Find dissolved nitrogen in pure liquid iron exposed to atmospheric air at 1873 K, in wt% (three decimal places). Sieverts constant satisfies $\log_{10}K_N=-518/T-1.063$, with $K_N=h_N/\sqrt{p_{N_2}}$ and activity relative to the 1 wt% Henrian standard. Assume $h_N=[\%N]$.
At 1873 K, log₁₀K=−518/1873−1.063=−1.33956, giving K≈0.04576. Atmospheric air supplies pN₂≈0.79 atm. Thus [%N]=K√0.79≈0.04067 wt%, or 0.041. The square-root pressure dependence comes from dissociation of molecular nitrogen.
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