Question

4 liter sealed container contains water vapour at 0.4 atm and 90°C. a) What is the dew point of the vapour? b) Determine the percentage of the vapour that will condense if the system is cooled to 50°C.

Answer 1

Partial pressure of water vapor =0.4*760= 304 mm Hg

Dew point is the temperature at which the partial pressure = vapor pressure

From vapor pressure data, it has to be seen what the temperature at which the vapor pressure correspond to 304mm Hg =40.53 Kpa

The saturation pressure connecting vapor pressure and temperature for water is given by Antoine equation

Log P =A- B/(C+T)

Log(304)= 8.07131

A= 8.07131

B= 1730.63 C= 233.426

Log (304) = 8.07131- 1730.63/ (233.426+C)

2.48= 8.07131- 1730.63/ (233.426+T)

1730.63/ (233.426+T)= 8.07131-2.48=5.5913

233.426+T =1730.63/ 5.5913= 309.5

T=309.5-233.426=76.07 deg.c

Total moles of water from PV=nRT

Therefore n= PV/RT= 0.4*4/{0.08206*(90+273.15)} =0.053961 moles

When the system is cooled to 50 deg.c

Partial pressure of water vapor = 92.5876 mm Hg =0.122 atm

Moles of water at 50 deg.c = 0.122*4/ 0.08206*(50+273.15)=0.0184 moles

Moles of water condensed = 0.053961-0.01843= 0.035288 moles

% of water vapor condensed = 100*(0.035288/0.053961)=65.395%