Question

In a binary gas-liquid system, \(N_{A,EMD}\) is the molar flux of a gas \(A\) for equimolar counter diffusion with a liquid \(B\). \(N_{A,UMD}\) is the molar flux of \(A\) for steady one-component diffusion through stagnant \(B\). Using the mole fraction of \(A\) in the bulk of the gas phase as 0.2 and that at the gas-liquid interface as 0.1 for both the modes of diffusion, the ratio of \(N_{A,UMD}\) to \(N_{A,EMD}\) is equal to __________ (rounded off to two decimal places).

Hint:

EMD uses a \(\Delta x_A\) (linear) driving force; diffusion through stagnant \(B\) introduces a logarithmic correction \(\ln\!\left(\dfrac{1-x_{A2}}{1-x_{A1}}\right)\).
Ratios are convenient—common factors like \(D_{AB},\,C,\) and \(L\) cancel.

Answer 1

Correct Answer: 1.17

Step 1: For equimolar counter diffusion (EMD) in a binary gas at constant \(T,P\), \[ N_{A,EMD} \;=\; \frac{D_{AB}C}{L}\,(x_{A1}-x_{A2}). \] Given \(x_{A1}=0.2\) (bulk gas) and \(x_{A2}=0.1\) (interface). Step 2: For one-component diffusion of \(A\) through stagnant \(B\) (UMD), \[ N_{A,UMD} \;=\; \frac{D_{AB}C}{L}\,\ln\!\left(\frac{1-x_{A2}}{1-x_{A1}}\right). \] Step 3: Take the ratio (note that \(D_{AB},\,C,\) and \(L\) cancel): \[ \frac{N_{A,UMD}}{N_{A,EMD}} = \frac{\ln\!\left(\dfrac{1-x_{A2}}{1-x_{A1}}\right)}{x_{A1}-x_{A2}} = \frac{\ln\!\left(\dfrac{0.9}{0.8}\right)}{0.2-0.1}. \] Step 4: Evaluate numerically: \[ \ln\!\left(\frac{0.9}{0.8}\right)=\ln(1.125)\approx 0.11778, \quad\Rightarrow\quad \frac{N_{A,UMD}}{N_{A,EMD}} \approx \frac{0.11778}{0.1}=1.1778 \approx \boxed{1.18}. \]