Storyboard

Water vapor interacts with the surface of the soil grains, leading to the formation of water layers on their surface. The degree of coverage depends on the existing water vapor pressure.

>Model

ID:(374, 0)

Concept

>Top

ID:(15209, 0)

Concept

>Top

Water molecules suspended in vapor form can be captured by the surface of grains. This absorption is a result of intermolecular forces.

On the other hand, due to thermal oscillations, an absorbed molecule may acquire enough energy to break free.

An equilibrium state is reached when the number of absorbed particles is equal to the number that can once again be released.

ID:(115, 0)

Concept

>Top

When comparing the Langmuir model with real systems, significant discrepancies are observed. For this reason, various models have emerged that rely on fitting curves to experimental data.

In the graph presented, the Langmuir models for one and two layers are shown on the left side, along with the Freundlich and Temkin models:

Graphs of Langmuir, Freundlich and Temkin models [1]

[1] "A comparison of the Langmuir, Freundlich and Temkin equations to describe phosphate adsorption properties of soil." - Mead, J.A., Aust. J. Soil Res. 19:333-342 (1981).

ID:(7977, 0)

Concept

>Top

The molecules within a liquid experience equal attractions towards all their neighbors. This results in the overall forces exerted on them cancelling each other out, causing the molecule to behave like a free particle.

However, the situation is different for molecules on the surface. Since there are more molecules within the liquid generating an effective inward force, this prevents the molecules on the surface from leaving the liquid.

The force described in the previous section gives rise to what is known as ($$). This surface tension creates a kind of membrane on the surface, allowing some insects to move across it without sinking. For example, the spider's leg in the image does not penetrate the surface, thus preventing it from sinking.

Surface tension is also responsible for the shape of water droplets. The attraction between molecules tends to make the droplet have the smallest possible surface area, which means it will seek to assume a spherical shape. This causes a stream of water to tend to break into droplets, and these droplets tend to be spherical or fluctuate around this shape.

ID:(1551, 0)

Concept

>Top

Capillarity causes water to accumulate between the soil grains. A meniscus forms around the union zone, as depicted by the red arrows in the image below:

the surface Tension ($\sigma$) generates an internal pressure the surface tension pressure ($p_c$), which depends on curvature radio ($r$), countering the pressure on the generated meniscus the water vapor pressure unsaturated ($p_v$).

The final effect of the meniscus is to generate a force that binds the soil grains, providing cohesion that is crucial for the mechanical properties of the soil.

On the other hand, the same tension hinders the removal of water from the soil during drying processes.

To illustrate, consider building sandcastles at the beach. If we add too much water to the sand, menisci do not form (the radius is larger than the grain), and the sand is not moldable. On the other hand, as the sun dries the sand, cohesion is lost, and the castle collapses.

ID:(10687, 0)

Concept

>Top

The geometry is defined by two grains assumed to be spheres of the radius of a generic grain ($r_0$) in contact at a point. In this region, a water region is formed, which, thanks to the surface Tension ($\sigma$), concentrates around the point of contact and creates a meniscus of the meniscus radius ($r_m$) in its vicinity, similar to if an imaginary torus existed, as shown in the image below:

&bull: the section ($S$) corresponds to the line of the imaginary torus between the two contact points between the grain and the torus, which must then be rotated around an axis passing through the two centers of the grains.
&bull: the volume ($V$) corresponds to the blue area, which must then be rotated around an axis passing through the two centers of the grains.

ID:(15151, 0)

Concept

>Top

Selected parameter

Calculations

ID:(15233, 0)

Equation

>Top, >Model

Water molecules can be adsorbed by the surface of grains, meaning they are captured and retained through intermolecular forces.

The amount retained depends on both these intermolecular forces and the concentration or pressure of water vapor present on the surface.

In 1918, Langmuir [1] studied the equilibrium between a gas of particles associated with the second component concentration ($[P]$), empty spaces associated with the first component concentration ($[S]$), and occupied spaces associated with the reacted component concentration ($[SP]$):

$S + P\rightleftharpoons SP$

He proposed that the concentrations obey an equation of the form:

$ K =\displaystyle\frac{ [SP] }{ [S] [P] }$

Conditions

Variables

$K$

Concentration quotient

$1/m^3$

$[S]$

First component concentration

$1/m^3$

$[SP]$

Reacted component concentration

$1/m^3$

$[P]$

Second component concentration

$1/m^3$

Relation

[1] "The Adsorption of Gases on Plane Surfaces of Glass, Mica, and Platinum", Irving Langmuir, Journal of the American Chemical Society, Volume 40, Issue 9, pages 1361-1403 (1918)

ID:(4745, 0)

Equation

>Top, >Model

Based on the assumption that concentrations in a chemical reaction can be compared to the 'components': water molecules in the air (the second component concentration ($[P]$)), empty spaces on the grain surface (the first component concentration ($[S]$)), and occupied sites (the reacted component concentration ($[SP]$)), we can model grain coverage with the relationship in the concentration quotient ($K$):

Thus, using the molar concentration ($c_m$), fraction of Water Coverage ($\theta$) and the langmuir constant ($ m^3/mol $) ($\alpha$) [1], we obtain the degree of coverage as a function of water vapor pressure:

$ \theta =\displaystyle\frac{ \alpha c_m }{1+ \alpha c_m }$

Conditions

Variables

$c_n$

Concentration

$1/m^3$

$\theta$

Fraction of Water Coverage

$-$

$\alpha$

Langmuir constant ($ m^3/mol $)

$m^3/mol$

Relation

Development

Considering the occupied spaces the second component concentration ($[P]$), the empty spaces associated with the first component concentration ($[S]$), and the occupied spaces associated with the reacted component concentration ($[SP]$) and the concentration quotient ($K$):

$ K =\displaystyle\frac{ [SP] }{ [S] [P] }$

If fraction of Water Coverage ($\theta$) represents the fraction of surface occupied by molecules:

$[SP] \propto \theta$ - occupied spaces on the grain

$[S] \propto 1-\theta$ - empty spaces on the grain, and

$[P] \propto c$ - vapor concentration of water in the intergranular space

Then, it can be expressed as

$\alpha=\displaystyle\frac{\theta}{(1-\theta)c}$

with the langmuir constant ($ m^3/mol $) ($\alpha$).

By solving for fraction of Water Coverage ($\theta$), we can obtain the degree of coverage as a function of water vapor pressure:

$ \theta =\displaystyle\frac{ \alpha c_m }{1+ \alpha c_m }$

The physical meaning of the Langmuir constant is an affinity coefficient of the grain surface for water. The higher its value, the more likely water molecules are to adhere to the surface.

[1] "The Constitution and Fundamental Properties of Solids and Liquids. Part I. Solids" - Irving Langmuir, Journal of the American Chemical Society, Volume 38, Issue 11, pages 2221-2295 (1916).

ID:(7973, 0)

Equation

>Top, >Model

The determination of fraction of Water Coverage ($\theta$) is based on the concentration ($c_n$), and therefore, it can be expressed in terms of the water vapor pressure unsaturated ($p_v$) using the universal gas constant ($R$) and the absolute temperature ($T$) through the equation:

In this case, the langmuir constant ($ m^3/mol $) ($\alpha$) can be replaced by the langmuir constant ($1/Pa$) ($\alpha_p$) using the equation:

$ \alpha_p \equiv\displaystyle\frac{ \alpha }{ R T }$

Conditions

Variables

$T$

Absolute temperature

$K$

$\alpha$

Langmuir constant ($ m^3/mol $)

$m^3/mol$

$\alpha$

Langmuir constant ($1/Pa$)

$1/Pa$

$R$

Universal gas constant

8.4135

$J/mol K$

Relation

An article describing how to estimate the Langmuir constant can be found in [1]. Another article investigates the adsorption of different ions on grain surfaces [2]. For a langmuir constant ($ m^3/mol $) ($\alpha$) on the order of 1000 L/mol, as described in the article, the langmuir constant ($1/Pa$) ($\alpha_p$) will be approximately 4.1E-4 1/Pa at room temperature.

[1] "Evaluation of thermodynamic parameters of cadmium adsorption on sand from Temkin adsorption isotherm" - Abdul Satter Ali Khan, Turk J. Chem 36, 437-443 (2012).

[2] "A Comparison of the Langmuir, Freundlich and Temkin Equations to Describe Phosphate Sorption Characteristics of Some Representative Soils of Bangladesh" - Mohammad Z. Afsar, Sirajul Hoque, K.T. Osman, International Journal of Soil Science 7 (3): 91-99 (2012).

ID:(7976, 0)

Equation

>Top, >Model

fraction of Water Coverage ($\theta$) is calculated based on the concentration ($c_n$), which allows it to be rewritten in terms of the water vapor pressure unsaturated ($p_v$) using the universal gas constant ($R$) and the absolute temperature ($T$) through the equation:

In this case, the equation for fraction of Water Coverage ($\theta$) can be expressed in terms of the water vapor pressure unsaturated ($p_v$) as follows:

$ \theta =\displaystyle\frac{ \alpha_p p_v }{1+ \alpha_p p_v }$

Conditions

Variables

$\theta$

Fraction of Water Coverage

$-$

$\alpha$

Langmuir constant ($1/Pa$)

$1/Pa$

$p_v$

Water vapor pressure unsaturated

$Pa$

Relation

Development

fraction of Water Coverage ($\theta$) can be calculated using the langmuir constant ($ m^3/mol $) ($\alpha$) and the molar concentration ($c_m$) with the equation:

$ \theta =\displaystyle\frac{ \alpha c_m }{1+ \alpha c_m }$

If we replace the molar concentration ($c_m$) with the water vapor pressure unsaturated ($p_v$) using the universal gas constant ($R$) and the absolute temperature ($T$) through the equation:

$ p = c_m R T $

it becomes evident that the langmuir constant ($ m^3/mol $) ($\alpha$) can be replaced by the langmuir constant ($1/Pa$) ($\alpha_p$) using the equation:

$ \alpha_p \equiv\displaystyle\frac{ \alpha }{ R T }$

resulting in:

$ \theta =\displaystyle\frac{ \alpha_p p_v }{1+ \alpha_p p_v }$

ID:(4444, 0)

Equation

>Top, >Model

In 1834, Émile Clapeyron recognized that the pressure ($p$), the volume ($V$), the absolute temperature ($T$), and the number of moles ($n$) are related by Boyle's law, Charles's law, Gay-Lussac's law, and Avogadro's law. These laws can be rewritten in the form:

$ p V = n R T $

Conditions

Variables

$T$

Absolute temperature

$K$

$n$

Número de Moles

$-$

$p$

Pressure

$Pa$

$R$

Universal gas constant

8.4135

$J/mol K$

$V$

Volume

$m^3$

Relation

Development

The pressure ($p$), the volume ($V$), the absolute temperature ($T$), and the number of moles ($n$) are related through the following physical laws:

• Boyle's law

$ p V = C_{pV} $

• Charles's law

$\displaystyle\frac{ V }{ T } = C_{VT}$

• Gay-Lussac's law

$\displaystyle\frac{ p }{ T } = C_{pT}$

• Avogadro's law

$\displaystyle\frac{ n }{ V } = C_{nV} $

These laws can be expressed in a more general form as:

$\displaystyle\frac{pV}{nT}=cte$

This general relationship states that the product of pressure and volume divided by the number of moles and temperature remains constant:

$ p V = n R T $

In this equation, the universal gas constant ($R$) assumes the value 8.314 J/K·mol.

ID:(3183, 0)

Equation

>Top, >Model

The number of moles ($n$) is determined by dividing the volume ($V$) of a substance by its the molar volume ($V_m$), which corresponds to the weight of one mole of the substance.

Therefore, the following relationship can be established:

$ n =\displaystyle\frac{ V }{ V_m }$

Conditions

Variables

Relation

The molar volume is expressed in cubic meters per mole ($m^3/mol$).

It's important to note that the molar volume depends on the pressure and temperature conditions under which the substance exists, especially in the case of a gas, so it is defined considering specific conditions.

ID:(15147, 0)

Equation

>Top, >Model

In general, the differential of the Gibbs free energy ($dG$) is equal to the entropy ($S$), the temperature variation ($dT$), the volume ($V$), and the pressure Variation ($dp$), which is expressed as follows:

During a phase change, it is necessary to consider the difference of this expression between the two phases. However, in this case, the absolute temperature ($T$) does not vary, so the term $SdT$ does not appear. On the other hand, when considering the variation of the terms the volume ($V$) and the pressure Variation ($dp$), it encompasses the volume variation in phase change ($\Delta V$). Finally, it is necessary to include the term the surface Tension ($\sigma$) and the surface variation ($dS$):

$ dG = \sigma dS + \Delta V dp $

Conditions

Variables

Relation

ID:(15146, 0)

Equation

>Top, >Model

In general, the differential of the Gibbs free energy ($dG$) is equal to the surface Tension ($\sigma$), the surface variation ($dS$), the volume ($V$), and the pressure Variation ($dp$), which is expressed as follows:

However, in the case of the water meniscus, the absolute temperature ($T$) does not vary, but the surface does change as more molecules condense within the volume ($V$) of water. Therefore, by integrating from the absence of water molecules to reach the equilibrium state, the gibbs free energy difference ($\Delta G$) is obtained in terms of the section ($S$), the water vapor pressure unsaturated ($p_v$), the pressure saturated water vapor ($p_s$), the molar volume ($V_m$), and the universal gas constant ($R$):

$ \Delta G = \sigma S - \displaystyle\frac{ V }{ V_m } R T \ln\left(\displaystyle\frac{ p_s }{ p_v }\right) $

Conditions

Variables

Relation

Development

The differential of the Gibbs free energy ($dG$) with the surface Tension ($\sigma$), the surface variation ($dS$), the volume ($V$), and the pressure Variation ($dp$) can be expressed as:

$ dG = \sigma dS + \Delta V dp $

Integrating the differential of the Gibbs free energy ($dG$) from the state where there are no water molecules in the meniscus to the state where they reach saturated pressure requires two trivial integrals that yield the gibbs free energy difference ($\Delta G$):

$\displaystyle\int dG = \Delta G$

and the integration of the entire the thermal conductivity ($\lambda$):

$\displaystyle\int \sigma dS = \sigma S$

To integrate $Vdp$, it is important to remember that the volume ($V$) results from the difference between the volume of the liquid $V_l$ and the gas $V_g$:

$\Delta V = V_l - V_g \sim - V$

since $V_l\ll V_g$.

Furthermore, if we consider the ideal gas equation with the universal gas constant ($R$)

$ p V = n R T $

the third integral becomes

$-nRT\displaystyle\int_{p_v}^{p_s}\displaystyle\frac{dp}{p}=-nRT\ln\left(\displaystyle\frac{p_s}{p_v}\right)$

With these considerations, the gibbs free energy difference ($\Delta G$) can be expressed as:

$ \Delta G = \sigma S - \displaystyle\frac{ V }{ V_m } R T \ln\left(\displaystyle\frac{ p_s }{ p_v }\right) $

ID:(15148, 0)

Equation

>Top, >Model

The relationship between the concentration of water vapor molecules ($c_v$) and saturated water vapor concentration ($c_s$) is called the relative humidity ($RH$):

and it can be expressed in terms of the water vapor pressure unsaturated ($p_v$) and the pressure saturated water vapor ($p_s$) as follows:

$ RH =\displaystyle\frac{ p_v }{ p_s }$

Conditions

Variables

$p_s$

Pressure saturated water vapor

$Pa$

$RH$

Relative humidity

$-$

$p_v$

Water vapor pressure unsaturated

$Pa$

Relation

Development

The relationship between the relative humidity ($RH$) with the concentration of water vapor molecules ($c_v$) and saturated water vapor concentration ($c_s$) is expressed as:

$ RH =\displaystyle\frac{ c_v }{ c_s }$

and by relating the pressure ($p$) with the molar concentration ($c_m$), the absolute temperature ($T$), and the universal gas constant ($R$), we obtain:

$ p = c_m R T $

This applies to the vapor pressure of water, where:

$p_v = c_v R T$

and the saturated vapor pressure of water:

$p_s = c_s R T$

resulting in the following equation:

$ RH =\displaystyle\frac{ p_v }{ p_s }$

ID:(4478, 0)

Equation

>Top, >Model

The gibbs free energy difference ($\Delta G$) with the surface Tension ($\sigma$), the section ($S$), the water vapor pressure unsaturated ($p_v$), the pressure saturated water vapor ($p_s$), the volume ($V$), the molar volume ($V_m$), the absolute temperature ($T$), and the universal gas constant ($R$) equals:

To calculate the gibbs free energy difference ($\Delta G$), it is necessary to calculate the surface Tension ($\sigma$) and the volume ($V$) for the radio mensic the meniscus radius ($r_m$), which is formed between the assumed values of the radius of a generic grain ($r_0$):

$ \Delta G = 4 \pi \sqrt{2 r_0 r_m ^3} \sigma - 2\pi r_m ^2 r_0 \displaystyle\frac{ R T }{ V_m }\ln\left(\displaystyle\frac{1}{ RH }\right) $

Conditions

Variables

Relation

Development

The sum of the gibbs free energy difference ($\Delta G$) with the surface Tension ($\sigma$), the section ($S$), the water vapor pressure unsaturated ($p_v$), the pressure saturated water vapor ($p_s$), the volume ($V$), the molar volume ($V_m$), the absolute temperature ($T$), and the universal gas constant ($R$) equals:

$ \Delta G = \sigma S - \displaystyle\frac{ V }{ V_m } R T \ln\left(\displaystyle\frac{ p_s }{ p_v }\right) $

The integral of the section ($S$) for two grains of the radius of a generic grain ($r_0$) with a water meniscus of the meniscus radius ($r_m$) in the limit $r_m\ll r_0$ is given by:

$S=4\pi \displaystyle\int_0^{r_0r_m/(r_0+r_m)}du[\sqrt{r_0^2+2r_mr_0}-\sqrt{r_m^2-u^2}]\sim 4\pi \sqrt{2r_0r_m^3}$

And for the volume ($V$),

$V=4\pi\displaystyle\int_0^{r_0r_m/(r_0+r_m)}du[r_m^2+r_0(r_m-u)-\sqrt{(r_m(r_m+2r_0)(r_m^2-u^2)}]\sim 2\pi r_0 r_m^2$

So, with the definition of the relative humidity ($RH$),

$ RH =\displaystyle\frac{ p_v }{ p_s }$

we obtain:

$ \Delta G = 4 \pi \sqrt{2 r_0 r_m ^3} \sigma - 2\pi r_m ^2 r_0 \displaystyle\frac{ R T }{ V_m }\ln\left(\displaystyle\frac{1}{ RH }\right) $

ID:(15149, 0)

Equation

>Top, >Model

The sum of the gibbs free energy difference ($\Delta G$) with the surface Tension ($\sigma$), the meniscus radius ($r_m$), the radius of a generic grain ($r_0$), the universal gas constant ($R$), the absolute temperature ($T$), and the relative humidity ($RH$) equals:

the meniscus radius ($r_m$) must be chosen such that the gibbs free energy difference ($\Delta G$) is minimized. To achieve this, we can differentiate the gibbs free energy difference ($\Delta G$) with respect to the meniscus radius ($r_m$), set it to zero, and solve for the meniscus radius ($r_m$). This results in:

$ r_m = \displaystyle\frac{3^2}{2^2}\displaystyle\frac{ \sigma ^2 V_m ^2}{ R ^2 T ^2 r_0} \displaystyle\frac{1}{\ln^2\left(\displaystyle\frac{1}{ RH }\right)}$

Conditions

Variables

Relation

Development

To minimize the gibbs free energy difference ($\Delta G$) with the surface Tension ($\sigma$), the meniscus radius ($r_m$), the radius of a generic grain ($r_0$), the universal gas constant ($R$), the absolute temperature ($T$), and the relative humidity ($RH$), we define the meniscus radius ($r_m$) as follows:

$ \Delta G = 4 \pi \sqrt{2 r_0 r_m ^3} \sigma - 2\pi r_m ^2 r_0 \displaystyle\frac{ R T }{ V_m }\ln\left(\displaystyle\frac{1}{ RH }\right) $

We can differentiate the gibbs free energy difference ($\Delta G$) with respect to the meniscus radius ($r_m$) and set the derivative to zero to find the value of the meniscus radius ($r_m$) that minimizes the gibbs free energy difference ($\Delta G$). This leads us to the following equation:

$\displaystyle\frac{\partial\Delta G}{\partial r_m} = 6 \pi \sqrt{2 r_0 r_m } \sigma - 4 \pi r_m r_0 \displaystyle\frac{ R T }{ V_m }\ln\left(\displaystyle\frac{1}{ RH }\right)$

Solving for the meniscus radius ($r_m$), we obtain:

$ r_m = \displaystyle\frac{3^2}{2^2}\displaystyle\frac{ \sigma ^2 V_m ^2}{ R ^2 T ^2 r_0} \displaystyle\frac{1}{\ln^2\left(\displaystyle\frac{1}{ RH }\right)}$

ID:(15150, 0)