Examples of Partial Vapor Pressures in Binary Mixture

Explore practical examples of partial vapor pressures in binary mixtures using Raoult's Law.
By Jamie

Introduction to Partial Vapor Pressures in Binary Mixtures

In chemistry, partial vapor pressure refers to the pressure exerted by an individual component of a mixture in the vapor phase. Raoult’s Law provides a framework for calculating these pressures in binary mixtures—systems composed of two components. Understanding these concepts is crucial in fields such as chemical engineering, environmental science, and materials science. Let’s delve into three practical examples to illustrate how partial vapor pressures can be calculated and applied.

Example 1: Ethanol and Water Mixture

Context

This example illustrates the vapor pressure of a common binary mixture: ethanol and water. It is relevant in distillation processes and beverage production.

To calculate the partial vapor pressures of ethanol and water at a given temperature, we can use the following data:

  • Vapor pressure of pure ethanol at 25°C: 78 mmHg
  • Vapor pressure of pure water at 25°C: 23 mmHg
  • Mole fraction of ethanol in the mixture: 0.6
  • Mole fraction of water in the mixture: 0.4

Using Raoult’s Law:

  • Partial vapor pressure of ethanol, P₁ = X₁ * P₁⁰ = 0.6 * 78 mmHg = 46.8 mmHg
  • Partial vapor pressure of water, P₂ = X₂ * P₂⁰ = 0.4 * 23 mmHg = 9.2 mmHg

Notes

The total vapor pressure of the mixture can be calculated by summing the partial pressures:

  • Total vapor pressure, P_total = P₁ + P₂ = 46.8 mmHg + 9.2 mmHg = 56 mmHg.
    This example can vary if the mole fractions change or if the mixture is at a different temperature, impacting the pure component vapor pressures.

Example 2: Benzene and Toluene Mixture

Context

Benzene and toluene are commonly used solvents in various chemical processes. Understanding their partial vapor pressures is essential for optimizing distillation and solvent recovery.

For this example, we will use the following data:

  • Vapor pressure of pure benzene at 30°C: 100 mmHg
  • Vapor pressure of pure toluene at 30°C: 28 mmHg
  • Mole fraction of benzene in the mixture: 0.75
  • Mole fraction of toluene in the mixture: 0.25

Applying Raoult’s Law:

  • Partial vapor pressure of benzene, P₁ = X₁ * P₁⁰ = 0.75 * 100 mmHg = 75 mmHg
  • Partial vapor pressure of toluene, P₂ = X₂ * P₂⁰ = 0.25 * 28 mmHg = 7 mmHg

Notes

The total vapor pressure can be calculated as:

  • Total vapor pressure, P_total = P₁ + P₂ = 75 mmHg + 7 mmHg = 82 mmHg.
    Changes in temperature or composition will alter the vapor pressures, demonstrating the dynamic nature of these mixtures.

Example 3: Acetone and Chloroform Mixture

Context

Acetone and chloroform are widely used in laboratory settings for extraction and purification processes. Their vapor pressures are critical for safe handling and efficient operation.

For this case, we have:

  • Vapor pressure of pure acetone at 20°C: 180 mmHg
  • Vapor pressure of pure chloroform at 20°C: 174 mmHg
  • Mole fraction of acetone in the mixture: 0.3
  • Mole fraction of chloroform in the mixture: 0.7

Using Raoult’s Law:

  • Partial vapor pressure of acetone, P₁ = X₁ * P₁⁰ = 0.3 * 180 mmHg = 54 mmHg
  • Partial vapor pressure of chloroform, P₂ = X₂ * P₂⁰ = 0.7 * 174 mmHg = 121.8 mmHg

Notes

The total vapor pressure is:

  • Total vapor pressure, P_total = P₁ + P₂ = 54 mmHg + 121.8 mmHg = 175.8 mmHg.
    The example is significant in understanding the behavior of volatile organic compounds in mixtures, as variations in composition can significantly affect their handling and reaction kinetics.