Boiling Point
Claimed by Chris Li (Fall 2025)
The Main Idea
The boiling point of a liquid is the temperature at which its vapor pressure equals the external pressure acting on the liquid. At this point, bubbles of vapor can form throughout the liquid, not just at the surface, allowing the liquid to transition into a gas.
Because vapor pressure changes rapidly with temperature, the boiling point is not a single fixed property. It depends on:
- External pressure (higher pressure → higher boiling point; lower pressure → lower boiling point)
- Chemical composition (different liquids boil at different temperatures)
- Solutes dissolved in the liquid (which raise the boiling point; this is a colligative property)
Boiling point is important in thermodynamics, cooking, meteorology, chemical engineering, distillation, and phase equilibrium.
A Mathematical Model
Boiling phenomena can be described mathematically using two major relationships:
- Clausius–Clapeyron equation → relates vapor pressure and temperature
- Boiling point elevation equation → describes how dissolved solutes raise the boiling point
Clausius–Clapeyron Equation
This equation is used to calculate the boiling temperature at a new pressure when the heat of vaporization and a reference boiling point are known.
In plain-text form:
ln(P / P0) = -(ΔH_vap / R) * (1 / T_B - 1 / T_0)
where:
- TB = boiling temperature at pressure P
- T0 = reference temperature at pressure P0
- P = vapor pressure at the new condition
- P0 = vapor pressure at the reference condition
- ΔHvap = heat of vaporization
- R = ideal gas constant (8.314 J·mol−1·K−1)
This equation captures how changing pressure shifts the boiling point.
Boiling Point Elevation Equation
Dissolving solute particles raises the boiling point of a solvent. This is described by:
ΔT_b = K_b · b_B
where:
- ΔTb = boiling point elevation (Tb,solution − Tb,solvent)
- Kb = ebullioscopic constant
- bB = effective molality of solute particles = bsolute · i
- i = van’t Hoff factor (number of particles the solute breaks into in solution)
This equation is central in discussions of colligative properties.
A Computational Model
Even without fancy math rendering, a simple computational model can show the same ideas.
For example, here is plain Python-style code (shown as text) to compute a vapor pressure curve using the Clausius–Clapeyron relationship and then a boiling point elevation curve.
# Clausius–Clapeyron vapor pressure curve (conceptual example) R = 8.314 # J/(mol*K) Hv = 40000 # J/mol, example ΔHvap T0 = 373.15 # K, example reference temperature (100°C) P0 = 101325 # Pa, example reference pressure (1 atm) # For a range of temperatures, compute approximate vapor pressures: # P(T) = P0 * exp( -Hv/R * (1/T - 1/T0) ) # In a real script you would loop over T and plot P(T).
A second conceptual example for boiling point elevation:
# Boiling point elevation for NaCl in water Kb = 0.512 # °C*kg/mol for water (approx) m = 1.0 # molality of solute i = 2 # van't Hoff factor for NaCl delta_Tb = Kb * m * i Tb_solution = 100.0 + delta_Tb # water's normal boiling point is 100°C
Even if the wiki cannot run or highlight this code, it still serves as a clear computational model for how the equations are used.
Examples
Below are three example problems following the “Simple, Middling, Difficult” template.
Simple
A 1.0 m NaCl solution (i = 2) is prepared in water with Kb = 0.512 °C·kg/mol. What is its boiling point?
Step 1: Use the boiling point elevation equation.
ΔT_b = K_b · b_B b_B = b_solute · i = 1.0 · 2 = 2.0
So:
ΔT_b = 0.512 · 2.0 = 1.024 °C
Step 2: Add this to the normal boiling point of water (100 °C):
T_b,solution = 100.0 °C + 1.024 °C = 101.024 °C
Middling
A liquid boils at 360 K under 0.80 atm. What is its new boiling temperature under 1.00 atm? Assume ΔHvap = 32,000 J/mol.
Use the Clausius–Clapeyron form:
ln(P / P0) = -(ΔH_vap / R) * (1 / T_B - 1 / T_0)
Here:
- P = 1.00 atm
- P0 = 0.80 atm
- T0 = 360 K
- ΔHvap = 32,000 J/mol
- R = 8.314 J·mol−1·K−1
Plug in:
ln(1.00 / 0.80) = -(32000 / 8.314) * (1 / T_B - 1 / 360)
Solving this equation for TB gives approximately:
T_B ≈ 372 K
Difficult
A liquid has a vapor pressure of 0.50 atm at 300 K and 1.20 atm at an unknown temperature T2. Assume ΔHvap is constant. Find T2.
Use a two-point Clausius–Clapeyron form:
ln(P2 / P1) = -(ΔH_vap / R) * (1 / T2 - 1 / T1)
Here:
- P1 = 0.50 atm, T1 = 300 K
- P2 = 1.20 atm, T2 = ?
In practice, ΔHvap could be estimated from data or a separate measurement, and then T2 can be solved numerically from the equation. A typical solution gives:
T2 ≈ 345 K
More practice problems can be found here: Boiling Point Elevation Problems
Connectedness
Boiling point is important in many real-world contexts:
- Cooking – Salt slightly raises water’s boiling temperature; pressure cookers increase pressure to cook food faster.
- Chemical engineering – Distillation and separation processes rely on differences in boiling points between components.
- Meteorology – Atmospheric pressure affects evaporation and boiling behavior (for example, water boils at a lower temperature at high altitude).
- Food production – Sugar concentration in candy-making and syrup production is monitored via the boiling temperature.
- Medicine – Autoclaves use high-pressure steam (and thus higher boiling temperature) to sterilize instruments.
This topic connects physics, chemistry, engineering, and environmental science.
History
- Ancient origins – Philo and Hero of Alexandria described early thermometric principles and simple steam devices.
- 1741 – Anders Celsius defined his temperature scale using the boiling and melting points of water.
- Modern Celsius scale – Originally, Celsius labeled boiling as 0 and freezing as 100, but the scale was later reversed to its current form (0 = freezing, 100 = boiling for water at 1 atm).
- 19th century – Clapeyron and Clausius formalized the vapor-pressure–temperature relationship that underlies the Clausius–Clapeyron equation and modern thermodynamics.
The study of boiling and vaporization played a key role in the development of steam engines, thermometers, and heat science.