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?

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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?

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Answer 1 · 2017-07-06T17:47:22

We assume a pressure of $\frac{800 \cdot m m \cdot H g}{760 \cdot m m \cdot H g \cdot a t m^{- 1}} = 1.052 \cdot a t m$ $(800*mm*Hg)/(760*mm*Hg*atm^-1)=1.052*atm$ . We finally get $P_{dioxygen} = 0.421 \cdot a t m$ $P_"dioxygen"=0.421*atm$

Explanation:

I am unhappy in quoting a pressure measurement of over 760 mm Hg, as it is something you would try to avoid in the lab.......

We have $P_{Total} = 1.052 \cdot a t m$ $P_"Total"=1.052*atm$ . But by definition, $P_{Total} = P_{dioxygen} + P_{dinitrogen} = 1.052 \cdot a t m \times 40 % + 1.052 \cdot a t m \times 60 % = 1.052 \cdot a t m$ $P_"Total"=P_"dioxygen"+P_"dinitrogen"=1.052*atmxx40%+1.052*atmxx60%=1.052*atm$

..............and thus $P_{dioxygen} = 0.421 \cdot a t m$ $P_"dioxygen"=0.421*atm$

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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?

1 Answer

Explanation:

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