Question

= carbon dioxide = oxygen • = neon Consider the mixture of gasses depicted above. If the parial pressure on neon in this mixture is 0.900 atm, what is the partial pressure of oxygen?

Answer 1

• n(CO2) = 4, n(O2) = 2 and n(Ne) = 6

mole fraction of Ne (XNe) = n(Ne) / (n(CO2) + n(O2) + n(Ne)) = 6/4+2+6 = 6/12 = 0.5

partial pressure of Ne = XNe* Ptotal  ⇒ 0.9 atm = 0.5*Ptotal  ⇒ Ptotal = 0.9/0.5 = 1.8 atm

mole fraction of O2 (XO2) = n(O2) / (n(CO2) + n(O2) + n(Ne)) = 2/4+2+6 = 2/12 = 0.167

partial pressure of O2 = XO2* Ptotal = 0.167*1.8 = 0.3 atm

• n(CO2) = 6, n(O2) = 10, n(N2) = 7 and n(Kr) = 3

mole fraction of CO2 (XCO2) = n(CO2) / (n(CO2) + n(O2) + n(N2) + n(Kr)) = 6/6+10+7+3 = 6/20 = 0.3

partial pressure of CO2 = XCO2* Ptotal   = 0.3*1775 = 532.5 torr