Determine Vapor Pressure of Pure Component in Ideal Solution

To find the vapor pressure of pure liquid $\mathbf{B}$ (denoted as $P_\mathbf{B}^\circ$), we apply Raoult's law for an ideal binary liquid solution of components $\mathbf{A}$ and $\mathbf{B}$.

According to Raoult's law, the total vapor pressure of an ideal solution ($P_{\text{total}}$) is given by: $P_{\text{total}} = P_\mathbf{A} + P_\mathbf{B} = P_\mathbf{A}^\circ x_\mathbf{A} + P_\mathbf{B}^\circ x_\mathbf{B}$

Since the sum of mole fractions is $x_\mathbf{A} + x_\mathbf{B} = 1$, we can express $x_\mathbf{B}$ as $1 - x_\mathbf{A}$: $P_{\text{total}} = P_\mathbf{A}^\circ x_\mathbf{A} + P_\mathbf{B}^\circ (1 - x_\mathbf{A})$ $P_{\text{total}} = P_\mathbf{B}^\circ + (P_\mathbf{A}^\circ - P_\mathbf{B}^\circ) x_\mathbf{A}$

We are given two different solution compositions:

Case 1: When $x_\mathbf{A} = 0.25$, the total vapor pressure is $P_{\text{total}} = 0.3\text{ bar}$. $0.3 = P_\mathbf{B}^\circ + 0.25 (P_\mathbf{A}^\circ - P_\mathbf{B}^\circ) \quad \text{--- (Equation 1)}$

Case 2: When $x_\mathbf{A} = 0.50$, the total vapor pressure is $P_{\text{total}} = 0.4\text{ bar}$. $0.4 = P_\mathbf{B}^\circ + 0.50 (P_\mathbf{A}^\circ - P_\mathbf{B}^\circ) \quad \text{--- (Equation 2)}$

Subtracting Equation 1 from Equation 2 gives: $(0.4 - 0.3) = (0.50 - 0.25)(P_\mathbf{A}^\circ - P_\mathbf{B}^\circ)$ $0.1 = 0.25 (P_\mathbf{A}^\circ - P_\mathbf{B}^\circ)$ $P_\mathbf{A}^\circ - P_\mathbf{B}^\circ = \frac{0.1}{0.25} = 0.4\text{ bar}$

Now, substituting $(P_\mathbf{A}^\circ - P_\mathbf{B}^\circ) = 0.4\text{ bar}$ back into Equation 1: $0.3 = P_\mathbf{B}^\circ + 0.25(0.4)$ $0.3 = P_\mathbf{B}^\circ + 0.1$ $P_\mathbf{B}^\circ = 0.3 - 0.1 = 0.2\text{ bar}$

Thus, the vapor pressure of pure liquid $\mathbf{B}$ is $0.2$ bar (or $0.20$ bar).