Dalton's Law Of Partial Pressure Example

Dalton's law of partial pressure states that the total pressure exerted by a mixture of non-reacting gases is equal to the sum of the partial pressures of individual gases within that mixture. This fundamental principle, formulated by John Dalton in 1801, serves as a cornerstone for understanding gas behavior in chemistry, physics, engineering, and even physiology. Whether you are calculating the pressure inside a scuba tank, analyzing atmospheric composition, or studying gas exchange in human lungs, mastering this concept requires working through practical scenarios that bridge theory and application.

Understanding the Core Concept

Before diving into specific calculations, Make sure you define the terminology. Still, it matters. The partial pressure of a specific gas in a mixture is the pressure that gas would exert if it occupied the entire volume of the mixture alone at the same temperature But it adds up..

P_total = P₁ + P₂ + P₃ + ... + P_n

Where P_total is the total pressure of the gas mixture, and P₁, P₂, P₃... P_n are the partial pressures of the component gases. Plus, this additive property holds true strictly for ideal gases and non-reacting mixtures. If gases react chemically (like hydrogen and oxygen forming water), the number of moles changes, and the simple additive relationship no longer applies directly to the initial components.

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A critical bridge between partial pressure and composition is the mole fraction (χ). The partial pressure of a gas is directly proportional to its mole fraction in the mixture:

P_i = χ_i × P_total

Where χ_i = moles of gas i / total moles of mixture. This relationship allows us to move without friction between pressure measurements and compositional analysis.

Example 1: Basic Calculation of Total Pressure

The most straightforward application involves summing known partial pressures to find the total pressure of a system.

Scenario: A rigid 10.0 L container at 298 K holds a mixture of three gases: nitrogen (N₂), oxygen (O₂), and carbon dioxide (CO₂). The partial pressures are measured as follows:

• P_N₂ = 0.75 atm
• P_O₂ = 0.20 atm
• P_CO₂ = 0.05 atm

Solution: Applying Dalton's law directly: P_total = P_N₂ + P_O₂ + P_CO₂ P_total = 0.75 atm + 0.20 atm + 0.05 atm P_total = 1.00 atm

Analysis: This example illustrates the additive nature of pressure. Note that the volume (10.0 L) and temperature (298 K) were extraneous information for this specific calculation. They would only be necessary if we needed to calculate the number of moles of each gas using the Ideal Gas Law (PV = nRT).

Example 2: Determining Partial Pressure from Mole Fraction

In many laboratory and industrial settings, the composition of a gas mixture is known in terms of percentages or mole fractions, and the total pressure is measured. The task is to find the partial pressure of a specific component.

Scenario: Synthetic air is prepared for a calibration gas standard. The mixture contains 78.08% N₂, 20.95% O₂, 0.93% Ar, and 0.04% CO₂ by volume (which corresponds to mole fraction for ideal gases). The total pressure of the cylinder is 1500 psi. Calculate the partial pressure of oxygen.

Solution:

1. Convert percentage to mole fraction (χ): χ_O₂ = 20.95% = 0.2095
2. Apply the mole fraction formula: P_O₂ = χ_O₂ × P_total P_O₂ = 0.2095 × 1500 psi P_O₂ = 314.25 psi

Significance: This calculation is vital for gas blending. If a specific process requires an oxygen partial pressure of exactly 300 psi for safety or reaction kinetics, the blender knows the total pressure must be adjusted, or the composition tweaked Simple as that..

Example 3: The "Water Displacement" Correction (Vapor Pressure)

One of the most common laboratory applications—and a frequent source of error for students—is collecting a gas over water. The gas collected is never pure; it is always saturated with water vapor at the temperature of the lab. The total pressure inside the collection vessel (usually equal to atmospheric pressure) is the sum of the dry gas pressure and the water vapor pressure.

Scenario: A student collects 250 mL of hydrogen gas over water at 25°C. The atmospheric pressure (barometric pressure) is 755 mmHg. The vapor pressure of water at 25°C is 23.8 mmHg. What is the partial pressure of the dry hydrogen gas? How many moles of H₂ were collected?

Solution:

1. Find Partial Pressure of Dry H₂: P_total = P_H₂ + P_H₂O 755 mmHg = P_H₂ + 23.8 mmHg P_H₂ = 731.2 mmHg

2. Calculate Moles (using Ideal Gas Law): Convert units for R = 0.08206 L·atm/mol·K.

   • P = 731.2 mmHg × (1 atm / 760 mmHg) = 0.9621 atm
   • V = 0.250 L
   • T = 25°C + 273.15 = 298.15 K
   • n = PV / RT
   • n = (0.9621 atm × 0.250 L) / (0.08206 L·atm/mol·K × 298.15 K)
   • n ≈ 0.00984 mol H₂

Critical Insight: Forgetting to subtract the water vapor pressure would result in a calculated mole value that is roughly 3% too high. At higher temperatures, where water vapor pressure is significantly greater (e.g., 47 mmHg at 37°C, 760 mmHg at 100°C), this error becomes massive. Always consult a vapor pressure table for the specific temperature of the experiment.

Example 4: Mole Fraction and Composition Analysis

This example reverses the process: using measured partial pressures to determine the composition of an unknown mixture Simple, but easy to overlook..

Scenario: A 5.00 L flask contains a mixture of helium (He) and neon (Ne) at 300 K. The