Question

\(A\) and \(B\) form an ideal solution at \(298 \mathrm{K},\) with \(x_{A}=0.320, P_{A}^{*}=84.3\) Torr, and \(P_{B}^{*}=41.2\) Torr. a. Calculate the partial pressures of \(A\) and \(B\) in the gas phase. b. A portion of the gas phase is removed and condensed in a separate container. Calculate the partial pressures of A and \(\mathrm{B}\) in equilibrium with this liquid sample at \(298 \mathrm{K}\).

Short answer

a. The partial pressures of A and B in the gas phase are calculated as follows: P_A = \(0.320 \times 84.3 \mathrm{Torr} = 26.976 \mathrm{Torr}\) P_B = \(0.680 \times 41.2 \mathrm{Torr} = 28.016 \mathrm{Torr}\) b. We do not have enough information to calculate the exact partial pressures of A and B after a portion of the gas phase is removed and condensed.

Step-by-step solution

a. Calculate the partial pressures of A and B in the gas phase

Given the following information: x_A = 0.320 (mole fraction of A) P_A* = 84.3 Torr (vapor pressure of pure A) P_B* = 41.2 Torr (vapor pressure of pure B) Now we can calculate the mole fraction of B: x_B = 1 - x_A = 1 - 0.320 = 0.680 Using Raoult's Law, we can find the partial pressures of A and B in the gas phase: P_A = x_A * P_A* = 0.320 * 84.3 Torr = 26.976 Torr P_B = x_B * P_B* = 0.680 * 41.2 Torr = 28.016 Torr

b. Calculate the partial pressures of A and B in equilibrium with the condensed liquid sample at 298 K

When part of the gas phase is removed and condensed, the mole fractions in the remaining gas phase may change. However, it is not possible to determine the new mole fractions without additional information. Therefore, we can only express the new partial pressures using the new mole fractions: P_A_new = x_A_new * P_A* P_B_new = x_B_new * P_B* We do not have enough information to find the exact values of x_A_new and x_B_new, and consequently, the new partial pressures of A and B.

Key concepts

Partial Pressures

Understanding partial pressures is crucial when studying gas mixtures. In the context of Raoult's Law, the partial pressure of a component in a gas phase is directly proportional to its mole fraction in the liquid phase and its vapor pressure when pure. Simply put, the partial pressure is the pressure each gas in a mixture would exert if it were alone in the container.

In the given exercise, the partial pressure of a component, say A, is calculated as the product of its mole fraction (\(x_A\text{, which is 0.320 for A}\)) and its pure vapor pressure (\(P_A^*\text{, which is 84.3 Torr for A}\)). This relationship reflects how the presence of other gases affects each gas's contribution to the total pressure in a mixture.

Vapor Pressure

Vapor pressure is an important concept in thermodynamics, indicating how a substance behaves when it enters the gas phase. It's defined as the pressure formed by the vapor of a compound above its liquid or solid form in a sealed container, at a specific temperature. Vapor pressure depends on the intrinsic properties of the substance and temperature.

In the exercise, the vapor pressures of pure components A and B (\(P_A^*\text{ and }P_B^*\text{, respectively}\)) are given. Together with their respective mole fractions, these values are essential for calculating the partial pressures using Raoult's Law. The vapor pressures in this context are used to reflect the tendency of A and B to evaporate from the liquid to the gas phase.

Mole Fraction

The mole fraction is a way of expressing the concentration of a component in a mixture. It is the ratio of the number of moles of that component to the total number of moles of all components in the mixture. The mole fraction is dimensionless and always ranges between 0 and 1.

When applying Raoult's Law as seen in the exercise, mole fractions help calculate partial pressures. For example, the mole fraction of A (\(x_A\text{, which is 0.320 for A}\)) is used to assess how much A contributes to the total pressure. The mole fraction is central to understanding how different components in a mixture interact with each other in terms of pressure contributions.

Ideal Solution

An ideal solution is a special type of solution where the intermolecular forces between unlike molecules are approximated to be equal to those between like molecules. This means that no significant change in enthalpy occurs on mixing—the heat of solution is essentially zero. Additionally, the volume of an ideal solution is the sum of the volumes of the pure components; no volume change occurs upon mixing.

For an ideal solution, Raoult's Law can be applied without corrections because it assumes that the components of the solution exhibit ideal behavior. In the exercise, the assumption that A and B form an ideal solution at 298 K allows us to use Raoult's Law straightforwardly to calculate the partial pressures.