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Unformatted text preview: Exercise 15.1 Subject: Estimation of adsorbent characteristics. Given: Porous particles of activated alumina. BET area = 310 m 2 /g = S g . Particle porosity = 0.48 = ε p . Particle density = 1.30 g/cm 3 = ρ p . Assumptions: Straight pores of circular crosssection with uniform diameter. Find: (a) V p = specific pore volume in cm 3 /g. (b) ρ s = true solid density, g/cm 3 . (c) d p = average pore diameter, angstoms. Analysis: (a) From Eq. (153), V p = ε p / ρ p = 0.48/1.30 = 0.369 cm 3 /g (b) From a rearrangement of Eq. (155), ρ s = ρ p /(1  ε p ) = 1.30/(1  0.48) = 2.50 g/cm 3 (c) From a rearrangement of Eq. (152), d p = 4 ε p / S g ρ g = 4(0.48)/[310 x 10 4 (1.30)] = 4.76 x 107 cm or 47.6 angstroms. Exercise 15.2 Subject: Estimation of surface area of two forms of molecular sieves. Given: Form A with V p = 0.18 cm 3 /g and average d p = 5 angstroms. Form B with V p = 0.38 cm 3 /g and average d p = 2.0 microns. Assumptions: Straight pores of circular crosssection with uniform diameter. Find: Surface area, S g , of each form. Analysis: By substitution of Eq. (153) into (152), S g = 4 V g / d p (1) Form A: From Eq. (1), S g = 4(0.18)/(5 x 108 ) = 14.4 x 10 6 cm 2 /g or 1,440 m 2 /g Form B: From Eq. (1), S g = 4(0.38)/(2 x 104 ) = 7,600 cm 2 /g or 0.760 m 2 /g This is a very large difference in surface area. Exercise 15.3 Subject: Consistency of the characteristics of a smallpore silica gel. Given: Smallpore silica gel. Pore diameter = 24 angstroms = d p . Particle porosity = 0.47 = ε p . Particle density = 1.09 g/cm 3 = ρ p . Specific surface area = 800 m 2 /g = S g . Assumption: Straight pores of circular crosssection with uniform diameter. Find: (a) Reasonableness of the above values. (b) Fraction of a monolayer adsorbed if the adsorption capacity for water vapor at 25 o C and 6 mmHg partial pressure is 18% by weight. Analysis: (a) From Eq. (152), based on the above assumption, S g = 4 ε p / d p ρ g = 4(0.47)/[(1.09 x 10 6 )(24 x 1010 )] = 719 m 2 /g This is reasonable compared to the given value of 800 m 2 /g. The true density of silica gel, assuming it is SiO 2 , depends on its crystalline form. From a handbook, the density is from 2.20 to 2.65 g/cm 3 , with 3 of 4 forms from 2.20 to 2.26 g/cm 3 . Therefore, using a value of ρ s = 2.20 g/cm 3 with Eq. (155), ε p = 1  ρ p / ρ s = 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 m 2 ....
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 Spring '11
 suh

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