Boiling Point & Elevation in Boiling Point – Comprehensive Notes

Boiling Point of Pure Liquid

• Operational definition
  • A liquid boils when the vapour pressure inside the bubble equals the external (atmospheric) pressure: $(P<em>{vap}=P</em>{ext})$.
• Evaporation vs. Boiling
  • Evaporation – takes place only at the surface and at any temperature.
  • Boiling – takes place throughout the bulk as well as the surface; requires bubble formation whose internal pressure equals $P_{ext}$.
• Vapour-pressure–temperature curve
  • Increasing $T$ → exponential rise in $P_{vap}$.
  • Intersection with a chosen $P_{ext}$ gives the corresponding boiling point.
  • Typical data for $\mathrm{H_2O}$:
  • $P<em>{ext}=0.5\,\text{atm}\;\Rightarrow\;T</em>b\approx60^\circ\text{C}$
  • $P<em>{ext}=1\,\text{atm}\;\Rightarrow\;T</em>b=100^\circ\text{C}$
  • $P<em>{ext}=2\,\text{atm}\;\Rightarrow\;T</em>b\approx120^\circ\text{C}$.
• Concept of a pressure cooker
  • Sealed vessel raises $P_{ext}$ (typically $\sim2–3\,\text{atm}$).
  • Water then boils near $120–130^\circ\text{C}$ ⇒ food cooks faster because rate of chemical change roughly doubles for every $10^\circ\text{C}$ rise.
  • Sample calculation for food initially at $25^\circ\text{C}$:
  • Normal pot: Temperature window $25\rightarrow100^\circ\text{C}$, $\Delta T=75^\circ\text{C}$.
  • Pressure cooker (3 atm): $25\rightarrow300^\circ\text{C}$ (idealised), $\Delta T=275^\circ\text{C}$.
• Boiling at high altitudes
  • Atmospheric pressure decreases with altitude (e.g., $P_{ext}\approx0.95\,\text{atm}$ on hill-top).
  • Lower $P<em>{ext}$ → lower $T</em>b$ → food takes longer to cook.
  • Energy requirement: heat needed is proportional to $\Delta H<em>{vap}$ plus sensible heat to reach the lower $T</em>b$.
• Clausius–Clapeyron (two-point form)$\ln\left(\frac{P<em>1}{P</em>2}\right)=\frac{\Delta H<em>{vap}}{R}\left(\frac{1}{T</em>2}-\frac{1}{T_1}\right)$
  • $R=8.314\,\text{J mol}^{-1}\text{K}^{-1}$ (or $2\,\text{cal mol}^{-1}\text{K}^{-1}$ in CGS).
  • Used to predict $T<em>b$ at any $P</em>{ext}$ (assuming $\Delta H_{vap}$ ≈ constant over small $T$ range).
• Illustrative problems
  1. BP of water at 2 atm (given $\Delta H<em>{vap}=9720\,\text{cal mol}^{-1}$): $T</em>b\approx393\,\text{K}\;(120^\circ\text{C}).$
  2. External pressure that allows water to boil at 0 °C:
     Using same equation, $P_{ext}\approx0.01\,\text{atm}$ (near vacuum).

Elevation of Boiling Point (Colligative Property)

• Conceptual basis
  • Addition of a non-volatile solute lowers the vapour pressure of the solvent (Raoult’s Law).
  • To reach $P_{ext}$ again, temperature must be increased ⇒ elevation in boiling point.
• Definitions & Symbols
  • $T_b^0$ – BP of pure solvent.
  • $T_b$ – BP of solution.
  • $\Delta T<em>b=T</em>b-T_b^0$ – elevation in boiling point.
  • $K_b$ – molal elevation (ebullioscopic) constant; depends only on the solvent.
  • $m$ – molality $\left(\dfrac{\text{moles solute}}{\text{kg solvent}}\right)$.
• Fundamental relation
  $\Delta T<em>b = K</em>b m$.
• Units
  $K<em>b: \dfrac{\text{K kg}}{\text{mol}}$ Example: $K</em>b(\mathrm{H_2O}) = 0.52\,\dfrac{\text{K kg}}{\text{mol}}$.
• Derivation (outline)
  1. Start with Clausius–Clapeyron for dilute solutions.
  2. Substitute Raoult’s law ($P<em>{soln}=\chi</em>A P_A^0$).
  3. Assume $m\ll1$ → obtain linear dependence on molality.
• Key implications
  • Property is colligative – depends only on number of solute particles not their nature (non-electrolyte assumption).
  • Valuable for molar-mass determination of non-volatile solutes.

Worked & Exam-Type Problems

• Kb determination (water)
  1 mol solute in 1 kg H₂O increases BP to 100.5 °C
  $K<em>b=\Delta T</em>b/m = \dfrac{0.5}{1}=0.5\,\dfrac{\text{K kg}}{\text{mol}}$ (≈ literature 0.52).
• Ratio problems
  1 g solute in 100 g solvents A & B, $K<em>{b,A}:K</em>{b,B}=1:5$
  $\frac{\Delta T<em>{b,A}}{\Delta T</em>{b,B}}=\frac{K<em>{b,A}}{K</em>{b,B}}=\frac{1}{5}$.
• Matching Kb with BP (three solvents X, Y, Z of equal molar mass)
  Higher normal BP ⇒ stronger intermolecular forces ⇒ larger $\Delta H<em>{vap}$ ⇒ larger $K</em>b$.
  Approx. order: Tb:Z>X>Y\;\Rightarrow\;Kb:Z>X>Y.
• Molar-mass from BP elevation (JEE 2023)
  Given: 2 g solute in 20 g water, $T<em>b=373.52\,\text{K}$. $\Delta T</em>b=0.52\,\text{K};\;m=1\,\text{mol kg}^{-1}$
  $M=\dfrac{\text{mass}}{n}=\dfrac{2\,\text{g}}{1\,\text{mol}}=180\,\text{g mol}^{-1}.$
• Elevation using CCl₄ (JEE 2021)
  $\Delta T<em>b=0.60\,\text{K};\;K</em>b=5.0$
  $m=0.12\,\text{mol kg}^{-1};\;M=\frac{3.0}{0.12}=25\,\text{g mol}^{-1}.$
• Vapour-pressure depression link (IIT 2012)Given: 2 °C elevation for 2.5 g solute in 100 g H₂O ($K_b=0.76$).
  1. Find $m=\dfrac{\Delta T<em>b}{K</em>b}=\dfrac{2}{0.76}=2.63\,\text{mol kg}^{-1}$.
  2. Mole fraction $X_B=\dfrac{m}{55.5+m}\approx0.045.$
  3. $P<em>{soln}=P</em>A^0(1-X_B)=760(1-0.045)\approx724\,\text{mmHg}.$

Typical Conceptual & Numerical Take-aways

• Raising $P<em>{ext}$ (pressure cooker) → raises $T</em>b$ of solvent (water) → faster cooking.
• Lowering $P<em>{ext}$ (high altitude, vacuum distillation) → lowers $T</em>b$.
• For dilute solutions of non-volatile, non-electrolytes: $(\Delta T_b)\propto m\propto\frac{1}{M}$; hence heavier solute causes smaller elevation for same mass percentage.
• $K<em>b$ is larger for solvents with larger $\Delta H</em>{vap}$ and lower $T_b^0$ denominator in derivation.
• Colligative properties allow experimental determination of molar masses and degree of ionisation (when extended to electrolytes).

Formulas & Constants Quick Sheet

• Boiling criterion: $P<em>{vap}=P</em>{ext}$.
• Clausius–Clapeyron (two-point): $\ln\frac{P<em>1}{P</em>2}=\frac{\Delta H<em>{vap}}{R}\left(\frac{1}{T</em>2}-\frac{1}{T_1}\right)$.
• Elevation relation: $\Delta T<em>b = K</em>b m$.
• Molality: $m=\frac{n_{solute}}{\text{kg solvent}}$.
• Mole fraction (for dilute soln): $X_B\approx\frac{m}{55.5}\;(\text{for water}).$
• Unit conversions:
  $1\,\text{atm}=760\,\text{mmHg};\;R=8.314\,\text{J mol}^{-1}\text{K}^{-1}=1.987\,\text{cal mol}^{-1}\text{K}^{-1}.$