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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.
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 fixed property—rather, it depends on:
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)
* External pressure (higher pressure → higher boiling point; lower pressure → lower boiling point)
* **Chemical composition** (different liquids boil at different temperatures)
* Chemical composition (different liquids boil at different temperatures)
* **Solutes dissolved in the liquid**, which raise the boiling point (a colligative property)
* 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.
Boiling point is important in thermodynamics, cooking, meteorology, chemical engineering, distillation, and phase equilibrium.
[[File:Boiling-water.jpg|center|350px|thumb|Boiling occurs when vapor pressure equals external pressure.]]
---


===A Mathematical Model===
===A Mathematical Model===
Line 21: Line 17:
Boiling phenomena can be described mathematically using two major relationships:
Boiling phenomena can be described mathematically using two major relationships:


* **Clausius–Clapeyron Equation** → relates vapor pressure and temperature   
* Clausius–Clapeyron equation → relates vapor pressure and temperature   
* **Boiling Point Elevation Equation** → describes how dissolved solutes raise the boiling point   
* Boiling point elevation equation → describes how dissolved solutes raise the boiling point   
 
----


'''Clausius–Clapeyron Equation'''
'''Clausius–Clapeyron Equation'''


Used to calculate the boiling temperature at a new pressure when ΔH<sub>vap</sub> and a reference boiling point are known.
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.


\[
In plain-text form:
\ln\left(\frac{P}{P_0}\right)
= -\frac{\Delta H_{vap}}{R} \left( \frac{1}{T_B} - \frac{1}{T_0} \right)
\]


[[File:ClausiusClapeyron.png|center|400px|thumb|Graph of vapor pressure vs temperature illustrating boiling point.]]
ln(P / P0) = -(ΔH_vap / R) * (1 / T_B - 1 / T_0)


''Variable definitions:''
where:


* '''T<sub>B</sub>''' – boiling temperature at pressure P   
* T<sub>B</sub> = boiling temperature at pressure P   
* '''T<sub>0</sub>''' – known temperature at pressure P<sub>0</sub>   
* T<sub>0</sub> = reference temperature at pressure P<sub>0</sub>   
* '''P''' – new vapor pressure   
* P = vapor pressure at the new condition  
* '''P<sub>0</sub>''' – reference vapor pressure   
* P<sub>0</sub> = vapor pressure at the reference condition  
* '''ΔH<sub>vap</sub>''' – heat of vaporization   
* ΔH<sub>vap</sub> = heat of vaporization   
* '''R''' – ideal gas constant (8.314 J·mol<sup>−1</sup>·K<sup>−1</sup>)
* R = ideal gas constant (8.314 J·mol<sup>−1</sup>·K<sup>−1</sup>)


----
This equation captures how changing pressure shifts the boiling point.


'''Boiling Point Elevation Equation'''
'''Boiling Point Elevation Equation'''


Dissolving solute particles raises the boiling point of a solvent:
Dissolving solute particles raises the boiling point of a solvent. This is described by:
 
\[
\Delta T_b = K_b \, b_B
\]
 
Where:


* '''ΔT<sub>b</sub>''' = T<sub>b,solution</sub> − T<sub>b,solvent</sub> 
  ΔT_b = K_b · b_B
* '''K<sub>b</sub>''' = ebullioscopic constant  
* '''b<sub>B</sub>''' = effective molality: b<sub>solute</sub> ·
* '''i''' = van’t Hoff factor 


[[File:BoilingPointElevation.png|center|350px|thumb|Boiling point elevation vs solute concentration.]]
where:


This equation is central in colligative property analysis.
* ΔT<sub>b</sub> = boiling point elevation (T<sub>b,solution</sub> − T<sub>b,solvent</sub>) 
* K<sub>b</sub> = ebullioscopic constant 
* b<sub>B</sub> = effective molality of solute particles = b<sub>solute</sub> · 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===
===A Computational Model===


We can simulate boiling point elevation or vapor-pressure curves numerically. The following VPython/GlowScript code models the Clausius–Clapeyron equation to generate a vapor pressure vs temperature plot:
Even without fancy math rendering, a simple computational model can show the same ideas.


<syntaxhighlight lang="python">
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.
import numpy as np
import matplotlib.pyplot as plt


R = 8.314
<pre>
Hv = 40000    # J/mol
# Clausius–Clapeyron vapor pressure curve (conceptual example)
T0 = 373.15  # K
P0 = 101325  # Pa


T = np.linspace(320, 400, 200)
R = 8.314          # J/(mol*K)
P = P0 * np.exp(-Hv/R * (1/T - 1/T0))
Hv = 40000        # J/mol, example ΔHvap
T0 = 373.15        # K, example reference temperature (100°C)
P0 = 101325        # Pa, example reference pressure (1 atm)


plt.plot(T, P)
# For a range of temperatures, compute approximate vapor pressures:
plt.xlabel("Temperature (K)")
# P(T) = P0 * exp( -Hv/R * (1/T - 1/T0) )
plt.ylabel("Vapor Pressure (Pa)")
plt.title("Vapor Pressure Curve via Clausius-Clapeyron")
plt.show()
</syntaxhighlight>


This simulation demonstrates how small temperature changes dramatically affect vapor pressure, explaining why boiling point shifts with altitude and pressure.
# In a real script you would loop over T and plot P(T).
</pre>


A second computational model simulates boiling point elevation:
A second conceptual example for boiling point elevation:


<syntaxhighlight lang="python">
<pre>
Kb = 0.512          # water
# Boiling point elevation for NaCl in water
m = np.linspace(0, 5, 200)  # molality range
i = 2              # NaCl (approx)
Tb = 100 + Kb * m * i


plt.plot(m, Tb)
Kb = 0.512      # °C*kg/mol for water (approx)
plt.xlabel("Molality (m)")
m = 1.0        # molality of solute
plt.ylabel("Boiling Point (°C)")
i  = 2          # van't Hoff factor for NaCl
plt.title("Boiling Point Elevation for NaCl in Water")
plt.show()
</syntaxhighlight>


These visual models help make the equations intuitive.
delta_Tb = Kb * m * i
Tb_solution = 100.0 + delta_Tb  # water's normal boiling point is 100°C
</pre>


---
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==
==Examples==


Below are expanded, step-by-step examples.
Below are three example problems following the “Simple, Middling, Difficult” template.


===Simple===
===Simple===


A 1.0 m NaCl solution (i = 2) is prepared in water with K<sub>b</sub> = 0.512 °C·kg/mol.
A 1.0 m NaCl solution (i = 2) is prepared in water with K<sub>b</sub> = 0.512 °C·kg/mol. What is its boiling point?
 
Step 1: Use the boiling point elevation equation.


\[
ΔT_b = K_b · b_B
\Delta T_b = (0.512)(1.0)(2) = 1.024^\circ C
b_B = b_solute · i = 1.0 · 2 = 2.0
\]


\[
So:
T_b = 100^\circ C + 1.024^\circ C = 101.024^\circ C
 
\]
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===
===Middling===


A liquid boils at 360 K under 0.80 atm. What is its boiling temperature under 1.00 atm?
A liquid boils at 360 K under 0.80 atm. What is its new boiling temperature under 1.00 atm? Assume ΔH<sub>vap</sub> = 32,000 J/mol.
Given: ΔH<sub>vap</sub> = 32 kJ/mol.


\[
Use the Clausius–Clapeyron form:
\ln\left(\frac{1.00}{0.80}\right)
= -\frac{32000}{8.314}\left(\frac{1}{T_B} - \frac{1}{360}\right)
\]


Solving gives:
ln(P / P0) = -(ΔH_vap / R) * (1 / T_B - 1 / T_0)
\[
 
T_B \approx 372\ \text{K}
Here:
\]
 
* P = 1.00 atm 
* P<sub>0</sub> = 0.80 atm 
* T<sub>0</sub> = 360 K 
* ΔH<sub>vap</sub> = 32,000 J/mol 
* R = 8.314 J·mol<sup>−1</sup>·K<sup>−1</sup> 
 
Plug in:
 
ln(1.00 / 0.80) = -(32000 / 8.314) * (1 / T_B - 1 / 360)
 
Solving this equation for T<sub>B</sub> gives approximately:
 
T_B 372 K


===Difficult===
===Difficult===


A liquid has vapor pressure 0.50 atm at 300 K and 1.20 atm at an unknown temperature T<sub>2</sub>.
A liquid has a vapor pressure of 0.50 atm at 300 K and 1.20 atm at an unknown temperature T<sub>2</sub>. Assume ΔH<sub>vap</sub> is constant. Find T<sub>2</sub>.
Use the two-point Clausius–Clapeyron form:


\[
Use a two-point Clausius–Clapeyron form:
\ln\left(\frac{1.20}{0.50}\right)
= -\frac{\Delta H_{vap}}{R}\left(\frac{1}{T_2} - \frac{1}{300}\right)
\]


Solving:
ln(P2 / P1) = -(ΔH_vap / R) * (1 / T2 - 1 / T1)


\[
Here:
T_2 \approx 345\ \text{K}
\]


More problems:  
* P<sub>1</sub> = 0.50 atm, T<sub>1</sub> = 300 K  
[http://www.chemteam.info/Solutions/BP-elevation-probs1-to-10.html Boiling Point Elevation]
* P<sub>2</sub> = 1.20 atm, T<sub>2</sub> = ? 


---
In practice, ΔH<sub>vap</sub> could be estimated from data or a separate measurement, and then T<sub>2</sub> can be solved numerically from the equation. A typical solution gives:
 
T2 ≈ 345 K
 
More practice problems can be found here: 
[http://www.chemteam.info/Solutions/BP-elevation-probs1-to-10.html Boiling Point Elevation Problems]


==Connectedness==
==Connectedness==


Boiling point matters in many real-world contexts:
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 relies entirely on different boiling points. 
* **Meteorology:** Atmospheric pressure affects evaporation and cloud formation. 
* **Food production:** Sugar concentration is monitored via boiling temperature in candy-making. 
* **Medicine:** Autoclaves use high-pressure steam to sterilize tools. 


Boiling point connects physics, chemistry, engineering, and environmental science.
* 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==
==History==


* **Ancient origins:** Philo and Hero of Alexandria described early thermometric principles and steam devices.   
* 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 (later reversed to the modern form).   
* 1741 Anders Celsius defined his temperature scale using the boiling and melting points of water
* **19th century:** Clapeyron and Clausius formalized the vapor‐pressure–temperature relationship, laying the foundation for phase diagrams and thermodynamics
* 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 was central to the development of thermometers, steam engines, and modern heat science.


---
The study of boiling and vaporization played a key role in the development of steam engines, thermometers, and heat science.


==See also==
==See also==
Line 193: Line 179:
* [[Melting Point]]   
* [[Melting Point]]   
* [[Vapor Pressure]]   
* [[Vapor Pressure]]   
* [[Clausius–Clapeyron Equation]] 
* [[Phase Diagram]]   
* [[Phase Diagram]]   
* [[Colligative Properties]]   
* [[Colligative Properties]]   
Line 199: Line 184:
===Further reading===
===Further reading===


* [https://www.chem.purdue.edu/gchelp/liquids/boil.html Purdue – Boiling]   
* [https://www.chem.purdue.edu/gchelp/liquids/boil.html Boiling – Purdue Chemistry]   
* [http://www.britannica.com/science/boiling-point Britannica – Boiling Point]   
* [http://www.britannica.com/science/boiling-point Boiling Point – Britannica]   


===External links===
===External links===
Line 210: Line 195:
* [http://www.ehow.com/info_8344665_uses-boiling-point-elevation.html Uses of Boiling Point Elevation]   
* [http://www.ehow.com/info_8344665_uses-boiling-point-elevation.html Uses of Boiling Point Elevation]   
* [http://www.chemteam.info/Solutions/BP-elevation.html Boiling Point Elevation]   
* [http://www.chemteam.info/Solutions/BP-elevation.html Boiling Point Elevation]   
* [https://www.chem.tamu.edu/class/majors/tutorialnotefiles/intext.htm Chemistry Basics]   
* [https://www.chem.tamu.edu/class/majors/tutorialnotefiles/intext.htm Chemistry Basics – TAMU]   
* [http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch14/melting.php Melting / Freezing / Boiling]   
* [http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch14/melting.php Melting, Freezing, Boiling – Purdue]   
* [http://didyouknow.org/celsius/ Boiling Point of Water]   
* [http://didyouknow.org/celsius/ Boiling Point of Water]   


[[Category:Properties of Matter]]
[[Category:Properties of Matter]]

Latest revision as of 22:08, 1 December 2025

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.

See also

Further reading

External links

References