An aluminum alloy, 1.0 cm thick, is subjected to gas diffusion. After 10 hours, the concentration at 1 mm deep is 2 at%. At what depth will the concentration be 2 at% after 15 hours?

\(t_{1}\) = 10 hrs \( \quad C_{x_{1}} \) = 2 at% \( \quad x_{1} \) = 1 mm

\(t_{2}\) = 15 hrs \( \quad C_{x_{2}} \) = 2 at%

\( x_{2} \)

\(C_{x} - C_{0} \over C_{s} - C_{0} \) \(= 1 - erf(\) \(x \over 2\sqrt {Dt} \)\()\)

\({C_{x_{1}} - C_{0} \over C_{s} - C_{0}} = {C_{x_{2}} - C_{0} \over C_{s} - C_{0}} \implies {x_{1} \over \sqrt {t_{1}}} = {x_{2} \over \sqrt {t_{2}}} \)

\( x_{2} = x_{1} \sqrt {t_{2} \over t_{1}}\)

\(x_{2} \) = \(1 [mm] \sqrt { 15 h \over 10 h} \)

\(x_{2} \) = 1.2 [mm]