How to calculate percentage of water vapour that condenses?

[Rhaegal]

Homework Statement

1kg of moist air of RH 70% at 21^oC is cooled at a constant pressure of 1 bar to 5^oC. The vapour pressure at 21^oC and 5^oC are 0.025 bar and 0.0087 bar. The percentage of water vapour that condenses into water at 5^oC is?

Want to check whether my intuition & logic are correct.

Homework Equations

RH = Vapour Pressure/ Saturation Pressure

The Attempt at a Solution

Before cooling, Saturation pressure = 0.025/0.7 = 1/28 bar

As cooling & dehumidification occurs at constant pressure, Saturation pressure before cooling = saturation pressure after cooling.
RH after cooling = 0.0087 * 28 = 0.2436

% decrease of RH = ((0.7-0.2436)*100)/0.7 = 65.2 %
which equals percentage of water vapour that condenses into water.

Am I missing anything here? Because I cannot find any text to support this type of calculation.
Thanks in advance!

 

[Chestermiller]

When they say that the vapor pressure at 21 C is 0.025 bars, what they mean is that this is the saturation vapor pressure at 21 C. This being the case, in the 1 kg of moist air under consideration, what is the partial pressure of water vapor at 21 C? What is the mole fraction of water vapor and what is the mole fraction of air? What is the mass fraction of water vapor and the mass fraction of air? How many kg of dry air are the in 1 kg of the moist air? How many kg or water are in the 1 kg of moist air?

 

[Rhaegal]

When they say that the vapor pressure at 21 C is 0.025 bars, what they mean is that this is the saturation vapor pressure at 21 C. This being the case, in the 1 kg of moist air under consideration, what is the partial pressure of water vapor at 21 C? What is the mole fraction of water vapor and what is the mole fraction of air? What is the mass fraction of water vapor and the mass fraction of air? How many kg of dry air are the in 1 kg of the moist air? How many kg or water are in the 1 kg of moist air?

So Humidity ratio = 0.622Pv/(Pa - Pv) = m_v / m_a
From this, mass of dry air m_a = 0.9890 kg & mass of water vapor m_v = 0.011 kg
Mass fraction of water vapor = 11/1000 & Mass fraction of dry air = 989/1000
Mole fraction of dry air = 0.9824 & Mole fraction of water vapor = 0.0176
I do not know what effect cooling & dehumidification has on these values. I feel I have to use the ideal gas equation but am confused about the P,V values for both the states.

 

[Chestermiller]

So Humidity ratio = 0.622Pv/(Pa - Pv) = m_v / m_a
From this, mass of dry air m_a = 0.9890 kg & mass of water vapor m_v = 0.011 kg
Mass fraction of water vapor = 11/1000 & Mass fraction of dry air = 989/1000
Mole fraction of dry air = 0.9824 & Mole fraction of water vapor = 0.0176
I do not know what effect cooling & dehumidification has on these values. I feel I have to use the ideal gas equation but am confused about the P,V values for both the states.

The amount of dry air in the cooled air would not change, and would still be 989.1 g, which is 989.1/29=34.11 moles. In the final air at 5 C, the partial pressure of water vapor in this air would be 0.0087 bars, and the mole fraction water vapor would be 0.0087. So the mole fraction of the dry air is 1-0.0087=0.9913. So the total number of moles of gas in the final sample would be 34.11/0.9913=34.41 moles. So the number of moles of water vapor would be (34.41)(0.0087)
=0.299 moles of water. The mass of this number of moles of water is (0.299)(18)=5.39 gm.

So, the amount of water that condenses from the moist air = 10.9 - 5.4 = 5.5 gm. So the fraction of water that condenses is 5.5/10.9=0.505 = 50.5%

 

[Rhaegal]

The amount of dry air in the cooled air would not change, and would still be 989.1 g, which is 989.1/29=34.11 moles. In the final air at 5 C, the partial pressure of water vapor in this air would be 0.0087 bars, and the mole fraction water vapor would be 0.0087. So the mole fraction of the dry air is 1-0.0087=0.9913. So the total number of moles of gas in the final sample would be 34.11/0.9913=34.41 moles. So the number of moles of water vapor would be (34.41)(0.0087)
=0.299 moles of water. The mass of this number of moles of water is (0.299)(18)=5.39 gm.

So, the amount of water that condenses from the moist air = 10.9 - 5.4 = 5.5 gm. So the fraction of water that condenses is 5.5/10.9=0.505 = 50.5%

I overlooked the fact that at the final state, vapor pressure is equal to saturation pressure. Thanks for helping me approach the problem correctly