Question #af457

The first thing that you need to do here is to figure out the number of moles of oxygen and the total number of moles of gas present in the mixture.

Use the molar mass of oxygen gas and the molar mass of helium to get

#2.0 color(red)(cancel(color(black)("g"))) * "1 mole O"_2/(32.0color(red)(cancel(color(black)("g")))) = "0.0625 moles O"_2#

and

#98.0color(red)(cancel(color(black)("g"))) * "1 mole He"/(4.003color(red)(cancel(color(black)("g")))) = "24.48 moles He"#

This means that the total number of moles present in the mixture is equal to

#"0.0625 moles O"_2 + "24.48 moles He" = "24.54 moles gas"#

Now, Dalton's Law of Partial Pressures tells you that the partial pressure of a gas that's part of a gaseous mixture is proportional to the mole fraction of the gas in the mixture and the total pressure of the mixture.

In your case, you will have

#P_ ("O"_ 2) = chi_ ("O"_ 2) * P_"total"#

Here

• #P_( "O"_ 2)# is the partial pressure of oxygen gas
• #chi_ ("O"_ 2)# is the mole faction of oxygen gas in the mixture
• #P_"total"# is the total pressure of the mixture

To find the mole fraction of oxygen gas, simply divide the number of moles of oxygen gas by the total number of moles of gas present in the sample.

#chi_( "O"_ 2) = (0.0625 color(red)(cancel(color(black)("moles"))))/(24.54color(red)(cancel(color(black)("moles")))) = 0.002547#

You can thus say that the partial pressure of oxygen gas in this mixture will be

#P_( "O"_ 2) = 0.002547 * "8.0 atm" = color(darkgreen)(ul(color(black)("0.020 atm")))#

The answer is rounded to two sig figs.