A solvent recovery system delivers a gas saturated with benzene (C6H6) vapor that analyzes on a benzene free basis as follows: CO2 - 15%; O2 - 4% and N2 - 81%. This gas is at 21oC and 750 mm Hg pressure. It is compressed to 5 atmospheres and cooled to 21oC after compression. How many kilograms of benzene are condensed by this process per 1000 m3 of the original mixture?
Data: Vapor pressure of benzene at 21oC = 75 mm Hg.
Calculations:
Basis: 100 moles of Benzene free vapor.
At saturation,
pA/(p - pA) = pS/(p - pS)
where pA is the partial pressure of benzene;
p is the total pressure of the system;
pS is the vapor pressure of benzene at system temperature.
Therefore,
pA = pS = 75 mm Hg
sum of the partial pressure of benzene free gases is = 750 - 75 = 675 mm Hg.
Moles of Benzene/mole of benzene free gases at a total pressure of 750 mm Hg= 75/675 = 0.1111
At a total pressure of 5 atm ( = 5 x 760 = 3800 mm Hg),
moles of benzene per mole of benzene free gases in the gas phase = 75/(3800 - 75) = 0.0201.
The remaining moles of benzene condense into liquid phase.
Initial mole fraction of benzene in the gas mixture = 75/750 = 0.1
Partial volume of benzene in the gas mixture at the initial conditions = 0.1 x 1000 = 100 m3
Number of moles of benzene in the gas mixture at the initial conditions:
Using ideal gas equation,
P1V1/n1T1 = P2V2/n2T2
760 x 22.4/(1 x 273) = 750 x 100/(n2 x 294)
n2 = 4.091 kmol.
i.e., no of moles of benzene in the gas mixture = 4.0909 kmol
And the number of moles of benzene free gas in the gas mixture = 40.909 - 4.0909 = 36.8181 kmol.
Number of moles of benzene condensed
= No of moles of benzene in the gas mixture at initial conditions - at final conditions
= 4.0909 - 0.0201 x 36.8181 = 3.3509 kmol.
Mass of benzene condensed
= 3.3509 x molecular weight of benzene = 3.3509 x 78 = 261.4 kgLast Modified on: 04-Feb-2022
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