Vapor Pressure Lowering Definition in Chemistry and Example

Now we will discuss about the vapor pressure lowering definition chemistry. Vapor pressure lowering occurs because of a very high level of solute and evaporation didn’t happen. Before we discuss vapor pressure lowering of a substance even further, let’s discuss about the evaporation process itself first.

Vapor pressure lowering

First, prepare a glass full of water. What would happen to the volume of the water if you left that glass of water outside for a couple of hours? That answer is obviously the water volume will reduce because of an evaporation process.
Because of the glass of water that’s left out without any cap and placed in an outdoor place, the water volume will reduce continuously. Different things would happen if the glass is left indoor.
Now fill the water inside a closed container and connect it with pressure gauge
In the beginning of our experiment, we can see that the height of both pipe will be equal, which is because the water molecule did not experience an evaporation yet. After we left it for a couple of hours, there will be a difference of the mercury height in the U pipe.
The occurrence of difference in the mercury height in that U pipe showed that there’s a pressure in water molecules that caused the evaporation process. Evaporation will occur in the molecule surface continuously until the water is in an equal state.
In an equal state, the number of water molecules that left the fluid caused by the evaporation process is the same as the number of water molecules that entered the liquid. Pressure that occurs in a liquid at an equal state with the liquid molecule’s vapor above it is what called as liquid vapor pressure.
Liquid, in this context, refers to certain kind of substances like water, ethanol, benzene, and other liquid substance that is usually used as solvent. Because of that, we will refer “liquid vapor pressure” as solvent vapor pressure.
The amount of solvent vapor pressure that shows up is not affected by the number of solvent that is used, but it is affected by temperature. So, in a different temperature, the solvent vapor pressure that is caused will be different. E.g. in a room temperature (25 degree celcius), the vapor pressure that we get is 20 mmHg.
What will happen if we dissolve a non-volatile substance of substance that does not evaporate easily? E.g. Glucose substance that is dissolved in a container filled with water and connected to a pressure gauge. If the temperature of both substance is equal, we will get glucose vapor pressure of 18,5 mmHg.
The existence of solute in a solvent will cause vapor pressure lowering of its solvent. As we can see in the example above that in a 25 degree celcius temperature, the acquired pure water vapor pressure is 20 mmHg while a glucose substance in a water with the same temperature, the vapor pressure lowered to 18,5 mmHg.
The vapor pressure that turned out to be lower between water and glucose indicates that the total solvent mole that evaporates in less than the total mole that evaporates in a pure solvent.

From several experiments done above, we can say that the occurrence of vapor pressure lowering of a solvent is bigger than solution vapor pressure. Therefore, this colligative property is called as solution vapor pressure lowering.

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Vapor Pressure of a Solvent is Bigger than Vapor Pressure of a Solution

Compound, like solution, has an entropy or energy bigger than single material or pure solvent. The entropy rising also rise the energy that is needed to transform solvent molecule from liquid phase to gas phase.

How to calculate the vapor pressure lowering of a solution? First, we emphasize the vapor pressure lowering occurs in a non-volatile solvent. Non-volatile is a property that makes a substance dies not evaporate easily while volatile means that a substance can evaporate easily.

Solute in a solvent will cause a lowering of total solvent molecule for every of its volume unit. The lower the total solvent molecule per volume unit goes that is inside in a solution if we compare it with the total of solvent molecule inside a pure solvent will reduce the total molecule that can evaporate. Therefore, the vapor pressure will also have lowered. To make it easier, If the volume is big, the surface area is big while if the volume is small, the surface area is small, so it caused the total difference of H2O molecule that evaporates.
In an energy form, the solute in a solvent substance will increase the irregularity in a substance

The correlation between saturated solution vapor pressure with its solvent vapor pressure already been explained by Francois M. Raoult where he explained below formula:

P = Xp.Po

Where:

P = Saturated solution vapor pressure
Po = Pure solvent vapor pressure
Xp = Solvent mole fraction

The pressure lowering from P0 to P is called vapor pressure lowering symbolized as ΔP, where ΔP is formulated as:

ΔP = Po–P
ΔP = Po.Xt
P = Xp. Po
= (Np/(Np+Nt)).Po

Where:

ΔP: Vapor pressure lowering
P: Saturated solution vapor pressure
Po: Pure solvent vapor pressure
Xp: Solvent mole fraction
np: Solvent mole
nt: Solute mole

Mole fraction (X), in this context, is explained as a ratio between a mole of a species and the total mole where the species lived. E.g. a solution made of water solvent and urea solute, then the mole fraction of each substance is as below:

X water = Water mole / water mole + urea mole
X urea = Urea mole / water mole + urea mole

The total mole fraction of each compound component if totaled is equal to 1. For urea mole fraction can be as explained above is written as:

X water + X urea = 1

If a solution only acquired from two components is solvent and solute. The correlation between both mole fractions can be expressed as:

Xp + Xt = 1
Xp = 1 – Xt (first expression)

If the first and second expression is combined, then you get the below expression:

P= Xp . Po
with Xp = 1 – Xt, then

P = (1 – Xt) Po
P = Po – Xt . Po
P – Po = Xt . Po

P = Xt.Po (third expression)

This third expression is what we can use to calculate how big the vapor pressure lowering number of a solution is. To find the vapor pressure lowering, we can see it from the first and third expression. Bear in mind that if we used the first expression, then the mole fraction that is used is Xp (solvent mole fraction), but if we use the third expression, the mole fraction that is used is Xt (solute mole fraction).

What if the Solute in a Solvent is Volatile?

The above explanation focused more on a solution with a non-volatile solute while what happened if a solution is made from a volatile solute?

Examples of these are Benzene – Toluene, Water – Ethanol, or Acetone – Ethyl acetate. Because of the volatile characteristic of the solute, this solute’s vapor will react to the total of solution’s vapor. Vapor that is inside this kind of solution is acquired from the solute’s mole and the solvent’s mole.

Based on above explanation, we can conclude that the total of solution’s vapor pressure is expressed with below formula:

P solution: P1+P2+P3+ … +Pn

Where:
P1 = Xt1.Po1
P2 = Xt2.Po2
P3 = Xt3.Po3
Pn= Xtn.Pon

Bear in mind that Raoult’s law only applied to an ideal solution or solution with low concentration level. Where the ideal solution can be acquired if an interaction between solute – solute, solvent – solvent, or solute – solvent got an almost equal number. Example of compound that meets the Raoult’s law is Benzene – Toluene. The mixing of those two results in an entalphy of almost “0”, so the compound is considered “ideal”.

When a dissolving of a solute inside a solvent is freed by heat (exoterm), then the eltalphy is negative. We can assume that there is a strong interaction between the solute and the solvent, this thing usually causes the solvent to have a lower tendency to evaporate and the vapor pressure of its solution will be lower that what’s determined by Raoult’s law. This occurrence is usually referred as negative deviation of Raoult’s law. E.g. dissolving acetone substance in a water or mixture between chloroform and acetone substance. A strong interaction between acetone – water or acetone chloroform is caused by a hydrogen bond between each substance.

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Vapor Pressure Lowering Exercises

The water vapor pressure in 25 0C temperature is 23,76 mmHg. With the same temperature, urea (Mr = 60 g/mole) is added in 900ml of water. The solution vapor pressure changed to 22,84 mmHg. What is the vapor pressure lowering of that solution?

Known:
T = 25 0C
P0 = 23,76 mmHg.
V = 900 mL
Mr = 60 g/mole.
P = 22,84 mmHg.
Asked:
ΔP = …
Solution:
ΔP = P0 – P = 23,76 mmHg – 22,84 mmHg = 0,92 mmHg

The water vapor pressure in 25 0C temperature is 23,76 mmHg. With the same temperature, urea (Mr = 60 g/mol) is added in 900ml of water. The solution vapor pressure changed to 22,84 mmHg. What is the mole fraction of that solution?

Known:
T = 25 ⁰C
P⁰ = 23,76 mmHg.
V = 900 mL
Mr = 60 g/mol.
P = 22,84 mmHg.
Asked: xp = …
Solution:
xp = ΔP / P⁰ = 0,92 mmHg / 23,76 mmHg
xp = 0,0387

Mole fraction of a urea inside water is 0,5. The water vapor pressure in 20°C temperature is 17,5 mmHg. What is the saturated vapor pressure of that solution with the same temperature?

Known:
xA = 0,5
P0 = 17,5 mmHg
Asked: P …?
Solution: ΔP = xA ⋅ P0
= 0,5 ⋅ 17,5 mmHg
= 8,75 mmHg
P = P0 – ΔP
= 17,5 mmHg – 8,75 mmHg
= 8,75 mmHg

That’s all about vapor pressure lowering definition chemistry. Examples of vapor pressure lowering in everyday life is on a floating pool in tourist destination that adopts the trait of Dead Sea that makes you float when swimming. This is because the dissolved solution in very high. Hope it’s helpful.