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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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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Step 1: Given Data. $$ y_i = 1.5 x_i \quad \text{(relation between interfacial gas and liquid mole fractions)} $$ $$ y_{\text{bulk}} = 0.02 \quad \text{(mole fraction of $A$ in the bulk gas)} $$ $$ x_{\text{bulk}} = 0.002 \quad \text{(mole fraction of $A$ in the bulk water)} $$ $$ \frac{k_y}{k_x} = 2 \quad \text{(ratio of mass transfer coefficients)} $$ 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).

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