Separation Process Principles- 2n - Seader & Henley - Solutions Manual

000886023 x 1023319 x 10 15 169 x 106 cm2g or 169 m2g

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Unformatted text preview: ysis: (a) From Eq. (15-2), based on the above assumption, Sg = 4 εp/dpρg = 4(0.47)/[(1.09 x 106)(24 x 10-10)] = 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. (15-5),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 water-free silica gel. Therefore, for 1.0 gram of dry silica gel, the surface area for adsorption = 800 m2. From Eq. (15-8) and Example 15.1, the projected surface area per molecule = α = 10.51 x 10-16 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 10-16 = 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 water-free 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. (15-8) and Example 15.1, the projected surface area per molecule = α = 10.51 x 10-16 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 10-16 = 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 least-squares 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. (15-7), S g = = = 3.205 × 1020 α V 22,400 Analysis: From Eq. (15-6), M From Eq. (15-8), 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 10-15) = 5....
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