A mixture of gases with a pressure of 800.0 mm hg contains 60% nitrogen and 40% oxygen by volume. What is the partial pressure of oxygen in this mixture?

The partial pressure of oxygen in the mixture is 320.0 mm Hg

Explanation:

1) Take a base of 100 liters of mixture:

N: 60% * 100 liter = 60 liter

O: 40 % * 100 liter = 40 liter.

2) Volume fraction:

At constant pressure and temperature, the volume of a gas is proportional to the number of molecules.

Then, the mole ratio is equal to the volume ratio. Callin n₁ and n₂, the number of moles of nitrogen and oxygen, respectively, and V₁, V₂ the volume of the respective gases you can set the proportion:

V₁ / V₂ = n₁ / n₂

That means that the mole ratio is equal to the volume ratio, and the mole fraction is equal to the volume fraction.

Then, since the law of partial pressures of gases states that the partial pressure of each gas is equal to the mole fraction of the gas multiplied by the total pressure, you can draw the conclusion that the partial pressure of each gas is equal to the volume fraction of the gas in the mixture multiplied by the total pressure.

Then calculate the volume fractions:

Volume fraction of a gas = volume of the gas / volume of the mixture

N: 60 liter / 100 liter = 0.6 liter

V: 40 liter / 100 liter = 0.4 liter

3) Partial pressures:

These are the final calculations and results:

Partial pressure = volume fraction * total pressure

Partial pressure of N = 0.6 * 800.0 mm Hg = 480.0 mm Hg

Partial pressure of O = 0.4 * 800.0 mm Hg = 320.0 mm Hg