Question

In an absorption column, the gas enters at y$_1$ = 0.10 and exits at y$_2$ = 0.01. The solvent enters pure (x$_2$ = 0). Using a liquid-gas ratio (L/G) of 2, the exit liquid mole fraction x$_1$ is:

• A. 0.045
• B. 0.09
• C. 0.05
• D. 0.02

Hint:

The equation \( G \cdot \Delta y = L \cdot \Delta x \) is the operating line equation for an absorption column.
Representing this line on a \( y \)-\( x \) plot helps visualize the driving force for mass transfer.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The question asks us to calculate the outlet liquid mole fraction (\( x_1 \)) of a solute leaving the bottom of an absorption column.
We are given the inlet and outlet gas-phase mole fractions (\( y_1 \) and \( y_2 \)), the inlet liquid mole fraction (\( x_2 \)), and the liquid-to-gas ratio (\( L/G \)).

Step 2: Key Formula or Approach:
The overall material balance for a counter-current absorption column is:
\[ G \cdot (y_1 - y_2) = L \cdot (x_1 - x_2) \]
where \( G \) is the gas flow rate, \( L \) is the liquid solvent flow rate, \( y \) is the gas mole fraction, and \( x \) is the liquid mole fraction.
Rearranging this equation yields:
\[ y_1 - y_2 = \frac{L}{G} \cdot (x_1 - x_2) \]

Step 3: Detailed Explanation:

• Identify the given values:
Inlet gas mole fraction (at the bottom), \( y_1 = 0.10 \)
Outlet gas mole fraction (at the top), \( y_2 = 0.01 \)
Inlet liquid mole fraction (at the top), \( x_2 = 0 \) (pure solvent)
Liquid-to-gas ratio, \( L/G = 2 \)

• Set up the material balance equation:
\[ 0.10 - 0.01 = 2 \cdot (x_1 - 0) \]

• Perform the arithmetic:
\[ 0.09 = 2 \cdot x_1 \]

• Solve for \( x_1 \):
\[ x_1 = \frac{0.09}{2} = 0.045 \]


Step 4: Final Answer:
The exit liquid mole fraction \( x_1 \) is 0.045.