Saurabh Koranglekar
Last Activity: 5 Years ago
To find the vapor pressures of the pure components X and Y, we can utilize Raoult's Law, which states that the vapor pressure of a solution is dependent on the vapor pressures of the pure components and their mole fractions in the solution. Let's break down the problem step by step.
Understanding the Initial Setup
We know the following:
- The solution at 300 K has 1 mol of X and 3 mol of Y, giving a total of 4 moles.
- The vapor pressure of this solution is 550 mmHg.
Calculating Mole Fractions
The mole fraction of each component can be calculated as follows:
- Mole fraction of X, \( X_X = \frac{1}{1 + 3} = \frac{1}{4} = 0.25 \)
- Mole fraction of Y, \( X_Y = \frac{3}{1 + 3} = \frac{3}{4} = 0.75 \)
Applying Raoult's Law
According to Raoult's Law, the vapor pressure of the solution (\( P_{solution} \)) can be expressed as:
P_solution = X_X * P^0_X + X_Y * P^0_Y
Where \( P^0_X \) and \( P^0_Y \) are the vapor pressures of pure X and Y, respectively. Plugging in the known values, we have:
550 mmHg = (0.25 * P^0_X) + (0.75 * P^0_Y)
After Adding More Y
Now, we add 1 mol of Y, which makes the new composition:
This gives a total of 5 moles. The new mole fractions are:
- Mole fraction of X, \( X_X = \frac{1}{1 + 4} = \frac{1}{5} = 0.20 \)
- Mole fraction of Y, \( X_Y = \frac{4}{1 + 4} = \frac{4}{5} = 0.80 \)
Using Raoult's Law again for the new solution, we have:
P_new = X_X * P^0_X + X_Y * P^0_Y
Since the vapor pressure increases by 10 mmHg, the new pressure is:
P_new = 550 mmHg + 10 mmHg = 560 mmHg
Substituting the mole fractions into the equation gives:
560 mmHg = (0.20 * P^0_X) + (0.80 * P^0_Y)
Setting Up Equations
Now we have a system of two equations:
- Equation 1: \( 550 = 0.25 * P^0_X + 0.75 * P^0_Y \)
- Equation 2: \( 560 = 0.20 * P^0_X + 0.80 * P^0_Y \)
Solving the Equations
We can solve these equations simultaneously. Start with Equation 1:
0.25 * P^0_X + 0.75 * P^0_Y = 550
From this, we can express \( P^0_Y \) in terms of \( P^0_X \):
P^0_Y = \frac{550 - 0.25 * P^0_X}{0.75}
Now substituting this expression into Equation 2:
560 = 0.20 * P^0_X + 0.80 * \left(\frac{550 - 0.25 * P^0_X}{0.75}\right)
Multiplying through by 0.75 to eliminate the fraction:
420 = 0.15 * P^0_X + 0.80 * (550 - 0.25 * P^0_X)
Expanding and simplifying gives:
420 = 0.15 * P^0_X + 440 - 0.20 * P^0_X
Combining like terms results in:
420 - 440 = -0.05 * P^0_X
-20 = -0.05 * P^0_X
Thus, we find:
P^0_X = 400 mmHg
Finding the Vapor Pressure of Y
Now substituting \( P^0_X \) back into the expression for \( P^0_Y \):
P^0_Y = \frac{550 - 0.25 * 400}{0.75} = \frac{550 - 100}{0.75} = \frac{450}{0.75} = 600 mmHg
Final Result
In conclusion, the vapor pressures of the pure components are:
- P^0_X = 400 mmHg
- P^0_Y = 600 mmHg