Reduction system in vapor pressure occurs due to very high solute and have no evaporation. In previous article described that one of the colligative characteristics of solutions is Boiling Point Increase. This time we will learn more about the other characteristics of Steam Pressure Reduction System. But before we discuss further about Steam Pressure Reduction System, it would be better if we first understand the process of evaporation first.
Steam Pressure Reduction System
- First provide a place / glass filled with water. What happens to the volume of water if the water-filled object is left open in the daytime for several hours? The answer is definitely the volume of water in the glass will be reduced due to the evaporation process.
- Because the glass condition is left uncovered and is in open space then the volume of water will decrease continuously. This will be different if we do it in a confined space.
- Now we fill the water into a closed container and connected with a pressure gauge:
- At the beginning of the experiment we can see the height in both parts of the foot of the pipe will be the same, it is because the water molecules that have not experienced evaporation. After we leave for a few hours there is a change of mercury height in the U pipe (Read Also: Hazardous Substance and Chemicals)
- The occurrence of alteration of mercury in the U pipe indicates that there is pressure on water molecules due to the evaporation process. Evaporation will occur on the surface of the molecule and continuously until the water state is in a balanced position.
- In this equilibrium position it is known that the number of water molecules leaving the liquid due to the evaporation process is equal to the number of water molecules entering into the liquid. The pressure that occurs when a liquid is in equilibrium with the liquid molecular vapor above it is referred to as Pressure of Liquid Steam or Solution.
- The purpose of the term liquid I use above is to lead to certain substances such as water, ethanol, bensena, and other compounds in the form of liquid which is usually used as a solvent, hence the term “liquid vapor pressure” for further discussion referred to as Solvent Vapor Pressure.
- The amount of solvent vapor pressure that is emerging is not affected by the amount of solvent used but is affected by the temperature. So at different temperatures will cause different solvent vapor pressure. For example, at room temperature (25 degrees Celsius) we get water vapor pressure is 20 mmHg.
- What would happen if we dissolved a nonvolatile or non-volatile substance, such as a glucose substance into a water-filled container and connected to a pressure gauge? Suppose that at the same temperature we find a vapor pressure of glucose solution of 18.5 mmHg.
- The presence of a solute in a solvent will cause a reduction in the vapor pressure of the solvent. We see in the example above is at a temperature of 25 degrees Celsius obtained pure water vapor pressure is equal to 20 mmHg while on glucose substances in water at the same temperature vapor pressure reducted to 18.5 mmHg. (Read Also: list of Organic Chemicals)
- The smaller differences in vapor pressure values between the water and glucose solutions indicate that the number of solvent molecules evaporating over the solution is less than when the number of molecules evaporates above the pure solvent.
From some experiments we do above it can be seen that there is a reduction in the vapor pressure of the solvent is greater when compared with the vapor pressure of the solution, therefore the colligative nature is commonly referred to as the reduction of Solvent Pressure.
The Solvent Vapor Pressure is greater than the Solvent Pressure
Mixtures such as solutions have greater entropy or energy than a single material or pure solvent. This increase of entropy will lead to an increase in energy as well, which is required to move solvent molecules from the liquid or liquid phase to the gas phase.
How Do We Calculate Reduced Solvent Pressure?
First we emphasize the reduction in vapor pressure that occurs for substances dissolved in non-volatile solvents. Non volatile is non volatile and volatile is volatile.
- The solute present in a solvent will cause a reduction in the number of solvent molecules for each volume unit, with the decreasing number of solvent molecules per unit volume present in a solution if we compare with the number of solvent molecules present in a pure solvent will reduce the number of molecules that can evaporate so that the vapor pressure will also decrease as well.
To make definition easier if large volumes then surface area is large, whereas when the volume is small then the surface area is small, causing differences in the number of molecules of H2O undergoing the evaporation process.
- In the form of energy, the presence of solutes in a solvent increases the irregularity in the substance. (Also Read: Factors Affecting pH solution)
The relationship between the saturated vapor pressure of the solution and the vapor pressure of the solvent was once described by Francios M. Raoult where he introduced the following formula:
P = Vacuum pressure saturated solution
Po = pure solvent vapor pressure
Xp = mole fraction of solvent
The pressure drop from P0 to P is called a reduction in vapor pressure, which is denoted by ΔP, which ΔP is formulated by:
By Description :
ΔP: Steam Pressure Reduction
P: Saturated Steam Pressure Solution
Po: Pure Solvent Steam Pressure
Xp: Fraction of Solvent Mol
np: Solvent Mol
nt: Mol Substances Dissolved
The mole fraction (X) in this case is expressed as a form of comparison between the moles of a species and the total moles in which they belong. So if it exemplified a solution made from a water solvent and a solute of urea.
Then the mole fraction of each substance is as follows:
X water = mole of water / water mol + mol urea and
X urea = mol urea / water mol + mol urea
The amount of mole fraction for each composition of the mixture when added will we obtain the value of = 1, for the mole fraction of the above urea solution can be expressed:
X water + X urea = 1
If the solution is obtained only from two components ie solvent (p) and a solute (t) then the relationship of the two mole fractions can be expressed as follows:
If equations 1 and 2 are combined then the following equation will be obtained:
P = Xp. Po
With Xp = 1 – Xt then obtained,
P = (1 – Xt) Po
P = Po – Xt. Po
P – Po = Xt. Po
P = Xt. Po …… (Equation 3)
Equation 3 is what we can use to know how much the steam reduction of a solution. To find the reduction in vapor pressure we can see through equation 1 or 3. We need to remember is that if we use formula 1 then the fraction of mole used is Xp (mole fraction of solvent) but if using formula 3 then the fraction of mole used is Xt ( mole fraction of solute). That is the fundamental difference of the two equations above. (Also Read: Differences of Electrolyte and Non-Electrolyte Solutions)
What if the solute in a solvent is volatile?
The above explanation is preferable to a solution whose solute nature is nonvolatile, whereas for solutions constructed from volatile solutes?
Examples of this mixture are bensene – toluene, water – ethanol, or acetone – ethyl acetate. Due to the volatile solute, the vapor of this solute will react to the total vapor of the solution. The steam present in this type of solution is obtained from solute molecules and solvent molecules.
After reading the above explanation then the total vapor pressure of the solution can be expressed by the formula:
It should be remembered that Raoult’s Law applies only to ideal solutions or low concentrations of solutions. Where the ideal solution is achieved when the interaction between solute – solute, solvent, solute – solvent is obtained almost the same value. The mixture that meets Raoult’s Law or is ideal for example is benzene-toluene. Mixing the two will produce enthalpy that is almost zero “0” so this mixture is “ideal”.
If when dissolving the solute into a solvent is released heat (exothermic) then the entalpinya will be negative value, we can assume that there is a strong interaction between the solvent and solute, this may cause the solvent to have a smaller tendency to evaporate and the value the vapor pressure of the solution would be much smaller than the value determined by Raoult’s own law, this event being called the “negative deviation of Raoult’s law”.
An example is to dissolve acetone with water or a mixture of chloroform with acetone. Strong interactions between acetone – water or aseto – chlorofomes are caused by the presence of hydrogen bonds between the two.
Example of Solvent Pressure Reduction Problem
1. Water vapor pressure at 25 ° C is 23.76 mmHg. If at the same temperature, into 900 mL water is added urea (Mr = 60 g / mol), the vapor pressure of the solution becomes 22.84 mmHg. Decreasing the vapor pressure of the solution is …
T = 25 0C
P0 = 23,76 mmHg.
V = 900 mL
Mr = 60 g / mol.
P = 22.84 mmHg.
Asked: ΔP = …
ΔP = P0 – P = 23.76 mmHg – 22.84 mmHg = 0.92 mmHg
2. Water vapor pressure at 25 ° C is 23.76 mmHg. If at the same temperature, into 900 mL water is added urea (Mr = 60 g / mol), the vapor pressure of the solution becomes 22.84 mmHg. The mole fraction of the solution is ….
T = 25 0C
P0 = 23,76 mmHg.
V = 900 mL
Mr = 60 g / mol.
P = 22.84 mmHg.
Asked: xp = …
xp = ΔP / P0 = 0.92 mmHg / 23,76 mmHg
xp = 0.0387
3. The fraction of water urea mole is 0.5. The vapor pressure at 20 ° C is 17.5 mmHg. What is the vapor pressure of the saturated solution at that temperature?
Is known :
xA = 0.5
P0 = 17.5 mmHg
Asked: P …?
Answer: ΔP = xA ⋅ P0
= 0.5 ± 17.5 mmHg
= 8.75 mmHg
P = P0 – ΔP
= 17.5 mmHg – 8.75 mmHg
= 8.75 mmHg
So that’s the explanation of the Steam Pressure Reduction System. An example of a reduction in the vapor pressure in everyday life is the floating pool at a tourist site that adopts the characteristic of the dead sea that when you swim you will not drown. This is because in the sea dead a very high solute. Hopefully this article will be useful for you.