How to Calculate Vapor Pressure

Have you ever left a water bottle out in the sun for a few hours, then opened the cap and heard a small “puff” sound? This sound is caused by the vapor pressure in the bottle. In chemistry, vapor pressure is the pressure exerted on the walls of a sealed vessel when the liquid in the vessel evaporates (turns into a gas). ^{[1] X Research Source} To find the vapor pressure at a known temperature, use the Clausius-Clapeyron equation: ln(P1/P2) = (ΔH _vap /R)((1/T2) – ( 1/T1)) .

Table of Contents

Using the Clausius-Clapeyron equation

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Write the Clausius-Clapeyron equation. When considering the change in vapor pressure over time, the formula for calculating vapor pressure is the Clausius-Clapeyron equation (named after physicists Rudpf Clausius and Benoît Paul Émile Clapeyron). ^{[2] X Research Source} This is a common formula used to solve common problems of vapor pressure in physics and chemistry. The formula is written as follows: ln(P1/P2) = (ΔH _vap /R)((1/T2) – (1/T1)) . In this formula, the variables represent:

ΔH _vap : Evaporation enthalpy of the liquid. This value can be found in the table at the end of the chemistry textbook.
R: Ideal gas constant and equals 8.314 J/(K × Mp).
T1: The temperature at which the vapor pressure is known (initial temperature).
T2: The temperature at which the vapor pressure is to be calculated (final temperature).
P1 and P2: The vapor pressure at temperature T1 and T2 respectively.

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Substitute known values into variables. The Clausius-Clapeyron equation looks quite complicated because it has so many variables, but it’s actually not too difficult if the problem provides enough information. The most basic vapor pressure problems will give you two values of temperature and one value of pressure, or two values of pressure and one value of temperature — once you have these data, solving is easy.

For example, suppose the problem is for a vessel that holds a liquid at 295 K and has a vapor pressure of 1 atmosphere (atm). The question is: What is the vapor pressure at 393 K? We have two values of temperature and one value of pressure, so we can solve for the remaining pressure using the Clausius-Clapeyron equation. Substituting values into the variables, we have ln(1/P2) = (ΔH _vap /R)((1/393) – (1/295)) .
For the Clausius-Clapeyron equation, we always have to use the Kelvin temperature value. You can use any pressure value, as long as the units are the same for both P1 and P2.

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Substitute the constants. The Clausius-Clapeyron equation has two constants: R and ΔH _vap . R is always equal to 8,314 J/(K × Mp). However, ΔH _vap (evaporation enthalpy) depends on the type of vaporizing liquid for which the problem is given. With that said, you can look up ΔH _vap values for a variety of substances at the end of chemistry or physics textbooks, or look them up online (here, for example.) ^{[3] X Research Sources}

In the above example, assume the liquid is pure water . If we look in the table for the value of ΔH _vap , we have that the ΔH _vap of pure water is approximately 40.65 kJ/mp. Since the value of H uses the unit of joul, we have to change it to 40,650 J/mp.
Substituting the constants into the equation, we get ln(1/P2) = (40.650/8,314)((1/393) – (1/295)) .

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Solve the equation. After you have substituted all the values for the variables of the equation, except for the variable we are calculating, continue to solve the equation according to the principles of ordinary algebra.

The hardest point when solving the equation ( ln(1/P2) = (40.650/8,314)((1/393) – (1/295)) ) is dealing with the natural logarithmic function (ln). To suppress the natural log function, use both sides of the equation as exponents for the mathematical constant e . In other words, ln(x) = 2 → e ^ln(x) = e ² → x = e ² .
Now we solve the example equation:
ln(1/P2) = (40,650/8,314)((1/393) – (1/295))
ln(1/P2) = (4,889,34)(-0.00084)
(1/P2) = e ^(-4,107)
1/P2 = 0.0165
P2 = 0.0165 ^-1 = 60.76 atm . This value is reasonable — in a closed vessel, when the temperature is increased by almost 100 degrees (to a temperature approximately 20 degrees above the boiling point of water) a lot of steam is produced, so the pressure will increase much.

Find the vapor pressure of the dissolved solution

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Write Raoult’s Law. In practice, we rarely work with pure liquids — often we have to work with mixtures of many different substances. Some common mixtures are made by dissolving a small amount of a chemical called a solute in a large amount of another chemical called a solvent to form a solution . In this case, we need to know the equation of Raoult’s Law (named after physicist François-Marie Raoult), ^{[4] X The research source} has the following form: P _solution = P _solvent X _solvent . In this formula, the variables represent:

_Solution P : Vapor pressure of the entire solution (all components constituting the solution)
_Solvent P : The vapor pressure of the solvent
X _solvent : The mp part of the solvent.
Don’t worry if you don’t know the term “mp part” — we’ll explain it in the next steps.

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Distinguish between solvents and solutes in solution. Before calculating the vapor pressure of a solution, you need to know the substances for which the problem is given. Note that a solution is formed when a solute dissolves in a solvent — the chemical being dissolved is always the solute, and the chemical doing the dissolution is the solvent.

In this section we will take a simple example to illustrate the above concepts. Suppose we want to find the vapor pressure of a syrup solution. Usually syrup is made from one part sugar dissolved in one part water, hence we say sugar is the solute and water is the solvent . ^{[5] X Research Sources}
Note: the chemical formula of sucrose (granulated sugar) is C ₁₂ H ₂₂ O ₁₁ . You will find this information very important.

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Find the temperature of the solution. As we saw in the above Clausius Clapeyron, the temperature of the liquid will affect its vapor pressure. In general, the higher the temperature, the higher the vapor pressure — as the temperature increases, more liquid evaporates and increases the pressure in the vessel.

In this example, assume the current temperature of the syrup is 298 K (about 25 C).

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Find the vapor pressure of the solvent. The chemical reference literature usually gives vapor pressure values for many common substances and mixtures, but usually only gives pressures at 25 C/298 K or at boiling point. If your solution is at this temperature you can use the lookup value from the literature, otherwise you need to find the vapor pressure at the solution’s initial temperature.

The Clausius-Clapeyron equation can help here, let’s use the traceable pressure and the temperature of 298 K (25 C) for P1 and T1.
In this example, the mixture has a temperature of 25 C so we can use a lookup table. We see that water at 25 C has a vapor pressure of 23.8 mmHg ^{[6] X Research Source}

Find the vapor pressure in special cases

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Recognize standard Temperature and Pressure conditions. Scientists often use a pair of temperature and pressure values as the “default” condition. These values are called Standard Pressure and Temperature (collectively, standard conditions or GCC). Vapor pressure problems often refer to the GCC, so you should memorize these values for your convenience. GSO is defined as: ^{[8] X Research Source}

Temperature: 273.15 K / 0 C / 32 F
Pressure: 760 mmHg / 1 atm / 101.325 kilopascal

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Transform the Clausius-Clapeyron equation to find other variables. In the example in Part 1, we see that the Clausius-Clapeyron equation is very effective when used to calculate the vapor pressures of pure substances. However, not all problems require finding P1 or P2, but many times they also require finding the temperature or even the value of ΔH _vap . In this case, to find the answer you just need to switch sides of the equation so that the variable you are looking for is on one side of the equation, and all the other variables are on the other side.

For example, suppose there is an unknown liquid that has a vapor pressure of 25 torr at 273 K and 150 torr at 325 K, and we want to find the evaporation enthalpy of this liquid (ΔH _vap ). We can solve it like this:
ln(P1/P2) = (ΔH _vap /R)((1/T2) – (1/T1))
(ln(P1/P2))/((1/T2) – (1/T1)) = (ΔH _vap /R)
R × (ln(P1/P2))/((1/T2) – (1/T1)) = ΔH _vap . Now we substitute the values in:
8.314 J/(K × Mp) × (-1.79)/(-0.00059) = ΔH _vap
8,314 J/(K × Mp) × 3,033.90 = ΔH _vap = 25,223.83 J/mp

Advice

To use the above Clausius Clapeyron equation, you must convert the temperature to Kevin (symbol K). If you have the temperature in Celsius, then convert it using the following formula: T _k = 273 + T _c
You can apply the above methods because the energy corresponds to the amount of heat supplied. The temperature of the liquid is the only environmental factor that affects vapor pressure.

Steps

Using the Clausius-Clapeyron equation

Find the vapor pressure of the dissolved solution

Find the vapor pressure in special cases

Advice

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