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.
Given Information
- Number of moles of liquid A (nA) = 10 moles
- Number of moles of liquid B (nB) = 10 moles
- Vapor pressure of pure A (P0A) = 200 mm Hg
- Vapor pressure of pure B (P0B) = 100 mm Hg
Understanding the Mixture
In our case, we have equal moles of two components, A and B. When the pressure is decreased isothermally, we want to find the pressure at which half of the liquid is converted into vapor. This means that at equilibrium, we will have 5 moles of liquid A and 5 moles of liquid B remaining, and 10 moles of vapor (5 moles of A and 5 moles of B).
Calculating Mole Fractions
Initially, the total number of moles in the liquid phase is:
Total moles (nTotal) = nA + nB = 10 + 10 = 20 moles
When half of the liquid is vaporized, the remaining moles in the liquid phase will be:
Remaining moles = 10 moles (5A + 5B)
The mole fractions of A and B in the liquid phase at this point are:
- X_A = nA / nTotal = 5 / 10 = 0.5
- X_B = nB / nTotal = 5 / 10 = 0.5
Applying Raoult's Law
According to Raoult's Law, the total vapor pressure (P) above the solution can be calculated as:
P = X_A * P0A + X_B * P0B
Substituting the values we have:
P = (0.5 * 200 mm Hg) + (0.5 * 100 mm Hg)
P = 100 mm Hg + 50 mm Hg = 150 mm Hg
Final Result
Therefore, the pressure at which half of the liquid mixture is converted into vapor is 150 mm Hg.
This calculation illustrates how the vapor pressures of the components and their mole fractions interact to determine the equilibrium pressure in a binary liquid mixture. By understanding these relationships, you can predict the behavior of mixtures under varying conditions.