Phase Equilibria
Transitions between solid, liquid, and gaseous phases; one-component phase diagrams; Raoult's law and deviations; colligative properties of solutions.
Complete Theory
4A phase transition occurs when a substance changes between solid, liquid, and gas. At the transition, two phases coexist at equilibrium.
The Clapeyron equation describes the pressure-temperature dependence of the phase boundary: For liquid-vapor transitions (assuming ideal gas and ), the Clausius-Clapeyron equation applies:
The Clapeyron equation describes the pressure-temperature dependence of the phase boundary: For liquid-vapor transitions (assuming ideal gas and ), the Clausius-Clapeyron equation applies:
A phase diagram plots vs. and shows the regions of stability for each phase. Key points:
- Triple point: three phases coexist
- Critical point: above and the liquid and gas phases become indistinguishable (supercritical fluid)
Raoult's law states that the partial vapor pressure of a component above an ideal solution is proportional to its mole fraction:
where is the vapor pressure of the pure component.
Positive deviation (): weaker A-B interactions than A-A/B-B (e.g. ethanol + water → azeotrope). Negative deviation (): stronger A-B interactions (e.g. chloroform + acetone).
Positive deviation (): weaker A-B interactions than A-A/B-B (e.g. ethanol + water → azeotrope). Negative deviation (): stronger A-B interactions (e.g. chloroform + acetone).
Properties that depend only on the number of solute particles (not their identity):
- Ebullioscopy (boiling point elevation):
- Cryoscopy (freezing point depression):
- Osmotic pressure: (where is the van't Hoff factor)
Worked Examples
2Example 1Boiling point elevation
Given
50.0 g glucose ( g/mol) in 500 g water
Find
of the solution
Step-by-step solution
1Calculate the molality: .
2Apply the boiling point elevation formula: .
✓ Final result: °C (boiling point = 100.285°C)
Example 2Osmotic pressure
Given
sucrose solution
Find
Osmotic pressure
Step-by-step solution
1Since sucrose is a non-electrolyte, the Van't Hoff factor . Use the osmotic pressure equation: .
2Substitute the values: .
3Calculate: .
✓ Final result: atm
Exercises with Solutions
3Exercise 1Freezing point depressionMedium
Problem to solve
Calculate the freezing point of a solution containing 10.0 g of NaCl ( g/mol) in 200 g of water. °C·kg/mol.
Given data
10.0 g NaCl in 200 g water °C·kg/mol
Step-by-step solution
1 mol/kg
2NaCl → Na⁺ + Cl⁻ (),
3°C → freezing point = °C
✓ Final answer: −3.18°C
Exercise 2Molar mass from cryoscopyMedium
Problem to solve
A 2.00 g sample of a non-electrolyte dissolved in 100 g water lowers the freezing point by 0.372°C. Find the molar mass ().
Given data
°Cm_solute = 2.00 gm_solvent = 0.100 kg °C·kg/mol
Step-by-step solution
1 mol/kg
2 mol
3
✓ Final answer: g/mol
Exercise 3Osmotic pressureHard
Problem to solve
A 0.50 L solution contains 9.0 g of an unknown protein. The osmotic pressure is 0.024 atm at 298 K. Calculate the molar mass.
Given data
V = 0.50 Lm = 9.0 g atmT = 298 KR = 0.0821 L·atm·mol⁻¹·K⁻¹
Step-by-step solution
1 M
2 mol
3
✓ Final answer: g/mol
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