Received 9 March 2016; accepted 3 May 2016; published 6 May 2016

1. Introduction
We derived the heat capacity equations of metals and then used consistent step multiplication of the appropriate binomial equations [1] . They are fitted to experimental data well [2] . The heat capacity equation (type V) and the adsorption equation (type II) draw sigmoid (S character) lines all together. And they are symmetrical with each other. The measurement gases of heat capacity are hydrogen and helium. The adsorption gases are vapor and nitrogen. The movements of their measurement gases are different. The formers are expansion and the later contraction.
The most important term in the derivation of heat capacity equation was the temperature increasing term,
. In case of adsorption it becomes
. Here
and
are the total molecules adsorbed at the layers of
and
. Let us put
which is unit in data fitting as the constant. The constant affects the adsorption equation from starting to ending like other constants. If
is transformed into
power of
in every derivation of adsorption such as
[3] , the more advanced adsorption equation than BET eq. comes out. Hence we get the surface area.
2. Statistical Modeling of Adsorption Isotherm
Suppose each layer has one binomial equation. And suppose
molecules are adsorbed on
localized sites of the unit surface layers of the adsorbent and
molecules on
sites made of 1st to
layers in sequence. Here m is quantization constant. The first adsorption layer has
adsorption energy and. the second to
layers,
. Hence the adsorption probability on the first layer is
and the non-adsorption probability
. Here
is the adsorption constant of the first adsorption layer.
is similar to
in heat capacity equation. The adsorption constant,
is calculable quantity. It is the expression to combine the rotation and vibration energy of the adsorbed molecules in the first layer.
is similar to
(Boltzmann’s constant) [2] . In the combination calculation of the first layer
can take from 0 to
as variables in sequence. Then the binomial equation of the first layer becomes [1]
(1)
Next the adsorption probability of from
to
layers is
and the non- adsorption probability
. The binomial equations for from
to
layers are
(2)
Let us multiply Equation (1) and Equation (2) side by side. Then for
![]()
(3)
In the above
(3.1)
In Equation (3) the largest term dominates the equation. So the total differential of Equation (3) becomes the zero which requires that the coefficients of all terms should be zero. Hence by using
Stirling
’s approximation we solve the equation,
. The first equation (Equation (1)) becomes
(4)
In [4]
value should be corrected as those in the present figure (Figure 1). The constants have three parts of
. As we see in Figure 1, bonding constants mean quantization. If
, bonding occurs in many directions. If
no quantization occur and if
, the adsorptions are interrupted in many directions. In Equation (4)
(4.1)
for
(4.2)
It is possible that
is put as unit.
![]()
Figure 1. Bonding constant and non-bonding constant in statistic quantization.
The next equations are
(5)
In Equations (4) and (5) let us put
, then
which is same as Equation (9) of [5] solved by using the chemical potential
(6)
In Equation (6) add side by side and rearrange
(7)
In Equation (6) multiple side by side and rearrange
(8)
We solve Equation (8) with
in order to eliminate ![]()
(9)
We solve Equation (7) with Equation (9) in order to eliminate ![]()
(10)
Equation (10) represents the adsorption amount of the first layer. It is Langmuir equation. The quantization values (
) and the numbers of the adsorption layer (
) are influential on the determination of
. All values of the parameters (
,
,
,
) directly participate in the determination of
and following
. The
value of
is much influential on the determination of the last term (
). But it does not much increase or decrease of
since it exists in the denominator and nominator. At
it becomes
(11)
Equation (11) is the same as Equation (10) obtained by using chemical potential in [1] . Therefore the total adsorption amount per unit surface (
), that is, the adsorption isotherm for from first layer to the last (
) layer becomes by using Equations (6) and (10) as follows
(12)
In the above
(12.1)
And the total adsorption rate is a linear function of
made of four constants (
,
,
,
). The equation draws BET-like lines and fits BET type experimental data well. At
it becomes
(13)
In the above
(13.1)
If the measurement gas is nitrogen, the general empirical formulae,
is introduced into.
to get the specific surface area of the adsorbate,
[6] . Then
is the mole-
cular weight of nitrogen and
density of nitrogen and
Avogadro number, Hence from empirical formulae
=Å2/number is used. The monolayer capacity,
should also be substituted by the value of the integration of Equation (12) subtracted by the surface adsorption isotherm (
).
3. Result and Discussion
The base of Equations ((10) and (12)) is
. So we may use the word, the rate without considering dimension. It affects the equation totally. Figure 2 shows the total adsorption rate according to the values of
. In accordance with the values of
approaching units, the total adsorption rates approach closely with one another. This seems to mean that the adsorption heat of the first layer is same as those of
layers. We call
the quantization values. It seems to have same notion as the quantization appears in quantum mechanism. We are dealing statistical quantization which should exist in statistics. We can discern them, three cases.
and
. The cases of
and
explain bonding and
explains the existence of the interruptions for bonding. Figure 3 explains the increase of the adsorption rate according to
values.
![]()
Figure 2. Theoretical adsorption isotherm curve, Equation (12) (m = 0.7, n = 4, ga = 0.5) with respect to βa values.
![]()
Figure 3. Theoretical adsorption isotherm curve, Equation (12) (n = 4, βa = 0.00901, ga = 1.0) with respect to m values.
Less
than unit seem to interrupt the bonding. Figure 4 represents the total adsorption rates according to
values. At less
than 0.5 the isotherms show the same adsorption rates. The constant
showed in Figure 5 should be positive. When
and
, the constants draw the type II isotherm, but at
the different type (type IV) of the isotherm appear. Figure 6 shows
with respect to the relative pressure like Langmuir’s lines which can’t become unit even if
is very large or
much smaller. The parameter values of Figure 6 come from Figure 7 and Figure 8. According to the above variations we optimized two kinds of the experimental adsorption isotherm data showed in Figure 7 [6] and Figure 8 [7] using trial and error method. The experimental data of Figure 7 are obtained from the Figures 2-10 of [6] . The
![]()
Figure 4. Theoretical adsorption isotherm curves, Equation (12) (βa = 0.009, ga = 0.5, m = 0.7) with respect to n values.
![]()
Figure 5. Theoretical adsorption isotherm curve, Equation (12) (m = 1.7, n = 8, βa = 0.009) with respect to ga values.
![]()
Figure 6. Theoretical adsorption isotherm of the first layer using Equation (10) including experimental constants of Figure 7 and Figure 8.
![]()
Figure 7. Theoretical adsorption isotherm, Equation (12) (m = 0.9, n = 3.28, βa = 0.000011, ga = 1.0) with BET equation (c = 150) and experimental nitrogen adsorption at −196˚ on non-porous samples of silica and aluminna [6] .
experimental data are fitted to Equation (12) well. As we see in Figure 7 and Figure 8, BET isotherms there can’t imitate the experimental data except for beginning.
Equation (13) can be used in the data fitting with
and
. Its quality is poor.
What is the catalyst? As we see Figure 9, 1, 2, 3, 4, 5, 6 and 7 molecules can function as the catalyst. That is, the molecules of the surface adsorption layer can’t function like the catalyst. Because they use much
![]()
Figure 8. Theoretical adsorption isotherm curve, Equation (12) (m = 2.4, n = 4.5, βa = 0.12, ga = 1.0) with BET equation (c = 130) curve and experimental water adsorption at 25˚ on cross-linked polystyrene sulfuric acid resin [7] .
![]()
Figure 9. Real adsorption molecules to get the surface area (numbered rings).
energy in order to hold the surface. So they are not active. The molecules which lie on the surface layer adsorption molecules which hold the surface, can function as the real catalyst. Therefore in Equation (12)
part must work in the adsorption reaction. Hence the real adsorption isotherm equation becomes
(14)
If this equation is integrated from the inflection point (
) to each relative pressure (
), we get the mono-area capacity (
) with respect to the relative pressure (
). Table 1 and Table 2 include those. Then
![]()
Table 1.
(mono-area capacity) and
(surface area) for Figure 7.
![]()
Table 2.
(mono-area capacity) and
(surface area) for Figure 8.
the surface area for the catalyst obtained from the equation becomes
with
per number for nitrogen and
per number for water (Table 2.12) [6] by subs-
tituting
. in the equation of (2.60) of the reference [6] by
in which
= inflection point
and
=optional
value after the inflection point. In the above equation
is the molecular weight (g) per g-mole and
Avogadro number per g-mole.
Then the inflection points are obtained by Secant method [8] through the program showed in Appendix 1. Specific surface areas are changed according to the relative pressures. These are showed in Table 1 and Table 2 precisely. The integrations with respect to z values give the total adsorption site numbers of the adsorbate. Before the inflection point the specific surface area of the adsorbent is not counted as a catalyst since it makes the strong surface film [9] . The adsorption rate increases consistently after the inflection point. The values of
of Figure 7 and Figure 8 match the range of the reference of BET [10] . The completion of
goes with the completion of
to the end as we see in Figure 6 and its equation. But from the inflection point the plugging of
may begin characteristically.
We have felt intimately that the adsorption molecules of more than 2nd layers lie down the first layer molecules softly since the small quantity of adsorption molecules control the drawing and the surface area of numerical number.
Our study have realized the saying that “After considerable work on the theory, Hill (1946) formed the opinion that any future improvement on it must be in the form of refinement rather than a modification on the basic theory” [11] [12] .
4. Conclusion
Acknowledgements
The author thanks for the encouragements of Yongduk Kim, an emeritus professor of Sogang University.
Appendix 1
/*--Finding zero by Secant method [8] to get inflection point from Eq. (12). with c++ and c languages */
#include
using namespace std;
#include
#include
#include
float eval_f(float x); //evaluation of f(x)
float fprime(float x1,float x2); //diff. of f(a)
void main()
{ float x1,x2,xn,e;
int i=0;
printf(“\nType initial point: “);
scanf(“%f”, &x1);
printf(“\nType second point: “);
scanf(“%f”,&x2);
printf(“\nType acceptable error interval in y: “);
scanf(“%f”, &e);
i=0;
do
{ xn=x1-eva;_f(x1)/fprime(x1,x2);
x1=x2;
x2=xn;
printf(“\n%3dth Iteration Root :%f
\n”,
++i,xn,fabs(eval_f(xn)));
getch();
} while(fabs(eval_f(xn))> e);
}
float eval_f(float x)
{ float b;
//cout<<”x=”<
double bc;
double bb,bb1,bb2,cc,dd,dd1,dd2,dd3,ee,ee1,ee2,ff,ff1,ff2;
double aa,aa1,aa2,ba1,ba2,ba3,ccc1,ccc2,cc1,cc2,cc3,cc4,cc11,cc12;
const double b1=.000091,ga=.983,an=10.0,am=.64;
double ad21,ad22,ad23;
bb=pow(x,(1./am))-pow(x,(an/am));
bb1=(1./am)*pow(x,(1./am)-1.))-(an/am)*pow(x,(an/am-1.));
bb2=(1./am)*(1./am-1.)*pow(x,(1./am)-2.))-(an/am)*(an/am-1.)*pow(x,(an/am-2.));
ccc1=-(1./am)*pow(x,,1./am-1.);
ccc2=-(1./am)*(1./am-1.)*pow(x,1./am-2.);
cc1=pow((1.-pow(x,1./am)),-1.);
cc2=pow((1.-pow(x,1./am)),-2.);
cc3=pow((1.-pow(x,1./am)),-3.);
cc4=pow((1.-pow(x,1./am)),-4.);
dd=pow(x,(an/am))/ga;
dd1=(an/am)*pow(x,((an/am)-1.))/ga;
dd2=(an/am)*(an/am-1.)*pow(x,((an/am)-2.))/ga;
ee=pow(x,2./am)-pow(x,an/am);
ee1=(2./am)*pow(x,(2./am)-1.)-(an/am)*pow(x,(an/am)-1.);
ee2=((2./am)*(2./am-1.)*pow(x,(2./am)-2.)-(an/am)*(an/am-1.)*pow(x,(an/am-2.);
ff=((an-1.)/ga)*pow(x,an/am);
ff1=((an-1.)/ga)*(an/am)*pow(x,(an/am)-1.);
ff2=((an-1.)/ga)*(an/am)*(an/am-1.)*pow(x,(an/am)-2.);
aa=bb*cc1+dd;
aa1=bb1*cc1+(-1)*cc2*ccc1*bb+dd1
aa2=bb2*cc1+(-1)*cc2*ccc1*bb1+(-1)*(-2)*cc3*ccc1*ccc1*bb+(-1)*ccc2*cc2*bb+(-1)*bb1*cc2*ccc1+dd2;
ba1=pow((b1+aa),-1);
ba2=pow((b1+aa),-2);
ba3=pow((b1+aa),-3);
ad21=aa2*ba1+(-1)*ba2*aa1*aa1+(-1)*(-2)*ba3*aa1*aa1*aa+(-1)*aa2*ba2*aa+(-1)*aa1*aa1*ba2;
ad22=ee2*cc2*ba1+(-2)*cc3*ccc1*ee1*ba1+(-1)*ba2*aa1*ee*cc2+(-2)*(-3)*cc4*ccc1*ccc1*ee*ba1+(-2)*ccc2*cc2*ee*ba1+(-2)*ee1*cc3*ccc1*ba1+(-2)*(-1)*ba2*aa1*cc3*ccc1*ee+(-1)*(-2)*ba3*aa1*aa1*ee*cc2+(-1)*aa2*ba2*ee*cc2+(-1)*ee1*ba2*aa1*cc2+(-1)*(-2)*cc3*ccc1*ba2*aa1*ee1;
ad23=ff2*ba1+(-1)*ba2*aa1*ff1+(-1)*(-2)*ba3*aa1*aa1*ff+(-1)*aa2*ff*ba2+(-1)*ff1*ba2*aa1;
b=ad21+ad22+ad23;
return(b);
}
float fprime (float x1,float x2)
{ float b6;
b6=(eval_f(x2)-eval_f(x1))/(x2-x1);
return(b6);
}