Answer :
Answer:
The mass of water present per 100 kg-mol of dry air is 3.17 kg-mol.
Explanation:
Given parameters:
The partial pressure of water vapor in the entering air = 8.7 kPa
The temperature of the entering air = 60°C
The pressure of entering air = 98.5 kPa
We are to calculate the number of kg-mol of water present per 100 kg-mol dry air
Solution: The mole fraction of water vapor in the air can be calculated as:
Y = Pv / PaPv = Partial pressure of water vapor in the entering air = 8.7 kPaPa = Total pressure of entering air = 98.5 kPaY = 8.7 / 98.5Y = 0.08817
Therefore, the mole fraction of water vapor in the air is 0.08817.
The total number of moles of air present can be calculated as:nA = PA * VA / RT... (i)
where, PA = Total pressure of entering air = 98.5 kPaVA = Volume of entering air = 1 kg-mol dry air
R = Universal gas constant = 8.314 J / mol K... (ii)
T = Temperature of entering air = 60 + 273 = 333 K
T = 333 K
Substituting the given values in equation (i)
nA = 98.5 × 1000 / (8.314 × 333) = 35.89 kg-mol
Therefore, the total number of moles of air present is 35.89 kg-mol. The mass of water present per 100 kg-mol of dry air can be calculated as:
nw = nA × Y
nw = 35.89 × 0.08817
nw = 3.17 kg-mol
Therefore, the mass of water present per 100 kg-mol of dry air is 3.17 kg-mol.
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