Molar Flux Ratio: EMD vs UMD Calculation
This solution calculates the ratio between the molar flux of gas A during unidirectional diffusion through stagnant liquid B ($N_{A,UMD}$) and equimolar counter diffusion ($N_{A,EMD}$) in a binary gas-liquid system.
Equimolar Counter Diffusion (EMD) Flux
For equimolar counter diffusion (EMD), the molar flux of A ($N_{A,EMD}$) is proportional to the difference in mole fractions across the diffusion path. The formula is:
$ N_{A,EMD} = C_{total} D_{AB} \frac{(y_{A,bulk} - y_{A,interface})}{\delta} $
Where:
- $C_{total}$ is the total molar concentration.
- $D_{AB}$ is the diffusivity of gas A in species B.
- $\delta$ is the diffusion path thickness.
- $y_{A,bulk}$ is the mole fraction of A in the bulk gas phase.
- $y_{A,interface}$ is the mole fraction of A at the gas-liquid interface.
Unidirectional Diffusion Through Stagnant B (UMD) Flux
For unidirectional diffusion through stagnant B (UMD), where the molar flux of B ($N_B$) is zero, the molar flux of A ($N_{A,UMD}$) is given by:
$ N_{A,UMD} = \frac{C_{total} D_{AB}}{\delta} \ln\left(\frac{1-y_{A,interface}}{1-y_{A,bulk}}\right) $
Calculating the Flux Ratio
The ratio $\frac{N_{A,UMD}}{N_{A,EMD}}$ is found by dividing the UMD flux by the EMD flux. Assuming $C_{total}$, $D_{AB}$, and $\delta$ are constant for both modes:
$ \frac{N_{A,UMD}}{N_{A,EMD}} = \frac{\frac{C_{total} D_{AB}}{\delta} \ln\left(\frac{1-y_{A,interface}}{1-y_{A,bulk}}\right)}{C_{total} D_{AB} \frac{(y_{A,bulk} - y_{A,interface})}{\delta}} $
This simplifies to:
$ \frac{N_{A,UMD}}{N_{A,EMD}} = \frac{\ln\left(\frac{1-y_{A,interface}}{1-y_{A,bulk}}\right)}{y_{A,bulk} - y_{A,interface}} $
Applying Given Values
The problem provides the following mole fractions:
- $y_{A,bulk} = 0.2$
- $y_{A,interface} = 0.1$
Substitute these values into the ratio formula:
$ \frac{N_{A,UMD}}{N_{A,EMD}} = \frac{\ln\left(\frac{1 - 0.1}{1 - 0.2}\right)}{0.2 - 0.1} $
$ \frac{N_{A,UMD}}{N_{A,EMD}} = \frac{\ln\left(\frac{0.9}{0.8}\right)}{0.1} $
$ \frac{N_{A,UMD}}{N_{A,EMD}} = \frac{\ln(1.125)}{0.1} $
Final Result
Calculate the natural logarithm:
$ \ln(1.125) \approx 0.117783 $
Now, compute the ratio:
$ \frac{N_{A,UMD}}{N_{A,EMD}} \approx \frac{0.117783}{0.1} \approx 1.17783 $
Rounding the result to two decimal places gives:
$ \frac{N_{A,UMD}}{N_{A,EMD}} \approx 1.18 $
Important Questions from Molecular Diffusion
Consider a non-reactive binary mixture of two ideal gases A and B in an isothermal horizontal closed tube. Due to the concentration difference at the two ends of the tube, equimolal counterdiffusion occurs along the axial direction. No concentration gradients exist in the radial direction. The profile for partial pressure of component A ($p_A$) as a function of axial distance $z$ is shown in the figure. The partial pressure of component B ($p_B$) approaches zero at $z = 0.8 \text{ m}$. At a location $z = z_1$ the partial pressures of the components A and B are equal. The value of $z_1$ (in m) is _____ (rounded off to two decimal places).
- For a binary mixture of components A and B, $N_A$ and $N_B$ denote the total molar fluxes of components A and B, respectively. $J_A$ and $J_B$ are the corresponding molar diffusive fluxes. Which of the following is true for equimolar counter-diffusion in the binary mixture?
- Consider the following two cases for a binary mixture of ideal gases A and B under steady state conditions. In Case 1, the diffusion of A occurs through non-diffusing B. In Case 2, equimolal counter diffusion of A and B occurs. In both the cases, the total pressure is 100 kPa and the partial pressures of A at two points separated by a distance of 10 mm are 10 kPa and 5 kPa. Assume that the Fick's first law of diffusion is applicable. What is the ratio of molar flux of A in Case 1 to that in Case 2?
- Consider two stationary spherical pure water droplets of diameters $d_1$ and $2d_1$. $CO_2$ diffuses into the droplets from the surroundings. If the rate of diffusion of $CO_2$ into the smaller droplet is $W_1$ mol $s^{-1}$, the rate of diffusion of $CO_2$ into the larger droplet is
- Consider steady-state diffusion in a binary A-B liquid at constant temperature and pressure. The mole-fraction of A at two different locations is $0.8$ and $0.1$. Let $N_{A1}$ be the diffusive flux of A calculated assuming B to be non-diffusing, and $N_{A2}$ be the diffusive flux of A calculated assuming equimolar counter-diffusion. The quantity $\frac{(N_{A1}-N_{A2})}{N_{A1}} \times 100$ is ________ (rounded off to one decimal place).
