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
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- \( \frac{P_A^\circ P_B^\circ}{2} + P_A^\circ + P_B^\circ \)
\(P_T = 3 \cdot \frac{P^\circ_A P^\circ_B}{2 P^\circ_A + P^\circ_B}\)
- \( \frac{P_A^\circ}{2} + P_B^\circ \)
- \( P_A^\circ + 2P_B^\circ \)
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The Correct Option is B
Approach Solution - 1
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:
\[ P_T = 3 \cdot \frac{P^\circ_A P^\circ_B}{2 P^\circ_A + P^\circ_B} \]
Approach Solution -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:
\[ P_T = 3 \cdot \frac{P^\circ_A P^\circ_B}{2 P^\circ_A + P^\circ_B} \]
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