3 Ways to Calculate Vapor Pressure

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3 Ways to Calculate Vapor Pressure
3 Ways to Calculate Vapor Pressure

Have you ever left a bottle of water in the blazing sun for a few hours, only to hear a slight “whistle” when you open it again? This phenomenon is caused by a principle called vapor pressure. In chemistry, vapor pressure is the pressure exerted on the walls of a closed container when the substance contained therein evaporates to a gas. To find the vapor pressure at a given temperature, use the Clausius-Clapeyron equation: ln(P1/P2) = (ΔHvap/R)((1/T2) - (1/T1)).


Method 1 of 3: Using the Clausius-Clapeyron Equation

Calculate Vapor Pressure Step 1

Step 1. Write the Clausius-Clapeyron equation

The formula used in calculating the vapor pressure, given a certain change in the existing pressure, is called the Clausius-Clapeyron equation (named after physicists Rudolf Clausius and Benoît Paul Émile Clapeyron). This is often the formula needed to discover the most common problems related to vapor pressure found in physics and chemistry textbooks. It is written as follows: ln(P1/P2) = (ΔHvap/R)((1/T2) - (1/T1)). In this formula, variables refer to the following variables:

  • ΔHvap:

    enthalpy of liquid vaporization. This value can usually be found in a table on the back cover of chemistry books.

  • A:

    the actual gaseous content, or 8.314 J / (K × mol).

  • T1:

    the temperature at which the vapor pressure is known (or the initial temperature).

  • T2:

    the temperature at which the vapor pressure is to be found (or the final temperature).

  • P1 / P2:

    steam pressures at temperatures T1 and T2, respectively.

Calculate Vapor Pressure Step 2

Step 2. Enter the known variables

The Clausius-Clapeyron equation looks challenging given the multitude of different variables, but it really isn't difficult when the right information is available. The most basic vapor pressure problems will give two values ​​relative to temperature and one relative to pressure, or two relative to pressure and one relative to temperature - once they are present, solving the problem will be easy.

  • For example, let's say that we have in front of us a container filled with liquid at a temperature of 295 K, whose vapor pressure is equal to 1 atm. The question is: What is the vapor pressure at a temperature of 393 K? We have two values ​​for temperature and one for pressure, so we can solve the problem with the Clausius-Clapeyron equation. Inserting the variables, we will have: ln(1/P2) = (ΔHvap/R)((1/393) - (1/295))
  • Note that in the Clausius-Clapeyron equations it is necessary to enter temperature values ​​in degrees Kelvin. You can use any pressure values ​​as long as they are in identical units in P1 and P2.
Calculate Vapor Pressure Step 3

Step 3. Enter the constants

The Clausius-Clapeyron equation contains two constants: R and ΔHvap. R is always equal to 8.314 J / (K × mol). The value of ΔHvap (enthalpy of vaporization), however, depends on the substance whose vapor pressure is being examined. As noted earlier, you can find values ​​for ΔHvap on various substances on the back cover of chemistry or physics books, or online (such as here).

  • In our example, let's say our liquid consists of pure liquid water. If we look in a table of ΔH valuesvap, we will find that the ΔHvap will be approximately equal to 40, 65 KJ / mol. Since our value for H uses joules, we can convert the found number to 40,650 J/mol.
  • Inserting the constants into our equation, we will have: ln(1/P2) = (40,650/8, 314) ((1/393) - (1/295)).
Calculate Vapor Pressure Step 4

Step 4. Solve the equation

Once you have all the variables entered into the equation except the one that must be discovered, proceed by solving it according to the rules of common algebra.

  • The only difficult part of the equation - ln(1/P2) = (40,650/8, 314) ((1/393) - (1/295)) - is dealing with the natural logarithm (ln). To cancel it, simply use both sides of the equation as the exponent for the mathematical constant e. In other words: ln(x) = 2 → andln(x) = and2 → x = and2.
  • Now, let's solve the equation:

    • ln(1/P2) = (40,650/8, 314)((1/393) - (1/295))
    • ln(1/P2) = (4,889, 34)(-0, 00084)
    • (1/P2) = and(-4, 107)
    • 1/P2 = 0.0165
    • P2 = 0.0165-1 = 60, 76 atm. This makes sense - in a closed container, increasing the present temperature by almost 100 degrees (up to almost 20 degrees above the boiling point of water) will create a huge amount of steam, considerably increasing the internal pressure.

Method 2 of 3: Finding Vapor Pressure with Dissolved Solutions

Calculate Vapor Pressure Step 5

Step 1. Write Raoult's Law

In real life, it's rare to work with a single pure liquid - we usually deal with liquids made up of mixtures of different substances. Some of the most common of these are created by dissolving a small amount of a certain chemical called a solute into large amounts of a chemical called a solvent, thus creating a solution. In such cases, it is useful to know an equation called Raoult's Law (after physicist François-Marie Raoult), which looks like the following: FORsolution =Psolvent × Xsolvent. In this formula, the variables refer to:

  • FORsolution:

    the vapor pressure of the entire solution (all component parts combined).

  • FORsolvent:

    the vapor pressure of the solvent.

  • Xsolvent:

    the molar fraction of the solvent.

  • Don't worry if you don't know terms like “mole fraction” - they will be explained in the next steps.
Calculate Vapor Pressure Step 6

Step 2. Identify the solvent and solute in the solution

Before calculating the vapor pressure of a mixed liquid, you need to identify the substances you are working with. It is important to remember that a solution is formed when a solute is dissolved in a solvent - the dissolved chemical is always the solute, and the chemical that dissolves is always the solvent.

  • We will work through a simple example to illustrate the concepts to be discussed. For example, let's say that we are aiming to find the vapor pressure of a common syrup. Traditionally, this substance consists of a part of sugar dissolved in a part of water, so that sugar is the solute and water the solvent.
  • Note that the chemical formula for sucrose (common sugar) is C12H22O11. It will be important soon.
Calculate Vapor Pressure Step 7

Step 3. Find out the temperature of the solution

As seen in the Clausius-Clapeyron section above, the temperature of a liquid will affect its vapor pressure. Generally speaking, the higher the temperature, the greater the vapor pressure - as the temperature increases, more of the liquid will evaporate, forming vapor and increasing the internal pressure of the vessel.

  • In our example, let's say the current temperature of the common syrup is equal to 298K (approximately 25°C).
Calculate Vapor Pressure Step 8

Step 4. Find out the solvent vapor pressure

Chemical Reference Materials generally have vapor pressure values ​​for a number of common compounds and substances, but they will generally be presented at a temperature of 25 °C (298 K) or their boiling point. If the solution is at one of these temperatures, you can use the reference value. If not, you will need to find out the vapor pressure at your current temperature.

  • The Clausius-Clapeyron relationship can help at this point - use the reference vapor pressure and 298 K (25 °C) for P1 and T1, respectively.
  • In our example the mixture is at 25 °C so we can use the reference tables. We found that water at 25 °C has a vapor pressure equal to 23, 8 mm Hg.
Calculate Vapor Pressure Step 9

Step 5. Find the molar fraction of the solvent

The last thing to do before solving the equation is to figure out the molar fraction of our solvent. Finding this value is easy: just convert the components to moles and then find the percentage of the total number of moles in the substance that are occupied by each component. In other words, each molar fraction is equal to: (moles of component) / (total number of moles in the substance).

  • Let's say our recipe for common syrup uses 1 liter (l) of water and 1 liter (l) of sucrose (sugar). In this case, we will need to find out the number of moles corresponding to each substance. To do this, it is necessary to find the mass of each of them and then use their molar mass to convert this value to moles.

    • Mass of 1 l of water: 1,000 grams (g).
    • Mass of 1 l of common sugar: approximately 1,056, 7 g.
    • Moles of water: 1,000 g × 1 mol / 18, 015 g = 55. 51 moles.
    • Moles of sucrose: 1056, 7 g × 1 mol / 342, 2965 g = 3.08 moles (note that it is possible to derive the molar mass of sucrose from its chemical formula, C12H22O11).
    • Total moles: 55, 51 + 3.08 = 58.59 moles.
    • Molar fraction of water: 55, 51 / 58, 59 = 0, 947.
Calculate Vapor Pressure Step 10

Step 6. Solve the equation

Finally, we have everything needed to solve Raoult's Law equation. This part is surprisingly easy: just enter the values ​​relative to the variables in the simplified equation at the beginning of the section: FORsolution =Psolvent × Xsolvent.

  • Replacing the present values, we have:

    • FORsolution = (23.8 mm Hg) (0.947).
    • FORsolution = 22, 54 mm Hg. This makes sense - in molar terms, there is only a little sugar dissolved in a lot of water (even though, in practical terms, both ingredients have the same volume), so the vapor pressure will decrease slightly.

Method 3 of 3: Finding Vapor Pressure in Special Cases

Calculate Vapor Pressure Step 11

Step 1. Be aware of the Normal Temperature and Pressure Conditions

Scientists often use, for convenience, a “standardized” set of values ​​for temperature and pressure. They are called Normal Temperature and Pressure Conditions, or CNTP. Vapor pressure problems generally refer to CNTP conditions, and it is very practical to have these values ​​always in memory. CNTP values ​​are defined as:

  • Temperature: 273, 15K / 0°C / 32°F.
  • Pressure: 760 mm Hg / 1 atm / 101, 325 kPa.
Calculate Vapor Pressure Step 12

Step 2. Rearrange the Clausius-Clapeyron equation to find other variables

In our example in Section 1, we observed that the Clausius-Clapeyron equation is quite useful for finding the vapor pressures for pure substances. However, not all questions will ask you to find the value of P1 or P2 - many want you to find a temperature value or even the value of ΔHvap. Fortunately, in these cases, to get the right answer, it is enough to rearrange the equation so that it leaves only the variable to be solved on one side of the equality.

  • For example, suppose we have an unknown liquid with a vapor pressure equal to 25 torr at 273 K and 150 torr at 325 K, and we want to find the enthalpy of vaporization of that liquid (ΔHvap). We could solve the problem as follows:

    • ln(P1/P2) = (ΔHvap/R)((1/T2) - (1/T1))
    • (ln(P1/P2))/((1/T2) - (1/T1)) = (ΔHvap/R)
    • R × (ln(P1/P2))/((1/T2) - (1/T1)) = ΔHvap
  • Now, we enter the values:

    • 8, 314 J/(K × Mol) × (-1, 79)/(-0, 00059) = ΔHvap
    • 8, 314 J/(K × Mol) × 3,033, 90 = ΔHvap = 25,223, 83 J/mol
Calculate Vapor Pressure Step 13

Step 3. Take into account the vapor pressure of the solute when it produces vapor

In our example of Raoult's Law above, the solute (sugar) does not produce any steam on its own at normal temperatures (think - when did you see a bowl of sugar evaporate on the kitchen table?). However, when the solute does evaporate, it will influence its vapor pressure. We will take this into account when using a modified version of Raoult's Law equation: FORsolution = Σ (Pcomponent × Xcomponent). The sigma (Σ) means that we need to add up all the vapor pressures of the different components to arrive at the answer.

  • For example, let's say you have a solution made up of two chemicals: benzene and toluene. The total volume of the solution is equal to 120 milliliters (ml): 60 ml of benzene and 60 ml of toluene. The temperature of the solution is equal to 25 °C, and the vapor pressure of each of these substances, at 25 °C, is equal to 95.1 mm Hg for benzene and 28.4 mm Hg for toluene. Given these values, find out the vapor pressure of the solution. We can solve the question as follows, using standard values ​​of density, molar mass and vapor pressure relative to the two substances:

    • Mass (benzene): 60 ml = 0.060 l × 876, 5 kg / 1,000 l = 0.053 kg = 53 g.
    • Mass (toluene): 0.060 l × 866, 9 kg / 1,000 l = 0.052 kg = 52 g.
    • Moles (benzene): 53 g × 1 mol / 78, 11 g = 0.679 mol.
    • Moles (toluene): 52 g × 1 mol / 92, 14 g = 0.564 mol.
    • Total moles: 0.679 + 0.564 = 1.243.
    • Molar fraction (benzene): 0.679/1,243 = 0.546.
    • Molar fraction (toluene): 0.564/1, 243 = 0.454.
  • Solve: Psolution =Pbenzene × Xbenzene + Ptoluene × Xtoluene.

    • FORsolution = (95.1 mm Hg)(0.546) + (28.4 mm Hg)(0.454).
    • FORsolution = 51.92 mm Hg + 12.89 mm Hg = 64, 81 mm Hg.


  • To use the Clausius-Clapeyron equation above, temperature must be measured in degrees Kelvin (expressed in K). If you have the temperature in degrees centigrade, you need to convert it with the following formula: TK = 273 + TÇ.
  • The above methods work because energy is directly proportional to the amount of heat delivered. The temperature of the liquid is the only environmental factor on which the vapor pressure depends.

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