It is proposed to absorb acetone from air using water as a solvent. Operation is at 10 atm and is isothermal at 20°C. The total flow rate of entering gas is 10 kmol /h. The entering gas is 1.2 mol% acetone. Pure water is used as the solvent. The water flow rate is 15 kmol/h. The desired outlet gas concentration should be 0.1 mol % acetone. For this system, Henry's law holds and Ye = 1.5 X where Ye is the mol fraction of acetone in the vapour in equilibrium with a mol fraction X in the liquid.
KGa = 0.4 kmol*m^-3*s^-1
1. Draw a schematic diagram to represent the process.
2. Determine the mole fraction of acetone in the outlet liquid.

The meole fraction of acetone in the outlet liquid is

Explanation:

1.

The schematic diagram to represent this process is shown in the diagram attached below:

2.

the mole fraction of acetone in the outlet liquid is determined as follows:

solute from Basis Gas flow rate

Let the entering mole be : % = 0.012

Let the outlet gas concentration be = 0.1% = 0.001

Thus; the mole fraction of acetone in the outlet liquid is:

The mole fraction of acetone in the outlet liquid is

answer:

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