Question

One hundred moles of moist air (100 moles) at 25oC and 100 kPa has a dew point of 16oC. If you want to remove 50% of the initial moisture in the air (at a constant pressure of 100 kPa).

1. Sketch the process.

2. Determine to what temperature should you cool the air? Assume that after removal of 50% of moisture from the air, the final air is saturated.

Answer 1

Vapor pressure of water at 16 deg.c = 13.6 mm Hg =13.6/760 atm =0.017895 atm

101. 3 Kpa= 1 atm 100 KPa =100/101.3= 0.987 atm

Moles of water vapor/ total moles= partial pressure of water vapor/ total pressure

Moles ofwater vapor/ 100= 0.017895/0.987

Moles of water vapor= 100*0.017895/0.987=1.813 moles of water vapor

Moles of dry air =100-1.813= 98.187 moles

Mosture to be removed= 1.813*0.5= 0.9065

Moles remaining = 0.9065

Total moles after removal of water vapor = 0.9065 ( moles of water vapor remaining )+98.187 ( moles of dry air )=99.0935

Partial presssure of water vapor/ Total pressure= moles of water vapor/ total moles

After removing some moisture by cooling

Partial pressure of water vapor/ 0.987= 0.9065/99.0935

Partial pressure of water= 0.987*0.9065/99.0935=0.009029 atm =6.86 mm

At saturation partial pressure= vapor pressure

So temperature correspoding to this pressure is the temperature the gas needs to be cooled to

And this temperature is 5 deg,c

The gas need s to be cooled to 5 deg.c