# Diffusion of Methane Through Helium ## Problem Statement (Paraphrased) A gas of CH₄ and He is contained in a tube at 101.32 kPa and 298 K. At one point, the partial pressure of methane, \( P_{A1} = 20.1 \) kPa, and at a point 0.02 m away, \( P_{A2} = 20.7 \) kPa. The total pressure is constant. Calculate the flux of CH₄ (methane) at steady state for equimolar counterdiffusion. Given: - Diffusion coefficient: \( D_{AB} = 6.52 \times 10^{-5} \) m²/s --- ## 1. List Knowns and Unknowns (2/10) ### **Knowns:** - \( P = 101.32 \) kPa (total pressure) - \( T = 298 \) K - \( P_{A1} = 20.1 \) kPa (partial pressure of CH₄ at point 1) - \( P_{A2} = 20.7 \) kPa (partial pressure of CH₄ at point 2) - \( x_2 - x_1 = 0.02 \) m (distance between points) - \( D_{AB} = 6.52 \times 10^{-5} \) m²/s (diffusion coefficient for CH₄ in He) - Steady-state, equimolar counterdiffusion ### **Unknowns:** - Flux of CH₄ (\( N_A \)), in mol/(m²·s) --- ## 2. Draw a Diagram Representing the Problem (2/10) ``` |--------------------- Tube ---------------------| Point 1 (x1) Point 2 (x2) P_A1 = 20.1 kPa P_A2 = 20.7 kPa |---------------------------------------------| Distance = 0.02 m ``` *(Hand-drawn diagram recommended for submission)* --- ## 3. Plan and List Assumptions (2/10) ### **Plan:** - Use Fick's law for equimolar counterdiffusion: \[ N_A = -D_{AB} \frac{C}{x_2 - x_1} (y_{A2} - y_{A1}) \] Or, in terms of partial pressures: \[ N_A = -\frac{D_{AB} P}{R T (x_2 - x_1)} (y_{A2} - y_{A1}) \] where \( y_{A} = \frac{P_A}{P} \). ### **Assumptions:** - Steady-state diffusion - Isothermal process - Total pressure is constant - Ideal gas behavior - No chemical reaction --- ## 4. Solve the Problem (4/10) ### **Step 1: Calculate mole fractions** \[ y_{A1} = \frac{P_{A1}}{P} = \frac{20.1}{101.32} = 0.1985 \] \[ y_{A2} = \frac{20.7}{101.32} = 0.2043 \] ### **Step 2: Substitute into Fick's Law** \[ N_A = - \frac{D_{AB} P}{R T (x_2 - x_1)} (y_{A2} - y_{A1}) \] - \( D_{AB} = 6.52 \times 10^{-5} \) m²/s - \( P = 101320 \) Pa - \( R = 8.314 \) J/(mol·K) - \( T = 298 \) K - \( x_2 - x_1 = 0.02 \) m - \( y_{A2} - y_{A1} = 0.2043 - 0.1985 = 0.0058 \) \[ N_A = - \frac{6.52 \times 10^{-5} \times 101320}{8.314 \times 298 \times 0.02} \times 0.0058 \] Calculate denominator: \[ 8.314 \times 298 \times 0.02 = 49.54 \] Calculate numerator: \[ 6.52 \times 10^{-5} \times 101320 = 6.6078 \] Plug in values: \[ N_A = - \frac{6.6078}{49.54} \times 0.0058 \] \[ N_A = -0.1335 \times 0.0058 = -0.000774 \text{ mol/(m}^2\text{·s)} \] ### **Final Answer:** \[ \boxed{N_A = -7.74 \times 10^{-4} \ \text{mol/(m}^2\cdot \text{s)}} \] (The negative sign indicates the direction of flux; you may report the magnitude if only the flux amount is needed.) --- **If you need a clean hand-drawn diagram or a step-by-step explanation for each part, let me know!**

# Step-by-Step Explanation of the Problem ## Problem Overview We need to calculate the flux of methane (CH₄) in a tube where it is diffusing through helium (He). The setup involves different partial pressures of CH₄ at two points separated by a distance. ## Step 1: Understanding Inputs and Outputs ### **Known Values** - **Total Pressure (P)**: \( 101.32 \) kPa - **Temperature (T)**: \( 298 \) K - **Partial Pressure at Point 1 (P_{A1})**: \( 20.1 \) kPa - **Partial Pressure at Point 2 (P_{A2})**: \( 20.7 \) kPa - **Distance (x₂ - x₁)**: \( 0.02 \) m - **Diffusion Coefficient (D_{AB})**: \( 6.52 \times 10^{-5} \) m²/s ### **Unknown Value** - **Flux of CH₄ (N_A)**: To be calculated in mol/(m²·s) --- ## Step 2: Calculate Mole Fractions **Mole Fraction at Point 1 (y_{A1})**: \[ y_{A1} = \frac{P_{A1}}{P} = \frac{20.1 \, \text{kPa}}{101.32 \, \text{kPa}} \approx 0.1985 \] **Mole Fraction at Point 2 (y_{A2})**: \[ y_{A2} = \frac{P_{A2}}{P} = \frac{20.7 \, \text{kPa}}{101.32 \, \text{kPa}} \approx 0.2043 \] --- ## Step 3: Use Fick's Law for Flux Calculation ### **Fick's Law for Equimolar Counterdiffusion** The formula for the flux \( N_A \) is given by: \[ N_A = -\frac{D_{AB} P}{R T (x_2 - x_1)} (y_{A2} - y_{A1}) \] ### **Substituting Known Values** - **Constants**: - \( D_{AB} = 6.52 \times 10^{-5} \) m²/s - \( P = 101320 \) Pa - \( R = 8.314 \) J/(mol·K) - \( T = 298 \) K - \( x_2 - x_1 = 0.02 \) m - \( y_{A2} - y_{A1} = 0.2043 - 0.1985 = 0.0058 \) ### **Calculate Denominator** \[ R T (x_2 - x_1) = 8.314 \times 298 \times 0.02 \approx 49.54 \] ### **Calculate Numerator** \[ D_{AB} P = 6.52 \times 10^{-5} \times 101320 \approx 6.6078 \] ### **Plug Values into the Flux Equation** \[ N_A = -\frac{6.6078}{49.54} \times 0.0058 \] ### **Perform Final Calculation** 1. Calculate the fraction: \[ \frac{6.6078}{49.54} \approx 0.1335 \] 2. Multiply by the mole fraction difference: \[ N_A = -0.1335 \times 0.0058 \approx -0.000774 \text{ mol/(m}^2\text{·s)} \] --- ## Step 4: Final Result ### **Flux of Methane** The calculated flux of CH₄ is: \[ \boxed{N_A = -7.74 \times 10^{-4} \ \text{mol/(m}^2\cdot \text{s)}} \] ### **Interpretation** - The negative sign indicates the direction of flux (from higher to lower concentration). - The magnitude of the flux is \( 7.74 \times 10^{-4} \) mol/(m²·s). If you have any further questions or need clarification on any part, feel free to ask!