Question

Solute \(A\) is absorbed from a gas into water in a packed bed operating at steady state. The absorber operating pressure and temperature are 1 atm and 300 K, respectively. At the gas-liquid interface, \(y_i = 1.5 x_i\),
where \(y_i\) and \(x_i\) are the interfacial gas and liquid mole fractions of \(A\), respectively. At a particular location in the absorber, the mole fractions of \(A\) in the bulk gas and in the bulk water are 0.02 and 0.002, respectively. If the ratio of the local individual mass transfer coefficients for the transport of \(A\) on the gas-side (\(k_y\)) to that on the water-side (\(k_x\)), \(\frac{k_y}{k_x} = 2\), then \(y_i\) equals __________ (rounded off to 3 decimal places).

Hint:

In absorption problems, the interfacial concentration can be derived from the bulk concentration using the given relationship and mass transfer coefficients.

Answer 1

Correct Answer: 0.014

Step 1: Given Data. \[ y_i = 1.5 x_i \quad {(relation between interfacial gas and liquid mole fractions)} \] \[ x_{{gas}} = 0.02 \quad {(mole fraction of \(A\) in the bulk gas)} \] \[ x_{{water}} = 0.002 \quad {(mole fraction of \(A\) in the bulk water)} \] \[ \frac{k_y}{k_x} = 2 \quad {(ratio of mass transfer coefficients)} \] Step 2: Apply the Relation Between \(y_i\) and \(x_i\). Using the relation \(y_i = 1.5 x_i\), we substitute: \[ y_i = 1.5 (x_{{water}}) \] Substitute \(x_{{water}} = 0.002\): \[ y_i = 1.5 (0.002) = 0.003 \] Final Answer: \(y_i = \boxed{0.003}\)