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).
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).
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Solution and Explanation
\[ y_i = 1.5 x_i \quad \text{(relation between interfacial gas and liquid mole fractions)} \] \[ y_{\text{bulk}} = 0.02 \quad \text{(mole fraction of \(A\) in the bulk gas)} \] \[ x_{\text{bulk}} = 0.002 \quad \text{(mole fraction of \(A\) in the bulk water)} \] \[ \frac{k_y}{k_x} = 2 \quad \text{(ratio of mass transfer coefficients)} \]
Step 2: Interfacial Composition Using Mass Transfer Coefficient Ratio.At steady state, the rate of mass transfer from gas to liquid is equal: \[ k_y (y_{\text{bulk}} - y_i) = k_x (x_i - x_{\text{bulk}}) \] Using the ratio: \[ \frac{k_y}{k_x} = 2 \Rightarrow k_y = 2k_x \] Substituting: \[ 2k_x (0.02 - y_i) = k_x (x_i - 0.002) \] Divide both sides by \(k_x\): \[ 2(0.02 - y_i) = x_i - 0.002 \tag{1} \]
Step 3: Use Interfacial Equilibrium Relation.From equilibrium: \[ y_i = 1.5 x_i \Rightarrow x_i = \frac{y_i}{1.5} \] Substituting into equation (1): \[ 2(0.02 - y_i) = \frac{y_i}{1.5} - 0.002 \] Multiply both sides by 1.5: \[ 3(0.02 - y_i) = y_i - 0.003 \] \[ 0.06 - 3y_i = y_i - 0.003 \] \[ 0.063 = 4y_i \Rightarrow y_i = \frac{0.063}{4} = 0.01575 \]
Final Answer: The interfacial mole fraction of A in the gas phase is \( \boxed{0.0158} \) (rounded to 4 decimal places).Top GATE CH Chemical, Thermal and Polymer Engineering Questions
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