Vapour Pressure of Liquid Solutions – Friendly Notes 😊
1 • Vapour Pressure in a Nutshell 💨
When a liquid sits in a closed container, some molecules shoot into the air and form vapour. The pressure these vapour molecules exert on the liquid surface at equilibrium is the vapour pressure. Higher temperature means more energetic molecules and a larger vapour pressure.:contentReference[oaicite:0]{index=0}
2 • Raoult’s Law for Two Volatile Liquids 🧪
Imagine a binary solution with components 1 and 2. Raoult said:
\( p_{1}=x_{1}p_{1}^{0} \) and \( p_{2}=x_{2}p_{2}^{0} \) (where \(p_{1}^{0}\) and \(p_{2}^{0}\) are the vapour pressures of the pure liquids):contentReference[oaicite:1]{index=1}
Total vapour pressure is the sum of the partials:
\( p_{\text{total}}=p_{1}+p_{2}=p_{1}^{0}+(p_{2}^{0}-p_{1}^{0})x_{2} \):contentReference[oaicite:2]{index=2}
- The graph of \(p_{\text{total}}\) versus mole fraction is a straight line.📈:contentReference[oaicite:3]{index=3}
- If \(p_{2}^{0} > p_{1}^{0}\), the line slopes upward as you add more of component 2.
Vapour-Phase Composition 💭
Using Dalton’s law:
\( y_{1}=\dfrac{p_{1}}{p_{\text{total}}} \quad\text{and}\quad y_{2}=\dfrac{p_{2}}{p_{\text{total}}} \):contentReference[oaicite:4]{index=4}
The vapour is richer in the component with the higher pure-liquid vapour pressure – that’s the more “eager” molecule! 😎
3 • Raoult × Henry = Best Buddies 🤝
For a very volatile gas dissolved in a liquid, Henry’s law says \( p = K_{H}x \). Compare this with \( p = x\,p^{0} \) above, and you’ll notice Raoult’s law is simply Henry’s law with \( K_{H}=p^{0} \).:contentReference[oaicite:5]{index=5}
4 • Non-Volatile Solute = Lower Vapour Pressure 🧂
Drop a pinch of a non-volatile solute into a solvent and some surface spots get occupied by solute particles. Fewer solvent molecules can now escape, so the vapour pressure falls. The bigger the solute fraction, the bigger the drop – the chemical identity of the solute hardly matters!:contentReference[oaicite:6]{index=6}
5 • Ideal vs Non-Ideal Solutions 🌈
Ideal Solutions (dream team) ✨
- Obey Raoult’s law across the whole composition range.:contentReference[oaicite:7]{index=7}
- Mixing gives \( \Delta_{\text{mix}}H = 0 \) and \( \Delta_{\text{mix}}V = 0 \) – no heat change, no volume change.:contentReference[oaicite:8]{index=8}
- A–A, B–B, and A–B intermolecular attractions are practically equal (think n-hexane + n-heptane).
Non-Ideal Solutions (real-world rebels) 😮
- Positive deviation: vapour pressure shoots above Raoult’s line because A–B forces are weaker than A–A or B–B ➜ molecules escape more easily.:contentReference[oaicite:9]{index=9}
- Negative deviation: vapour pressure sinks below Raoult’s line because A–B forces are stronger ➜ molecules cling tighter, escape less.:contentReference[oaicite:10]{index=10}
6 • High-Yield NEET Points 🎯
- Raoult’s law equations and straight-line vapour-pressure plots are must-know.📊
- When solute is non-volatile, vapour-pressure lowering depends only on mole fraction, not solute type.
- Vapour phase is richer in the more volatile component; use \( y_{i}=\dfrac{p_{i}}{p_{\text{total}}} \) to prove it easily.
- Ideal solution criteria: \( \Delta_{\text{mix}}H = 0 \) and \( \Delta_{\text{mix}}V = 0 \).
- Positive vs negative deviation concepts help predict azeotrope formation – a favourite exam twist! 😉
Keep practicing, and let the chemistry concepts evaporate into your memory! 🚀