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Vapor pressure is a fundamental concept in chemistry and physics that measures the tendency of a substance to evaporate and form a gas phase. It is a crucial property in various scientific fields, including material sciences, environmental studies, and thermodynamics. Understanding how to calculate vapor pressure is essential for predicting the behavior of substances in different conditions, such as boiling points, phase transitions, and equilibrium vaporization. This article aims to provide a comprehensive guide on the principles and methods used to calculate vapor pressure, covering both theoretical aspects and practical applications. By the end of this article, readers will have a solid understanding of the factors influencing vapor pressure and the mathematical strategies involved in its calculation.
This article was co-written by Bess Ruff, MA. Bess Ruff is a graduate student in geography at Florida. She received her Master’s degree in Environmental Science and Management from the Bren School of Environmental Science & Management, UC Santa Barbara in 2016. She has conducted survey work for marine spatial planning projects in the coastal area. Caribbean and support research as a contributor to the Sustainable Fisheries Group.
There are 8 references cited in this article that you can see at the bottom of the page.
This article has been viewed 28,640 times.
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)) .
Steps
Using the Clausius-Clapeyron equation
- 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.
- 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.
- 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)) .
- 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
- 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.
- 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.
- In this example, assume the current temperature of the syrup is 298 K (about 25 C).
- 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
- Assume the recipe for the syrup is 1 liter (L) of water and 1 liter of sucrose (sugar) . Then we need to find the mp number of each component. To do this, we will find the mass of each component, then use the mass mp of those components to calculate the mp.
- Weight (1 L water): 1,000 grams (g)
- Weight (1 L raw sugar): Approx. 1,056.7 g [7] X Research Source
- Number of mp (water): 1,000 grams × 1 mp/18,015 g = 55.51 mp
- Number of mp (sugar): 1,056.7 grams × 1 mp/342.2965 g = 3.08 mp (Note that you can find the mp mass of sugar from its chemical formula, C 12 H 22 O 11 .)
- Total mp: 55.51 + 3.08 = 58.59 mp
- MP part of water: 55.51/58.59 = 0.947
- Substituting the values in, we have:
- Solution P = (23.8 mmHg)(0.947)
- P solution = 22.54 mmHg. This result makes sense — in mp terms only a little sugar dissolves in a lot of water (even though the two components are in fact the same volume), so the vapor pressure will only drop a little.
Find the vapor pressure in special cases
- Temperature: 273.15 K / 0 C / 32 F
- Pressure: 760 mmHg / 1 atm / 101.325 kilopascal
- 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
- For example, suppose we have a solution prepared from two chemicals: benzene and tpuene. The total volume of the solution is 120 mL; 60 mL benzene and 60 mL tpuene. The solution temperature is 25 °C and the vapor pressure of each chemical component at 25 °C is 95.1 mmHg for benzene, and 28.4 mmHg for tpuene. With the values given, find the vapor pressure of the solution. We can solve the problem using the density, mass mp, and vapor pressure of those two chemicals:
- Mass (benzene): 60 mL = 0.06 L × 876.50 kg/1,000 L = 0.053 kg = 53 g
- Weight (tpuen): 0.06 L × 866.90 kg/1,000 L = 0.052 kg = 52 g
- Number of mp (benzene): 53 g × 1 mp/78.11 g = 0.679 mp
- Number of mp (tpuen): 52 g × 1 mp/92.14 g = 0.564 mp
- Total mp: 0.679 + 0.564 = 1.243
- Part mp (benzene): 0.679/1.243 = 0.546
- Part mp (tpuen): 0.564/1.243 = 0.454
- Solve for the result: P solution = P benzene X benzene + P tpuene X tpuene
- P solution = (95.1 mmHg)(0.546) + (28.4 mmHg)(0.454)
- P solution = 51.92 mmHg + 12.89 mmHg = 64.81 mmHg
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.
This article was co-written by Bess Ruff, MA. Bess Ruff is a graduate student in geography at Florida. She received her Master’s degree in Environmental Science and Management from the Bren School of Environmental Science & Management, UC Santa Barbara in 2016. She has conducted survey work for marine spatial planning projects in the coastal area. Caribbean and support research as a contributor to the Sustainable Fisheries Group.
There are 8 references cited in this article that you can see at the bottom of the page.
This article has been viewed 28,640 times.
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)) .
In conclusion, calculating vapor pressure is an important aspect of understanding the behavior of a substance and its ability to undergo phase changes. By utilizing the Clausius-Clapeyron equation or the Antoine equation, one can determine the vapor pressure of a substance at a given temperature. Additionally, the degree of intermolecular forces, volatility, and molecular weight impacts the vapor pressure, allowing for a comparison between different substances. Overall, understanding the concept of vapor pressure and its calculation can aid in various applications, such as predicting evaporation rates, determining boiling points, and designing efficient chemical processes.
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