3 Ways to Calculate Vapor Pressure

2024 Author: Samantha Chapman | [email protected]. Last modified: 2024-01-15 13:47

Have you ever left a bottle of water exposed to the sun for a few hours and hear a "hiss" when opening it? This phenomenon is caused by a principle called "vapor pressure" (or vapor pressure). In chemistry it is defined as the pressure exerted by an evaporating substance (which turns into gas) on the walls of an airtight container. To find the vapor pressure at a given temperature, you need to use the Clausius-Clapeyron equation: ln (P1 / P2) = (ΔH_vap/ R) ((1 / T2) - (1 / T1)).

Steps

Method 1 of 3: Using the Clausius-Clapeyron Equation

Step 1. Write the Clausius-Clapeyron formula

This is used to calculate the vapor pressure from a pressure change over a period of time. The name of the equation derives from the physicists Rudolf Clausius and Benoît Paul Émile Clapeyron. The equation is typically used to solve the most common vapor pressure problems faced in physics and chemistry classes. The formula is: ln (P1 / P2) = (ΔH_vap/ R) ((1 / T2) - (1 / T1)). Here is the meaning of the variables:

• ΔH_vap: the enthalpy of vaporization of the liquid. You can find this data in a table on the last pages of the chemistry texts.
• R.: the universal gas constant, i.e. 8, 314 J / (K x Mol).
• T1: the temperature corresponding to the known vapor pressure value (initial temperature).
• T2: the temperature corresponding to the vapor pressure value to be calculated (final temperature).
• P1 and P2: the vapor pressure at temperatures T1 and T2 respectively.

Step 2. Enter the known variables

The Clausius-Clapeyron equation looks complex because it has many different variables, but it's not difficult at all when you have the right information. The basic problems concerning the vapor pressure, in general, provide the two values of temperature and a datum for the pressure, or a temperature and the two pressures; once you have this information, the process of finding the solution is elementary.

• For example, consider a container filled with liquid at a temperature of 295 K, whose vapor pressure is 1 atmosphere (atm). The problem asks to find the vapor pressure at the temperature of 393 K. In this case we know the initial, final temperature and a vapor pressure, so we just have to insert this information into the Clausius-Clapeyron equation and solve it for the 'unknown. We will therefore have: ln (1 / P2) = (ΔH_vap/ R) ((1/393) - (1/295)).
• Remember that in the Clausius-Clapeyron equation the temperature must always be expressed in degrees Kelvin (K). The pressure can be expressed in any unit of measurement, as long as it is the same for P1 and P2.

Step 3. Enter the constants

In this case we have two constant values: R and ΔH_vap. R is always equal to 8, 314 J / (K x Mol). ΔH_vap (the enthalpy of vaporization), on the other hand, depends on the substance in question. As stated earlier, it is possible to find the values of ΔH_vap for a wide range of substances in the tables on the last pages of chemistry, physics or online books.

• Suppose the liquid in our example is pure water in the liquid state. If we look for the corresponding value of ΔH_vap in a table, we find that it is equal to about 40.65 KJ / mol. Since our constant R is expressed in joules and not kilojoules, we can convert the vaporization enthalpy value into 40,650 J / mol.
• By inserting the constants into the equation we get that: ln (1 / P2) = (40.650 / 8, 314) ((1/393) - (1/295)).

Step 4. Solve the equation

Once you have replaced the unknowns with the data at your disposal, you can start solving the equation to find the missing value, respecting the basic rules of algebra.

• The only difficult part of the equation (ln (1 / P2) = (40.650 / 8, 314) ((1/393) - (1/295)) is to find the natural logarithm (ln). To eliminate it, simply use both sides of the equation as the exponent of the mathematical constant e. In other words: ln (x) = 2 → e^{ln (x)} = and² → x = e².

• At this point you can solve the 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 makes sense because in a sealed container, increasing the temperature by at least 100 degrees (20 degrees above the boiling value of the water), a lot of steam is generated and consequently the pressure increases considerably.

Method 2 of 3: Finding the Vapor Pressure of a Solution

Step 1. Write Raoult's law

In the everyday world it is very rare to deal with a single pure liquid; usually you have to work with liquids that are the product of the mixing of different substances. One of these common liquids originates from dissolving a certain amount of a chemical, called a "solute," in a large amount of another chemical, called a "solvent." In this case, the equation known as Raoult's law comes to our aid, which owes its name to the physicist François-Marie Raoult. The equation is represented as follows: P._solution= P_solventX_solvent. In this formula the variables refer to:

• P._solution: the vapor pressure of the entire solution (with all the "ingredients" combined).
• P._solvent: the vapor pressure of the solvent.
• X_solvent: the mole fraction of the solvent.
• Don't worry if you don't know the term "mole fraction"; we will address the topic in the next steps.

Step 2. Identify the solvent and solute of the solution

Before calculating the vapor pressure of a liquid with multiple ingredients, you need to understand which substances you are considering. Remember that the solution consists of a solute dissolved in a solvent; the chemical substance that dissolves is always called "solute", while that which allows dissolution is always called "solvent".

• Let's consider a simple example to better illustrate the concepts discussed so far. Suppose we want to find the vapor pressure of a simple syrup. This is traditionally prepared with one part of sugar dissolved in one part of water. We can therefore affirm that sugar is the solute and water the solvent.
• Remember that the chemical formula of sucrose (common table sugar) is C.₁₂H.₂₂OR₁₁. This information will soon prove very useful.

Step 3. Find the temperature of the solution

As we saw in the Clausius-Clapeyron equation, in the previous section, the temperature acts on the vapor pressure. Generally speaking, the higher the temperature, the higher the vapor pressure, since as the temperature increases, the amount of liquid that evaporates also increases, consequently increasing the pressure inside the container.

In our example, suppose we have a simple syrup at a temperature of 298 K (about 25 ° C).

Step 4. Find the vapor pressure of the solvent

Chemistry textbooks and teaching materials generally report the vapor pressure value for many common substances and compounds. However, these values refer only to the temperature of 25 ° C / 298 K or the boiling point. If you are dealing with a problem where the substance is not at these temperatures, then you will need to make some calculations.

• The Clausius-Clapeyron equation can help in this step; replace P1 with the reference pressure and T1 with 298 K.
• In our example, the solution has a temperature of 25 ° C, so you can use the reference value we find in the tables. The vapor pressure of water at 25 ° C is equal to 23.8 mm Hg.

Step 5. Find the mole fraction of the solvent

The last piece of information you need to solve the formula is the mole fraction. It is a simple process: you just need to convert the solution into moles and then find the percentage "dosage" of the moles of each element that compose it. In other words, the mole fraction of each element is equal to: (moles of element) / (total moles of solution).

• Suppose the recipe for syrup plans to use 1 liter of water and the equivalent of 1 liter of sucrose. In that case you need to find the number of moles in each of them. To do this, you need to find the mass of each substance and then use the molar mass to find the number of moles.
• Mass of 1 l of water: 1000 g.
• Mass of 1 l of raw sugar: approximately 1056.7 g.
• Moles of water: 1000 g x 1 mol / 18.015 g = 55.51 moles.
• Moles of sucrose: 1056.7 g x 1 mol / 342.2965 g = 3.08 moles (you can find the molar mass of sugar from its chemical formula, C₁₂H.₂₂OR₁₁).
• Total moles: 55.51 + 3.08 = 58.59 moles.
• Molar fraction of water: 55.51/58.59 = 0, 947.

Step 6. Solve the equation

You now have everything you need to solve Raoult's law equation. This step is incredibly simple - just enter the known values into the simplified formula that was described at the beginning of this section (P._solution = P_solventX_solvent).

• By replacing the unknowns with values, we obtain:
• P._solution = (23.8 mm Hg) (0.947).
• P._solution = 22.54 mm Hg. This value makes sense, in terms of moles; there is little sugar dissolved in a lot of water (even if the two ingredients have the same volume), so the vapor pressure only increases slightly.

Method 3 of 3: Finding the Vapor Pressure in Special Cases

Step 1. Know the standard pressure and temperature conditions

Scientists use set values of pressure and temperature as a sort of "default" condition, which is very convenient for calculations. These conditions are called Standard Temperature and Pressure (abbreviated to TPS). Vapor pressure issues often refer to TPS conditions, so it's worth memorizing them. TPS values are defined as:

• Temperature: 273, 15K / 0 ° C / 32 ° F.
• Pressure: 760 mm Hg / 1 atm / 101, 325 kilopascals

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

In the example of the first section of the tutorial this formula was very useful in finding the vapor pressure of pure substances. However, not all problems require finding P1 or P2; it is often necessary to find the temperature value and in other cases even that of ΔH_vap. Fortunately, in these cases the solution can be found simply by changing the arrangement of the terms within the equation, isolating the unknown to one side of the equality sign.

• For example, consider that we want to find the vaporization enthalpy of an unknown liquid that has a vapor pressure of 25 torr at 273 K and 150 torr at 325 K. We can solve the problem in this way:
• ln (P1 / P2) = (ΔH_vap/ R) ((1 / T2) - (1 / T1)).
• (ln (P1 / P2)) / ((1 / T2) - (1 / T1)) = (ΔH_vap/ R).
• R x (ln (P1 / P2)) / ((1 / T2) - (1 / T1)) = ΔH_vap. At this point, we can enter the values:
• 8, 314 J / (K x Mol) x (-1, 79) / (- 0, 00059) = ΔH_vap.
• 8.314 J / (K x Mol) x 3.033.90 = ΔH_vap = 25,223.83 J / mol.

Step 3. Consider the vapor pressure of a solute that produces vapor

In the section dealing with Raoult's law, the solute (sugar) produces no steam at normal temperature (think, when was the last time you saw a bowl of evaporating sugar?). However, when you use a solute that "evaporates" then it interferes with the vapor pressure value. We need to take this into account using a modified formula for Raoult's law: P._solution = Σ (P_componentX_component). The sigma symbol (Σ) indicates that you have to add all the pressure values of the various components to find the solution.

• For example, consider a solution made up of two chemicals: benzene and toluene. The total volume of the solution is 120 ml, 60 ml of benzene and 60 ml of toluene. The temperature of the solution is 25 ° C and the vapor pressure of each substance at 25 ° C is 95.1 mm Hg for benzene and 28.4 mm Hg for toluene. From this information, the vapor pressure of the solution must be derived. You can do this using the standard value of density, molar mass and vapor pressure of the two substances:
• Benzene mass: 60ml = 0.060l & times 876.50kg / 1000l = 0.053kg = 53 g.
• Toluene mass: 60 ml = 0.060 l & times 866.90 kg / 1000 l = 0.052 kg = 52 g.
• Moles of benzene: 53 g x 1 mol / 78.11 g = 0.679 mol.
• Moles of Toluene: 52 g x 1 mol / 92.14 g = 0.564 moles.
• Total moles: 0, 679 + 0, 564 = 1, 243.
• Molar fraction of benzene: 0, 679/1, 243 = 0, 546.
• Molar fraction of toluene: 0, 564/1, 243 = 0, 454.
• Resolving: P._solution = P_benzeneX_benzene + P_tolueneX_toluene.
• P._solution = (95, 1 mm Hg) (0, 546) + (28, 4 mm Hg) (0, 454).
• P._solution = 51.92 mm Hg + 12.89 mm Hg = 64, 81 mm Hg.

Advice

• To use the Clausius-Clapeyron equation described in the article, the temperature must be expressed in degrees Kelvin (denoted by K). If this is given in degrees centigrade, you need to convert using the formula: T._k = 273 + T_c.
• The methods shown work because the energy is directly proportional to the amount of heat applied. The temperature of a liquid is only an environmental factor on which the pressure depends.