Question

43 lb. moles of CO2, 6 lb. moles of water vapor and 195 lb. moles of N2 and 10 lb. moles of oxygen flow from a stack at 121 C. What is the partial pressure of water vapor in the flue gas in mm Hg.

Answer 1

43 lb moles of CO2

6 lb moles water

195 lb moles N2

10 lb moles O2

to calculate partial Pressure of water vapour in flue gas in mm Hg

total lb moles of gaseous mixture = 43 + 6+195 +10 = 254 lb moles

to calculate partial pressure of water vapour we should know total pressure of flue gases first

but as we aren't provided with sufficient information to calculate total Pressure nor we are provided with pressure data

they should give you volume so you can use ideal gas law to get Pressure of gaseous mixture

so we can assume that flue gases are at atmospheric pressure i.e 1 atm

1 atm = 101.325kPa

the partial pressure p_i= y_iP_T

y_i= mole fraction of species i in gaseous mixture &

P_T = total pressure of flue gas

p_i = partial pressure of species i

for water y_water = 6/(43+6+195+10) = 0.02362204724

therefore

p_water = 0.02362204724*101.325 =2.393503937 kPa

we can convert it into mm of Hg

1 kPa = 7.50062 mm of Hg

so ,

2.393503937 kPa =17.95275334779 mm of Hg

so partial pressure of water vapour in flue gas = 17.95275334779 mm of Hg