A gas mixture contains 1.5 mol% CO2, 18 mol% O2 and 80.5 mol% Ar. Calculate the partial pressures (in mm Hg) of each gas if the total pressure is 1045 mm Hg?
The correct answer and explanation is:
To find the partial pressure of each gas in the mixture, we apply Dalton’s Law of Partial Pressures, which states:
Partial Pressure of a gas = Mole fraction of the gas × Total pressure
Given:
- Total pressure = 1045 mm Hg
- CO₂ = 1.5 mol% = 0.015 mole fraction
- O₂ = 18 mol% = 0.18 mole fraction
- Ar = 80.5 mol% = 0.805 mole fraction
Calculations:
1. Partial pressure of CO₂:
PCO2=0.015×1045=15.675 mm HgP_{CO₂} = 0.015 × 1045 = 15.675 \, \text{mm Hg}
2. Partial pressure of O₂:
PO2=0.18×1045=188.1 mm HgP_{O₂} = 0.18 × 1045 = 188.1 \, \text{mm Hg}
3. Partial pressure of Ar:
PAr=0.805×1045=841.225 mm HgP_{Ar} = 0.805 × 1045 = 841.225 \, \text{mm Hg}
Final Answers:
- CO₂: 15.675 mm Hg
- O₂: 188.1 mm Hg
- Ar: 841.225 mm Hg
Explanation (Approx. 300 Words):
Dalton’s Law of Partial Pressures is foundational in understanding gas behavior in mixtures. It states that the total pressure of a gas mixture is equal to the sum of the partial pressures of each individual component gas. The partial pressure of each gas depends solely on its mole fraction and the total pressure of the system.
In this problem, we’re given the mole percent (or mol%) of each gas, which is directly convertible to mole fractions by dividing by 100. These mole fractions represent the proportion of the total number of moles that each gas contributes to the mixture.
Using Dalton’s Law, we multiply each mole fraction by the total pressure (1045 mm Hg) to find the partial pressure of that gas. For example, CO₂ makes up only 1.5% of the mixture, so its contribution to the total pressure is small — just 15.675 mm Hg. In contrast, Argon is the major component at 80.5%, resulting in a much larger partial pressure of 841.225 mm Hg.
The sum of all the calculated partial pressures should equal the total pressure (1045 mm Hg), validating the accuracy of the method.
This approach is crucial in chemistry and fields like respiratory physiology or environmental science, where understanding the behavior of individual gases in a mixture is essential — such as calculating oxygen delivery in medical gases or analyzing atmospheric components.