Chemistry

Lowering of vapor pressure

Lowering of vapor pressure


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Area of ​​Expertise - thermodynamics

If a non-volatile substance is dissolved in a volatile solvent, the vapor pressure drops according to Raoult's law (Raoult's law). This process is known as steam pressure lowering:

Δp=p1x2Δp=Lowering of vapor pressurep1=Vapor pressure of the pure solventx2=Mole fraction (mole fraction) of the solute in the solution

This effect can be explained physically as follows: If, for example, sodium chloride is dissolved in water, the dissolved salt is evenly distributed in it. Some water molecules on the surface are displaced and replaced by salt molecules. Fewer water molecules can evaporate at the same temperature. The condensation of already evaporated water molecules, however, is not hindered. Consequence: At the same temperature, a lower saturation vapor pressure is set.

See also: Boiling point increase, freezing point lowering

Learning units in which the term is dealt with

Binary gas / liquid equilibria - Raoult's case30 min.

ChemistryPhysical chemistrythermodynamics

In this chapter we consider binary systems in which there is equilibrium between a liquid and a gaseous phase. The target variable is the vapor pressure of a solvent as a function of the concentration of a less volatile solute.


Lowering of vapor pressure

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Articles on the topic

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Vapor pressure of a solution with a vaporizable component

Without going into the theory again in detail, let's go straight to the measuring principle. For the sake of simplicity, water is usually used as the solvent (inexpensive and harmless), which is the vaporizable component. In the water (as a solvent) we add a non-evaporable component (e.g. salts or sugar). This is the basis of the later evaluation: the non-evaporable component has a negligible vapor pressure above the solution (the component hardly evaporates at the temperatures).

In the first step we calculate the & # 8220 substance amount & # 8221 in the solution. Even if the use of the & # 8220 Molehill & # 8221 is out of date, it is still often used in this calculation.

We now consider water W (as a vaporizable solvent) and glucose G (as a not easily vaporizable component that is dissolved in water). Since water is the vaporizable component, we calculate the mole fraction of water:

We also derive the mole fraction for glucose: mole fraction (G) = n (G): [n (W) + n (G)]

Now we still have to consider how the vapor pressure behaves above pure water and above the sugar solution. Sugars are polar organic substances, just like water is a polar substance. This creates an attractive interaction between the water molecules and the sugar molecules. In a descriptive sense, the sugar molecules hold the water molecules more strongly in solution, so that the tendency of the water molecules to evaporate is reduced.

Therefore, the vapor pressure of a vaporizable component of a solution (the other component is not easily vaporizable) is less than the vapor pressure above the liquid of the pure vaporizable component.

  • For pure water: state of equilibrium of the system water: water vapor at a given temperature. The vapor pressure above the water is p (W0).
  • For the solution, the equilibrium state of the glucose solution system: water vapor. The vapor pressure above the solution is p (W).

Now we can use the previous formulas to derive the relationship between the vapor pressure and the concentration of a solution (with a vaporizable component):

Let's start with the first part of the equation: [p (W0) & # 8211 p (W)], this expression is the decrease in vapor pressure caused by the interaction with the non-evaporable component. This expression is usually abbreviated as follows: Δp (W).

Now the second part of the equation: mole fraction (G): [mole fraction (W) + mole fraction (G)] insert the amounts of substance and shorten = & gt n (G): [n (W) + n (G)] = mole fraction (G )

Now we have an equation that is: Δp (W): p (W) = mole fraction (G)
This equation, which we have just derived, is nothing other than Raoult's 1st law. This means that the relative decrease in vapor pressure: Δp: p is equal to the mole fraction of the dissolved component (which is not highly volatile). This relative decrease in vapor pressure is independent of the temperature.

We can therefore determine the concentration or the mole fraction of a dissolved substance by measuring the vapor pressure.

Now the question is whether the lowering of the vapor pressure can be used elsewhere & # 8220 & # 8221.

We can answer this question with & # 8220Yes & # 8221 & # 8211 from the theoretical chapters at Lernort-Mint we know that the lowering of the vapor pressure of solutions with a non-evaporable substance leads to an increase in the boiling temperature and a decrease in the freezing temperature. We can use this increase or decrease to determine the molar mass. The corresponding procedures are called cryoscopy (lowering the freezing temperature) and ebullioscopy (increasing the boiling temperature).

The constants E (g) and E (s) are determined by dissolving one mole of the substance in 1000 g of solvent and then determining the temperature differences or looking up the corresponding lexicons.


Lowering of vapor pressure

Lowering of vapor pressure, Raoult's law.

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In a closed, thermostated room saturated with solvent vapor (vapor pressure osmometer) there are two highly sensitive semiconductor temperature sensors whose electrical resistance changes with temperature. These are wetted differently. One thermistor with pure solvent, the other with the polymer solution to be measured. The vapor pressures of solution and pure solvent are different. However, they are in contact with one another via the vapor phase and strive for a state of equilibrium. Over time, equilibrium between the drops is established, as solvent vapor condenses on the thermistor with the sample drop, since the vapor pressure of the solvent in the polymer solution is lower than that of the pure solvent. When the solvent condenses, heat of condensation is released, as a result of which the temperature of the solution drop rises. The temperature difference between the sample and solvent drops is measured after equilibrium has been established (certain waiting time). Since the temperature differences are small (<0.1 & # 160K) and therefore a high level of measurement accuracy is required, the two thermistors, when wetted with the same solvent, are compared using a Wheatstone bridge before the actual measurement. This means that even small changes in resistance of 5 & # 160 × & # 16010 -4 & # 160% can be measured.

The temperature difference between the two drops is a measure of the decrease in vapor pressure caused by the dissolved substance and, if the concentration is known, also for its molar mass. The device constant of the measuring arrangement must be determined beforehand using calibration substances (with known concentration and molar mass) in order to be able to calculate the molar mass from the temperature difference measured for the sample.


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