Question

For two pure volatile liquids X and Y, attractive intermolecular interactions of both X-X and Y-Y are weaker than those of X-Y. The total vapour pressure of an equimolar solution of X and Y is p$_{\text{total}$. The vapour pressure of pure X and pure Y are p$^0_{\text{X}}$ and p$^0_{\text{Y}}$, respectively. Which one of the following relations is correct?

• A. p$_{\text{total}} < $ (p$^0_{\text{X}}$ + p$^0_{\text{Y}}$)/2
• B. p$_{\text{total}} = $ (p$^0_{\text{X}}$ + p$^0_{\text{Y}}$)/2
• C. p$_{\text{total}} = $ p$^0_{\text{X}}$ + p$^0_{\text{Y}}$
• D. p$_{\text{total}} > $ (p$^0_{\text{X}}$ + p$^0_{\text{Y}}$)/2

Hint:

Intermolecular forces rules:
- Stronger $\text{X}-\text{Y}$ interactions $\rightarrow$ negative deviation $\rightarrow$ actual $P < P_{\text{ideal}}$ $\rightarrow$ $p_{\text{total}} < \frac{p_{\text{X}}^0 + p_{\text{Y}}^0}{2}$.
- Weaker $\text{X}-\text{Y}$ interactions $\rightarrow$ positive deviation $\rightarrow$ actual $P > P_{\text{ideal}}$ $\rightarrow$ $p_{\text{total}} > \frac{p_{\text{X}}^0 + p_{\text{Y}}^0}{2}$.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
This question asks us to identify the correct relationship for the total vapour pressure ($p_{\text{total}}$) of an equimolar mixture of two volatile liquids, X and Y, where the adhesive intermolecular forces ($\text{X}-\text{Y}$) are stronger than the cohesive forces ($\text{X}-\text{X}$ and $\text{Y}-\text{Y}$).

Step 2: Key Formula or Approach:
According to Raoult's Law, for an ideal liquid solution, the total vapour pressure is:
\[ p_{\text{ideal}} = x_{\text{X}} p_{\text{X}}^0 + x_{\text{Y}} p_{\text{Y}}^0 \] For an equimolar solution, the mole fractions are:
\[ x_{\text{X}} = x_{\text{Y}} = 0.5 \] Thus, the ideal vapour pressure would be:
\[ p_{\text{ideal}} = 0.5 p_{\text{X}}^0 + 0.5 p_{\text{Y}}^0 = \frac{p_{\text{X}}^0 + p_{\text{Y}}^0}{2} \]

Step 3: Detailed Explanation:
Let's analyze how the relative strength of intermolecular interactions affects the actual vapour pressure:

• We are given that the attractive interactions between unlike molecules ($\text{X}-\text{Y}$) are stronger than those between like molecules ($\text{X}-\text{X}$ and $\text{Y}-\text{Y}$).
- This means that when X and Y are mixed together, they form stronger attractive bonds with each other than they did in their pure liquid states.
- Because the intermolecular forces are stronger in the solution, the molecules of both X and Y are held more tightly in the liquid phase.
- This significantly reduces the tendency of both components to escape into the gas phase (i.e., less evaporation occurs).
- Consequently, the partial vapour pressures of both X and Y will be lower than predicted by Raoult's Law:
\[ p_{\text{X}} < x_{\text{X}} p_{\text{X}}^0 \quad \text{and} \quad p_{\text{Y}} < x_{\text{Y}} p_{\text{Y}}^0 \] - This behavior represents a negative deviation from Raoult's Law.
- Therefore, the actual total vapour pressure ($p_{\text{total}}$) will be lower than the ideal total vapour pressure:
\[ p_{\text{total}} < p_{\text{ideal}} \] \[ p_{\text{total}} < \frac{p_{\text{X}}^0 + p_{\text{Y}}^0}{2} \]

Step 4: Final Answer:
Since the solution shows a negative deviation, the total vapour pressure $p_{\text{total}}$ is less than the average of the two pure vapour pressures, matching option (A).