126989931-Solution-Manual-for-Stoichiometry-Bhatt-and-Thakore.pdf - Dimensions and Units EXERCISE 1.1(a Wave length 5500 = 5500 108 cm 1 m 10 9 nm 10 2

126989931-Solution-Manual-for-Stoichiometry-Bhatt-and-Thakore.pdf

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Unformatted text preview:  Dimensions and Units EXERCISE 1.1 (a) Wave length 5500 Å = 5500 ´ 10–8 cm ´ 1 m 10 +9 nm 10 +2 cm 1 m = 550 nm Ans. (b) Moisture content of humid air 175 grain = 1 lb dry air = 175 grain 1 lb ´ 1 lb dry air 7000 grains = 1 lb moisture 40 lb dry air or kg moisture kg dry air 1 kg moisture 1000 g ´ 1 kg 40 kg dry air = 25 g moisture/kg dry air Ans. Vacuum = 475 Torr Absolute pressure = 760 – 475 = 285 Torr º 285 ´ 101.325/760 º 38 kPa or 0.38 bar 1 Pa = 1.450 377 ´ 10–4 psi Pressure = 38 000 ´ 1.450 377 ´ 10–4 = 5.511 psia Ans. = EXERCISE 1.2 EXERCISE 1.3 F (5)2 4 = 19.635 cm2 Pressure = Force/Area = 192.6 ´ 104/(19.635 ´ 106) = 0.0981 MPa º 0.981 bar Cross sectional area of piston = 2 Solutions Manual—Stoichiometry In FPS units, pressure = 0.0981 ´ 106 ´ (1.450 377 ´ 10–4) = 14.228 psi Ans. EXERCISE 1.4 Weight = 500 lb º 500 ´ 0.4536 º 226.8 kg Volume = 29.25 L = 0.029 25 m3 Density of Fe, r = 226.8/0.029 25 = 7753.85 kg/m3 º 7.754 t/m3 Ans. EXERCISE 1.5 Diameter = 5 ft = 1.524 m Height of tank = 6 ft 6 inch = 1.981 m Volume of tank = (p/4)(Dia)2(Height) = (3.1416/4)(1.524)2(1.981) = 3.614 m3 Volume of CCl4 liquid in the tank = 3.614 ´ 0.75 = 2.71 m3 Density, r = 1600 kg/m3 Mass of CCl4 in the tank = 2.71 ´ 1600 = 4336 kg Ans. EXERCISE 1.6 Pressure drop = 0.05 ´ lbf × min (Pa. in 2 ) ´ 6894.759 (in 2 × US gal) lbf m3 1 US gal 1 h ´ ´ 10 3 –3 h m 60 min 3.785 412 ´ 10 1 kPa 1000 Pa = 15.178 kPa ´ Ans. EXERCISE 1.7 Total exposed area = 2 ´ 7.595 ´ 1.276 + 2 ´ 7.595 ´ 0.1535 + 2 ´ 1.276 ´ 0.1535 = 22.106 cm2 Weight loss = 14.9412 – 14.6254 = 0.3158 g Density = 7753.85 kg/m3 0.3158 1 10 6 365 1 ´ ´ ´ ´ 1000 7753.85 50 22.106 1 = 0.0134 cm/a º 0.0053 in/year º 5.3 mpy Corrosion rate = Ans. Dimensions and Units 3 EXERCISE 1.8 Let p¢ be the vapour pressure in kPa and T be the temperature in K 101 325kPa = 760 Torr or 1 kPa = 7.500 62 Torr p = 7.500 62 p¢ T = t + 273.15 or t = T – 273.15 Substitute the values in the equation, 1211.0 log10 (7.500 62 p¢) = 6.9057 – (T - 27315 . + 220.8) log10 p¢ + 0.8751 = 6.9057 – 1211.0 (T - 52.35) log10 p¢ = 6.0306 – 1211.0 (T - 52.35) Ans. EXERCISE 1.9 Let Cmo p¢ be the heat capacity of n-butane in SI units. 1 Btu/(lb mole × °R) = 1 kcal/(kmol × K) = 4.1868 kJ/(kmol × K) Cmo p¢ /4.1868 = Cmo p T = 1.8 T ¢ Substituting the values in the equation, ( Cmo p¢ /4.1868) = 4.429 + 40.159 ´ 10–3 ´ 1.8 T ¢ – 68.562 ´ 10–7 (1.8 T¢)2 Simplifying, Cmo p¢ = 18.5433 + 302.6479 ´ 10–3 T ¢ – 93.006 ´ 10–6 T ¢2 Ans. EXERCISE 1.10 Let Dp¢, v¢ and L¢ be pressure drop in cm WC, gas velocity is m/s and liquid flow rate in m3/m3 gas flow. Dp¢ = Dp ´ 2.54 v = v¢ ´ 3.281 L¢ = (L/264.172) 35.314 67 = 0.133 68 L Substitute above values in the Calvert equation. Dp ¢ L¢ = 5 ´ 10–5 ´ (v¢ ´ 3.281)2 ´ 0.133 68 2.54 Dp¢ = 1.0227 ´ 10–2 v¢2 L¢ Ans. 4 Solutions Manual—Stoichiometry EXERCISE 1.11 Let h¢ = heat transfer coefficient, kW/(m2 × K) G¢ = mass velocity of fluid, kg/(m2 × s) Cp¢ = specific heat, kJ/(kg × K) k¢ = thermal conductivity of fluid, kW/(m × K) D¢ = diameter of tube, m m¢ = viscosity of fluid, kg/(m × s) h= 3412.142 ´ 10-3 h¢ = 0.1761 h¢ 10.7639 ´ 1.8 G= 2.204 62 ´ 3600 G¢ = 737.28 G¢ 10.7639 Cp = 9.478 172 ´ 10 -4 = 2.2885 ´ 10–4 Cp¢ 2.204 62 ´ 1.8 k= 3412.142 ´ 10-3 k¢ = 0.5778 k¢ 3.2808 ´ 1.8 D = 3.2808 D¢ m¢ = 2.204 62 ´ 3600 m¢ = 2419.11 m¢ 3.2808 Substituting the values, 0.1761 h¢ = 0.023( 737.28G ¢ )0.8 (0.5778k ¢ )0.67 (2.3885 ´ 10 -4 C p¢ )0.33 (3.2808 D¢ )0.2 (2419.11m ¢ )0.47 or h¢ = 0.023 G¢0.8 k¢0.67 Cl¢0.33 / (D¢0.2 m¢0.47) Thus the equation does not change when consistent SI units are used. This is because the equation is a simplified form of Sieder-Tate equation which is madeup of three dimensionless numbers. Ans. Basic Chemical Calculations EXERCISE 2.1 Molar mass of oxygen = 2 ´ 16 = 32 g/mol Amount of oxygen = 500/32 = 15.625 mol Ans. EXERCISE 2.2 1 mol C is present in 1 mol CO2. 12 g C º 44 g CO2 Carbon content in 264 g CO2 = (12 ´ 264/44) = 72 g Ans. EXERCISE 2.3 Molar mass of KMnO4 = 39 + 55 + 4 ´ 16 = 158 Ans. EXERCISE 2.4 Molar mass of HNO3 = 63 Molar mass of H2SO4 = 98 100 g HNO3 = 100/63 = 1.5873 mol 100 g H2SO4 = 100/98 = 1.0204 mol Excess = 1.5873 – 1.0204 = 0.5669 mol HNO3 Excess atoms = 0.5669 ´ 6.022 142 ´ 1023 = 3.414 ´ 1023 atoms of HNO3 EXERCISE 2.5 Molar mass of CS2 = 12 + 2 ´ 32 = 76 76 g CS2 º 12 g C Ans. 6 Solutions Manual—Stoichiometry 1 mol CS2 = 1 mol C 3.5 kmol C = 3.5 kmol CS2 º 3.5 ´ 76 º 266 kg CS2 Ans. EXERCISE 2.6 Molar mass of Al2(SO4)3 = 2 ´ 27 + 3(32 + 4 ´ 16) = 342 Valence of Al2(SO4)3 = 6 Equivalent mass of Al2(SO4)3 = 342/6 = 57 Ans. EXERCISE 2.7 Molar mass of KMnO4 = 158 Valence of KMnO4 = 5 (based on oxidation number) Equivalent mass of KMnO4 = 158/5 = 31.6 500 g KMnO4 = 500/31.6 = 15.82 g eq Ans. EXERCISE 2.8 Basis: 100 kg magnesite ore Compound Molar mass kg kmol mole % MgCO3 SiO2 H2O 84.3 60 18 81 14 5 0.961 0.233 0.278 65.28 15.83 18.89 Total — 100 1.472 100.00 Ans. EXERCISE 2.9 Basis: 100 kg glass Compound Molar mass Na2O MgO ZnO Al2O3 B2O3 SiO2 62 40.3 81.4 102.0 69.6 60.0 Total — kg kmol mole % 7.8 7.0 9.7 2.0 8.5 65.0 0.1258 0.1737 0.1192 0.0196 0.1221 1.0833 7.65 10.57 7.25 1.19 7.43 65.91 1.6437 100.00 100 Ans. Basic Chemical Calculations 7 EXERCISE 2.10 Concentration of solids = 35 000 ppm or mg /L º (35 000 ´ 100)/106 º 3.5% For watery solutions, 10 000 ppm = 1% by mass Ans. EXERCISE 2.11 Basis: 100 kg limestone 1 kmol CaO = 1 kmol CaCO3 Molar mass of CaCO 3 ´ 54.5 CaCO3 in limestone = Molar mass of CaO = 100 ´ 54.5/56 = 97.32% (by mass) Ans. EXERCISE 2.12 (a) In one mole of ammonium sulphate, two atoms (or one mole) nitrogen are present. Nitrogen content of commercial ammonium sulphate = Molar mass of N 2 ´ 96 28 ´ 96 = Molar mass of (NH 4 ) 2 SO 4 132 = 20.36% Ans. (b) One mole of sodium nitrate contains half mole of nitrogen. Nitrogen content of pure sodium nitrate = 0.5 ´ Molar mass of N 2 ´ 100 0.5 ´ 28 ´ 100 = Molar mass of NaNO 3 85 = 16.47% Ans. EXERCISE 2.13 Chemical equation: 2 NaOH ¾ ¾¾® Na2O + H2O NaOH content of flakes = 2 ´ Molar mass of NaOH ´ 74.6 Molar mass of Na 2 O 2 ´ 40 ´ 74.6 62 = 96.26% = Ans. 8 Solutions Manual—Stoichiometry EXERCISE 2.14 Basis: 1 kmol azeotropic mixture Compound kmol Molar mass kg % by mass H2O HNO3 0.622 0.378 18 63 11.196 23.814 31.98 68.02 Total 1.000 — 35.010 100.00 Ans. EXERCISE 2.15 Basis: 100 kg saline solution Compound kg Molar mass kmol mole % NaCl H2O 25 75 58.5 18 0.4274 4.1667 9.3 90.7 Total 100 — 4.5941 100.0 Ans. EXERCISE 2.16 Basis: 100 g water Component mass, g HCl KCl H2O 4.00 19.61 100.0 Total 123.61 mass % Molar mass mole mole % 3.24 15.86 80.90 36.4609 74.5513 18.0153 0.1097 0.2630 5.5508 1.85 4.44 93.71 100.00 — 5.9235 100.00 Ans. EXERCISE 2.17 Basis: 100 kg solution Compound kg Nitrogen content kg/kmol comp. Molar mass Nitrogen Content kg NH3 NH4NO3 NH2CONH2 19 65.6 6 14 28 28 17 80 60 15.647 22.960 2.800 Total 90.6 — — 41.407 Nitrogen content of the solution = 41.41% Ans. Basic Chemical Calculations 9 EXERCISE 2.18 In one mole of ethanol, 2 moles of carbon are present. TOC of the solution = 2 ´ 12 ´ 1000/46 = 522 mg/L Oxidation (combustion) reaction: C2H5OH + 3 O2 ¾ ¾¾® 2 CO2 + 3 H2O 46 3 ´ 32 2 ´ 44 3 ´ 18 ThOD of the solution = 3 ´ 32 ´ 1000/46 = 2087 mg/L Ans. EXERCISE 2.19 Basis: 100 kg Phosphoric acid of 35% P2O5 strength 2 H3PO4 ¾ ¾¾® P2O5 + 3 H2O 2 ´ 98 142 3 ´ 18 H3PO4 content of the acid = 2 ´ 98 ´ 35/142 = 48.31% Ans. EXERCISE 2.20 Basis: 100 kg spent acid Free acid (H2SO4) content = 20 kg NH4HSO4 content = 45 kg + H ¾® H SO + NH + NH4HSO4 ¾ ¾¾® NH4+ + HSO–4 ¾ ¾¾ 2 4 4 115 18 97 98 Chemically bound acid = 98 ´ 45/115 = 38.35 kg Total acid content = 20 + 38.35 = 58.35 kg or 58.35% 18 Ans. EXERCISE 2.21 Basis: 1 m3 of aqueous TEA solution Mass of TEA = 0.47 ´ 1.00 ´ 1125 = 528.75 kg Mass of water = 0.53 ´ 1.00 ´ 1000 = 530 kg mass % TEA = 528.75 ´ 100/(528.75 + 530) = 49.94% Ans. Note: Volumes are strictly not additive but for an ideal solution, they can be considered additive. 10 Solutions Manual—Stoichiometry EXERCISE 2.22 Basis: 100 L wine Mass of alcohol = 20 ´ 0.79 = 15.80 kg Mass of alcohol-free liquid = (100 – 20) ´ 1.0 = 80 kg mass % alcohol = 15.80 ´ 100 = 16.49 (15.80 + 80) Ans. EXERCISE 2.23 Component CaCO3 Na2CO3 MgSO4 (a) Concentration of Na2CO3 = = (b) Concentration of MgSO4 = = Equivalent mass 100/2 = 50 106/2 = 53 120.3/2 = 60.15 50 ´ 800/53 754.7 ppm or mg/L as CaCO3 50 ´ 85/60.15 70.7 ppm or mg/L as CaCO3 Ans. EXERCISE 2.24 (a) Equivalent mass of H2SO4 = 98/2 = 49 Concn. of H2SO4 in solution = 294/49 = 6 mol/L º6N Ans. (b) Equivalent mass of CaCl2 = 112/2 = 55.5 Concn. of CaCl2 in solution = 4.8 ´ 1000/(1000 ´ 55.5) = 0.0865 mol/L º 0.0865 N (c) Equivalent mass of H3PO4 = 98/3 = 32.67 Concn. of H3PO4 in solution = 5 N º 5 ´ 32.67 º 163.35 g/L (d) Equivalent mass of HCl = 36.5 Concn. of HCl in solution = 54.75/36.5 = 1.5 g/L º 1.5 N º 1.5 M (e) Molar mass of K2SO4 = 174 Concn. of K2SO4 in solution = 174 ´ 3 = 522 g/L Ans. Ans. Ans. Basic Chemical Calculations 11 EXERCISE 2.25 Basis: 100 kg aqueous acetic acid solution Molar mass of acetic acid = 60 Acetic acid content = 35/60 = 0.5833 kmol Volume of 100 kg solution = 100/1.04 = 96.15 L Molarity = 0.5833 ´ 1000/96.15 = 6.066 M Since acetic acid is monovalent. Normality of the solution = 6.066 N Molality = 0.5833 ´ 1000/65 = 8.974 Ans. EXERCISEE 2.26 Basis: 1 litre solution CO2 dissolved (in gas form) = 40 L at NTP = 40/22.414 = 1.785 mol MEA content = 1 ´ 1.011 ´ 0.2 = 0.2022 kg º 202.2 g Molar mass of MEA = 61 Moles of MEA in solution = 202.2/61 = 3.315 mol CO2 concentration = 1.785/3.315 = 0.5385 mol/mol MEA Ans. EXERCISE 2.27 Basis: 100 L of 60 volume H2O2 solution O2 liberated in total = 60 ´ 100 = 6000 L Sp. volume of ideal gas at STP = 23.69 L/mol O2 liberated = 6000/23.69 = 253.27 mol H2O2 ¾ ¾¾® H2O + 1/2 O2 H2O2 decomposed = 2 ´ 253.27 = 506.54 mol Mass of H2O2 decomposed = 506.54 ´ 34 = 17 222.36 g Mass of H2O2 solution = 100 ´ 1.075 = 107.5 kg mass % H2O2 in solution = (17 222.36 ´ 100)/(107.5 ´ 1000) = 16.02 Ans. EXERCISE 2.28 Boiling point of 40% solution = 402.2 K(129.2°C). (Ref. Fig. 2.3) Boiling point of pure water at atm. pressure = 100°C (373.15 K) Boiling point elevation = 402.2 – 373.2 = 29 K or 29°C Ans. 12 Solutions Manual—Stoichiometry EXERCISE 2.29 Since solution is an ideal solution, it follows Raoult's law. Vapour pressure of solution, pvs = 0.5 ´ 37.2 + 0.5 ´ 12.3 = 24.75 kPa Ans. EXERCISE 2.30 Basis: 100 kg urea solution containing 25 % urea Urea present in solution = 25 = 0.4167 kmol 60 75 = 4.1667 kmol 18 Quantum of solution = 0.4167 + 4.1667 = 4.5834 kmol Water present in solution = 4.1667 4.5834 = 0.9091 Since vapour pressure of urea at 60°C (333.15 K) is negligible, Vapour pressure of solution = 0.9091 ´ 19.92 = 18.11 kPa Mole fraction of water in solution = Ans. EXERCISE 2.31 1 bar = 10.1972 m H2O 1000 m H2O = 98.066 bar According to Henry's law, mole fraction of solute = Partial pressure of solvent Henry's constant Assuming partial pressure of solvent (water) to be vapour pressure (as its mole fraction is near unity), mole fraction of nitrogen in water at 1000 m depth = 98.066 = 1.119 ´ 10–3 87 650 mole fraction of helium in water at 1000 m depth = 98.066 = 0.775 ´ 10–3 126 600 Since helium is much less dissolved in water, it is preferred as a mixture with oxygen. Ans. Basic Chemical Calculations 13 EXERCISE 2.32 Mole fraction of NH3 in solution = 0.02 Roult's law: yi p = xi pi yi = = [Ref. Eq. (2.18)] xi pi p 11.58 ´ 0.02 1.01 325 = 0.2286 Henry's law: pi = = = = yi = Hi xi 0.861 ´ 0.02 0.01722 bar 1.722 kPa [Ref. Eq. (2.44)] pi 0.01722 = p 1.013 25 = 0.017 Ans. EXERCISE 2.33 Basis: 100 kmol gas Gas Formula Molar mass kmol kg mass % Ethylene Benzene Oxygen Methane Ethane Nitrogen C2H4 C6H6 O2 CH4 C2H6 N2 28 78 32 16 30 28 30.6 24.5 1.3 15.5 25.0 3.1 856.8 1911.0 41.6 248.0 750.0 86.8 22.00 49.07 1.07 6.37 19.26 2.23 Total — — 100.00 3894.2 100.00 Average molar mass = 3894.2/100 = 38.942 Ideal gas occupies 22.414 m3/kmol at NTP. Density of gas mixture = 38.942/22.414 = 1.737 kg/m3 Ans. 14 Solutions Manual—Stoichiometry EXERCISE 2.34 Basis: 100 kmol sewage gas Gas Formula Molar mass kmol kg mass % Methane Carbon dioxide Ammonia CH4 CO2 16 44 68 30 1088 1320 44.55 54.05 NH3 17 2 34 1.40 Total — — 100 2442 100.00 Average molar mass = 2442/100 = 24.42 Density of sewage gas = 24.42/22.414 = 1.09 kg/m3 Ans. EXERCISE 2.35 Basis: 1.10 kg CO2 Moles of CO2 = 1.10/44 = 0.025 kmol Volume occupied = 33 L = 0.033 m3 V = 0.033/0.025 = 1.32 m3/kmol T = 300 K van der Waals equation: F p + a I (V – b) = RT H VK F p + 3.6 I (1.32 – 0.043) = 0.008 314 ´ 300 ´ 1000 H 1.32 K 2 2 Solving the equation, p = 1951.104 kPa º 19.51 bar Ans. EXERCISE 2.36 For chlorine gas: (a) Ideal gas law: p = 15.2 MPa a, T = 503.15 K (230°C) V = RT/p = 0.008 314 ´ 503.15/15.2 = 0.2752 m3/kmol 71 Molar mass = = 258.0 kg/m3 0.2752 V (b) van der Waals equation: For chlorine, pc = 79.77 bar a and Tc = 416.90 K a = 27 R2 T2/64 pc Density = Ans. Basic Chemical Calculations 15 = 27(0.083 14)2(416.90)2/(64 ´ 79.77) = 6.353 72 (m3 )2× bar/(kmol)2 b = RTc/8pc = 0.083 14 ´ 416.9/(8 ´ 79.77) = 0.054 314 m3/kmol Substituting in the van der waals equation 6.353 72 152 + (V – 0.054 314) = 0.083 14 ´ 503.15 = 41.831 89 V2 Simplifying, 152 V3 – 5.0075 V2 + 0.6354 V – 0.03451 = 0 Solving the equation by Newton-Raphson method. V = 0.152 m3/kmol Density = 71/0.152 = 467.1 kg/m3 Mathcad Solution: FG H I K  Density = 71/0.1536 = 462.24 kg/m3 Ans. EXERCISE 2.37 Equation: Pressure p = 73 bar, Temperature T = 423.15 K p = [RT(1 – e)/ v2] [V + b] – A/v2 a A = A0 1 = (5.88) [1 – (0.058 61/V)] V b = 0.094 [1 – (0.019 15/V)] B = B0 1 V F H F H I K I K 90 ´ 104 V (423.15)9 0.9 0.011 89 = = 75.687V V e = c/VT3 = Substituting the values. é æ 0.011 89 ö 2 ù /V ú 73 = ê0.08314 ´ 423.15 ç1 è ø÷ V ë û [V + 0.094 – (0.0018/V)] – [5.88 – (0.344 63/V)]/V2 16 Solutions Manual—Stoichiometry Simplifying and solving the equation by Newton-Raphson method, V = 0.4 m3/kmol Molar mass of ethane = 30.0704 Density of ethane = 30.0704/0.4 = 75.176 kg/m3 Mathcad Solution: Density = 30.0704/0.40045 = 75.092 kg/m3 Ans. Ans. EXERCISE 2.38 For dimethyl ether (DME): p = 15 bar a T = 353.15 K pc = 53.4 bar a Tc = 400.2 K Acentric factor w = 0.192 Tr = f ° = 0.1445 – T 353.15 = = 0.8824 Tc 400.2 0.607 ´ 10 –3 0.1385 0.0121 0.330 – – – (0.8824)8 0.8824 (0.8824) 2 (0.8824)3 = – 0.4262 f 1 = 0.0637 + 0.8 ´ 10 –2 0.331 0.423 – – (0.8824)8 (0.8824) 2 (0.8824)3 = – 0.148 B= 0.08314 ´ 400.2 [– 0.4262 + 0.192 (– 0.148)] 53.4 = – 0.283 26 Basic Chemical Calculations 17 Z=1– ( - 0.283 26)15 0.08314 ´ 353.15 = 1.1447 V= ZRT 1.1447 ´ 0.08314 ´ 353.15 = p 15 = 2.241 m3/kmol Ans. EXERCISE 2.39 (a) Ideal gas law: p = 7.09 + 1.013 25 = 8.103 25 T = 923.15 K (650 °C) V= 0.08314 ´ 923.15 RT = 8.103 25 p = 9.472 m3/kmol Density = 27.587/9.472 = 2.912 kg/m3 (b) van der Waals equation: a= 27 R 2Tc2 64 pc and b = RTc / 8pc where pc and Tc are pseudo critical properties. Gas Mole fraction ni Molar mass Mi Mass kg n × Mi N2 O2 H2O NH3 0.705 0.188 0.012 0.095 28 32 18 17 19.740 6.016 0.216 1.615 Total 1.000 — 27.587 Critical temp Tc, K Critical pressure pc, bar Tc ni × Tc pc 88.893 29.061 7.718 38.523 33.94 50.42 221.2 113.50 126.09 154.58 643.30 405.50 164.195 ni × pc 23.93 9.48 2.65 10.78 46.84 Pseudo critical temperature, Tc = 164.195 K Pseudo critical pressure, pc = 46.84 bar a = [27(0.083 14)2(164.195)2]/(64 ´ 46.84) = 1.678 (m3)2× bar/(kmol)2 b = (0.083 14 ´ 164.195)/(8 ´ 46.84) = 0.036 42 m3/kmol 18 Solutions Manual—Stoichiometry Substituting the values in the equation, 1.679 (V – 0.036 43) = 0.083 14 ´ 923.15 = 75.7507 8.1 + V2 Mathcad Solution: F H I K Average molar mass = 27.587 Density = 27.587/9.3626 = 2.9465 kg/m3 Specific gravity of the gas mixture = 27.587/28.96 = 0.952 Ans. EXERCISE 2.40 Concentration of water vapours = 0.216 kg/kmol º (0.216/9.3626)106 º 23 071 mg/m3 23.071 ´ 27.587 2.9465 ´ 18 = 12000 ppm In terms of ppm concentration = Ans. EXERCISE 2.41 Basis: 9.082 g furfural-n-butane mixture Partial pressure of water = 12.5 Torr at 295.2 K Partial content of n-butane = 763.2 – 12.5 = 750.7 torr n-butane content of vapours = pV/RT 750.7 ´ 105.7 = 760 ´ 1000 ´ 0.082 06 ´ 295.2 = 0.004 314 mol Mass of n-butane = 0.004 314 ´ 58 = 0.2501 g Mass of furfural = 9.082 – 0.2501 = 8.8319 g mass % n-butane = 0.2501 ´ 100/9.082 = 2.75% Moles of furfural = 8.8319/96 = 0.092 mol Total moles = 0.092 + 0.004 314 = 0.096 314 mol mole % n-butane = 0.004 314 ´ 100/0.096 314 = 4.48 Ans. Basic Chemical Calculations 19 EXERCISE 2.42 Basis: 100 kmol flue gases Gas CO2 O2 N2 SO2 Molar mass 44 32 28 64 Total — Avg. molar mass = Partial pressure of water = = Absolute humidity = = EXERCISE 2.43 kmol 10.00 7.96 82.00 0.04 kg 440.00 254.72 2296.00 2.56 100.00 2993.28 29.933 (of dry gas) Vapour pressure of water at DP 10.612 kPa (ref. Chapter 6) [10.612 ´ 18]/[(100 – 10.612) 29.933] 0.071 39 kg/kg º 71.39 g/kg Ans. Concentration of SO2 in flue gases = (0.04/100) ´ 106 = 400 ppm (v/v) = 2.56/2993.28 = 0.8552 ´ 10–3 kg/kg º 855.2 mg/kg or ppm (by mass) Volume of gas = 0.082 06 ´ 463.15 ´ 760/750 = 38.513 m3/kmol mg kmol 2.56 kg 1 ´ 106 ´ kg 100 kmol 38.513 m 3 3 = 664.7 mg/m Ans. Concentration of SO2 = EXERCISE 2.44 Basis: 733 kg mixture Component kg Molar mass CH3CH Steam Oxygen (from air) Nitrogen (from air) 200 133 32 18 Component O2 N2 CH3OH H2O Total kmol 6.25 7.39 0.21 ´ 400/29 = 2.897 0.79 ´ 400/29 = 10.897 Total pressure = 68.6 kPa g Absolute Pressure = 68.6 + 101.325 = 169.925 kPa kmol mole % Partial pressures, kPa 2.897 10.56 17.94 10.897 39.72 67.49 6.250 22.78 38.71 7.390 26.94 45.78 27.434 100.00 169.92 Ans. 20 Solutions Manual—Stoichiometry EXERCISE 2.45 Basis: 1 kmol LPG mixture at 313.15 K (40 °C). Gas kmol Molar mass kg C3H8 n-C4H10 i-C4H10 0.30 0.45 0.25 44.0956 58.1222 58.1222 13.229 26.155 14.531 Total 1.00 53.915 Average molar mass = 53.915 Specific gravity = Gas kmol C3H8 n-C4H10 i-C4H10 0.30 0.45 0.25 Total 1.00 Ans. (a) 53.915 = 1.861 28.97 Vapour pressure bar 13.975 3.773 5.290 Pressure in LPG cylinder = 7.2129 bar Ans. (b) Partial pressure bar 4.1925 1.6979 1.3225 7.2129 Ans. (c) EXERCISE 2.46 Basis: 5 m3 solution Total pressure = 6.77 MPa g = 6.77 + 0.1013 = 6.8713 MPa a Solubility of N2 at 0.1013 MPa and 305.75 K = 1.35 ´ 5/100 = 0.0675 Nm3/5 m3 soln. Partial pressure of N2 at 6.77 MPa g total pressure = (6.8713 ´ 0.206)/(1 – 0.022) = 1.4473 MPa (on ammonia-free basis) Solubility of N2 at 6.77 MPa g and 305.75 K = 0.0675 ´ 1.4473/0.1013 = 0.9644 Nm3/5 m3 solution In a similar manner, solubility can be calculated for all the gases as shown in the following, table, Basic Chemical Calculations 21 Gas Solubility at 0.1013 MPa a and 305.75 K/5 m3 soln. N2 H2 Ar CH4 Total Partial pr. at 6.77 MPa g Solubility at 305.75 K and 6.77 MPa g/5 m3 solution 0.0675 0.0800 0.1375 0.1400 1.4473 4.3560 0.2881 0.7800 0.964 3.440 0.391 1.078 0.4250 m3 6.8714 MPa a 5.873 Nm3 Ans. EXERCISE 2.47 Refer Example 2.25 pi = 7 bar g = 8.013 bar a , pf = 1.013 bar a T f = Ti – m (pi – pf) = 313.15 – 0.21 (8.013 – 1.013) = 311.68 K or 38.53°C Ans. EXERCISE 2.48 p1 = 101 atm a p2 = 1 atm a Ti = 308.15 K T f = 308.15 – 0.169 (101 – 1) = 291.25 K or 18.1°C Ans. EXERCISE 2.49 Assume ideal gas law. Basis: Receiver of 2 m3 capacity Volume of air in the receiver at 7.5 bar g and 313.15 K, 8.513 25 ´ 2 ´ 27315 . v1 = 31315 . ´ 1.013 25 = 14.657 Nm3 Volume of air in the receiver at 1 bar g and 313.15 K, v2 = 2.013 25 ´ 2 ´ 27315 . 31315 . ´ 1.013 25 = 3.466 Nm3 Volume of air pressurized = 14.657 – 3.466 = 11.191 Nm3 in 4 min 11191 . ´ 60 4 = 167.9 Nm3/h Capacity of air compressor = Ans. 22 Solutions Manual—Stoichiometry EXERCISE 2.50 qL = (1 ´ 10-3 - 2 ´ 10-4 )25 7.5 ´ 60 = 4.444 ´ 10–5 (mbar × L)/s Ans. 3 Material Balances without Chemical Reaction EXERCISE 3.1 Basis: 4 kg feed water Product water quantity = 3 kg Material balance of dissolved solids (DS): Feed water quantity ´ ...
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