Liquids A and B form ideal solution for all compositions of A and B at 25°C. Two such solutions with

Q: Why do we use $X_B = 1 - X_A$?

In a binary mixture, there are only two components. The sum of all fractional parts must equal the whole, which is 1. Therefore, if you know the fraction of one component, the other is simply the remainder.

Q: Can we solve this by equating the slopes?

Yes! The graph of $P_T$ versus $X_A$ is a straight line for an ideal solution. You can find the slope $m = \frac{0.4 - 0.3}{0.50 - 0.25} = 0.4$. The equation of the line is $P_T = m X_A + c$. Since $P_T = (P_A^\circ - P_B^\circ)X_A + P_B^\circ$, the y-intercept $c$ is exactly $P_B^\circ$. Using a point $(0.25, 0.3)$, $0.3 = 0.4(0.25) + c \implies c = 0.2$. So $P_B^\circ = 0.2$.

Q: What if the solution was non-ideal?

If the solution was non-ideal, it would show either positive or negative deviation from Raoult's Law. The total pressure would not be a simple linear combination of the pure vapor pressures, and we would need activity coefficients to solve it.

Q: Why did we simplify the second equation first?

Simplifying $0.4 = 0.50 P_A^\circ + 0.50 P_B^\circ$ gives a very clean relation $P_A^\circ + P_B^\circ = 0.8$. This makes the subsequent substitution much easier and less prone to calculation errors compared to isolating a variable from the first equation.

Q: Is it possible for $P_B^\circ$ to be negative?

No, vapor pressure is a physical pressure exerted by a gas in equilibrium with its liquid phase. It must always be a positive value (or zero for a completely non-volatile substance).

Question

Liquids A and B form ideal solution for all compositions of A and B at 25°C. Two such solutions with 0.25 and 0.50 mole fractions of A have the total vapor pressures of 0.3 and 0.4 bar, respectively. What is the vapor pressure of pure liquid B in bar?

Sigma Percentile · Accepted Answer

The problem of finding the vapor pressure of pure components from the total vapor pressure of their ideal mixtures is a classic application of Raoult's Law. It tests your ability to translate physical chemistry principles into a solvable system of linear equations.

### Understanding the Physical Setup

Imagine you have two volatile liquids, A and B. When you mix them, they form an **ideal solution**. This means the molecules of A and B interact with each other exactly as they would with their own kind. There is no preference, no extra attraction, and no repulsion. Because of this perfect harmony, the solution obeys **Raoult's Law** across all concentrations.

Raoult's Law states that the partial vapor pressure of any volatile component in an ideal solution is equal to the vapor pressure of the pure component multiplied by its mole fraction in the solution.

Mathematically, for our two components:
$$P_A = P_A^\circ X_A$$
$$P_B = P_B^\circ X_B$$

The total vapor pressure ($P_T$) above the solution is simply the sum of these partial pressures, according to Dalton's Law of Partial Pressures:
$$P_T = P_A + P_B = P_A^\circ X_A + P_B^\circ X_B$$

### The Master Equation

We are given data for two different mixtures. To use our total pressure equation, we need to remember a crucial constraint: in a binary mixture, the sum of the mole fractions must always equal 1.
$$X_A + X_B = 1 \implies X_B = 1 - X_A$$

This allows us to express the total pressure entirely in terms of the mole fraction of A:
$$P_T = P_A^\circ X_A + P_B^\circ (1 - X_A)$$

### Setting Up the System of Equations

The problem provides two specific scenarios. Let's translate them into mathematical equations.

**Scenario 1:**
We are told that when the mole fraction of A ($X_A$) is $0.25$, the total vapor pressure ($P_T$) is $0.3 	ext{ bar}$.
Since $X_A = 0.25$, the mole fraction of B must be $X_B = 1 - 0.25 = 0.75$.
Substituting these into our master equation:
$$0.3 = P_A^\circ (0.25) + P_B^\circ (0.75) \quad 	ext{--- (Equation 1)}$$

**Scenario 2:**
In the second mixture, the mole fraction of A ($X_A$) is $0.50$, which means the total vapor pressure ($P_T$) is $0.4 	ext{ bar}$.
Here, $X_B = 1 - 0.50 = 0.50$.
Substituting these values:
$$0.4 = P_A^\circ (0.50) + P_B^\circ (0.50) \quad 	ext{--- (Equation 2)}$$

### Solving the Linear Equations

We now have a neat system of two linear equations with two variables, $P_A^\circ$ and $P_B^\circ$. Our goal is to find $P_B^\circ$.

Let's look at Equation 2. It can be simplified beautifully. Notice that $0.50$ is a common factor on the right side.
$$0.4 = 0.50 (P_A^\circ + P_B^\circ)$$

Dividing both sides by $0.50$:
$$P_A^\circ + P_B^\circ = \frac{0.4}{0.50} = 0.8$$

This simplified relation is incredibly useful. We can express $P_A^\circ$ in terms of $P_B^\circ$:
$$P_A^\circ = 0.8 - P_B^\circ$$

Now, we substitute this expression for $P_A^\circ$ back into Equation 1:
$$0.3 = (0.8 - P_B^\circ)(0.25) + P_B^\circ (0.75)$$

Let's expand the brackets carefully:
$$0.3 = (0.8 	imes 0.25) - 0.25 P_B^\circ + 0.75 P_B^\circ$$
$$0.3 = 0.2 + 0.50 P_B^\circ$$

### The Final Calculation

We are just one step away from the answer. Let's isolate the term with $P_B^\circ$:
$$0.3 - 0.2 = 0.50 P_B^\circ$$
$$0.1 = 0.50 P_B^\circ$$

Finally, solving for $P_B^\circ$:
$$P_B^\circ = \frac{0.1}{0.50} = 0.2 	ext{ bar}$$

The vapor pressure of pure liquid B is **$0.2 	ext{ bar}$**.

*(As a bonus, if we plug this back into our simplified Equation 2, we find that $P_A^\circ = 0.8 - 0.2 = 0.6 	ext{ bar}$. This confirms that liquid A is more volatile than liquid B.)*

Animated Solution for Chemistry - Solutions: Liquids A and B form ideal solution for all compositions of A and B at 25°C. Two such solutions with 0.25 and 0.50 mole fractions of A have the total vapor pressures of 0.3 and 0.4 bar, respectively. What is the vapor pressure of pure liquid B in bar?

Enter Numerical Value:

Visualized Solution

$P_{T} = P_{A}^{\circ} X_{A} + P_{B}^{\circ} X_{B}$

$X_{A} + X_{B} = 1$

$0.3 = P_{A}^{\circ} (0.25) + P_{B}^{\circ} (0.75)$

$0.4 = P_{A}^{\circ} (0.50) + P_{B}^{\circ} (0.50)$

$P_{A}^{\circ} + P_{B}^{\circ} = 0.8$

$0.3 = (0.8 - P_{B}^{\circ}) (0.25) + P_{B}^{\circ} (0.75)$

$P_{B}^{\circ} = 0.2 bar$

\text{Conclusion}

The Sigma Insight: Henry's Law and Raoult's Law

Understanding the Physical Setup

The Master Equation

Setting Up the System of Equations

Solving the Linear Equations

The Final Calculation

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