Question

Liquid A has vapor pressure $x$ , and liquid $\mathrm{B}$ has vapor pressure $y.$ What is the mole fraction of the liquid mixture if the vapor above the solution is $30.\mathrm{\%}\mathrm{A}$ by moles? $50.\mathrm{\%}\mathrm{A}?80.\mathrm{\%}\mathrm{A}$ ? (Calculate in terms of $x$ and $y.)$ Liquid A has vapor pressure $x,$ liquid $\mathrm{B}$ has vapor pressure y. What is the mole fraction of the vapor above the solution if the liquid mixture is $30.\mathrm{\%}\mathrm{A}$ by moles? $50.\mathrm{\%}\mathrm{A}?80.\mathrm{\%}\mathrm{A}$ ? (Calculate in terms of $x$ and $y.)$

Short answer

For given mole fractions of vapor, the mole fractions of A and B in the liquid phase are as follows: Case 1: $30\mathrm{\%}A$ $x_A=\frac{0.3\times (x+y)}{x}$, $x_B=\frac{0.7\times (x+y)}{y}$ Case 2: $50\mathrm{\%}A$ $x_A=\frac{0.5\times (x+y)}{x}$, $x_B=\frac{0.5\times (x+y)}{y}$ Case 3: $80\mathrm{\%}A$ $x_A=\frac{0.8\times (x+y)}{x}$, $x_B=\frac{0.2\times (x+y)}{y}$ For given mole fractions of the liquid mixture, the mole fractions of A and B in the vapor phase are as follows: Case 1: $30\mathrm{\%}A$ $y_A=\frac{0.3\times x}{x+y}$, $y_B=\frac{0.7\times y}{x+y}$ Case 2: $50\mathrm{\%}A$ $y_A=\frac{0.5\times x}{x+y}$, $y_B=\frac{0.5\times y}{x+y}$ Case 3: $80\mathrm{\%}A$ $y_A=\frac{0.8\times x}{x+y}$, $y_B=\frac{0.2\times y}{x+y}$

Step-by-step solution

Write down Raoult's Law for each component

According to Raoult's Law, the partial pressure of a component i in a mixture is equal to its mole fraction times its vapor pressure. We write the partial pressure of A and B with respect to their mole fraction in the liquid phase and vapor pressures as: $P_A=x_A\times x$ $P_B=x_B\times y$

Calculate the mole fractions of each component in the vapor phase

We are given the mole fraction of A in the vapor phase for each case (30%, 50%, and 80%). We can find the mole fraction of B in the vapor phase simply by subtracting the mole fraction of A from 1: Case 1: $30\mathrm{\%}$ $y_A=0.3$ $y_B=1-y_A=0.7$ Case 2: $50\mathrm{\%}$ $y_A=0.5$ $y_B=1-y_A=0.5$ Case 3: $80\mathrm{\%}$ $y_A=0.8$ $y_B=1-y_A=0.2$

Use the given mole fractions of the vapor phase to find the mole fractions of A and B in the liquid phase

We can use the partial pressure equations from Step 1 to find the mole fractions of A and B for each case. We will solve for $x_A$ and $x_B$ in terms of the given vapor pressures x and y: Case 1: $30\mathrm{\%}A$ $P_A=x_A\times x\Rightarrow x_A=\frac{0.3\times (x+y)}{x}$ $P_B=x_B\times y\Rightarrow x_B=\frac{0.7\times (x+y)}{y}$ Case 2: $50\mathrm{\%}A$ $P_A=x_A\times x\Rightarrow x_A=\frac{0.5\times (x+y)}{x}$ $P_B=x_B\times y\Rightarrow x_B=\frac{0.5\times (x+y)}{y}$ Case 3: $80\mathrm{\%}A$ $P_A=x_A\times x\Rightarrow x_A=\frac{0.8\times (x+y)}{x}$ $P_B=x_B\times y\Rightarrow x_B=\frac{0.2\times (x+y)}{y}$ Part 2: Finding the mole fraction of the vapor for given mole fractions of the liquid mixture In this part, we will follow the same steps as before, but we will calculate the mole fraction of the vapor for given mole fractions of the liquid mixture rather than vice versa.

Calculate the mole fractions of liquid A

We are given the mole fraction of A in the liquid phase for each case (30%, 50%, and 80%). We can find the mole fraction of B in the liquid phase simply by subtracting the mole fraction of A from 1: Case 1: $30\mathrm{\%}$ $x_A=0.3$ $x_B=1-x_A=0.7$ Case 2: $50\mathrm{\%}$ $x_A=0.5$ $x_B=1-x_A=0.5$ Case 3: $80\mathrm{\%}$ $x_A=0.8$ $x_B=1-x_A=0.2$

Use the given mole fractions of the liquid phase to find the mole fractions of A and B in the vapor phase

We can use the partial pressure equations from Step 1 to find the mole fractions of A and B for each case. We will solve for $y_A$ and $y_B$ in terms of the given vapor pressures x and y: Case 1: $30\mathrm{\%}A$ $P_A=x_A\times x\Rightarrow y_A=\frac{0.3\times x}{x+y}$ $P_B=x_B\times y\Rightarrow y_B=\frac{0.7\times y}{x+y}$ Case 2: $50\mathrm{\%}A$ $P_A=x_A\times x\Rightarrow y_A=\frac{0.5\times x}{x+y}$ $P_B=x_B\times y\Rightarrow y_B=\frac{0.5\times y}{x+y}$ Case 3: $80\mathrm{\%}A$ $P_A=x_A\times x\Rightarrow y_A=\frac{0.8\times x}{x+y}$ $P_B=x_B\times y\Rightarrow y_B=\frac{0.2\times y}{x+y}$ In conclusion, we have calculated the mole fraction of the liquid mixture and of the vapor for the given cases in terms of the vapor pressures x and y.

Key concepts

Vapor Pressure

Vapor pressure is an essential concept when talking about Raoult's Law, as it directly relates to the behavior of liquid mixtures. In simple terms, vapor pressure is the pressure exerted by the vapor when a liquid is in equilibrium with its vapor phase. This value depends on several factors, including the temperature and the nature of the liquid.

Some key points about vapor pressure:

• Every pure liquid has a specific vapor pressure at a given temperature.
• A higher vapor pressure indicates that a liquid evaporates more readily.
• In a mixture, the total vapor pressure is influenced by the vapor pressure of each component and their respective mole fractions.

Understanding vapor pressure helps in analyzing and predicting how components in a solution will behave under various conditions, paving the way for applications in chemistry and industry.

Mole Fraction

The mole fraction is a way to express the concentration of a component in a mixture. It is the ratio of the number of moles of a specific component to the total moles of all components in the mixture. This value is crucial in applying Raoult's Law to determine the partial pressures of the components in a liquid-vapor mixture.

Here’s why mole fraction is important:

• The mole fraction has no units, making it a convenient way of expressing concentrations.
• It's used in Raoult's Law to calculate the partial vapor pressure of each component in the mixture: $P_i=x_i\times P_i^0$.
• Mole fractions of all components in a mixture add up to 1.

By knowing the mole fraction and vapor pressure of a component, you can deduce how a mixture will behave in different scenarios, helping to predict and control chemical processes.

Liquid-Vapor Equilibrium

Liquid-vapor equilibrium is a state in which a liquid and its vapor exist together in balance at a certain temperature and pressure. This concept is vital in systems where phase changes occur, like distillation or refrigeration, and directly ties back to Raoult's Law and vapor pressure.

Important aspects of liquid-vapor equilibrium include:

• At equilibrium, the rate of evaporation equals the rate of condensation.
• The vapor pressure of the mixture governs the equilibrium state.
• Using Raoult's Law, you can deduce how changes in composition or temperature will affect the vapor phase.

Studying this equilibrium helps in understanding how mixtures can be separated into components, which is particularly valuable in industries like petrochemicals or pharmaceuticals.