Question

Air at 26 degrees C that is polluted with benzene vapor (1.80% benzene, 98.2% dry air) needs to be cleaned before being released to the atmosphere for safety reasons. One technique that can be done is by compressing the gas mixture so that 95% of the benzene condenses and is separately recovered as a liquid. Assuming constant temperature (isothermal) compression, to what pressure must you compress the polluted air to recover the required amount of benzene?

The following constants are used for benzene (Antoine equation):

A = 6.87987

B = 1196.76

C = 219.161

Please help me solve in details. Thank you!

Answer 1

Antonie constants of benzene

A: 6.87987

B: 1196.76

C: 219.161

Antonie equation is

P* = 10^{(A-(B/(T+C) )}

P* - mmHg

T = °C

P* = 10^{(6.87987-(1196.76/(26+219.161) )}

P* = 99.6192 mmHg

1 atm = 760 mmHg

99.6192 = 0.13107 atm

Saturation or vapor pressure of benzene = 0.13107 atm

The given data is assumed to be in mole %

Initial mole fraction of benzene in air = 0.018

Basis : 1 mole feed

Amount of benzene in feed = 1(0.018) =

0.018 mole

L- condensed benzene

95% of benzene is to be condensed

L = 0.95(0.018) = 0.0171 moles

V - outlet gas stream

Doing overall material balance

1 = L+ V

1 = 0.0171+ V

V = 0.9829 moles

Doing benzene balance we get

1(0.018) = 0.0171(1) + 0.9829(y)

y = 0.00091565

Since liquid and outlet gas is in equilibrium

Saturation pressure = partial pressure =

0.13107 atm

y = partial pressure /(total pressure)

0.00091565= 0.13107/P

P = 143.144 atm

I mole is Condenser 343 1h atm, 26 el

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