Asample of gas isolated from unrefined petroleum contains 90.0% ch4, 8.9% c2h6, and 1.1% c3h8 at a total pressure of 307.2 kpa. what is the partial pressure of each component of this gas?

The partial pressure of each component of the gas are:

1. The partial pressure of CH₄ is 276.48 KPa

2. The partial pressure of C₂H₆ is 27.34 KPa

3. The partial pressure of C₃H₈ is 3.38 KPa

We'll begin by calculating the mole fraction of each gas.

Percentage of CH₄ = 90%

Total = 100%

Mole fraction = mole / total

Mole fraction CH₄ = 90 / 100

Percentage of C₂H₆ = 8.9%

Total = 100%

Mole fraction = mole / total

Mole fraction C₂H₆ = 8.9 / 100

Percentage of C₃H₈ = 1.1%

Total = 100%

Mole fraction = mole / total

Mole fraction C₃H₈ = 1.1 / 100

Finally, we shall determine the partial pressure of each gas. This can be obtained as follow:

1. Determination of the partial pressure of CH₄

Mole fraction CH₄ = 0.9

Total pressure = 307.2 KPa

Partial pressure = mole fraction × Total pressure

Partial pressure of CH₄ = 0.9 × 307.2

Partial pressure of CH₄ = 276.48 KPa

2. Determination of the partial pressure of C₂H₆

Mole fraction C₂H₆ = 0.089

Total pressure = 307.2 KPa

Partial pressure = mole fraction × Total pressure

Partial pressure of C₂H₆ = 0.089 × 307.2

Partial pressure of C₂H₆ = 27.34 KPa

3. Determination of the partial pressure of C₃H₈

Mole fraction C₃H₈ = 0.011

Total pressure = 307.2 KPa

Partial pressure = mole fraction × Total pressure

Partial pressure of C₃H₈ = 0.011 × 307.2

Partial pressure of C₃H₈ = 3.38 KPa

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276.48 atm → CH₄

27.3 atm → C₂H₆

3.38 atm → C₃H₈

Explanation:

Percentages of each gas, are the mole fraction

0.9 CH₄

0.089 C₂H₆

0.011 C₃H₈

Mole fraction = Partial pressure each gas/ Total pressure

0.9 = Partial pressure CH₄ / 307.2 kPa

307.2 kPa . 0.9 = 276.48 atm

0.089 = Partial pressure C₂H₆ / 307.2 kPa

307.2 kPa . 0.089 = 27.3 atm

0.011 = Partial pressure C₃H₈ / 307.2 kPa

307.2 kPa . 0.011 = 3.38 atm