Understanding Vapor Composition at 60°C

What is the composition of the vapor above the solution at 60°C?

At 60°C, the vapor pressure of ethyl alcohol is 352.7 mm Hg, and that of methyl alcohol is 652 mm Hg. A mixture of the two, assuming ideal behavior, contains 50% by weight of each constituent. What will be the composition of the vapor above the solution at 60°C?

Composition of Vapor Above the Solution at 60°C

The vapor above the solution at 60°C, assuming ideal behavior, will have a composition of approximately 35.3% ethyl alcohol and 64.7% methyl alcohol by mole fraction.

To determine the composition of the vapor above the solution at 60°C, we can use Raoult's law, which states that the partial pressure of each component in an ideal solution is directly proportional to its mole fraction in the solution.

Given that the mixture contains 50% by weight of each constituent (ethyl alcohol and methyl alcohol), we need to convert the weight percentages to mole fractions.

First, calculate the mole fractions of each component:

Mole fraction of ethyl alcohol (X₁) = 0.50

Mole fraction of methyl alcohol (X₂) = 0.50

Next, we can use Raoult's law to determine the partial pressures of each component in the vapor phase.

Partial pressure of ethyl alcohol (P₁) = X₁ * vapor pressure of ethyl alcohol at 60°C

Partial pressure of methyl alcohol (P₂) = X₂ * vapor pressure of methyl alcohol at 60°C

Using the given vapor pressure values:

Partial pressure of ethyl alcohol (P₁) = 0.50 * 352.7 mm Hg

Partial pressure of methyl alcohol (P₂) = 0.50 * 652 mm Hg

Finally, we can calculate the mole fractions of each component in the vapor phase using the partial pressures and the total pressure of the vapor above the solution.

Total pressure of the vapor = P₁ + P₂

Mole fraction of ethyl alcohol in the vapor phase = P₁ / (P₁ + P₂)

Mole fraction of methyl alcohol in the vapor phase = P₂ / (P₁ + P₂)