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? What is the mole fraction of Ar in the mixture?

Respuesta :

Answer:

Mole Fraction of O2 --> 0.42

Mole Fraction of Ar --> 0.037

Explanation:

The mole fraction of Oxygen has been 0.41. The mole fraction of Argon has been 0.037.

According to the Dalton's law of vapor pressure, the total pressure has been the sum of the partial pressure of all the gases in the mixture.

The total pressure (P)of the solution has been:

[tex]P=P_{N_2}\;+\;P_{O_2}\;+\;P_{Ar}\;+\;P_{He}[/tex]

Where, Partial pressure of nitrogen, [tex]P_{N_2}=3.5\;\rm atm[/tex]

Partial pressure of oxygen, [tex]P_{O_2}=2.8\;\rm atm[/tex]

Partial pressure of argon, [tex]P_{Ar}=0.25\;\rm atm[/tex]

Partial pressure of helium, [tex]P_{He}=0.15\;\rm atm[/tex]

Substituting the values, total pressure (P) has been given as:

[tex]P=3.5\;+\;2.8\;+\;0.25\;+\;0.15\;\rm atm\\\textit P=6.7\;atm[/tex]

The total pressure of the gas has been 6.7 atm.

The mole fraction (m) of a gas has been given by:

[tex]m=\dfrac{P'}{P}[/tex]

Where, P' has been the partial pressure.

P has been the total pressure

  • The mole fraction of Oxygen ([tex]m_{O_2}[/tex]) has been given as:

[tex]m_{O_2}=\dfrac{2.8}{6.7}\\m_{O_2}=0.41[/tex]

The mole fraction of Oxygen has been 0.41.

  • The mole fraction of Argon ([tex]m_{Ar}[/tex]) has been given as:

[tex]m_{Ar}=\dfrac{0.25}{6.7}\\m_{Ar}=0.0.37[/tex]

The mole fraction of Argon has been 0.037.

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