Ch. 3 a - 1 PGE 312 Physical and Chemical Behavior of...

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1 PGE 312 Physical and Chemical Behavior of Fluids Home Work 5 – Solutions Problem 3-2 Given: Density = 0.103lb/cu ft Pressure = 14.7 psia Temperature = 100°F = 560°R Molecular formula = C n H 2n Solution: For an ideal gas, we have PV = nRT ------ (1) The number of moles n of any compound is given by n = m/M where m is the mass of the compound in lb and M is the molecular weight in lb/lbmoles. Thus, rewriting equation (1) by substituting for number of moles we get, PV = mRT/M ------- (2) Taking volume to the other side, we have P = (m/V)*(RT/M) -------- (3) We know that density = mass/volume. Therefore, replacing (m/V) in equation (3) by density we get, P = ρRT/M -------- (4) Therefore, the Molecular weight M is given by M = ρRT/P -------- (5) Substituting the values for ρ, R and T in equation (5), we have M = 0.103*10.732*560/14.7 which gives M = 42lb/lbmol. From the general formula of the given compound, we find that 12*n + 1*2n = 42, where n is the number of carbon atoms. Solving the above equation we get n = 3. Therefore, the molecular formula of the given compound is C 3 H 6 Propene .
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2 Problem 3-4 Solution: Component Molecular Weight (M j ) Compostion, mole fraction (y j ) Methane 16 0.6904 Ethane 30 0.0864 Propane 44 0.0534 i-Butane 58 0.0115 n-butane 58 0.0233 i-pentane 72 0.0093 n-pentane 72 0.0085 Hexanes 86 0.0173 Heptanes plus 158 (given) 0.0999 The mole fraction of a compound in a mixture is given by: y j = n j /n ------ (1) and the number of moles of any substance is given by n j = m j /M ------ (2) The mass fraction is given by j = m j /m ------ (3) where, m j is the mass of each component and m is the total mass of the mixture. Rewriting equation 3 in terms of mole fraction we get j = (n j *M j )/(n*M) ------ (4) Thus, from equation (4) we have j = y j *M j /M ------ (5) The apparent Molecular weight M of the mixture is given by the equation M = Σy j *M j ------ (6) Thus, M = 16*0.6904+30*0.0864+44*0.0534+58*0.0115+ 58*0.0233+72*0.0093+72*0.0085+86*0.0173+158*0.0999 = 36.56lb/lbmol The mass fraction is shown in the table below for each of the components in the mixture.
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3 Component Compostion, mole fraction (y j ) Molecular Weight (M j ) Mass fraction ( j ) Volume fraction (v j ) Methane 0.6904 16 0.3021 0.6904 Ethane 0.0864 30 0.0709 0.0864 Propane 0.0534 44 0.0642 0.0534 i-Butane 0.0115 58 0.0182 0.0115 n-butane 0.0233 58 0.0370 0.0233 i-pentane 0.0093 72 0.0183 0.0093 n-pentane 0.0085 72 0.0167 0.0085 Hexanes 0.0173 86 0.0407 0.0173 Heptanes plus 0.0999 158 (given) 0.4317 0.0999 Total 1.0000 0.9998 1.0000 The volume fraction
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This note was uploaded on 11/27/2011 for the course PGE 312 taught by Professor Peters during the Spring '08 term at University of Texas at Austin.

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Ch. 3 a - 1 PGE 312 Physical and Chemical Behavior of...

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