To find the vapor pressure of the solution after one mole of the mixture has been vaporized, we can use Raoult's Law, which states that the vapor pressure of each component in an ideal solution is proportional to its mole fraction in the solution. Let's break this down step by step.
Initial Setup
We start with one mole of liquid A and one mole of liquid B. Therefore, the total number of moles in the solution is:
- 1 mole of A + 1 mole of B = 2 moles total
Calculating Mole Fractions
The mole fractions of A and B in the solution can be calculated as follows:
- Mole fraction of A, \( X_A = \frac{n_A}{n_A + n_B} = \frac{1}{2} = 0.5 \)
- Mole fraction of B, \( X_B = \frac{n_B}{n_A + n_B} = \frac{1}{2} = 0.5 \)
Applying Raoult's Law
According to Raoult's Law, the vapor pressure of the solution, \( P_{solution} \), can be calculated using the formula:
P_{solution} = X_A \cdot P^0_A + X_B \cdot P^0_B
Where:
- \( P^0_A = 900 \, \text{torr} \) (vapor pressure of pure A)
- \( P^0_B = 100 \, \text{torr} \) (vapor pressure of pure B)
Calculating the Initial Vapor Pressure
Substituting the values into the equation:
P_{solution} = (0.5 \cdot 900) + (0.5 \cdot 100)
P_{solution} = 450 + 50 = 500 \, \text{torr}
After Vaporization
Now, when one mole of the mixture is vaporized, we need to consider how this affects the remaining amounts of A and B. Since we started with 1 mole of each, and we vaporized 1 mole of the mixture, we can assume that the vaporized mixture consisted of both components in their respective mole fractions.
Calculating Moles Vaporized
Since the mole fractions are equal, the vaporized mixture will contain:
- Moles of A vaporized = \( 0.5 \, \text{moles} \)
- Moles of B vaporized = \( 0.5 \, \text{moles} \)
Remaining Moles in the Solution
After vaporization, the remaining moles of A and B are:
- Remaining moles of A = \( 1 - 0.5 = 0.5 \, \text{moles} \)
- Remaining moles of B = \( 1 - 0.5 = 0.5 \, \text{moles} \)
The total remaining moles in the solution is now:
- Total remaining moles = \( 0.5 + 0.5 = 1 \, \text{mole} \)
New Mole Fractions
The new mole fractions after vaporization are:
- New mole fraction of A, \( X_A' = \frac{0.5}{1} = 0.5 \)
- New mole fraction of B, \( X_B' = \frac{0.5}{1} = 0.5 \)
Calculating New Vapor Pressure
Using Raoult's Law again for the remaining solution:
P_{solution}' = X_A' \cdot P^0_A + X_B' \cdot P^0_B
P_{solution}' = (0.5 \cdot 900) + (0.5 \cdot 100)
P_{solution}' = 450 + 50 = 500 \, \text{torr}
Final Thoughts
Interestingly, even after vaporizing one mole of the mixture, the vapor pressure of the solution remains at 500 torr. This is due to the fact that the mole fractions of A and B did not change after the vaporization process, as they were initially equal and vaporized in equal amounts. Thus, the vapor pressure of the solution at this point is 500 torr.
