# A mixture of gases with a pressure of 800.0 Hg contains 60% nitrogen and 40% oxygen by volume. What is the partial pressure of oxygen in this mixture?

We assume a pressure of $\frac{800 \cdot m m \cdot H g}{760 \cdot m m \cdot H g \cdot a t {m}^{-} 1} = 1.052 \cdot a t m$. We finally get ${P}_{\text{dioxygen}} = 0.421 \cdot a t m$
We have ${P}_{\text{Total}} = 1.052 \cdot a t m$. But by definition, P_"Total"=P_"dioxygen"+P_"dinitrogen"=1.052*atmxx40%+1.052*atmxx60%=1.052*atm
..............and thus ${P}_{\text{dioxygen}} = 0.421 \cdot a t m$