Vapor Pressure Calculation Examples for Mixtures

Understanding Vapor Pressure in Mixtures

Vapor pressure is a crucial concept in chemistry that describes the pressure exerted by a vapor in thermodynamic equilibrium with its liquid or solid phase. When dealing with mixtures, Raoult’s Law provides a valuable framework for calculating the vapor pressure of each component in a solution. Here, we will explore three diverse, practical examples of vapor pressure calculation for mixtures.

Example 1: Calculating Vapor Pressure of an Ethanol-Water Mixture

In a laboratory setting, a chemist needs to determine the vapor pressure of a mixture containing ethanol and water at 25°C. This is pertinent for applications such as distillation, where knowing the vapor pressures aids in optimizing separation processes.

• Given Data:
  • Mole fraction of ethanol (X₁) = 0.6
  • Mole fraction of water (X₂) = 0.4
  • Vapor pressure of pure ethanol (P₁°) at 25°C = 5.8 kPa
  • Vapor pressure of pure water (P₂°) at 25°C = 3.2 kPa

Using Raoult’s Law:

• P₁ = X₁ * P₁°
• P₂ = X₂ * P₂°

Calculating the vapor pressures:

• P₁ = 0.6 * 5.8 kPa = 3.48 kPa
• P₂ = 0.4 * 3.2 kPa = 1.28 kPa

Total vapor pressure (P_total):

• P_total = P₁ + P₂ = 3.48 kPa + 1.28 kPa = 4.76 kPa

Notes:

This example illustrates how the vapor pressure of a mixture can be influenced by the composition of the components. In this case, the higher mole fraction of ethanol significantly increases the total vapor pressure.

Example 2: Vapor Pressure of a Hydrocarbon Mixture

In the petroleum industry, understanding the vapor pressure of hydrocarbon mixtures is essential for refining processes. Consider a mixture of hexane and heptane where the engineers need to calculate the vapor pressure at a given temperature to ensure safety during storage.

• Given Data:
  • Mole fraction of hexane (X₁) = 0.7
  • Mole fraction of heptane (X₂) = 0.3
  • Vapor pressure of pure hexane (P₁°) at 60°C = 20.0 kPa
  • Vapor pressure of pure heptane (P₂°) at 60°C = 12.0 kPa

Applying Raoult’s Law:

• P₁ = X₁ * P₁°
• P₂ = X₂ * P₂°

Calculating the vapor pressures:

• P₁ = 0.7 * 20.0 kPa = 14.0 kPa
• P₂ = 0.3 * 12.0 kPa = 3.6 kPa

Total vapor pressure (P_total):

• P_total = P₁ + P₂ = 14.0 kPa + 3.6 kPa = 17.6 kPa

Notes:

This calculation helps engineers design equipment that can handle the pressures generated by the vapor phase, ensuring the safety and efficiency of the refining process.

Example 3: Vapor Pressure of a Saltwater Mixture

In environmental science, calculating the vapor pressure of saltwater is vital for understanding evaporation rates in oceans. A researcher wants to determine how the presence of salt affects the vapor pressure of a water solution.

• Given Data:
  • Mole fraction of water (X₁) = 0.95
  • Mole fraction of salt (X₂) = 0.05
  • Vapor pressure of pure water (P₁°) at 100°C = 101.3 kPa
  • Vapor pressure of pure salt (P₂°) is negligible (assumed to be 0)

Using Raoult’s Law:

• P₁ = X₁ * P₁°
• P₂ = X₂ * P₂°

Calculating the vapor pressures:

• P₁ = 0.95 * 101.3 kPa = 96.24 kPa
• P₂ = 0.05 * 0 kPa = 0 kPa

Total vapor pressure (P_total):

• P_total = P₁ + P₂ = 96.24 kPa + 0 kPa = 96.24 kPa

Notes:

In this scenario, the addition of salt lowers the overall vapor pressure of the solution, demonstrating the colligative properties of solutions, which is crucial for understanding phenomena like ocean evaporation.

These examples illustrate the practical application of Raoult’s Law in calculating vapor pressures of various mixtures, providing insights for industries ranging from chemical manufacturing to environmental science.