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1 August, 10:34

Consider a mixture of 3 gases (Helium, Argon, and Nitrogen). Helium has mole fraction of 0.28. Argon has a partial pressure of 450 kPa. There are 7.5 moles of Nitrogen. The total pressure is 1470 kPa. Determine:

a. Mole fraction of Argon and Nitrogen

b. Partial pressure of Helium and Nitrogen

c. Number of moles of Helium and Nitrogen

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  1. 1 August, 12:21
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    a. Mole fraction of Argon = 0.28

    Mole fraction of N₂ = 0.414

    b. Partial pressure He = 411.6 kPa

    Partial pressure N₂ = 608.58 kPa

    c. Mole of He = 5.54 mole

    Mole of N₂ = 7.5 mole

    Explanation:

    There is a formula for mole fraction which has a relation with partial pressure in a mixture of gases

    Mole fraction = Fraction for partial pressure

    Mole of Ar / Total moles = Partial pressure Ar / Total pressure

    Partial pressure Ar = 450kPa

    450kPa / 1470kPa = 0.306

    If mole fraction of He is 0.28 we can assume this:

    Sum of mole fraction = 1

    0.28 (He) + 0.306 (Ar) + Mole fraction N₂ = 1

    1 - 0.28 - 0.306 = Mole fraction N₂ → 0.414

    Now we can know partial pressure of He and N₂

    0.28 = Partial pressure He / 1470kPa

    1470kPa. 0.28 = 411.6 kPa

    0.414 = Partial pressure N₂ / 1470kPa

    1470kPa. 0.414 = 608.58 kPa

    Mole fraction: Mole of gas / Total mole

    0.414 = Mole of N₂ / Total Mole

    0.414. Total mole = 7.5 mole

    7.5 mole / 0.414 = 18.11 Total mole

    18.11. 0.28 (Ar) = 5.07 mole

    18.11. 0.306 (He) = 5.54 mole
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