Let's break down problem and solve it step by step. ### **Given Data** - **Thickness of sheet, \( x \):** .5 mm = .0005 m - **Diffusion coefficient, \( D \):** \( 1.8 \times 10^{-11} \) m²/s - **Concentration on high-pressure side, \( C_1 \):** 3 kg/m³ - **Diffusion flux, \( J \):** \( 1.7 \times 10^{-10} \) kg/m²·s - **Depth for low-pressure side concentration:** .0005 m (same as thickness) ### **Required** Determine the concentration on the low-pressure side (\( C_2 \)). --- ## **Step 1: Use Fick's First Law for Steady-State Diffusion** \[ J = -D \frac{dC}{dx} \] For a thin sheet with concentration difference across it: \[ J = D \frac{C_1 - C_2}{x} \] --- ## **Step 2: Plug in the Values** \[ 1.7 \times 10^{-10} = 1.8 \times 10^{-11} \cdot \frac{3 - C_2}{.0005} \] --- ## **Step 3: Rearranging for \( C_2 \)** \[ 1.7 \times 10^{-10} \times .0005 = 1.8 \times 10^{-11} \cdot (3 - C_2) \] \[ 8.5 \times 10^{-14} = 1.8 \times 10^{-11} \cdot (3 - C_2) \] \[ \frac{8.5 \times 10^{-14}}{1.8 \times 10^{-11}} = 3 - C_2 \] \[ 4.72 \times 10^{-3} = 3 - C_2 \] \[ C_2 = 3 - 4.72 \times 10^{-3} \] \[ C_2 \approx 2.995 \text{ kg/m}^3 \] --- ## **Step 4: Match with Options** Closest option is: **3.2 kg/m³** --- ## **Final Answer** ### **The correct option is:** \[ \boxed{3.2 \text{ kg/m}^3} \]