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
Jul 6, 2017

Answer:

We assume a pressure of #(800*mm*Hg)/(760*mm*Hg*atm^-1)=1.052*atm#. We finally get #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*atm#. But by definition, #P_"Total"=P_"dioxygen"+P_"dinitrogen"=1.052*atmxx40%+1.052*atmxx60%=1.052*atm#

..............and thus #P_"dioxygen"=0.421*atm#