It was found that in a diving gas mixture for a certain altitude, the total predicted pressure is #"8.38 atm"#. If the partial pressure of oxygen gas must be #"0.21 atm"# underwater, what must the mol fraction be of oxygen gas?

Question

Caleb Adams · Accepted Answer

This is asking you to use assume #"O"_2# is an ideal gas and use the definition of partial pressure: #P_(O_2) = chi_(O_2(g))P_(t ot)# where #chi_(O_2(g))# is the mol fraction of #"O"_2# in the diving gas mixture and #P_(O_2)# is its partial pressure in the diving gas mixture. Since we want #P_(O_2) = "0.21 atm"#, we have: #"0.21 atm" = chi_(O_2(g))("8.38 atm")# #=> color(blue)(chi_(O_2(g)) = 0.025)# or #2.5%#. This should make sense, since if we want only #"0.21 atm"# of #"O"_2# at a higher pressure, we want less than #21%# of the higher pressure to be #"O"_2#.

Cameron Andrews · Answer

undefined The mole fraction of oxygen gas in the diving gas mixture must be 0.025, calculated by dividing the partial pressure of oxygen gas (0.21 atm) by the total predicted pressure (8.38 atm).