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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anfabba15 anfabba15 · Answer 1 · 2019-12-24T20:23:29+01:00

Answer:

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