4.4 Monolayers and multilayers

For open surfaces, adsorption consists of a layer-by-layer loading process, where the first layer is filled as in the case when <sup>θ</sup> <sup>¼</sup> na Nm = 1, where θ is the surface recovery and Nm is the monolayer capacity; as a result, it is understood that we have monolayer adsorption when θ < 1 and multiplayer adsorption when θ > 1 [31].

#### 4.5 Porous adsorbents

They are characterized by their specific surface area, denoted by S, and measured in (m2 /g), in which the surface area is the outer surface, concretely, the area external to the micropores. If the adsorbents do not have micropores, the surface area and the outer surface area match; furthermore, the micropore volume is represented by WMP, measured in (cm3 /g), whereas the total pore volume labeled W is the sum of the micropores and mesopore volumes [32] of the adsorbent, also measured in (cm3 /g). Lastly, we have the pore size distribution (PSD), i.e., a plot of ΔvP <sup>Δ</sup>Dp versus Dp, where Vp is the pore volume accumulated up to the pore width Dp measured in (cc-STP/g Å) [26, 27]. Now it is necessary to state that cc-STP is a unit denoting the amount adsorbed, measured in cubic centimeters at standard temperature and pressure (STP), that is, 273.15 K and 760 Torr, that is, 1.01325 � <sup>10</sup><sup>5</sup> Pa.

#### 4.6 Magnitude of adsorption

The interfacial layer is the non-homogeneous section of an adsorption system, i.e., between two neighboring bulk phases (Figure 2), where the properties typifying this region are radically dissimilar from, but associated to, the properties of the bulk phases; in which, to deal with this system is assumed that in the ideal reference system, the concentration must be constant up to the Gibbs dividing surface (GDS) (see Figure 2); been, in an actual system [32]

Figure 2. Gibbs dividing surface.

$$n^a = \int\_0^{V^a} c dV = \int\_0^t c dz \tag{2}$$

$$V^\infty = At$$

In Figure 11, a schematic representation of volumetric gas adsorption measure-

<sup>2</sup> X, <sup>i</sup> þ l0

Initially Vm is filled with air and then the valves are opened allowing air up to equilibrium with the atmospheric pressure of the room, which can be measured with a barometer; now given that Vm was calibrated, filling it with water, then Xi

Pi½ �¼ mmHg 2Xi½ � mmHg Pat ¼ 2lat½ � mmHg where Pat is the atmospheric pressure and P1 is the measurement of pressure

Vm � Vm þ πr

760 mmHg ½ � <sup>¼</sup> Pi½ � Atm

Xi mm<sup>3</sup> x103 <sup>þ</sup> Vd½ � ml <sup>¼</sup> ni½ � mmol <sup>R</sup> atm:ml

T K½ � <sup>¼</sup> ni½ � mmol

2 Xi 

(6)

mmol:K T K½ �

<sup>a</sup> and the numbers of mole gas

T K½ � <sup>¼</sup> ni

Xi mm<sup>3</sup> x10<sup>3</sup> <sup>þ</sup> Vd½ � ml <sup>¼</sup> Va

R atm:ml mmol:K

2

ment equipment is shown, where using the scheme, it is possible to show that:

Vx ¼ πr

Synthesis, Characterization, and Adsorption Properties of Nanoporous Materials

DOI: http://dx.doi.org/10.5772/intechopen.83355

X, <sup>i</sup> þ l0 <sup>¼</sup> Xi

Substituting Eq. (6) on (5), the dose volume is calculated as:

2Xi½ � mmHg x1 Atm ½ �

Pi½ � Atm <sup>2</sup>πr2Xi mm<sup>3</sup> ½ �x10<sup>3</sup> <sup>þ</sup> Vd½ � ml R atm:ml mmol:K

Number of moles in the dose volume and the volume provided by the

When the valve of the sample holder is opened, the gas is expanded; hence:

T K½ � <sup>¼</sup> Pj½ � Atm <sup>2</sup>πr2Xj mm<sup>3</sup> ½ �x10<sup>3</sup> <sup>þ</sup> Vd½ �þ ml Va

Vd <sup>¼</sup> lat Xi

In this sense, using the proportions rule, we get:

Consequently applying the ideal gas law:

5. Experimental determination of Vaand V<sup>0</sup>

Xj mm3 x10<sup>3</sup> <sup>þ</sup> Vd½ �þ ml Va � ½ � Atm <sup>2</sup>π<sup>r</sup>

2

Pi½ � Atm 2πr

displacement of the manometer.

Pi½ � Atm <sup>2</sup>πr2Xi mm<sup>3</sup> ½ �x103 <sup>þ</sup> Vd½ � ml R atm:ml mmol:K

adsorbed naj

where:

½ � Atm 2πr

2

Pj Pi

75

where

allow us to calculate:

after expansion.

Then the total amount of gas molecules in the system measured in (mol/g) is

$$
\mathfrak{n} = \mathfrak{n}^a + V^{\mathfrak{g}} c\_{\mathfrak{g}}^{\mathfrak{g}}
$$
 
$$
\mathfrak{n}^a = \mathfrak{n} - V^{\mathfrak{g}} c\_{\mathfrak{g}}^{\mathfrak{g}}
$$

where

V<sup>∝</sup> is the volume of the adsorption space, V<sup>g</sup> is the volume of the gas phase, c g <sup>g</sup> is the concentration of the gas phase, c g <sup>s</sup> is the concentration of the gas in the solid phase, A is the adsorbent surface area, t is the thickness of the surface layer, and n is the total amount of gas molecules in the system.

Finally the isotherm is calculated as follows:

$$m\_a = \frac{\mathfrak{n}^a}{m\_s} = F(P)\_T$$

where ms is the degassed adsorbent mass in grams. Then if the ideal gas equation is applied,

$$\text{PV} = \text{nRT} \tag{3}$$

Been, n and T, kept constants, then [29]:

$$\mathbf{P\_iV\_i = P\_jV\_j = nRT = const} \tag{4}$$

where P is the pressure in atmosphere, V is the volume in milliliters; 1 ml <sup>¼</sup> 1 cm3 <sup>¼</sup> <sup>10</sup>3mm3 <sup>¼</sup> <sup>10</sup>�3dm3 <sup>¼</sup> <sup>10</sup>�3l, n is the number of gas moles, T is the temperature in kelvin (K), and R is the gas constant.

Thereafter, using Eq. (4), based in the scheme of the volumetric adsorption equipment shown in Figure 11, we get:

$$\mathbf{P}\_{at}\mathbf{V}\_{\mathbf{m}} = \mathbf{P}\_{\mathbf{i}}(\mathbf{V}\_{\mathbf{m}} + \mathbf{V}\_{\mathbf{d}} + \mathbf{V}\_{\mathbf{x}})\tag{5}$$

$$\mathbf{P}\_{at} = \mathbf{2}l\_{at}$$

Synthesis, Characterization, and Adsorption Properties of Nanoporous Materials DOI: http://dx.doi.org/10.5772/intechopen.83355

In Figure 11, a schematic representation of volumetric gas adsorption measurement equipment is shown, where using the scheme, it is possible to show that:

$$\mathbf{V\_x = \pi r^2 (X\_i^\circ + l\_0)}\tag{6}$$

$$\left(\mathbf{X\_i^\circ + l\_0}\right) = \mathbf{X\_i}$$

Initially Vm is filled with air and then the valves are opened allowing air up to equilibrium with the atmospheric pressure of the room, which can be measured with a barometer; now given that Vm was calibrated, filling it with water, then Xi allow us to calculate:

$$\mathbf{P\_i[mmHg]} = 2\mathbf{X\_i[mmHg]}$$

$$\mathbf{P\_{at} = 2l\_{at}[mmHg]}$$

where Pat is the atmospheric pressure and P1 is the measurement of pressure after expansion.

Substituting Eq. (6) on (5), the dose volume is calculated as:

$$\mathbf{V\_{d}} = \frac{\mathbf{l\_{at}}}{\mathbf{X\_{i}}} \mathbf{V\_{m}} - \left(\mathbf{V\_{m}} + \pi \mathbf{r^{2}} \mathbf{X\_{i}}\right)^{2}$$

In this sense, using the proportions rule, we get:

$$\frac{2\mathbf{X}\_i[\mathbf{mmHg}]\mathbf{x1}[\mathbf{Atm}]}{760[\mathbf{mmHg}]} = \mathbf{P}\_i[\mathbf{Atm}]$$

Consequently applying the ideal gas law:

$$\mathbf{P\_i[Attm]} \left( 2\pi r^2 \mathbf{X\_i[mm^3]} \mathbf{x10^3} + \mathbf{V\_d[ml]} \right) = \mathbf{n\_i[mmol]} \mathbf{R} \left[ \frac{\text{atm.ml}}{\text{mmol.K}} \right] \mathbf{T[K]}$$

where

<sup>n</sup><sup>a</sup> <sup>¼</sup>

where

Figure 2.

Gibbs dividing surface.

Applied Surface Science

is applied,

74

the concentration of the gas phase, c

the total amount of gas molecules in the system. Finally the isotherm is calculated as follows:

Been, n and T, kept constants, then [29]:

temperature in kelvin (K), and R is the gas constant.

equipment shown in Figure 11, we get:

ð<sup>V</sup><sup>a</sup> 0

cdV ¼

<sup>V</sup><sup>∝</sup> <sup>¼</sup> At Then the total amount of gas molecules in the system measured in (mol/g) is

<sup>n</sup> <sup>¼</sup> <sup>n</sup><sup>a</sup> <sup>þ</sup> <sup>V</sup><sup>g</sup>

na <sup>¼</sup> <sup>n</sup> � <sup>V</sup><sup>g</sup>

V<sup>∝</sup> is the volume of the adsorption space, V<sup>g</sup> is the volume of the gas phase, c

phase, A is the adsorbent surface area, t is the thickness of the surface layer, and n is

¼ F Pð Þ<sup>T</sup>

where ms is the degassed adsorbent mass in grams. Then if the ideal gas equation

g

na <sup>¼</sup> <sup>n</sup><sup>a</sup> ms

where P is the pressure in atmosphere, V is the volume in milliliters;

1 ml <sup>¼</sup> 1 cm3 <sup>¼</sup> <sup>10</sup>3mm3 <sup>¼</sup> <sup>10</sup>�3dm3 <sup>¼</sup> <sup>10</sup>�3l, n is the number of gas moles, T is the

Thereafter, using Eq. (4), based in the scheme of the volumetric adsorption

Pat ¼ 2lat

ðt 0

c g g

c g g

<sup>s</sup> is the concentration of the gas in the solid

PV ¼ nRT (3)

PiVi ¼ PjVj ¼ nRT ¼ const (4)

PatVm ¼ Pið Þ Vm þ Vd þ Vx (5)

cdz (2)

g <sup>g</sup> is

$$\frac{\mathbf{P}\_{\mathrm{i}}[\mathrm{Atm}]\left(2\pi\mathrm{r}^{2}\mathbf{X}\_{\mathrm{i}}[\mathrm{mm}\mathrm{n}^{3}]\mathbf{x}\mathbf{10}^{3} + \mathbf{V}\_{\mathrm{d}}[\mathrm{ml}]\right)}{\mathbf{R}\left[\frac{\mathrm{atm}\mathrm{,}\mathrm{ml}}{\mathrm{mmol}\,\mathrm{K}}\right]\mathbf{T}[\mathrm{K}]} = \mathbf{n}\_{\mathrm{i}}[\mathrm{mmol}]$$

Number of moles in the dose volume and the volume provided by the displacement of the manometer.

#### 5. Experimental determination of Vaand V<sup>0</sup> <sup>a</sup> and the numbers of mole gas adsorbed naj

When the valve of the sample holder is opened, the gas is expanded; hence:

$$\frac{\mathbf{P}\_{\mathrm{i}}[\mathrm{Atm}](2\pi^{2}\mathbf{X}\_{\mathrm{i}}[\mathrm{mm}\mathrm{m}^{3}]\mathbf{x}10^{3} + \mathbf{V}\_{\mathrm{d}}[\mathrm{m}\mathrm{l}])}{\mathbf{R}\_{\mathrm{i}\mathrm{m}\mathrm{mol}\mathrm{K}}\mathrm{I}\mathrm{T}[\mathrm{K}]} = \frac{\mathbf{P}\_{\mathrm{i}}[\mathrm{Atm}](2\pi^{2}\mathbf{X}\_{\mathrm{i}}[\mathrm{mm}\mathrm{s}^{3}]\mathbf{x}10^{3} + \mathbf{V}\_{\mathrm{d}}[\mathrm{m}\mathrm{l}] + \mathbf{V}\_{\mathrm{a}})}{\mathbf{R}\_{\mathrm{i}\mathrm{m}\mathrm{mol}\mathrm{K}}\mathrm{I}\mathrm{T}[\mathrm{K}]} = \mathbf{n}\_{\mathrm{i}}$$

where:

$$\frac{P\_j}{P\_i} \left[ \text{Atm} \right] \left( 2\pi \text{r}^2 \text{X}\_{\text{j}} \left[ \text{mm}^3 \right] \text{x} \text{10}^3 + \text{V}\_{\text{d}} \left[ \text{ml} \right] + \text{V}\_{\text{a}} \right) - \left[ \text{Atm} \right] \left( 2\pi \text{r}^2 \text{X}\_{\text{i}} \left[ \text{mm}^3 \right] \text{x} \text{10}^3 + \text{V}\_{\text{d}} \left[ \text{ml} \right] \right) = \text{V}\_{\text{a}}$$

This allows the calculation of Va, the so-called dead volume. Nevertheless the really expanded volume should consider the volume occupied by the adsorbent sample, i.e.:

μ<sup>g</sup> ¼ μ<sup>L</sup> þ ε<sup>i</sup> ¼ μ<sup>a</sup> (7)

Pi

(8)

(9)

(10)

(11)

(12)

In which, μ<sup>g</sup> is the chemical potential of the gas phase adsorbate, μ<sup>L</sup> is the chemical potential of the pure liquid adsorbate, μ<sup>a</sup> is the adsorbed phase chemical potential, while ε<sup>i</sup> is the potential energy of the adsorption field. Consequently,

Synthesis, Characterization, and Adsorption Properties of Nanoporous Materials

<sup>ε</sup><sup>i</sup> <sup>¼</sup> RT ln <sup>P</sup><sup>0</sup>

where Po is the vapor pressure of the adsorptive at the temperature,T is the adsorption experiment, while Pi is the equilibrium adsorption pressure (ε could be also designed as the differential work of adsorption). Now, following the so-called Gurvich rule, it is possible to obtain the subsequent relation Vi <sup>¼</sup> <sup>V</sup>Lna between the volume of the adsorption space, Vi, and the amount adsorbed, where VL is the molar volume of the liquid phase that conforms to the adsorbed phase. Combining Eq. (8) with the characteristic function, the relation between the volume of the adsorption

F Vð Þ¼<sup>i</sup> f nð Þ¼ <sup>a</sup> <sup>ε</sup><sup>i</sup> <sup>¼</sup> RT ln <sup>P</sup><sup>0</sup>

Now, applying the Weibull distribution function, the relation between the amount adsorbed, na, and the differential work of adsorption, ε, is defined by the

na <sup>¼</sup> Na exp � <sup>ε</sup>

while, Na is the maximum amount adsorbed in the volume of the micropore, n (1 < n < 5) being an empirical parameter. Now combining Eqs. (9) and (10) is feasible to construe the Dubinin adsorption isotherm equation [32] as follows:

na <sup>¼</sup> Na exp � RT

ln ð Þ¼ na ln ð Þ� Na

been evidently the experimental data correctly fitted by Eq. (12)

y ¼ ln ð Þ¼ na ln ð Þ� Na

where E is a parameter termed the characteristic energy of adsorption; mean-

It is possible, as well, to express the Dubinin adsorption isotherm equation in

which is a very powerful tool for the description of the experimental data of

In Figure 4 is shown the Dubinin plot of N2 adsorption data at 77 K in the pressure range: 0.001 < P/Po < 0.03, in a high silica commercial H-Y zeolite, precisely, the acid Y zeolite labeled CBV-720, manufactured by PQ corporation; where, adsorption data was gathered in an Autosorb-1 automatic volumetric gas adsorption system [26];

Recapitulating, the concrete form to made the linear Dubinin plot as represented

RT E <sup>n</sup> ln <sup>P</sup><sup>0</sup> P <sup>n</sup>

¼ b � mx

E <sup>n</sup>

<sup>E</sup> ln <sup>P</sup><sup>0</sup> P

RT E <sup>n</sup> ln <sup>P</sup><sup>0</sup> P <sup>n</sup>

<sup>n</sup>

Pi

applying of Eq. (7), it is possible to demonstrate that:

DOI: http://dx.doi.org/10.5772/intechopen.83355

space, and the amount adsorbed, we will get [32]:

following relation [33]:

linear form:

77

adsorption in microporous material.

in Figure 4 was as follows:

$$\mathbf{V}\_{\text{muestra}}\left[\text{cm}^3\right] = \frac{\mathbf{m}|\mathbf{g}|}{\rho\left[\frac{\mathbf{g}}{\text{cm}^3}\right]}$$

And ρ is the apparent density of the tested adsorbent material. Consequently:

$$\mathbf{V\_a - V\_{muestra} = V\_a'}$$

According ideal gas low:

$$\begin{aligned} \mathbf{P}\_{\mathrm{i}}[\mathrm{Atm}] \left(2\pi \mathbf{r}^{2} \mathbf{X}\_{\mathrm{i}}[\mathrm{mm}\mathbf{3}^{3}] \mathbf{x} \mathbf{10}^{3} + \mathbf{V}\_{\mathrm{d}}[\mathrm{ml}]\right) + \mathbf{P}\_{\mathrm{j}-1} \mathbf{V}\_{\mathrm{a}}' \\ \mathrm{R} \left[\frac{\mathrm{atm}, \mathrm{ml}}{\mathrm{mmol}.\mathrm{K}}\right] \mathbf{T}[\mathrm{K}] \\ -\frac{\mathbf{P}\_{\mathrm{j}}[\mathrm{Atm}] \left(2\pi \mathbf{r}^{2} \mathbf{X}\_{\mathrm{j}}[\mathrm{mm}\mathbf{3}^{3}] \mathbf{x} \mathbf{10}^{3} + \mathbf{V}\_{\mathrm{d}}[\mathrm{ml}] + \mathbf{V}\_{\mathrm{a}}'\right)}{\mathrm{R} \left[\frac{\mathrm{atm}, \mathrm{ml}}{\mathrm{mmol}.\mathrm{K}}\right] \mathbf{T}[\mathrm{K}]} = \mathbf{n}\_{\mathrm{i}} - \mathbf{n}\_{\mathrm{j}} \end{aligned}$$

where:

$$\mathbf{n}\_{\mathbf{i}} - \mathbf{n}\_{\mathbf{j}} = \mathbf{n}\_{\mathbf{a}\mathbf{j}}$$

which is the amount adsorbed in the tested sample, which plotted against the equilibrium pressure Pj provides the adsorption isotherm.
