Steady-State Diffusion Flux Comparison
The problem involves comparing two diffusive flux scenarios for a binary A-B liquid under steady-state conditions:
- $N_{A1}$: Flux of A assuming B is non-diffusing ($N_B = 0$).
- $N_{A2}$: Flux of A assuming equimolar counter-diffusion ($N_A + N_B = 0$).
We are given the mole fractions of A at two locations, $x_{A1} = 0.8$ and $x_{A2} = 0.1$. The goal is to calculate $\frac{(N_{A1}-N_{A2})}{N_{A1}} \times 100$.
Flux $N_{A1}$ Derivation (B Non-Diffusing)
When B is non-diffusing ($N_B=0$), the flux of A ($N_{A1}$) relative to stationary coordinates is given by $N_{A1} = \frac{-D_{AB} C_{total}}{1-x_A} \frac{dx_A}{dz}$. Assuming constant total molar concentration ($C_{total}$), the integrated form over a distance $\Delta z$ is:
$N_{A1} = \frac{D_{AB} C_{total}}{\Delta z} \ln\left(\frac{1-x_{A2}}{1-x_{A1}}\right)$
Substituting the given mole fractions $x_{A1} = 0.8$ and $x_{A2} = 0.1$:
- $1 - x_{A1} = 1 - 0.8 = 0.2$
- $1 - x_{A2} = 1 - 0.1 = 0.9$
The flux $N_{A1}$ becomes:
$N_{A1} = \frac{D_{AB} C_{total}}{\Delta z} \ln\left(\frac{0.9}{0.2}\right) = \frac{D_{AB} C_{total}}{\Delta z} \ln(4.5)$
Flux $N_{A2}$ Derivation (Equimolar Counter-Diffusion)
Under equimolar counter-diffusion ($N_A + N_B = 0$), the flux of A ($N_{A2}$) relative to stationary coordinates is given by $N_{A2} = -D_{AB} C_{total} \frac{dx_A}{dz}$. The integrated form is:
$N_{A2} = \frac{D_{AB} C_{total}}{\Delta z} (x_{A1} - x_{A2})$
Substituting the given mole fractions $x_{A1} = 0.8$ and $x_{A2} = 0.1$:
$N_{A2} = \frac{D_{AB} C_{total}}{\Delta z} (0.8 - 0.1) = \frac{D_{AB} C_{total}}{\Delta z} (0.7)$
Percentage Difference Calculation
The quantity to compute is $\frac{(N_{A1}-N_{A2})}{N_{A1}} \times 100$. This can be rewritten as:
$\left(1 - \frac{N_{A2}}{N_{A1}}\right) \times 100$
Let $K = \frac{D_{AB} C_{total}}{\Delta z}$ for simplicity. The ratio of the fluxes is:
$\frac{N_{A2}}{N_{A1}} = \frac{K \times 0.7}{K \times \ln(4.5)} = \frac{0.7}{\ln(4.5)}$
Using the approximate value $\ln(4.5) \approx 1.504077$:
$\frac{N_{A2}}{N_{A1}} \approx \frac{0.7}{1.504077} \approx 0.46540$
Now, calculate the required percentage difference:
$\left(1 - 0.46540\right) \times 100 = 0.53460 \times 100 = 53.460\%$
Rounding the result to one decimal place gives $53.5\%$.
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?
- 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).
- 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
