Askiitians Tutor Team
Last Activity: 4 Months ago
To determine the pressure at which half of the liquid in an ideal binary mixture is converted into vapor, we can apply Raoult's Law. This law states that the partial vapor pressure of each component in an ideal solution is proportional to its mole fraction in the liquid phase. Let's break down the problem step by step.
Understanding the Components
We have a binary mixture of two liquids, A and B, with the following properties:
- Liquid A: Vapor pressure, PA = 200 mm Hg
- Liquid B: Vapor pressure, PB = 100 mm Hg
- Moles of A = 10
- Moles of B = 10
Calculating Initial Conditions
Initially, the total number of moles in the mixture is:
Total moles = Moles of A + Moles of B = 10 + 10 = 20 moles
The mole fractions of A and B can be calculated as follows:
- Mole fraction of A, XA = Moles of A / Total moles = 10 / 20 = 0.5
- Mole fraction of B, XB = Moles of B / Total moles = 10 / 20 = 0.5
Applying Raoult's Law
According to Raoult's Law, the partial pressures of A and B in the vapor phase can be calculated as:
- Partial pressure of A, PA = XA * PA = 0.5 * 200 mm Hg = 100 mm Hg
- Partial pressure of B, PB = XB * PB = 0.5 * 100 mm Hg = 50 mm Hg
Finding Total Vapor Pressure
The total vapor pressure, Ptotal, when half of the liquid is converted into vapor is the sum of the partial pressures:
Ptotal = PA + PB = 100 mm Hg + 50 mm Hg = 150 mm Hg
Final Result
Thus, the pressure at which half of the liquid mixture is converted into vapor is 150 mm Hg. This pressure reflects the balance between the vapor pressures of the two components in the mixture at the given conditions.
In summary, by applying Raoult's Law and calculating the mole fractions and partial pressures, we can effectively determine the pressure at which half of the liquid is vaporized in an ideal binary mixture. This approach is fundamental in understanding phase equilibria in solutions.