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Four gases were combined in a gas cylinder with these partial pressures: 3.5 atm N2, 2.8 atm O2, 0.25 atm Ar, and 0.15 atm He.

What is the mole fraction of O2 in the mixture?

___ atm

What is the mole fraction of Ar in the mixture?

___atm

Answer :

Answer:

Mole fraction of [tex]O_{2}[/tex] = 0.42

Moles fraction of Ar = 0.037

Explanation:

Let's assume all the four gases and their mixture behave ideally.

According to Dalton's law of partial pressure for mixture of ideal gases-

                                    [tex]P_{i}=x_{i}P_{total}[/tex]

where [tex]P_{i}[/tex] and [tex]x_{i}[/tex] are partial pressure and mole fraction of "i"-th gas respectively. [tex]P_{total}[/tex] is total pressure of gaseous mixture.

Here [tex]P_{total}[/tex] = [tex]P_{N_{2}}+P_{O_{2}}+P_{Ar}+P_{He}[/tex]

                  = (3.5+2.8+0.25+0.15) atm

                   = 6.7 atm

So [tex]x_{O_{2}}=\frac{P_{O_{2}}}{P_{total}}[/tex]  =  [tex]\frac{2.8atm}{6.7atm}[/tex]  = 0.42

    [tex]x_{Ar}=\frac{P_{Ar}}{P_{total}}[/tex]  = [tex]\frac{0.25atm}{6.7atm}[/tex]  = 0.037

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