Unformatted text preview: ysis: (a) From Eq. (152), based on the above assumption,
Sg = 4 εp/dpρg = 4(0.47)/[(1.09 x 106)(24 x 1010)] = 719 m2/g
This is reasonable compared to the given value of 800 m2/g.
The true density of silica gel, assuming it is SiO2, depends on its crystalline form. From a
handbook, the density is from 2.20 to 2.65 g/cm3, with 3 of 4 forms from 2.20 to 2.26 g/cm3.
Therefore, using a value of ρs = 2.20 g/cm3 with Eq. (155),εp = 1  ρp/ ρs = 1  1.09/2.20 = 0.505.
This is reasonable compared to the given value of 0.47.
(b) From the problem statement, it is not clear whether the 18 wt% refers to a dry basis or a wet
basis, so consider both possibilities.
Dry basis:
Adsorb 0.18 grams water per 1.0 gram of waterfree silica gel. Therefore, for 1.0
gram of dry silica gel, the surface area for adsorption = 800 m2.
From Eq. (158) and Example 15.1, the projected surface area per molecule =
α = 10.51 x 1016 cm2/molecule.
The number of water molecules adsorbed = 0.18(6.023 x 1023)/18.02 = 6.02 x 1021.
The number of molecules that could be adsorbed to cover 800 m2 = 800(104)/10.51 x 1016 =
7.61 x 1021.
Therefore, the fraction of the area covered = 6.02/7.61 = 0.79
Adsorb 0.18 grams water per 0.82 gram of waterfree silica gel (1.00 gram total).
Wet basis:
Therefore, for 0.82 gram of dry silica gel, the surface area for adsorption = (0.82)800 = 656 m2.
From Eq. (158) and Example 15.1, the projected surface area per molecule =
α = 10.51 x 1016 cm2/molecule.
The number of water molecules adsorbed = 0.18(6.023 x 1023)/18.02 = 6.02 x 1021.
The number of molecules that could be adsorbed to cover 656 m2 = 656(104)/10.51 x 1016 =
6.89 x 1021.
Therefore, the fraction of the area covered = 6.02/6.89 = 0.87 Exercise 15.4
Subject: Estimation of the specific surface area from BET data. Given: Data for adsorption equilibrium of nitrogen on silica gel (SG) at 195.8oC:
υ , Volume of N2 adsorbed,
P, N2 partial
pressure, torr cm3 (0oC, 1 atm) per gram SG
6.0
6.1
24.8
12.7
14.3
17.0
230.3
19.7
285.1
21.5
320.3
23.0
430
27.7
505
33.5
Find: Specific surface area in m2/g of SG. Compare to Table 15.2. P
1
(c − 1) P
=
+
υ P0 − P υ mc υ mc P0
where: P = total pressure = partial pressure of N2 in the above table in mmHg
P0 = vapor pressure of N2 at 195.8oC = 760 mmHg
υ = volume of gas adsorbed at STP
υ m = volume of monomolecular layer of gas adsorbed at STP
c = constant
Eq. (1) is of the form of a straight line, y = mx + b, where,
P
P
(c − 1)
1
y=
(3)
, x=
, m=
(2) and
b=
υ P0 − P
P0
υ mc
υ mc
If a leastsquares fit of the data is made, using a spreadsheet or POLYMATH,
m = 0.08557 and
b = 0.0016905
Combining Eqs. (2) and (3) to eliminate c,
υ m = 1/(m + b) = 1/(0.08557  0.0016905) = 11.92 cm3/g
c = 49.62
αυ m N A α(1192)(6.023 × 1023 )
.
From Eq. (157), S g =
=
= 3.205 × 1020 α
V
22,400
Analysis: From Eq. (156), M
From Eq. (158), α = 1.091
N Aρ L 2/3 28
= 1.091
6.023 × 1023 (0.808) (1) 2/3 = 1.626 × 10 −15 cm2 Therefore, Sg = 3.205 x 1020 (1.626 x 1015) = 5....
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.
 Spring '11
 Levicky
 The Land

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