Question

Two liquids A and B are mixed in such a proportion that they form an ideal solution whose total vapor pressure is exactly three times that of the partial pressure of A. If \( P_A^\circ \) and \( P_B^\circ \) are the vapor pressures of pure A and B respectively, then the total vapor pressure of the solution is given by

• A. \( \frac{P_A^\circ P_B^\circ}{2} + P_A^\circ + P_B^\circ \)
• B. \(P_T = 3 \cdot \frac{P^\circ_A P^\circ_B}{2 P^\circ_A + P^\circ_B}\)
• C. \( \frac{P_A^\circ}{2} + P_B^\circ \)
• D. \( P_A^\circ + 2P_B^\circ \)
• E. More data needed to solve the problem

Hint:

Raoult’s Law helps in calculating the vapor pressures of components in an ideal solution. Make sure to consider the mole fractions and partial pressures for each component.

Answer 1

The Correct Option is B

Given:

• Total Pressure: \( P_T = P_A + P_B \), where \( P_A \) and \( P_B \) are the partial pressures of A and B, respectively.
• Partial Pressure of A: \( P_A = P^\circ_A X_A \), where \( P^\circ_A \) is the vapor pressure of A and \( X_A \) is the mole fraction of A.
• Partial Pressure of B: \( P_B = P^\circ_B X_B \), where \( P^\circ_B \) is the vapor pressure of B and \( X_B \) is the mole fraction of B.
• The sum of the mole fractions: \( X_A + X_B = 1 \).

Step 1: Given that the total pressure \( P_T \) is 3, we can write: 

\[ P_T = P_A + P_B = 3 \]

Substituting \( P_A \) and \( P_B \) in terms of their vapor pressures and mole fractions:

\[ 3 = P^\circ_A X_A + P^\circ_B X_B \]

Step 2: We are also given that \( 2P_A = P_B \), so:

\[ 2P^\circ_A X_A = P^\circ_B X_B \]

Step 3: From Step 2, we can express \( X_B \) as:

\[ X_B = \frac{2P^\circ_A X_A}{P^\circ_B} \]

Step 4: Substitute this expression for \( X_B \) into the mole fraction sum equation:

\[ X_A + \frac{2P^\circ_A X_A}{P^\circ_B} = 1 \]

Now, solve for \( X_A \):

\[ X_A \left( 1 + \frac{2P^\circ_A}{P^\circ_B} \right) = 1 \]

So, we get:

\[ X_A = \frac{P^\circ_B}{2P^\circ_A + P^\circ_B} \]

Step 5: Finally, substitute this value of \( X_A \) back into the total pressure equation:

\[ P_T = 3 \cdot \frac{P^\circ_A P^\circ_B}{2 P^\circ_A + P^\circ_B} \]

Final Answer:

The total pressure \( P_T \) is calculated as:

\[ P_T = 3 \cdot \frac{P^\circ_A P^\circ_B}{2 P^\circ_A + P^\circ_B} \]

Answer 2

Total Pressure Calculation for Two Liquid Mixture 

We know that the total pressure \( P_T \) of a mixture is given by:

\[ P_T = P_A + P_B \]

Where \( P_A \) and \( P_B \) are the partial pressures of components A and B, respectively.

Partial pressures are related to their vapor pressures and mole fractions:

\[ P_A = P^\circ_A X_A \]

Where \( P^\circ_A \) is the vapor pressure of A, and \( X_A \) is the mole fraction of A.

Similarly, for component B:

\[ P_B = P^\circ_B X_B \]

Also, the sum of the mole fractions must be equal to 1:

\[ X_A + X_B = 1 \]

Step-by-Step Calculation

Step 1: We are given that the total pressure \( P_T \) is 3, so:

\[ P_T = P_A + P_B = 3 \]

Substitute the expressions for \( P_A \) and \( P_B \) in terms of vapor pressures and mole fractions:

\[ 3 = P^\circ_A X_A + P^\circ_B X_B \]

Step 2: We are also given that \( 2P_A = P_B \), so:

\[ 2P^\circ_A X_A = P^\circ_B X_B \]

Step 3: Rearranging this equation, we get:

\[ X_B = \frac{2P^\circ_A X_A}{P^\circ_B} \]

Step 4: Substitute this expression for \( X_B \) into the equation \( X_A + X_B = 1 \):

\[ X_A + \frac{2P^\circ_A X_A}{P^\circ_B} = 1 \]

Solve for \( X_A \):

\[ X_A \left( 1 + \frac{2P^\circ_A}{P^\circ_B} \right) = 1 \]

So we get:

\[ X_A = \frac{P^\circ_B}{2P^\circ_A + P^\circ_B} \]

Step 5: Finally, substitute this value of \( X_A \) back into the total pressure equation:

\[ P_T = 3 \cdot \frac{P^\circ_A P^\circ_B}{2 P^\circ_A + P^\circ_B} \]

Final Answer:

The total pressure \( P_T \) is given by:

\[ P_T = 3 \cdot \frac{P^\circ_A P^\circ_B}{2 P^\circ_A + P^\circ_B} \]