# Vapor Pressure vs Boiling Point: Several Graphs And Insights

In this article, we are going to learn what is the difference between the vapor pressure and the boiling point with detailed insights.

The vapor pressure and boiling point graph shows an exponential curve and denotes the saturation of vapor pressure too. Here is a table below differentiating the vapor pressure vs boiling point:-

## Vapor Pressure And Boiling Point Graph

The boiling point is nothing but the temperature at which the phase change occurs and the vapor pressure attains the highest value at that fixed atmospheric pressure. Hence let us plot a graph of vapor pressure v/s the temperature for a liquid boiling at a constant pressure condition.

The graph of vapor pressure v/s temperature shows the exponential curve as the number of vapors escaping from the liquid overcoming the attractive intermolecular bonds doubles at every rise in temperature of the liquid.

The point TBP denotes the boiling point of the particular liquid on the x-axis, beyond the boiling point of the liquid, the temperature of the liquid does not rise further but only the phase change from liquid to vapors takes place. The point on the y-axis Vsat represents the saturation point of the vapor pressure. As the vapors evaporated cool down and condense back into the liquid form. The vapor pressure is maintained constant after reaching the boiling point of the liquid.

## How to Calculate Boiling Point from Heat of Vaporization?

The heat of vaporization is the amount of heat energy required to transform the liquid state of matter into the gaseous state.

The boiling point of a liquid can be calculated from heat of vaporization by using the Clausius – Clapeyron equation given as $ln\left ( \frac{P_2}{P_1} \right )=\frac{-\Delta H_{vap}}{R}\left ( \frac{1}{T_2}-\frac{1}{T_1} \right )$.

## What is the boiling point of water in the pressure cooker which functions at 1.8 bar if the heat of vaporization of water is 45k J/mol?

Given: P2= 1.8 bar

The water at a normal atmospheric condition that is at 1 atm, boils at 1000C, hence

P1= 1 bar

T1 =1000C=373.2K

$\Delta H_{vap}=45k\ J/mol$

Using the Clasius – Clapeyron equation

$ln\left ( \frac{P_2}{P_1} \right )=\frac{-\Delta H_{vap}}{R}\left ( \frac{1}{T_2}-\frac{1}{T_1} \right )$

$ln\left ( \frac{1.8}{1} \right )=\frac{-45000}{8.314}\left ( \frac{1}{T_2}-\frac{1}{373.2} \right )$

$ln\left ( 1.8 \right )=-5412.56\left ( \frac{1}{T_2}-0.0027 \right )$

$0.5878=-5.412\left ( \frac{1}{T_2}-0.0027 \right )$

$-10.86 \times 10^{5}=\frac{1}{T_2}-0.0027$

$\frac{1}{T_2}=-10.86 \times 10^{5}-0.0027$

$\frac{1}{T_2}=0.00257$

$T_2=\frac{1}{0.00257}=389.1 K$

And 389.1K = 115.90C

Hence the boiling point of water inside the pressure cooker is 115.90C.

## How to find Boiling Point from Vapor Pressure?

The boiling point can be found by measuring the saturated vapor pressure developed at that temperature.

The liquid can have varied boiling points at different pressure in the system. The vapor pressure can be found using the Clausius – Clapeyron equation, also from the phase diagrams, and from the graph of vapor pressure v/s temperature too.

## What is the boiling point of methane at vapor pressure equal to 2 atm? Given the heat of vaporization of methane is 8.20k J/mol.

At normal atmospheric pressure, the boiling point of methane is -161.50C.

P1 =1atm

P2 =2atm

T1 =-161.50C =-161.5+273.2 =111.7K

$\Delta H_{vap}=8.2k\ J/mol$

Using Clausius – Clapeyron equation

$ln\left ( \frac{P_2}{P_1} \right )=\frac{-\Delta H_{vap}}{R}\left ( \frac{1}{T_2}-\frac{1}{T_1} \right )$

$ln\left ( \frac{2}{1} \right )=\frac{-8200}{8.314}\left ( \frac{1}{T_2}-\frac{1}{111.7} \right )$

$ln\left ( 2 \right )=-986.3\left ( \frac{1}{T_2}-\frac{1}{111.7} \right )$

$0.6931=-986.3\left ( \frac{1}{T_2}-0.00895 \right )$

$70.3\times 10^{-5}=-\left ( \frac{1}{T_2}-0.00895 \right )$

$70.3\times 10^{-5}=0.00895-\frac{1}{T_2}$

$\frac{1}{T_2}=0.00895-70.3\times 10^{-5}$

$\frac{1}{T_2}=0.00825$

$T_2=\frac{1}{0.00825}$

$T_2=121.2K$

This is equal to -1520 C.

Hence the boiling point of the methane at the vapor pressure of 2 atm increases to -1520 C.

## What are the factors affecting the vapor pressure of the liquid?

The vapor pressure is due to the pressure felt on the area by the vapors evaporated from the system into the surrounding.

The most vital factor on which the vapor pressure depends is the temperature and the heat energy supplied to the liquid. Also, the chemical composition and the impurities added will vary the vapor pressure.

## How vapor pressure depends upon the intermolecular bonding between the atoms?

On supplying heat energy to the liquid, the intermolecular bonding between the atoms breaks, and particles move in random motion.

If this intermolecular bonding between the atom in case of a certain liquid is low, which means there is a weak force of attraction between the atoms then these bonds will easily break down will even a small amount of energy supplied to the liquid and thus vapor pressure will be high at small temperature.

## How are the boiling point and the vapor pressure related to each other?

The vapors are the result of the rising temperature of the liquid supplying heat.

At a boiling point, the liquid phase is converted to the gaseous phase and at this temperature, the vapor pressure formed becomes equal to the atmospheric pressure.

AKSHITA MAPARI

Hi, I’m Akshita Mapari. I have done M.Sc. in Physics. I have worked on projects like Numerical modeling of winds and waves during cyclone, Physics of toys and mechanized thrill machines in amusement park based on Classical Mechanics. I have pursued a course on Arduino and have accomplished some mini projects on Arduino UNO. I always like to explore new zones in the field of science. I personally believe that learning is more enthusiastic when learnt with creativity. Apart from this, I like to read, travel, strumming on guitar, identifying rocks and strata, photography and playing chess. Connect me on LinkedIn - linkedin.com/in/akshita-mapari-b38a68122