Density is defined as the weight of a mixture per unit volume and is independent of temperature. As the volume of liquids and gases is a strong function of temperature, density also varies significantly with temperature for a specified composition. However, in the case of solids the variation of density with temperature is not very significant.
WORKED EXAMPLES
3.1 Calculate the volume of 15 kg of Chlorine at a pressure of 0.9 bar and 293 K.
Basis: 15 kg Cl2 = 15/71.0 = 0.2113 kmole Its volume is 0.2113 ¥ 22.414 ¥ 1.0
0.9 ¥ 293
273 = 5.643 m3
3.2 Calculate the volume occupied by 6 lb of chlorine at 743 mm Hg and 70 °F Basis: 6 lb of Cl2 ∫ 6/71 = 0.0845 lb mole of chlorine
Volume at standard condition = (0.0845 ¥ 359) = 30.34 ft3 Volume at given condition = 0 0 1
0 1
P V T
T P
Ê ˆ Ê ˆ
¥ Á ˜
Á ˜ Ë ¯
Ë ¯
= 760 530
3.3 Calculate the weight of 200 cu.ft. of water vapour at 15.5 mm Hg and 23 °C
Basis: 200 ft3 of gas at given condition Volume at standard condition = 1 1 0
0 1
Moles of water = 3.76
359
Find the pressure of the gas stored (required)?
Basis: 30 lb CO2 = 30 Volume at standard condition = 0.3102 ´ 22.414 = 6.9528 m3 Volume at given condition =20 ´ 0.02832 = 0.5664 m3
3.5 Calculate the maximum temperature to which 10 lb of nitrogen enclosed in a 30 ft3 chamber may be heated without exceeding 100 psi pressure.
Basis: 10 lb of N2 = 10/28 = 0.357 lb mole
Volume at standard condition = 0.357 ´ 359 = 128.21 ft3
\ Temperature ‘T1’ = 0 1 1
3.6 When heated to 100 °C and 720 mm Hg, 17.2 g of N2O4 gas occupies a volume of 11,450 cc. Assuming that the ideal gas law applies, calculate the percentage dissociation of N2O4 to NO2?
2 4 2
Parameter Given condition (1) Standard condition (0)
Pressure 720 mm Hg 760 mm Hg
Temperature 373 K 273 K
Volume 11,450 cc ?
Volume at standard condition = 1 1 0
0 1 Therefore, no. of g moles remaining = 7939.2
22, 414 = 0.354
\ (0.187 + x) = 0.354
\ x = 0.167
Percentage dissociation = 0.167 0.187
È Ø
É Ù
Ê Ú ´ 100 = 89.42%
3.7 Calculate the average molecular weight of a gas having the following composition by volume. CO2: 13.1%, O2: 7.7% and N2: 79.2%
Basis: 1 g mole of the gas
Component Volume % = Molecular g mole Weight, g mole % weight
CO2 13.1 44 0.131 0.131 ´ 44 = 5.764
O2 7.7 32 0.077 0.077 ´ 32 = 2.464
N2 79.2 28 0.792 0.792 ´ 28 = 22.176
Total 100.0 1.000 30.404
Average molecular weight is 30.404.
3.8 Calculate the density in lb/ft3 at 29.0 inches of Hg and 30 °C for a mixture of hydrogen and oxygen that contains 11.1% of hydrogen by weight.
Basis: 1 lb of gas mixture
Component Weight % Molecular weight lb mole
Hydrogen 0.111 2 0.111/2 = 0.0555
Oxygen 0.889 32 0.889/32 = 0.0278
Total 0.0833 lb moles Temperature = 30 °C = 86° F = 546° R
Volume of the gas at the given condition = 0.0833 ´ 359 ´ (29.92/29) ´ (546/492) = 34.24 ft3 Density = (1/34.24) = 0.0292 lb/ft3
3.9 Calculate the density in g/litre at 70 °F and 741 mm Hg of air Basis: 1 g mole of air
Component Volume % = mole % Molecular weight Weight, g
Oxygen 0.21 32 6.72
Nitrogen 0.79 28 22.12
Total 28.84 g
Volume of air = 1 ´ 22.414 ´ 530 760 492 741 È Ø È Ø
É Ù É Ù
Ê Ú Ê Ú = 24.8 litres
Density of air = 28.84 24.8
È Ø
É Ù
Ê Ú = 1.162 g/litre
3.10 In 1000 ft3 of a mixture of hydrogen, nitrogen and carbon-dioxide at 250 °F, the partial pressures are 0.26, 0.32 and 1.31 atm. Assuming
‘Ideal Gas’ behaviour, find the following:
(a) lb moles of H2; (b) mole fraction and mole % H2; (c) pressure fraction of H2 (d) partial volume of H2; (e) volume fraction and volume % of H2; (f) weight of H2; (g) weight fraction and weight % of H2; (h) average molecular weight; (i) density of gas mixture;
(j) density at standard condition
Also, show that volume % = pressure % = mole % Basis: 1000 ft3 of gas mixture
(a) Partial pressure of hydrogen = 0.26 atm V = 1000 ft3, T = 710 °R
Volume of H2 at standard condition = 1000 ´ 0.26 492
Total moles = 1000 1.89 492 359 1.00 710 È Ø È Ø È Ø
É Ù É Ù É Ù
Ê Ú Ê Ú Ê Ú = 3.648 lb moles
mole fraction of hydrogen = 0.502
3.648 = 0.138 (c) Pressure fraction of hydrogen = 0.26
1.89
È Ø
É Ù
Ê Ú = 0.138
(d) Partial volume of hydrogen is the volume occupied by 0.502 lb moles of it at 1.89 atm and 710 °R (e) Volume fraction of hydrogen = 138
1000 = 0.138 Thus volume % = pressure % = mole % (f) Weight of hydrogen = 0.502 ´ 2 = 1.004 lb (g) Basis 100 lb moles of gas mixture
Mole fraction of nitrogen = 0.32
189 = 0.169 Mole fraction of carbon dioxide = 1.31
1.89 = 0.693 Mole fraction of hydrogen = 0.26
1.89 = 0.138
Component Molecular mole % lb mole Weight, lb Weight % weight
Hydrogen 2 13.8 13.8 13.8 ´ 2 = 27.6 0.78 Nitrogen 28 16.9 16.9 16.9 ´ 28 = 473.2 13.33 Carbon dioxide 44 69.3 69.3 69.3 ´ 44 = 3049.2 85.89
Total 100.0 100.0 3550.0 100.00
(h) Average molecular weight = 3550 Number of moles = 1309.7
359 = 3.648 lb moles º 3.648 ´ 35.5 = 129.55 lb.
\ Density at given condition = 129.55 1000
È Ø
É Ù
Ê Ú = 0.12955 lb/ft3 ( j) Density at standard condition:
Volume at standard condition = 1000 ´ 1.89 492 1.00 710
3.11 A certain gaseous hydrocarbon is known to contain less than 5 carbon atoms. This compound is burnt with exactly the volume of oxygen required for complete combustion. The volume of reactants (all gases) is 600 ml and the volume of products (all gases) under the same condition is 700 ml. What is the compound?
Basis: 1 mole of hydrocarbon.
Let the hydrocarbon be CxHy
Total moles of reactants = 1 4 x y È Ø
É Ù
Ê Ú
Total moles of products = 2 x y È Ø
É Ù
Ê Ú
Reactants (1 /4) 600 6
Products ( /2) 700 7
(7x – 6x) + 7
Since the hydrocarbon has carbon atoms less than 5, we have x < 5;
solving above equation assuming carbon atoms as 1, 2, up to 5 we get the values of y as indicated below:
x = 1; –5
3.12 Combustion gases having the following molal composition are passed into an evaporator at 200 °C and 743 mm Hg (N2: 79.2%, O2: 7.2%, CO2: 13.6 %) Water is added to the stream as vapour and the gases leave at 85 °C and 740 mm Hg with the following composition N2: 48.3%, O2: 4.4% and H2O : 39%. Calculate (a) volume of gases leaving evaporator per 100 litres of gas entering and (b) weight of water added per 100 litres of gas entering.
(N2, O2, CO2) ® Evaporator ® (N2, O2, CO2) + H2O
Water Basis: 100 g moles gas entering
Volume = 100 ´ 22.414 ´ 473 760 273 743 È Ø È Ø
É Ù É Ù
Ê Ú Ê Ú = 3972.31 litres This 100 g moles of entering gas º 61% of gases leaving.
\ g moles of gases leaving = 100/0.61 = 164 g moles.
(a) Volume of gas leaving 4950.0 100 100 litres gas-entering 3972.31
t
= 124.6 litres (b) Volume of water added 1152.0 100
29 g. on taking T1 as 273.16 K the g moles of nitrogen will be 33,677) 3.14 In the manufacture of hydrochloric acid, a gas is obtained that
contains 25% HCl and 75% air by volume. This gas is passed through an absorption system in which 98% of the HCl is removed. The gas enters the system at 48.8 °C and 743 mm Hg and leaves at 26.7 °C and 738 mm Hg. Applying the pure component volume method, calculate:
(a) The volume of gas leaving per 1000 litres entering the absorption column.
(b) The weight of HCl removed per 100 litres entering.
Basis: In 100 litres of entering gas volume of air = 75 litres Pure component volume of HCl = 25 litres
Pure component volume of HCl absorbed = (25 ´ 0.98) = 24.5 litres
Pure component volume of HCl remaining = 0.5 litres Volume of gas leaving = 75 + 0.5 = 75.5 litres
(a) Parameter Entering condition Leaving condition
Pressure 743 mm Hg 738 mm Hg
Temperature 48.8 °C 26.7 °C
Volume 75.5 litres ?
Volume of (entering) gas at leaving condition
= 75.5 ´ 743 299.7
738321.8 = 70.8 lit.
Component Litre Volume % (or) mole %
(b) Composition: HCl 0.5 0.66
Air 75.0 99.34
Total: 75.5 100.00
Volume of HCl absorbed at standard condition = 24.5´ 743 492
760580 = 20.3 litres Weight of HCl absorbed = 20.3
22.414 ´ 36.5 = 33.057 g.
3.15 Absorbing chlorine in milk of lime produces calcium hypochlorite. A gas produced by the Deacon process enters the absorption apparatus at 740 mm Hg and 75 °F. The partial pressure of Cl2 is 59 mm Hg and the remainder being inert gas. The gas leaves at 80 °F and 743 mm Hg with Cl2 having a partial pressure of 0.5 mm Hg. Calculate, by applying the partial pressure method:
(a) volume of gas leaving per 100 litres entering (b) weight of Cl2 absorbed.
Basis: 100 litres of gases entering
Partial pressure of inert gas entering = 740 – 59 = 681 mm Hg Partial pressure of inert gas leaving = 743 – 0.5 = 742.5 mm Hg Volume of inert gases entering = 100 litres (681 mm Hg)
Volume of inert gas leaving = 100 ´ 681 299.7 742.5 297 = 92.5 litres
(a) Volume of gas leaving = 92.5 litres (743 mm Hg)
Volumes of Cl2 entering and leaving are 100 litres and 92.5 litres Entering condition:
Parameter Given condition (1) Standard condition (0)
Pressure 59 mm Hg 760 mm Hg
Temperature 297 K 273 K
Volume 100 litres ?
Volume at standard condition of Cl2 entering
= 100 ´ 59 273
760297 = 7.14 litres Volume at standard condition of Cl2 leaving
= 92.5 ´ 0.5 273
760299.7 = 0.055 litre Volume at standard condition of Cl2 absorbed
= 7.14 – 0.055 = 7.085 litres (b) Weight of Cl2 absorbed = 7.285
22.414 ´ 71 = 22.45 g.
3.16 Nitric acid is produced by the oxidation of ammonia with air. In the first step of the process, ammonia and air are mixed together and passed over a catalyst at 700 °C. The following reaction takes place.
4NH3 + 5O2 ® 6H2O + 4NO. The gases from this process are passed into towers where they are cooled and the oxidation is completed according to the reactions:
2NO + O2 ® 2NO2
3NO2 + H2O ® 2HNO3 + NO
The NO liberated is re-oxidized in part and forms more nitric acid in successive repetitions of the above reactions. The ammonia and the air enter the process at 20 °C and 755 mm Hg. The air is present in such proportion that the oxygen will be 20% in excess of that required for complete oxidation of ammonia to nitric acid. The gases leave the catalyst at 700 °C and 743 mm Hg. Given the overall reaction:
2NO + 1.5O2 + H2O ® 2HNO3, calculate the following:
(a) The volume of air to be used per 100 litres of NH3 entering the process (b) The composition of gases entering the catalyzer
(c) The composition of gases leaving (assuming the reaction in catalyzer is 85%)
(d) The volume of gases leaving the catalyzer for 100 litres ammonia entering
(e) Weight of acid produced per 100 litres NH3, assuming 90% of the nitric oxide entering the tower is oxidized to acid.
Basis: 1 g mole of NH3
overall reaction is NH3 + 2O2 ® HNO3 + H2O O2 required is 2 g moles
But O2 supplied is 2 ´ 1.2 = 2.4 g moles i.e. air supplied is 2.4
0.21
= 11.42 g moles
Thus, N2 = 9.02 g moles.
(a) Volume of air = 11.42´ 22.414 ´ 293 760
273755 = 276.4 litres Volume of NH3 = 1 ´ 22.414 ´ 293 760
273755 = 24.2 litres Volume of air/100 litres of NH3 = 276.4 100
24.2
= 1142.2 litres
(b)
Component g mole mole % = volume %
NH3 1.00 8.0
O2 2.40 19.3
N2 9.02 72.7
Total 12.42 100.0
(c) Gases leaving catalyzer are nitrogen, oxygen, ammonia, nitric oxide and water:
N2 (all that enters) = 9.02 g moles
NH3 (85% conversion) = (1 – 0.85) = 0.15 g mole O2 consumed = 5
0.85 4
È Ø
É Ù
Ê Ú = 1.06 g moles
NH3
Air Catalyzer Absorber
HNO3
Exit gas
O2 leaving = (2.4 – 1.06) = 1.34 g moles NO formed = 0.85 g mole
H2O formed = 6 0.85 4
Ê ¥ ˆ
Á ˜
Ë ¯ = 1.275 g moles
Component N2 O2 NH3 NO H2O Total
g moles 9.02 1.34 0.15 0.85 1.275 12.635
mole % = volume % 71.40 10.60 1.20 6.70 10.10 100.000 (d) Volume of gases leaving the catalyzer
= 12.635 ¥ 22.414 ¥ 973 760
273¥243 = 1031.8 litres
This is the volume of gases leaving for 24.2 litres of ammonia entering
Therefore, for 100 litres of NH3 entering
= (1031.8 ¥ 100)/24.2 = 4264 litres of gas leaves (e) NO entering the tower = 0.85 g mole
NO converted = 0.85 ¥ 0.9 = 0.765 g mole
HNO3 produced = 0.765 g mole = (0.765 ¥ 63) = 48 g For 100 litres of NH3 weight of HNO3 produced
= 48 100 24.2
¥ = 199 g
3.17 1000 kg/h of an organic ester C19H36O2 is hydrogenated to C19H38O2 in a process. The company purchases its H2 in cylinders of 1 m3 capacity. The pressure in cylinder is initially 10 kg/cm2 (abs.) and drops to 2 kg/cm2 (abs.) after use. The company works 24 h/day, 7 days a week. How many cylinders are needed per week?
Basis: Ester being processed in one week = (1000 ¥ 24 ¥ 7) = 1,68,000 kg
C19H36O2 + H2 Æ C19H38O2
296 2 298
296 kg of ester reacts with 2 kg of H2 168000 kg of ester reacts with 2 168000
296
¥ = 1135 kg of hydrogen
1 kmole of any gas occupies 22.414 m3 at NTP
H2 initially present in the cylinder (10 m3) = (1)´ 10 (by reducing to NTP condition)
H2 after use = 1 2 22.414 1 È Ø È Ø
É Ù É Ù
Ê Ú Ê Ú = 0.09 kmole
\ H2 available from one cylinder = 0.36 kmole H2 needed = 1135 kg = 1135
2
È Ø
É Ù
Ê Ú = 567.5 kmoles
The number of cylinders required should be full number, rounded to next highest value
\ Cylinders required per week = 567.5 0.36
È Ø
É Ù
Ê Ú = 1577
3.18 A mixture of toluene and air is passed through a cooler where some toluene is condensed. 1000 ft3 of gases enter the cooler per hour at 100 °C and 100 mm Hg (gauge). The partial pressure of toluene is 300 mm Hg. 740 ft3 of gases leave cooler per hour at 40 mm Hg and 50 °C. Calculate the weight of toluene condensed per hour. Vapour pressure of toluene at 50 °C = 90 mm Hg.
Basis: One hour
100 mm Hg (gauge) = 860 mm Hg (abs.) Toluene entering = 300
1000 860
At exit, air is saturated with toluene.
Toluene leaving = 740 ´ 90
3.19 Air is dried from a partial pressure of 50 mm of water vapour to a partial pressure of 10 mm. The temperature of entering air is 500 °F and the pressure remains constant at 760 mm Hg. How much water is removed per 1000 ft3 of entering air?
Basis: 1000 ft3 of entering air.
Moles of air entering = 1000 492
359 960 Moles of water leaving = 10
1.336 750
È Ø
É Ù
Ê Ú = 0.0178 lb mole Water condensed = (0.094 – 0.0178) = 0.0762 lb mole = 1.37 lb 3.20 Chimney gas has the following composition:
CO2: 9.5%, CO : 0.2%, O2: 9.6% and N2: 80.7%. Using ideal gas law, calculate:
(a) its weight percentage
(b) volume occupied by 0.5 kg of gas at 30 °C and 760 mm Hg.
(c) density of the gas in kg/m3 at condition of (b) (d) specific gravity of the gas mixture.
(Density of air may be taken as 1.3 g/cc) Basis: 100 kmoles of chimney gas (a)
Component Mol. weight Weight, kmole Weight, kg Weight % CO2 44 9.5 (9.5 ´ 44) = 418.0 13.978
29.904 = 0.01672 kmole
Volume at 30 °C, 760 mm Hg = 0.01672 ´ 303
(d) Specific gravity = density of gas/density of air = 1.202
1.300 = 0.925
3.21 A producer gas has the following composition CO2: 4.4%, CO : 23%, O2: 2.6% and N2: 70%. Calculate the following:
(a) volume of air at 25 °C and 750 mm Hg required for the combustion of 100 m3 of gas at the same condition if 25% excess air is used
(b) the composition and volume of gases leaving the burner at 350 °C and 750 mm Hg per 100 m3 of gas burnt.
Basis: 100 kmoles of producer gas.
CO = 23 kmoles Composition of gases leaving (volume % = mole %):
CO2: 19.37%, O2: 1.57% and N2: 79.06%
Volume of gases leaving = 141.475´ 22.414 ´ 623 760 273 750 È Ø È Ø
É Ù É Ù
Ê Ú Ê Ú
= 7332.92 m3
Volume of gases leaving/100 m3 of feed = 7332.92 100 2479.28
= 295.76 m3
3.22 Natural gas has the following composition: CH4: 94.1%, C2H6: 3%
and N2: 2.9%. This gas is piped from the well at 80°F and 80 psi.
Calculate the following.
(a) Partial pressure of N2
(b) Pure component volume of N2 per 100 ft3 of the gas
Component mole % lb mole Weight, lb
CH4 94.1 1.38 ´ 0.941 = 1.298 (1.298 ´ 16) = 20.777 C2H6 3.0 1.38 ´ 0.03 = 0.042 (0.042 ´ 30) = 1.260 N2 2.9 1.38 ´ 0.029 = 0.040 (0.040 ´ 28) = 1.120
Total 100.0 1.380 23.157
Density of gas = 23.157 100
È Ø
É Ù
Ê Ú = 0.23157 lb/ft3 Density at standard condition = 23.157
497
È Ø
É Ù
Ê Ú = 0.0466 lb/ft3
3.23 Compare pressures given by the ideal gas and van der Waals equation for 1 mole of CO2 occupying a volume of (381 ´ 10–6) m3 at 40 °C (a) Ideal gas law P = RT
nV = 6.831 ´ 106 N/m2
where, R = 8.314 N.m/g mole K (J/g mole K); T = 313 K;
V = 381 ´ 10–6 m3 (b) van der Waals equation
2
P = 6 5 6 2
3.24 It is desired to market O2 in small cylinders having volumes of 0.5 ft3 and exactly containing 1 lb of gas at 120 °F. What is the pressure?
Basis: 1 lb of O2 = 1/32 = 0.031 lb mole; R = 0.73 ft3 atm/lb mole °R T = 120 + 460 = 580 °R
P = nRT
V = 0.031 ´ 0.73 ´ 580
0.5 = 26.5 atm
3.25 A tire is inflated to 35 psig at 0 °F. To what temperature it can be heated up to a pressure of 50 psig, volume remaining same?
P1 = 35 + 14.67 = 49.67 psia; P2 = 50 + 14.67 = 64.67 psia.
\ Temperature = 140 °F
3.26 Calculate densities of C2H6 and air at NTP.
At NTP 1 g mole occupies 22,414 cc.
Density of C2H6 = 1 30
3.27 Acetylene gas is produced according to the reaction CaC2 + 2H2O ® C2H2 + Ca(OH)2
Calculate the number of hours of service that can be got from 1 lb of carbide in lamp burning 2 ft3 of gas/hour at 75 °F and 743 mm Hg.
A similar problem has been solved in MKS system also:
3.28 By electrolyzing a mixed brine a mixture of gases is obtained at the cathode having the following composition by weight: Cl2: 67%, Br2: 28%, O2: 5%. Calculate:
(a) composition by volume
(b) density at 25 °C and 740 mm Hg (c) specific gravity of the gas mixture.
Basis: 100 g of gas mixture (a)
Component Molecular Weight, g Weight, Volume %
weight g mole = mole %
Cl2 71 67 (67/71) = 0.945 74.0
Br2 160 28 (28/160) = 0.175 13.7
O2 32 5 (5/32) = 0.156 12.3
Total 100 1.276 100
(b) Volume of gases = 1.276 ´ 22.414 ´ 298 760
(c) Density of air = 1.293 g/litre (at standard condition) Density of air at 25 °C and 740 mm Hg Alternatively, volume of air at 25 °C and 740 mm Hg
= 22.414 ´ 760 298 740 273 È Ø È Ø
É Ù É Ù
Ê Ú Ê Ú = 25.13 litres
Density of air = 28.84
25.13 = 1.15 g/litre Specific gravity of gas mixture = 3.12
1.15 = 2.7
3.29 A mixture of NH3 and air at 730 mm Hg and 30 °C contains 5.1 NH3 by volume. This gas is passed at a rate of 100 ft3/min. through an absorption tower in which NH3 is removed. The gases leave the tower at 725 mm Hg and 20 °C having 0.05% NH3 by volume. Calculate:
(a) the rate of flow of gases leaving the tower and (b) weight of NH3 absorbed.
Basis: 100 ft3 of entering gases.
Volume of NH3 = 5.1 ft3 and of air = 94.9 ft3 (730 mm Hg, 30 °C)
3.30 1000 ft3 of moist air at 740 mm Hg and 30 °C contains water vapour in such proportions that its partial pressure is 22 mm Hg. Without the total pressure being changed, the temperature is reduced to 15 °C and some water condenses. After that the partial pressure of water is 12.7 mm Hg. Using partial pressure method, find the following:
(a) Volume of gas after cooling and (b) Weight of water condensed.
Basis: 1000 ft3 of moist air at 740 mm Hg and 30 °C Partial pressure of water = 22 mm Hg
Partial pressure of air = (740 – 22) = 718 mm Hg
Partial pressure of air after cooling = (740 – 12.7) = 727.3 mm Hg (a) Volume of air after cooling = 1000 ´ 718 288
727.3 303
È Ø È Ø
É Ù É Ù
Ê Ú Ê Ú = 938 ft3 After cooling volume of gases = Volume of water vapour + Dry air (740 mm, 15 °C) (12.7 mm, 15 °C) Volume of air leaving at inlet condition = 938 ´ 12.7 303
22 288
È Ø È Ø
É Ù É Ù
Ê Ú Ê Ú
= 570 ft3
Volume of water vapour condensed = 1000 – 570 = 430 ft3 (b) Water condensed = 430 22 273
359 760 303
È Ø È Ø È Ø
É Ù É Ù É Ù
Ê Ú Ê Ú Ê Ú ´ 18 = 0.562 lb 3.31 A producer gas has the following composition by volume CO : 23%,
CO2: 4.4%, O2: 2.6%and N2: 70%
(a) Calculate the ft3 of gas at 70 °F and 750 mm Hg per lb of carbon present.
(b) Calculate the volume of air required for the combustion of 100 ft3 of the gas if 20% excess air is used.
(c) Calculate the volumetric composition of gases leaving assuming complete combustion.
(d) Calculate the volume of gases leaving at 600 °F and 750 mm Hg per 100 ft3 gas burnt.
Basis: 100 lb moles of gas
i.e. carbon in the feed = 23 + 4.4 = 27.4 atoms
(Volume of gas/lb of carbon) = 39200 328.8
Air supplied = 10.68 ´ 100
Total = 139.35 lb moles
Composition of CO2 O2 N2
Volume % 19.66 1.28 79.06
(d) Volume of gases leaving
= 139.35 ´ 359 ´ 760 1060
3.32 A furnace is to be designed to burn coke at the rate of 200 lb/h having a composition C : 89.1% and ash : 10.9%. The grate efficiency of the furnace is such that 90% of the carbon present in the coke charged is burnt. Air supplied is 30% in excess of that required for complete combustion. It may be assumed that 97% of the carbon burnt is oxidized to carbon dioxide and the rest to carbon monoxide.
(a) Calculate the composition of the flue gases.
(b) If the flue gases leave the furnace at 550 °F and 743 mm Hg, calculate the rate of flow of gases in ft3/min.
Basis: 100 lb of coke.
C + O2 ® CO2
C : 89.1 lb. Ash : 10.9 lb.
Carbon burnt = 89.1 ´ 0.9 = 80.19 lb
Carbon burnt to CO2 = 80.19 ´ 0.97 = 77.78 lb = 6.48 lb moles
Component lb mole mole %
CO2 6.48 14.07 Hence, for 200 lb/h of coke charge, volume of flue gases is
= (34,722 ´ 2) = 69,444 ft3/h
i.e. Volumetric flow rate of flue gases = 1157.4 ft3/min.
3.33 In the fixation of nitrogen by the arc process, air is passed through a magnetically flattened electric arc. Some of the nitrogen is oxidized to NO, which on cooling, oxidizes to NO2. Of the NO2 formed, 66% will be associated to N2O4 at 26 °C. The gases are then passed into water washed absorption towers where nitric acid is formed.
3NO2 + H2O ® 2HNO3 + NO NO liberated in this reaction will be reoxidized in cooler.
In the operation of such a plant it is possible to produce gases from the arc furnace in which the NO is 2% by volume while hot. The gases are cooled to 26 °C at 750 mm Hg before entering the absorption column.
Basis: 100 g moles of air contain N2: 79 g moles and O2: 21 g moles.
(a) Reaction in arc furnace is given as N2 + O2 ® 2NO
(a) Calculate the composition of hot gases leaving the furnace assuming air is at NTP.
(b) Calculate the partial pressure of NO2 and N2O4 in the gas entering the absorption apparatus.
(c) Calculate the weight of acid formed per 1000 litres of gas entering the absorption system if the combustion to HNO3 of the combined nitrogen in the furnace gases is 85% complete.
Partial pressure of NO2 = 0.68 750
Total moles of gas in 1000 litres of entering gas
= PV
3.34 The gas leaving a gasoline stabilizer has the following analysis by volume C3H8: 8%, CH4: 78%, C2H6: 10% and C4H10: 4%
This gas leaving at 90 oF and 16 psia at the rate of 70,000 ft3/h is fed to gas reforming plant where the following reaction takes place.
CnH2n + 2 + nH2O ® nCO + (2n + 1)H2 (1)
CO + H2O ® CO2 + H2 (2)
Reaction (1) is 95% complete and Reaction (2) is 90% complete. Find (a) Average molecular weight of the gas leaving stabilizer;
(b) weight of gas fed to reforming plant (lb/h) (c) weight of H2 leaving (lb/h) and (d) composition of gases leaving (weight %) Basis: One hour = 70,000 ft3 of gas.
Components C3H8 CH4 C2H6 C4H10 Total
Volume % 8 78 10 4 100
lb mole 15.22 148.39 19.02 7.61 190.24
Molecular weight 44 16 30 58
Weight, lb 669.68 2374.24 570.6 441.38 4055.9 (a) Average molecular weight of the gas leaving stabilizer
= 4055.9
From Reaction (1) (hydrocarbons undergo 95% conversion), we have
(15.22 ´ 0.95 ´ 7) + (148.39 ´ 0.95 ´ 3) + (19.02 ´ 0.95 ´ 5) + (7.61 ´ 0.95 ´ 9) = 679.535 lb moles
Total CO formed = (262.53 ´ 0.95) = 249.40 lb moles Hydrogen from CO (90% conversion) = (249.40 ´ 0.90)
= 224.46 lb moles
\ Total hydrogen leaving reformer = (679.535 + 224.46) ´ 2 = 1808 lb/h
(d) Gases leaving unreacted = HC, H2O, CO, CO2, H2 H2 = 1808.000 lb
C3H8 15.22 ´ 0.05 ´ 44 = 33.484 lb CH4 148.39 ´ 0.05 ´ 16 = 118.712 lb C2H6 19.0 ´ 0.05 ´ 30 = 28.530 lb C4H10 7.61 ´ 0.05 ´ 58 = 22.069 lb CO 249.4 ´ 0.1 ´ 28 = 698.320 lb CO2 249.4 ´ 0.9 ´ 44 = 9876.240 lb H2O 262.53 ´ 0.05 ´ 18 = 236.277 lb
(from Reaction 1) H2O [4,725.54 – (224.46´ 18)] = 685.260 lb
(from Reaction 2) 13,506.892 lb
Component C3H8 CH4 C2H6 C4H10 CO CO2 H2O H2 Total Weight, lb 33.484 118.712 28.53 22.069 698.32 9,876.24 921.537 1808 13,506.892 Weight % 0.248 0.879 0.212 0.163 5.17 73.12 6.823 13.385 100.000
3.35 Analysis of a sewage gas sample from municipal sewage plant is given. CH4 : 68%, CO2: 30% and NH3: 2%. 600 m3/h of this gas at 30 oC and 2 atmospheres is flowing through a pipe. Find (a) the average molecular weight of the sewage gas and (b) the mass rate of flow of gas in kg/h and (c) density of the gas.
Basis: 100 kmoles of the sewage gas.
(a) the average molecular weight can be calculated from the following table.
Gas Molecular weight Weight, kmole Weight, kg
CH4 16 68 68 ´ 16 = 1,088
CO2 44 30 30 ´ 44 = 1,320
NH3 17 2 2 ´ 17 = 34
Total — 100 2,442
Average molecular weight = 2, 442
100 = 24.42 (b) Volumetric flow rate at standard condition
= 600 ´ 273 2
303 1
È Ø È Ø É Ù É Ù
Ê Ú Ê Ú = 1081.2 m3/h
Molar flow rate of gases = 1081.2 22.414
È Ø
É Ù
Ê Ú = 48.24 kmoles/h We know that 100 kmoles of this gas weighs = 2,442 kg Therefore, the weight of 48.24 kmoles
= 2, 442 48.24 100
= 1,178 kg.
Hence, the mass flow rate of gas in kg/h = 1,178 kg/h (c) Density of the gas is 2442
1081.2 = 2.26 kg/m3.
3.36 In the process of manufacturing Cl2, HCl gas is oxidized with air as follows:
4HCl + O2 ® 2Cl2 + 2H2O
If the air used is 30% excess and oxidation is 80% complete, find the composition of dry gases leaving.
Basis: 4 kmoles HCl gas.
O2 needed = 1 kmole O2 supplied = 1.3 kmoles N2 entering = 1.3 ´ 79
21 È ØÉ Ù
Ê Ú = 4.89 kmoles O2 remaining = (1.3 – 0.8) = 0.5 kmole HCl un-reacted = (4 ´ 0.2) = 0.8 kmole Cl2 formed = (2 ´ 0.8) = 1.6 kmoles
Composition of dry gases leaving is presented in the following table
Gas mole mole %
HCl 0.80 10.27
O2 0.50 6.42
N2 4.89 62.77
Cl2 1.60 20.54
Total 7.79 100.00
3.37 A mixture of NH3 and air at 730 mm Hg and 30 °C contains 5.1%
NH3. The gas is passed through an absorption tower at the rate of 100 m3/h where NH3 is removed. The gases leave the tower at 725 mm Hg and 20 °C having 0.05% NH3. Calculate (a) the rate of flow of gas leaving the tower and (b) weight of NH3 absorbed in kg/h.
Basis: One hour of operation
Moles of gas entering per hour = 100 730 273 22.414 760 303
3.663 kmoles of air contains 0.05% NH3 while leaving the absorber
3.663 kmoles of air contains 0.05% NH3 while leaving the absorber