Question: Calculate the diffusion flux for a 3.1 mm-thick sheet of palladium having an area of $0.53$ m$^2$ at $500^\circ$C. Assume a diffusion coefficient of $4.8 \times 10^{-8}$ m$^2$/s, that the concentrations at the high- and low-pressure sides of the plates are $3.9$ and $0.57$ kg/m$^3$ of hydrogen per cubic meter of palladium, and that steady-state conditions have been attained.
1. **State the problem:** Calculate the diffusion flux $J$ through a palladium sheet of thickness $L = 3.1$ mm, area $A = 0.53$ m$^2$, diffusion coefficient $D = 4.8 \times 10^{-8}$ m$^2$/s, with concentration difference between sides $C_1 = 3.9$ kg/m$^3$ and $C_2 = 0.57$ kg/m$^3$ at steady state.
2. **Formula used:** Fick's first law of diffusion for steady state:
$$J = -D \frac{\Delta C}{L} = -D \frac{C_2 - C_1}{L}$$
where $J$ is the diffusion flux (kg/m$^2$/s), $D$ is the diffusion coefficient, $\Delta C = C_2 - C_1$ is the concentration difference, and $L$ is the thickness.
3. **Convert thickness to meters:**
$$L = 3.1 \text{ mm} = 3.1 \times 10^{-3} \text{ m}$$
4. **Calculate concentration difference:**
$$\Delta C = C_2 - C_1 = 0.57 - 3.9 = -3.33 \text{ kg/m}^3$$
5. **Calculate diffusion flux magnitude:**
$$J = -D \frac{\Delta C}{L} = -4.8 \times 10^{-8} \times \frac{-3.33}{3.1 \times 10^{-3}}$$
6. **Simplify the fraction:**
$$\frac{-3.33}{3.1 \times 10^{-3}} = \frac{\cancel{-3.33}}{\cancel{3.1 \times 10^{-3}}} = -1074.19 \text{ kg/m}^4$$
7. **Calculate $J$:**
$$J = -4.8 \times 10^{-8} \times (-1074.19) = 5.156 \times 10^{-5} \text{ kg/m}^2/\text{s}$$
8. **Interpretation:** The positive sign indicates diffusion occurs from the high concentration side to the low concentration side.
**Final answer:**
$$\boxed{J = 5.16 \times 10^{-5} \text{ kg/m}^2/\text{s}}$$