A sheet of steel 1.5 mm thick has nitrogen atmospheres on both sides at \(1200^{\circ} \)C and is permitted to achieve a steady-state diffusion condition. The diffusion coefficient for nitrogen is steel at this temperature is \(6x10^{-11} m^2/s\), and the diffusion flux is found to be \(1.2x10^{-7} {kg \over m^2 s}\). Also, it is known that the concentration of nitrogen in the steel at the high-pressure surface is 4 \(kg \over m^3\). How far into the sheet from this high-pressure side will the concentration be 2.0 \(kg \over m^3\)? Assume a linear concentration profile.
50Cr: 4.34% A = 49.9460 amu
52Cr: 83.79% A = 51.9405 amu
53Cr: 9.50% A = 52.9407 amu
54Cr: 2.37% A = 53.9389 amu
Prove that \(A_{Cr}\) = 51.9963 amu
\(A = f_{1} * A_{1} + f_{2} * A_{2} + f_{3} * A_{3} + f_{4} * A_{4} \)
4.34% * 49.9460 + 83.79% * 51.9405 + 9.50% * 52.9407 + 2.37% * 53.9389 = 51.9963 amu
51.9963 amu = 51.9963 amu \( \implies \) correct