Let's solve **Question 1** step by step. --- ## Question 1 Statement Material A diffuses in the +z direction inside a room, and material B diffuses in the opposite direction with a flux being equal to half of the molecular flux of A. If the total pressure and temperature inside the room are 1 atm and 25°C, and there is no reaction between A and B, derive the equation for calculating: a) The concentration of A b) The concentration of B The process is at steady-state. --- ## **Step 1: Write the General Flux Equations** For a binary mixture: - \( J_A \): Flux of A relative to stationary coordinates (total flux) - \( J_B \): Flux of B relative to stationary coordinates - \( N_A \): Molar flux of A relative to stationary coordinates - \( N_B \): Molar flux of B relative to stationary coordinates The general equation for diffusion (Fick's law, steady-state, one-dimensional): \[ N_A = -D_{AB} \frac{dC_A}{dz} + y_A (N_A + N_B) \] \[ N_B = -D_{BA} \frac{dC_B}{dz} + y_B (N_A + N_B) \] Where: - \( D_{AB} \): Diffusion coefficient of A in B - \( D_{BA} \): Diffusion coefficient of B in A - \( C_A, C_B \): Concentration of A, B - \( y_A, y_B \): Mole fractions of A, B --- ## **Step 2: Problem Data and Relationship** Given: - The flux of B is half (but in the opposite direction) the molecular flux of A: \[ N_B = -\frac{1}{2}N_A \] - Total pressure and temperature are constant. - No reaction. **From stoichiometry:** \[ N_A + N_B = N_A - \frac{1}{2}N_A = \frac{1}{2}N_A \] --- ## **Step 3: Express the Diffusion Equation** From Fick's law for steady-state and one-dimensional diffusion: \[ N_A = -D_{AB} \frac{dC_A}{dz} + y_A (N_A + N_B) \] Plug in \( N_A + N_B = \frac{1}{2}N_A \): \[ N_A = -D_{AB} \frac{dC_A}{dz} + y_A \left(\frac{1}{2}N_A\right) \] \[ N_A - \frac{1}{2}y_A N_A = -D_{AB} \frac{dC_A}{dz} \] \[ N_A (1 - \frac{1}{2}y_A) = -D_{AB} \frac{dC_A}{dz} \] \[ N_A = \frac{-D_{AB}}{1 - \frac{1}{2}y_A} \frac{dC_A}{dz} \] --- ## **Step 4: Express Everything in Terms of Mole Fraction** Remember: - \( C_A = y_A C \), where \( C \) is the total concentration (constant). - \( \frac{dC_A}{dz} = C \frac{dy_A}{dz} \). So, \[ N_A = \frac{-D_{AB} C}{1 - \frac{1}{2}y_A} \frac{dy_A}{dz} \] --- ## **Step 5: Final Equation for Concentration of A** \[ \boxed{ N_A = \frac{-D_{AB} C}{1 - \frac{1}{2}y_A} \frac{dy_A}{dz} } \] This is the equation to use for finding the concentration profile of A. --- ## **Step 6: Concentration of B** Since \( y_B = 1 - y_A \), the concentration of B is: \[ C_B = (1 - y_A) C \] --- ## **Final Answers** ### a) **Equation for Calculating Concentration of A**: \[ N_A = \frac{-D_{AB} C}{1 - \frac{1}{2}y_A} \frac{dy_A}{dz} \] ### b) **Equation for Calculating Concentration of B**: \[ C_B = (1 - y_A) C \] --- **Explanation:** - We used the relationship between fluxes, Fick's law, and the constraint that the flux of B is half (and opposite) that of A. - The total pressure is constant, so the sum of mole fractions is 1. - You can solve for \( y_A \) as a function of \( z \) by integrating the equation with proper boundary conditions as needed. - No logarithms appear in the final answer, as required. --- Let me know if you want the step-by-step integration for the profile!

Let's solve **Question 1** step by step. ## Question 1 Statement Material A diffuses in the +z direction inside a room, while material B diffuses in the opposite direction, with a flux equal to half of the molecular flux of A. Given that the total pressure and temperature inside the room are 1 atm and 25°C, and there is no reaction between A and B, we need to derive equations for calculating the concentrations of A and B. The process is at steady-state. ## Step 1: Write the General Flux Equations In a binary mixture, we define several parameters. The total flux of material A is denoted as \( J_A \) and for material B as \( J_B \). The molar fluxes \( N_A \) and \( N_B \) represent the respective flows of A and B. Under steady-state conditions, Fick's law can be applied, leading us to the following equations: \[ N_A = -D_{AB} \frac{dC_A}{dz} + y_A (N_A + N_B) \] \[ N_B = -D_{BA} \frac{dC_B}{dz} + y_B (N_A + N_B) \] Where \( D_{AB} \) and \( D_{BA} \) are the diffusion coefficients, \( C_A \) and \( C_B \) are the concentrations of A and B, respectively, and \( y_A \) and \( y_B \) are their mole fractions. ## Step 2: Problem Data and Relationship We know that the flux of B is half that of A but in the opposite direction: \[ N_B = -\frac{1}{2} N_A \] Because there is no reaction, the total flux can be represented as: \[ N_A + N_B = N_A - \frac{1}{2}N_A = \frac{1}{2}N_A \] ## Step 3: Express the Diffusion Equation Inserting the relationship for \( N_B \) into Fick's law gives us: \[ N_A = -D_{AB} \frac{dC_A}{dz} + y_A \left(\frac{1}{2}N_A\right) \] Rearranging this, we find: \[ N_A (1 - \frac{1}{2}y_A) = -D_{AB} \frac{dC_A}{dz} \] ## Step 4: Express Everything in Terms of Mole Fraction Using the relationship \( C_A = y_A C \), where \( C \) is the total concentration (a constant), we can modify our expression for \( \frac{dC_A}{dz} \): \[ \frac{dC_A}{dz} = C \frac{dy_A}{dz} \] Thus, we can rewrite the equation for \( N_A \): \[ N_A = \frac{-D_{AB} C}{1 - \frac{1}{2}y_A} \frac{dy_A}{dz} \] ## Step 5: Final Equation for Concentration of A The final equation for the concentration of A becomes: \[ \boxed{N_A = \frac{-D_{AB} C}{1 - \frac{1}{2}y_A} \frac{dy_A}{dz}} \] ## Step 6: Concentration of B For material B, since \( y_B = 1 - y_A \), the concentration can be expressed as: \[ C_B = (1 - y_A) C \] ## Final Answers ### a) Equation for Calculating Concentration of A: \[ N_A = \frac{-D_{AB} C}{1 - \frac{1}{2}y_A} \frac{dy_A}{dz} \] ### b) Equation for Calculating Concentration of B: \[ C_B = (1 - y_A) C \] In summary, we derived the equations needed to calculate the concentrations of materials A and B under the given steady-state diffusion conditions. If you need further assistance with the next question or any specific integrations, feel free to ask!