Question

In an absorber the equilibrium curve is always:

• A. Parallel to the operating line
• B. Above the operating line
• C. Below the operating line
• D. Irrelevant to the operating line

Hint:

Graphical positions on a \(y-x\) mass transfer diagram: - Absorption (Gas \(\rightarrow\) Liquid): Solute enters the liquid, so \(y_{\text{actual}} > y^* \rightarrow\) The Operating Line lies ABOVE the Equilibrium Curve (Equilibrium curve is below). - Stripping Desorption (Liquid \(\rightarrow\) Gas): Solute leaves the liquid, so \(y_{\text{actual}} < y^* \rightarrow\) The Operating Line lies BELOW the Equilibrium Curve (Equilibrium curve is above).

Answer 1

The Correct Option is C

Concept: Gas absorption is a mass transfer operation where a soluble solute present in a gas mixture is dissolved into a liquid solvent. For a solute to transfer spontaneously from the gas phase into the liquid phase, there must be a positive mass transfer driving force. This means that at any point inside the absorption column, the actual partial pressure or mole fraction of the solute in the bulk gas phase ($y$) must be greater than the equilibrium vapor pressure or mole fraction ($y^*$) that corresponds to the solute concentration in the liquid phase ($x$).

Step 1: Examining the mathematical operating line equation.
Let us perform a steady-state material balance around one end of a countercurrent absorption column. The operating line equation, which maps the actual gas composition ($y$) against the actual liquid composition ($x$) at any cross-section in the column, is given by: \[ y = \frac{L_s}{V_s} \cdot x + \left( y_1 - \frac{L_s}{V_s} \cdot x_1 \right) \] Where $L_s$ and $V_s$ represent the constant molar flow rates of the solute-free solvent and carrier gas, respectively.

Step 2: Comparing the operating line and equilibrium line on a y-x plot.
Let us plot both relationships on a standard coordinate diagram where the solute mole fraction in the gas phase ($y$) is plotted on the vertical axis and the solute mole fraction in the liquid phase ($x$) is plotted on the horizontal axis:
• Operating Line: Represents the actual, real-time operating concentrations ($x, y$) present within the column.
• Equilibrium Curve: Represents the theoretical limit of mass transfer ($y^* = f(x)$), where the gas and liquid phases are in thermodynamic equilibrium. Because absorption requires the solute to transfer from the gas phase into the liquid phase, the actual gas phase concentration $y$ must always exceed the equilibrium concentration $y^*$ at every position along the column: \[ y_{\text{operating}} > y^*_{\text{equilibrium}} \quad \text{for a given value of } x \] On a standard $y-x$ plot, this inequality means that the coordinates of the operating line lie at a higher vertical position than those of the equilibrium curve. Therefore, in an absorption column, the operating line is situated entirely above the equilibrium curve. Flip this perspective around to evaluate the position of the equilibrium curve relative to the operating line: the equilibrium curve is always located below the operating line.