provides the mole fraction of water in the air, and not the mole fraction of air itself. As a result, we cannot accurately calculate the volume ratio without this information.
In summary:
- No water is condensed from the air during this process.
- The relative humidity of the outlet air at 95.0°C is 100%.
- The volume ratio of the outlet air at 95.0°C to the feed air at 85.0°C cannot be calculated accurately without the mole fraction of air.
Step-by-step explanation: