katom of carbon =
wt .in kg atomic weight of carbon
=
36 12
=
3
(2) Calculate the kilograms of ‘Na’ of which the amount is specified as 3 katom. Solution: 3 katom Na Atomic weight of Na = 23
katom of Na =
∴
wt .in kg of Na atomic weight of Na
Kg of Na = katom of Na x Atomic weight of Na = 3 x 23 = 69
(3) How many kilograms of ethane are there in 210 kmol? Solution: Basis: 210 kmol ethane. Atomic weights: C=12, H=1, Chemical formula of ethane = C 2H6 Molecular weight of ethane = 2x12+1x6 = 30
kmol
∴
of C 2 H 6
=
wt .in kg of C 2 H 6 mol . weight of C 2 H 6
kg of ethane (C 2H6) = kmol of ethane x Molecular weight of ethane = 210 x 30 = 6300 kg
∴ 210
kmol of ethane = 6300 kg ethane
(4) Convert 88 kg of carbon dioxide into its amount in molar units. Solution: Basis: 88 kg of carbon dioxide Molecular formula of carbon dioxide = CO 2 Atomic weights: C=12, O=16 Molecular weight of CO 2 = (1x12) + (2x16) = 44
kmol of CO 2 =
88 g of CO2 = 2 kmol CO2
wt .in kg of C O 2 mol . weight of CO 2
=
88 44
= 2
katom of carbon =
wt .in kg atomic weight of carbon
=
36 12
=
3
(2) Calculate the kilograms of ‘Na’ of which the amount is specified as 3 katom. Solution: 3 katom Na Atomic weight of Na = 23
katom of Na =
∴
wt .in kg of Na atomic weight of Na
Kg of Na = katom of Na x Atomic weight of Na = 3 x 23 = 69
(3) How many kilograms of ethane are there in 210 kmol? Solution: Basis: 210 kmol ethane. Atomic weights: C=12, H=1, Chemical formula of ethane = C 2H6 Molecular weight of ethane = 2x12+1x6 = 30
kmol
∴
of C 2 H 6
=
wt .in kg of C 2 H 6 mol . weight of C 2 H 6
kg of ethane (C 2H6) = kmol of ethane x Molecular weight of ethane = 210 x 30 = 6300 kg
∴ 210
kmol of ethane = 6300 kg ethane
(4) Convert 88 kg of carbon dioxide into its amount in molar units. Solution: Basis: 88 kg of carbon dioxide Molecular formula of carbon dioxide = CO 2 Atomic weights: C=12, O=16 Molecular weight of CO 2 = (1x12) + (2x16) = 44
kmol of CO 2 =
88 g of CO2 = 2 kmol CO2
wt .in kg of C O 2 mol . weight of CO 2
=
88 44
= 2
(5) Find the moles of oxygen present in 500 grams Solution: Basis: 500 g of oxygen Molecular weight of O 2 = 2 x 16 = 32 kmoles
of O 2 =
500 32
=
15 . 625
(6) Convert 499 g of CuSO 4.5H2O into moles. Solution: Basis: 499 g of CuSO 4.5H2O Atomic weights: Cu=63.5, S=32, O= 16 and H = 1 Molecular weight of CuSO 4.5H2O = (1x63.5)+(1x32)+(4x16)+5(2x1+1x16) = 249.5 Moles of CuSO 4.5H2O = 499 = 2 mol 249.5 kmoles
of CuSO 4 .5H 2 O =
499 249 . 5
=
2
The relationship of a compound and its constituents is given for some compounds as follows: Each mole of NaOH contains one atom of Na 1 mol of NaOH≡ 1 atom of Na ≡ 1 atom of H Each mole of NaOH contains 1 atom of Na. The sign ≡ refers to ‘equivalent to’ and not ‘equal to’. Similarly for H 2SO4 and ‘S’ 1 mol of H2SO4 = 1 atom of S (atom is written for gram-atom). 1 mol of H2SO4 = 1 atom of S (atom is written for gram-atom) 1 kmol of H2SO4 = 1 katom of S (katom is written for kilogram-atom) i.e., each mole of H2SO4 contains 1 atom of S. For CuSO4.5H2O and CuSO4 1 mol of CuSO 4.5H2O ≡ 1 mol CuSO4 1 kmol of CuSO 4 . 5H2O ≡ 1 kmol of CuSO 4
(7) How many kmoles of H 2SO4 will contain 64 kg S
Solution: Basis: 64 kg of S Atomic weight of S = 32
atoms
of S =
kg of S katom
=
64 32
=
2
Each moles of H 2SO4 contains one atom of S. 1 kmol of H2SO4≡ 1 katom of S moles of H 2 SO 4 =
1 1
x 2 = 2 kmol
(8) Find kmoles of K 2CO3 that will contain 117 kg of K? Solution: Basis: 117 kg of K Atomic weight of K = 39 atoms of K =
117 39
=
3 katom
Each mole of K 2CO3 contains 2 atom of K 2 atom of K ≡ 1 mole of K 2CO3 2 katom of K ≡ a kmol of K2CO3 moles
of K 2 CO 3 =
1 2
x 3 = 1 . 5 kmol
(9) How many kilograms of carbon are present in 64 kg of methane? Solution: Basis: 64 kg of methane Atomic weight of C = 12 Molecular weight of CH 4 = 16 1 katom of carbon ≡ 1 kmol of CH4 ∴ 12
kg of carbon ≡ 16 kg of CH 4 i.e., in 16 kg of CH 4, 12 kg of carbon are present.
So amount of carbon present in 64 kg of methane = 12 x 64 16
=
48
(10) Find equivalent moles of Na 2SO4 in 644 kg of Na 2SO4.10H2O crystals Solution: Basis: 644 kg of Na 2SO4.10H2O crystals Molecular weight of Na 2SO4 = 2x23+1x32+4x16 = 142 Molecular weight of Na 2SO4.10H2O = (2x23)+1x32+4x16+10 (2x1+1x16) = 322
moles of Na 2 SO 4 .10H 2 O =
644 322
=
2 mol
The relationship betweenNa 2SO4 and Na2SO4.10H2O is : 1 mol Na2SO4.10H2O ≡ 1 mol Na2SO4 ∴
2 mol of Na2SO4.10H2O ≡ ?
Moles of Na2SO4 = 2 mol In the formula of Na 2SO4.10H2O, the moles of Na 2SO4 are equal (one in each) and hence, the equivalent moles of Na 2SO4 in crystals are 2 mol. (11)
Find nitrogen (N) content of 100 kg urea sample containing 96.43% pure urea. Solution: Basis: 100 kg of urea sample It contains 96.43 kg of urea Molecular weight of urea [NH 2CONH2] = 60 1 kmol of NH 2CONH2≡ 2 katom of N 60 kg of NH2CONH2≡ 28 kg of N ∴96.43
kg of NH 2CONH2≡ ?
Nitrogen content of 100 kg sample = 28 x 96 . 43 60
=
45 kg
(12) Find kilograms of C 2H6 that contain 4 katom of carbon. Solution: Basis: 4 katom carbon Molecular weight of C 2H6 = 30 Relationship between C 2H6 and ‘C’ is 2 katom of C ≡ 1 kmol of C2H6 ∴ 4
katom of C ≡ ? Moles of C 2H6 = 1 x 4 = 2 kmol Weight of C 2H6 = kmol of C 2H6 x Molecular weight of C 2H6 = 2 x 30 = 60 kg (13) How many grams of carbon are present in 264 g of CO 2? Solution: Basis: 264 g of CO 2 Atomic weight of C = 12, Molecular weight of CO 2 = 44 1 mol of CO2≡ 1 atom of C 44 g of CO 2≡ 12 g of C ∴264
g of CO2≡ ?
Amount of carbon present in 264 g of CO 2 = 12 x 264 44
=
72 g
EQUIVALENT WEIGHT
It is defined as the ratio of the atomic weight or molecular weight to its valence.The valence of an element or a compound depends on the numbers of hydroxyl ions (OH ) donated or the + hydrogen ions (H ) accepted for each atomic weight or molecular weight. Equivalent weight = Molecular weight Valence equivalent
weight
=
molecular
weight
valence
NORMALITY
It is designated by the symbol ‘N’. It is defined as the number of gram equivalents of solute dissolved in one litre of solution.
Normality (N) = gram-equivalents of solute= gram of solute_____________________ Volume of solution in Litre eq wt. of solute x Volume of solution in Litre
Normality =
where
gequivalen ts of solute volume of solution in L
concentrat ion (g / L ) =
=
g of solute eqwt . of solute x volume of solution in L
g of solute volume of solution in L
By using definition of normality, it is possible to find out the concentration of solute in g/l. Concentration (g/L) = Normality x Equivalent weight
MOLARITY It is defined as the number of gram moles of solute dissolved in one litre of solution.
Designated by the symbol ‘M’.
Molarity ( M ) =
g of solute volume of solution in L
MOLALITY It is defined as the number of gram-moles (mol) of solute dissolved in one kilogram of solvent.
Molality ( M ) =
gmoles of solute mass of solvent in kg
Ppm and ppb
1ppm =
1g solute 1 g solute 1 mg solute = = 10 6 g solution 10 6 L solution 1 L solution
since density of solution(very dilute) = 1 g/cc
1ppb =
1g solute 1 g solute = 10 9 g solution 10 9 L solution
(14) Find the equivalent weights of (1) HCl, (2) NaOH, (3) Na 2CO3 and (4) H2SO4 Solution: (1) HCl: Molecular weight of HCl = (1x1)+(1x35.5) = 36.5 Valence of HCl = 1
equivalent weight of HCl =
mol wt of HCl valence of HCl
=
36 .5 1
(2) NaOH: Molecular weight of NaOH = (1x23)+(1x16)+(1x1) = 40 Valence of NaOH = 1
equivalent weight of NaOH =
40 1
= 40
= 36 .5
1 (3) Na2CO3 : Molecular weight of Na 2CO3=(2x23)+(1x12)+(3x16) = 106 Valence of Na2CO3 = 2
equivalent weight of Na 2 CO 3 =
mol wt of Na 2 CO 3 valence of Na 2 CO 3
=
106 2
= 53
(4) H2SO4 Molecular weight of H 2SO4 = (2x1) + (1x32) + (4x16) = 98 Valence of H 2SO4 = 2
equivalent weight of H 2 SO 4 =
98 2
= 49
(15) Calculate the equivalent weights of the following compounds: (1) H3PO4, (2) CaCl2, (3) FeCl3, (4) Al2(SO4)3, and (5) KMnO4 Solution: (1) H 3PO4 Molecular weight of H 3PO4 = (3x1)+(1x31)+(4x16) = 98 Valence of H 3PO4 = 3
equivalent weight of Na 2 CO 3 =
98 3
= 32 . 67
(2) CaCl2 Molecular weight of CaCl 2 = (1x40)+(2x35.5) = 111 Valence of CaCl2 = 2
equivalent weight of CaCl 2 =
mol wt of CaCl 2 valence of CaCl 2
=
111 2
= 55 . 5
(3) FeCl3: Molecular weight of FeCl 3 = (1x56)+(3x35.5) = 162.5 Valence of FeCl 3 = 3 ∴ Equivalent
weight of FeCl 3 = 162.5 = 54.17 3
equivalent weight of FeCl 3 =
(4) Al2(SO4)3:
mol wt of FeCl 3 valence of FeCl 3
=
162 .5 3
= 54 .17
Molecular weight of Al 2(SO4)3 = (2x27)+(3x32)+(12x16) = 342 Valence of Al 2(SO4)3 = 6
equivalent weight of Al 2 (SO4) 3 =
mol wt of Al 2 (SO4) 3 valence of Al 2 (SO4) 3
=
342 6
= 57
(5) KMnO4: Molecular weight of KMnO 4 = (1x39)+(1x55)+(4x16) = 158 Valence of KMnO4 = 5,
equivalent weight of Al 2 (SO4) 3 = (16)
mol wt of KMnO 4 valence of KMnO 4
=
158 5
= 31.6
98 grams of sulphuric acid (H 2SO4) are dissolved in water to prepare one litre of solution. Find normality and molarity of the solution. Solution: Basis: One litre of solution Amount of H2SO4 dissolved = 98 g Molecular weight of H 2SO4 = 98
equivalent weight of H 2SO 4 =
mol wt of H 2SO 4 valence of H 2SO 4
=
98 2
= 49
Gram-equivalents (g eq) of H 2SO4 = 98 = 2 49 g equivalent
s of H 2 SO 4 =
Normality (N) =
g eq of H 2SO 4 vol.of H 2 SO 4 sol. in L
Molarity (M) =
(17)
wt of H 2 SO 4 98 = =2 eq wt of H 2 SO 4 49
=
2 1
=2
moles of H 2 SO 4 1 = =1 vol.of H 2SO 4 sol. in L 1
Find grams of HCl needed to prepare 1 litre 2N HCl solution. Solution: Basis: 1 litre 2N HCl solution.
Normality
(N)
=
2 =
g eq of HCl vol .of H Cl sol . in L wt of HCl in g eq wt . of HCl x volume
of solution
in L
wt of HCl in g = 2 x 36 . 5 x 1 = 73 g
(18)
The concentration of an aqueous solution of acetic acid is specified as 30% on weight basis. Find the molarity of the solution. Solution: Basis: 100 kg of solution. Amount of acetic acid in it = 30 kg Amount of water (solvent) in it = 70 kg 3
Amount of acetic acid = 30 x 10 g Molecular weight of CH 3COOH = 60 moles molality
(19)
of acetic =
acid
moles kg
=
x 10 3 60 acid
30
of acetic of solvent
= 500 =
500 70
mol = 7 . 142
m
A solution of caustic soda contains 20% NaOH by weight. Taking density of the solution as 1.196 kg/L find the normality, molarity, and molality of the solution. Solution: Basis: 100 kg of solution. The solution contains 20 kg of NaOH and 80 kg water (solvent). Density of solution = 1.196 kg/L 100 = = 83.62L volume of solution 1.196 Molesof NaOHin solution=
Molarity(M) =
20 40
g molesof NaOH volumeof solution in L
=
= 0.5 kmol = 500 mol
500 83.52
= 5.98 M
For NaOH as valence = 1 Equivalent weight = Molecular weight ∴ Normality
(N) = Molarity (M) = 5.98
Molality(m) =
(20)
g molesof NaOH kg of solvent
500 80
= 6.25 m (mol/kg)
A chemist is interested in preparing 500 ml of 1 normal, 1 molar and 1 molal solution 3 of H2SO4. Assuming the density of H 2SO4 solution to be 1.075 g/cm , calculate the quantities of H 2SO4 to be taken to prepare these solutions. Solution: Basis: 500 ml of solution. Volume of solution = 500 ml = 0.5 L Normality (N) =
∴
=
g equivalent s of H 2 SO 4 volumw of solution in L
gram-equivalents of H 2SO4 = Normality x Volume of solution = 1 x 0.5 = 0.5 g eq Molecular weight of H 2SO4 = 98 equivalent weight of H2SO4 =
98 2
= 49
Amount of H2SO4 required for 1 normal solution = 0.5 x 49 = 24.5 g
molarity =
gmoles of H 2 SO 4 vol. of solution in L
Moles of H2SO4 = Molarity x Volume of solution = 1 x 0.5 = 0.5 mol Amount of H2SO4 required to make 1 molar solution = 0.5 x 98 = 49 g Let x be the quantity in grams of H 2SO4 required for making 1 molal solution. Density of solution = 1.075 g/cc Quantity of solution = 500 x 1.075 = 537.5 g grams of solvent = grams of solution – grams of solute = 537.5 – x -3 Weight of solvent = (537.5 - x) x 10 kg Molecular weight of H 2SO4 = 98 moles of H2SO4 =
x 98
Molalityof H 2SO 4 =
1=
gmoles of solute wt of solventin kg
x / 98 (537.5 - x) .10 -3
Solving we get, x = 47.97 g Amount of H2SO4 required for preparing 1 molal solution = 47.97 g. (21)
2000 ml solution of strength 0.5 N H 2SO4 is to be prepared in laboratory by adding 98% H2SO4 (sp.gr.1.84) to water. Calculate the volume in ml of 98% to be added to get solution to required strength. Solution: Basis: 2000 ml of 0.5 N H 2SO4 solution. Normality (N) =
g equivalent s of H 2 SO 4 volume of solution in L
Volume of solution = 2000 ml = 2 L gram equivalents of H 2SO4 = Normality x Volume of solution in L = 0.5 x 2 = 1 g eq Amount of H2SO4 required = 1 x 49 = 49 g amount of 98% H 2 SO 4 required =
49 0.98
= 50 g
Specific gravity of 98% H 2SO4 = 1.84 volume of 98% H 2 SO 4 required =
(22)
g of 98% H 2 SO 4 50 = = 2 . 72 ml specific gravity 1 . 84
A H2SO4 solution has a molarity of 11.24 and molality of 94. Calculate the density of solution. Solution: Basis: 1 litre of solution Molarity = 11.24 and Molality = 94 Now,
Molalityof H 2SO 4 =
gmoles of solute wt of solventin kg
11.24 =
gmoles of H 2SO 4 1
∴Moles
of H2SO4 = 11.24 x 1 = 11.24 mol Amount of H2SO4 = 11.24 x 98 = 1101.52 g = 1.101 kg Molality = mol H2SO4 / kg of solvent
amount of solution =
amount of solvent =
gmoles of H 2SO 4 1
11.24
= 0.1195 kg 94 Amount of solution = 1.101 + 0.1195 = 1.2205 kg
density of H 2SO 4 solution =
(23)
1.2205
= 1.2205 kg / L 1 0 2 litres of NH 3 at 303 K (30 C) and 20.265 kPa is neutralized by 135 ml of solution of H2SO4. Find the normality of the acid. Solution: Basis: 2 litres of NH 3 PV = nRT PV n = moles of NH 3 = RT Where P = 20.265 kPa, T= 303 K, V = 2 l 3 R = 8.31451 m .kPa/(kmol.K) = 8.31451 kPa/(mol.K)
n = moles of NH 3 =
20.265.(2) 8.314.(303)
=
0.0161 m
2 NH3 + H2SO4 (NH4)2SO4 For neutralization of 2 moles of NH 3, 1 mole of H2SO4 is required. 1 moles of H 2SO 4 required = .(0.0161) = 8.045 x10 − 3 2 Amount of H2SO4 required -3 = 8.045 x 10 x98= 0.788 g 0.788 = 0.0161 gequivalents of H 2SO 4 required = 49
Volume of H2SO4 solution = 135 ml = 0.135 L
N of H 2SO 4 =
(24)
0.0161 0.135
=
0.12 N
A sample of Na2CO3. H2O weighing 0.62 grams is added to 100 ml of 0.1 H2SO4solution. Will the resulting solution be acidic, basic or neutral? (Get the answer by numerical method.) Solution: Basis: 100 ml of 0.1 N H 2SO4solution and 0.62 g of Na 2CO3. H2O sample. Normality of H 2SO4 solution = 0.1 N Volume of H2SO4 solution = 100 ml = 0.1 l Gram equivalents of H 2SO4 = Normality x Volume of solution = 0.1 x 0.1 = 0.01 Amount of H2SO4 in solution = 0.01 x 49 = 0.49g Molecular weight of Na 2CO3 = 124, Molecular weight of H 2SO4 = 98 Na2CO3. H2O + H2SO4 Na2SO4 + 2H2O + CO2 1 mol Na2CO3.H2O = 1 mol H2SO4 124 g Na2CO3.H2O = 98 g H2SO4 ∴ 0.62
Na2CO3.H2O = ?
∴ Amount
of H2SO4 required for 0.63 g Na 2CO3.H2O =
98
(0.62) = 0.49 g 124 For neutralizing, we need 0.49 g H 2SO4 as per reaction and we have added solution containing 0.49 g H 2SO4. None of the components is in excess. Hence, resulting solution is neutral.
(25)
Do the following conversions: a. 294 g/L H 2SO4 to normality, b. 5N H 3PO4 to g/L, c. 54.75 g/L HCl to molarity, d. 3M K2SO4 to g/L,e. 4.8 mg/ml CaCl 2 to normality Sol: (a) 294 g/l H 2SO4 to normality Basis: 1 litre of solution. H2SO4 in solution = 294 g 294 = 6 geq gequivalen ts of H 2 SO 4 required = 49
N=
6 1
=
6N
(b) 5N H3PO4 to g/L Molecular weight of H 3PO4 = 98 Valence of H 3PO4 = 3 equivalent weight of H 3 PO 4 required =
98
= 32.67 3 Concentration in grams per litre of solution = Normality x Equivalent weight Concentration of solution = 5 x 32.67 = 163.35 g/L (c) 54.75 g/L HCl to molarity Basis: 1 l of solution
Amount of HCl in g/l = 54.75 g
moles of HCl =
54.75 36.5
= 1.5 mol
molarity HCl solution
=
1 .5 1
=
1 .5 M
(d) 3M K2SO4 to g/L Basis: 1 l of solution. Moles of solution = 3 x 1 = 3 mol Molecular weight of K 2SO4 = 174 ∴ Amount
of K2SO4 = 3 x 174 = 522 g
molarity HCl solution =
522
=
1
522 g / L
(e) 4.8 mg/ml CaCl2 to normality Basis: 1 L of solution 4.8 mg/ml CaCl2 = 4.8 g/l CaCl2 CaCl2 in solution = 4.8 g 111 = 55 . 5 Molecular weight of CaCl 2 = 2 48 = 0 . 0865 Gram equivalents of CaCl 2 = 55 . 5 0 . 0865 = 0 . 0865 Normality = = 1 (26)
An aqueous solution of K 2CO3 is prepared by dissolving 43 kg of K 2CO3 in 100 kg of 0 water at 293 K (20 C). Calculate molarity, normality, molality of solution. Density of solution is 1.3 kg/L. Solution: Basis: 43 kg of K 2CO3 and 100 kg of water Weight of K 2CO3 solution = 43 + 100 = 143 kg 143 = 110 . L Volume of solution = 1 .3 Molecular weight of K 2CO3 = 138
∴
Equivalent weight of K 2CO3 =
∴ Moles
of K2CO3 in solution =
Molarity of solution =
43 138
311 . 6 110
138 2
=
69
=
0 . 3116 kmol = 311 . 6 mol
=
2 . 832 M
Gram equivalents of K 2CO3
=
43 .(1000 )
Normality of solution = = Molality of solution =
69 623 . 19
110 311 . 6 100
=
=
=
623 . 19 eq
5 . 665 N
0 . 3116 N
METHODS OF EXPRESSING THE COMPOSITION OF MIXTURES AND SOLUTIONS There are various methods used in expressing the composition of mixtures and solutions.The methods are being explained by considering the system composed of two components namely, A and B and is also used for more than two components. The composition of a mixture or solution are expressed in weight percent, volume percent, mole percent. WEIGHT PERCENT The weight of any component expressed as a percentage of the total weight of the system. weight of A WA = ∴ Weight % of A = x 100 − − − − − − (1) total weight of system W
Where, WA = weight of the component A W= WA + WB = weight of the system …. for a binary system of A and B. Weight percent of component A present in system is defined as the weight of the component A expressed as a percentage of the total weight of the system. VOLUME PERCENT The pure component volume of any component is expressed as a percentage of the total volume of the system. VA pure component volume of A = x 100 − − − − − ( 2 ) Volume % of A = total volume of system V Where, VA = pure component volume of A, V B = total volume of the system VA+VB … for a binary system of A and B. MOLE FRACTION The ratio of the number of moles of a particular component to the total moles of a system. ∴ For
a binary system of A and B : moles of A Mole fraction of A = total moles of system
xA =
W A M A W A M A
+
W B
− − − − − − − − − − ( 3)
M B
Where, XA = mole fraction of A XB = mole fraction of B.
xB =
W B MB W A M A
+
W B
− − − − − − − − − −(4)
M B
From equation (3) and (4), we have Mole % of A = Mole fraction of A x 100 Adding equations (3) and (4), we get
xA =
W A M A W A M A
+
W B
M B
+
W B MB W A M A
+
W B
= 1 − − − − − − − (5)
M B
∴the sum of the mole fractions of all the components present in a given system is equal
to unity, or
NOTE: The sum of all the mole percentage for a given system totals to 100. WEIGHT FRACTION The ratio of the weight of a particular component to the total weight of the system. ∴In a binary system of A and B :
x 'A =
WA − − − − − − − − − −(6) WA + WB
Where xA is the weight fraction of A. Similarly for B,
x 'B =
WB − − − − − − − − − −(7) WA + WB
From equations (6) & (7), we get
x 'A =
WA WB + =1 WA + WB WA + WB
Also, Weight % of A = Weight fraction of A x 100 ∴The sum of the weight fractions of all the components present in a given system is
equal to unity, or
NOTE: The sum of all the weight percentages for a given system totals to 100.
(28) An aqueous solution of sodium chloride is prepared by dissolving 25 kg of sodium chloride in 100 kg of water. Determine (a) weight % and (b) mole % composition of solution. Solution:Basis: 25 kg of sodium chloride and 100 kg of water Amount of solution = 25 + 100 = 125 kg Weight % NaCl in solution =
weight of NaCl in kg total weight of solution in kg
=
25 125
x 100 = 20 %
Weight % H2O = 100 – Weight % of NaCl = 100 – 20 = 80 % Molecular weight of NaCl = 58.5, Molecular weight of H2O = 18. ∴
Moles of NaCl =
Moles of H2O =
10 18
= 5 . 56
25 58 . 5
= 0 . 427
kmol
kmol
Total moles of solution = 0.427 + 5.56 = 5.987 kmol Mole % NaCl in solution =
kmol NaCl kmol of solution
=
0 . 427 5 . 987
x 100 = 7 . 13 %
Mole % H2O in solution = 100 – Mole % of NaCl = 100 – 7.13 = 92.87 % (29) Sodium chloride weighing 200 kg is mixed with 600 kg potassium chloride. Calculate the composition of the mixture in (i) weight % and (ii) mole % Solution: Basis: 200 kg NaCl and 600 kg KCl Weight of NaCl in the mixture = 200 kg Weight of KCl in the mixture = 600 kg Total weight of the mixture = 600 + 200 = 800 kg ∴
Weight % of NaCl in the mixture =
weight of NaCl in kg total weight of mixture in kg
200
=
800
x 100 = 75 %
Weight % KCl = 100 – weight % of NaCl = 100 – 25 = 75% Molecular weight of NaCl = (1 x 23) + (1 x 35.5) = 58.5 Molecular weight of KCl = 1 x 39 + 1 x 35.5 = 74.5 Moles of NaCl =
Moles of KCl=
200 58 . 5 600 74 . 5
=
=
3 . 419 kmol
8 . 05 kmol
Total moles of the mixture = 3.419 + 8.05
Mole % NaCl =
kmol of NaCl kmol of mixture
=
= 11.469 kmol
3 . 419 11 . 469
x 100 = 29 . 81 %
Mole % KCl = 100 – mole % NaCl 100 – 29.81 = 70.19% (30) A saturated solution of salicylic acid (HOC6H4COOH) in methanol (CH3OH) contains 0 64 kg salicylic acid per 100 kg methanol at 298 K (25 C). Find the composition of solution in (a) weight % and (b) mole % Sol: Basis: 100 kg of methanol in the saturated solution Amount of salicylic acid corresponding to 100 kg methanol in saturated solution Weight of solution = 100 + 64 = 164 kg ∴
Weight % salicylic acid solution =
64 164
x 100 = 39 . 02 %
Weight % methanol = 100 – 39.02 = 60.98% Molecular weight of CH3OH = 32, Molecular weight of HOC6H4COOH = 138 Mole of methanol = 100/32 = 3.125 kmol Moles of Salicylic acid = 64/138 = 0.464 kmol
Total amount of solution = 3.125 + 0.464 = 3.589 kmol Mole % of methanol =
3 . 125
x 100 = 87 . 07 % 3 . 589 Mole % of salicylic acid = 100 – 87.07 = 12.93% 0
(31) At 298 K (25 C) the solubility of methyl bromide in methanol is 44 kg per 100 kg. Calculate (i) the weight fraction and (ii) the mole fraction of methanol in the saturated solution. Solution: Basis : 100 kg of methanol in the saturated solution Solution contains 44 kg of methyl bromide Weight of the saturated solution = 100 + 44 = 144 kg Weight fraction of methanol in the saturated solution =
100
= 0 . 694 144 Molecular weight of CH3OH = 32, Molecular weight of CH3Br = 94.91 Moles of CH3OH in solution = 100/32 = 3.125 kmol
Moles of CH3Br in solution = =
44 94 . 91
= 0 . 4636
kmol
∴ Total moles of solution = 3.125 + 0.4636 = 3.5886 kmol
Mole fraction of methanol in saturated solution=
3 . 125 3 . 5886
=
0 . 871
(32) Ethanol and water forms aazeotrope containing 96% ethanol by weight. Find the composition of azeotrope by mole %. Solution: Basis: 100 kg of ethanol-water mixture It contains 96 kg of ethanol and 4 kg of water Molecular weight of H2O = 18. Molecular weight of C2H5OH = 46 Moles of ethanol = 96/46 = 2.087 kmol Moles of water = 4/18 = 0.222 kmol Moles of azeotrope mixture = 2.087 + 0.222 = 2.309 kmol Mole % ethanol in azeotrope mixture = =
2 . 087 2 . 309
x 100 = 90 . 38 %
(33) Nitric acid and water forms maximum boiling azeotrope containing 62.2 % water by mole. Find the composition of the azeotrope in weight %. Solution: Basis: 100 kmol of HNO3 + water azeotrope It contains 62.2 kmol of H2O and 37.88 kmol of HNO 3 Molecular weight of HNO3 = 63. Molecular weight of H2O = 18 Amount of HNO 3 in azeotrope = 37.8 x 63 = 2381.4 kg Amount of H2O in azeotrope = 62.2 x 18= 1119.6 kg Amount (weight) of azeotrope = 2381.4 + 1119.6 = 3501 kg
Weight % HNO3 in azeotrope = =
2381 . 4 3501
x 100 = 68 . 02 %
(34) An aqueous solution contains 15% ethanol by volume. Find the weight % ethanol, if 3 densities of ethanol and water are 0.79 g/cm respectively. 3 Solution: Basis: 100 cm of aqueous solution. 3 Volume of ethanol in solution = 0.15 x 100 = 15 cm 3 Volume of water in solution = 0.85 x 100 = 85 cm Amount of ethanol in solution = Volume x Density = 15x 0.79 = 11.85 g Amount of water in solution = 85 x 1 = 85 g ∴ Total amount of solution = 11.85 + 85 = 96.85 g ∴
Weight % ethanol in solution = =
11 . 85 96 . 85
x 100 = 12 . 23 %
(35) The available nitrogen (N) in the urea sample is found to be 45 % by weight. Calculate the actual urea content in the sample. Solution: Basis: 100 kg of urea sample. It contains 45 kg of nitrogen as N Atomic weight of N = 14. Molecular weight of NH2CONH2 = 60 1 kmol NH2CONH2≡ 2 katom N ∴ 60 kg NH2CONH2 ≡ 28 kg N ∴ Actual urea in sample = =
60 28
x 45 = 96 . 43 kg
Weight % urea content of sample =
kg urea kg sample
x 100 =
96 . 43 100
x100 = 96 . 43 %
(36) The nitrogen content of NH4NO3 sample is given as 34.5 % by weight. Find the actual ammonium nitrate content of the sample. Solution: Basis: 100 kg of sample It contains 34.5 kg of nitrogen as N Molecular weight of NH4NO3 = 80, Atomic weight of N = 14 1 kmol NH4NO3≡ 2 katom N ∴ 80 kg NH4NO3≡ 28 kg N ∴ Amount of NH 4NO3 in sample =
% NH4NO3 content of sample =
80 28
98 . 57 100
x 34 . 5 = 98 . 57 kg
x 100 =
∴ Actual NH4NO3 content of sample is 98.57 %
98 . 57 100
x 100 = 98 . 57 %
(37) Calculate the available nitrogen content of solution having 30 % urea (NH2CONH2), 20 % ammonium sulphate and 20 % ammonium nitrate. Solution: Basis: 100 kg of solution. It contains 30 kg urea, 20 kg of ammonium sulphate and 20 kg ammonium nitrate. Molecular weight of urea (NH2CONH2) = 60 1 kmol NH2CONH2 ≡ 2 katom N 60 kg NH2CONH2≡ 28 kg N ∴ Nitrogen available from urea =
28
x 30 = 14 kg
60
Molecular weight of (NH4)2SO4 = 132 1 kmol (NH4)2SO4 ≡ 2 katom N 132 kg (NH4)2SO4≡ 28 kg N (on weight basis) ∴ Nitrogen available from (NH4)2SO4 =
28 132
x 20 = 4 . 24 kg
Molecular weight of NH4NO3 = 80 1 kmol NH4NO3≡ 2 katom N 80 kg NH4NO3≡ 28 kg of N (on weight basis) ∴ Nitrogen available from NH4NO3 =
28 80
x 20 = 7 kg
∴ Total nitrogen available from the solution= 14 + 4.24 + 7 = 25.24 kg ∴ Nitrogen available from the solution =
25 . 24 100
x 30 = 25 . 24 %
(38)
Spent acid from fertilizer plant has the following composition by weight: H2SO4 = 20%, NH4SO4 = 45%, H2O = 30% and organic compounds = 5%. Find the total acid content of the spent acid in terms of H2SO4 after adding the acid content, chemically bound in ammonium hydrogen sulphate. Solution: Basis: 100 kg of spent acid It contains 20 kg of H2SO4 and 45 kg of NH4SO4 1 kmol NH4SO4 ≡ 1 kmol H2SO4 115 kg NH4SO4 ≡ 98 kg H2SO4 ∴ H2SO4 chemically bound in NH4SO4 =
98 115
x 45 = 38 . 35 kg
H2SO4 from the spent acid = free H 2SO4 + H2SO4 chemically bound in NH4SO4 = 20 + 38.35 = 58.35 kg Total acid content of the spent acid as weight % H2SO4
58 . 35 100
x 100 = 58 . 35 %
(39) What will be % Na2O content of lye containing 73% (by weight) caustic soda ? Solution: Basis: 100 kg of lye It contains 73 kg of caustic soda (NaOH) 2 NaOH = Na2O = Na2O + H2O Molecular weight of NaOH = 40, Molecular weight of Na2O = 62 2 kmol NaOH ≡ 1 kmol Na2O 80 kg NaOH ≡ 62 kg Na2O ∴ Amount of Na 2O in the lye =
Weight % Na2O in the lye =
62 80
x 73 = 56 . 575 kg
56.575 100
x 100 = 56 . 575 %
(40) An aqueous solution of soda ash contains 20 % soda ash by weight. Express the composition as weight % Na2O. Solution: Basis: 100 kg of an aqueous solution of soda ash Amount of soda ash in the solution = 0.2x100= 20 kg Na2CO3 = Na2O + CO2 Molecular weight of Na2CO3 = 106, Molecular weight of Na2O = 62 1 kmol Na2CO3≡ 1 kmol Na2O 106 kg Na2CO3≡ 62 kg of Na2O Weight of Na2O in the solution =
Weight % Na2O in the solution =
62 106
x 100 = 11 . 7 kg
11.7
x 100 = 11 . 7 % 100 (41) A sample of caustic soda flakers contains 74.6 % Na2O by weight. Determine the purity of the flakes. Solution: Basis: 100 kg of caustic soda flakes It contains 74.6 kg of Na2O 2 NaOH = Na2O + H2O 2 kmol of NaOH ≡ 1 kmol of Na2O 80 kg of NaOH ≡ 62 kg of Na2O Amount of NaOH in the flakes =80 x74.6=96.26 kg 62 % purity of the flakes = kg NaOH x 100 kg flakes =
96.26 100
x 100 = 96 . 26 %
(42) The strength of a phosphoric acid sample is found to be 35% P2O5 by weight. Determine the actual concentrations of H3PO4 (by weight) in the acid.
Solution: Basis: 100 kg of phosphoric acid sample It contains 35 kg of P2O5. 2H3PO4 = P2O5 + 3H2O Molecular weight of H3PO4 = 96, Molecular weight of P2O5 = 142 2 kmol H3PO4≡ 1 kmol P2O5 196 kg of H3PO4≡ 142 kg of P2O5 ∴Amount of H3PO4 in sample =
196 142
x 35 = 48 . 31 kg
Weight % of H3PO4 in phosphoric acid sample= (43)
48.31
x 100 = 48 . 31 % 100 Caustic soda flakes received from a manufacturer are found to contain 60 ppm silica (SiO2), Convert this impurity into weight %. Sol: Basis: 106 kg of caustic soda flakes Silica in flakes is given as 60 ppm ∴ Amount of silica in flakes =
Weight % of silica in flakes =
60 x 10 6 = 60 kg 10 6
60 x 100 = 0 . 006 6 10
(44) A sample of water contains 2000 ppm solids. Express the concentration of solids in the sample in weight percent. 6 Solution: Basis: 10 kg of sample of water. Solids in water = 2000 ppm Amount of solids in water = Weight % of solids =
2000 x 10 6 = 2000 kg 10 6
2000 x 100 = 0 . 2 % 6 10
GASES The composition are expressed in terms of volume percent, as the volume can be measured easily. The density can be calculated if the parameters such as temperature and pressure are known, which in turn gives mass of the gas. The relationship among mass, temperature and volume should be a known one, if it that needed to dealing with substances existing in the gaseous state.
IDEAL GAS LAW BOYLE’S LAW Given mass of an ideal gas, the product of the pressure and volume is constant at a constant temperature i.e., P x V = Constant --------- (A) Where P is the absolute pressure and V is the volume occupied by the gas. CHARLE’S LAW : Given mass of an ideal gas, the ratio of the volume to temperature is constant at a given pressure, i.e., V T
= cons tan t − − − − − − − − − (B)
Where T is the absolute temperature By combining the above two laws, it can be formulated as an ideal gas laws as PxV T
= cons
tan t − − − − − − − − − ( C )
The constant of the above equation is designated by the symbol R and is known as Universal gas constant Therefore, PV = RT ---- (D) When V is the volume in cubic meters of n kmol of gas, then the ideal gas is written as PV = nRT ------ (E) The ideal gas law states three facts : (i) volume of gas is directly proportional to numberof moles. (ii) volume is directly proportional to the absolutetemperature (iii) volume is inversely proportional to the pressure. When P (absolute pressure) is in kPa, V is in kmol and T is in K, then the units of R will be 3. . m kPa/(kmol K). 3 When P is in Mpa, V is in m , T is in K and n is in kmol, then numerical value of R is 3. . 0.008314 with the units m Mpa/(kmol K). 3 When P is in kPa, V is in m , T is in K and n is in kmol, then numerical value of R is 3. . 8.31451 with units m kPa/(kmol K) When the mass of a gas is not known, and if we know the volume occupied at specified temperature and pressure and conditions are changed and we know two of the three variables in final state, then the third one can be calculated by means of proportionality indicated by the gas law. Let V1, T1, and P1 be the volume, temperature and pressure of n kgmol of gas at conditions1 Let V2, T2, and P2 be the volume, temperature and pressure of n kgmol of gas at conditions -2 then, P1V1 = nRT1 ---- (F) and P2V2 = nRT2 ---- (G) Combining the above two (F & G) equations, we get
P1 V1 P V = 2 2 − − − − − − − − − − − − − (H ) T1 T2 In the ideal gas law given by equation (5), V is called the molar volume. At 273.15 K* 0 3 (0 C) and 101.325 kPa (1 atm), the volume occupied by 1 kmol of a gas is 22.4136 m i.e., 3 V = 22.4136 m /kmol (or 22.4136 l/mol). These conditions are said to be normal temperature and pressure (NTP). 0 In USA, 101.325 kPa (1 atm) and 288.7 K (15.6 C) are considered to be standard temperature and pressure conditions (STP). If two of three variables are known(temperature, pressure and volume) at one specified condition and the third one can be calculated, thus are calculated with the help of equation 3 (8) where the other situation may be taken as NTP i.e., P2= 101.325 kPa; V2 = 22.4 m per kmol and T2 = 273 K. GASEOUS MIXTURE The composition of component gases are expressed in terms of volume percent. The molecules of the each component gas are distributed throughout the entire volume of the container in a closed vessel.The total pressure exerted by the entire mixture is equal to the sum of the pressure exerted by each component gas molecules. PARTIAL PRESSURE The pressure exerted by that component gas, if that alone is present in the same volume and at the same temperature . PURE COMPONENT VOLUME The volume occupied by that component gas, if that alone is present at the same pressure and temperature as the gas mixture. DALTON’S LAW It states that the total pressure exerted by a gaseous mixture is equal to the sum of the partial pressures of the component gases present in gas mixture P = pA+pB+pC ---------------------(I) AMAGAT’S LAW The total volume occupied by the gaseous mixture is equal to the sum of the pure component volumes of component gases. Mathematically, V = VA + VB+VC + ………… (J) Where V is the total volume and VA, VB, and VC etc., are the pure component volumes of component gases A,B,C etc., respectively. RELATIONSHIP BETWEEN PARTIAL PRESSURE, MOLE FRACTION OF COMPONENT GAS TO TOTAL PRESSURE The gas mixture consisting of component gases A,B,C are considered.
Let V be the total volume of the gas mixture and P be the total pressure exerted by the gas mixture. VA,VB,VC etc are the pure component volumes of A,B,C etc., respectively. PA,PB,PC are the partial pressures of component gases A,B,C etc., respectively. The ideal gas law for the component gas A is pAV = nART n A RT
pA =
V
− − − − − − − − − − − −(K )
Where R = gas constant T = temperature of gas mixture Similarly, for component B:
n B RT
pB =
V
− − − − − − − − − − − − (L )
Similarly, for component C : n C RT
pC =
V
− − − − − − − − − − − −(M )
Adding Equations (J), (K), and (L) we get
p
A
+
p
B
+
p
(n C
=
+n + ........) RT A + nB C − − − − − − − − − − − −(N ) V
According to the Equation (1) P = pA+pB+pC+ ------- (O) Dividing the equation (J) by Equation (M) we get
pA pA + pB + pC
=
nA nA + nB + nC
nA+nB+nC + ------- = n = total moles ∴pA = x A P
−− −− −− − −−− −
(P )
The mixture of ideal gases (i.e., one which follows an ideal gas law): The partial pressure of a component gas mixture is equal to the product of the total pressure and the mole fraction of that component. Multiplying both sides of Equation (O) by 100, pA pA + pB + pC
x 100 =
nA nA + nB + nC
x 100 − − − − − − − − − − − ( Q )
i.e., pressure % of A = mole % of A --- (Q) This result can be proved for other components of the gas mixture ∴ For gases behaving ideally, Pressure % = Mole % ------ (R) When an ideal gas law is applicable, it is written for any component gas A as : PVA = nA RT ------ (S) Similarly, for component B : PVB = nB RT ------ (T) For component C : PVC = nCRT ------ (U) Let VA,VB,VC etc., are the pure component volumes of components A,B,C …. etc., V = Total volume of the gas mixture Adding Equations (S), (T), and (U), we get P (VA+VB+VC + ------) = (nA+nB+nC+-----) RT – (V) Dividing Equation (S) by Equation (V) gives VA VA + V B + V C
=
nA nA + nB + nC
− − − − − − − − − − − (W )
According to Amagat’s law, V = VA+VB+VC + ------W/X gives :
VA
= xA
V
i.e., Volume fraction of A = mole fraction of A ---- (Y) Where, x A
∴
=
nA nA + nB + nC
Volume % of A = Mole % of A --- (Z)
----- (X)
This result can be proved for other components of the gas mixture.Combining results obtained, i.e., equations (Q) and (Z) for gases behaving ideally/for an ideal gas mixture, we have: Pressure % = Mole % = Volume % AVERAGE MOLECULAR WEIGHT OF GAS MIXTURE Required for calculating the weight of the gas mixture.It is calculated by assuming the quantity of the gas mixture as one mole. The weight of the one mole of the gas mixture represents its average molecular weight. The composition of the gas mixture is specified in terms of volume % (on a volume basis). Assuming the applicability of ideal gas, thus the volume analysis treated same as the mole analysis. The mole analysis (i.e., volume % = mole %) of each component is calculated which is needed for calculating the average molecular weight of the gas mixture. The gas mixture consisting of components A, B, C etc., is considered. Let MA, MB, MC etc., and XA,XB,XC, etc., are the molecular weights and mole fractions of components gases respectively. Let Mavg be the average molecular weight of the gas mixture. Then, Mavg = MA.xA+MB.xB+MC.xC +…….. (1) Average molecular weight of the gas mixture is the sum of the product of molecular weight and mole fraction of all the individual component gases present in the gas mixture. In general, Mavg = th
th
Where, Mi = molecular weight of i component gas, Xi = mole fraction of i component gas DENSITY OF GAS MIXTURE The density of a gas mixture at a given temperature and pressure is very easily calculated by using the ideal gas law.For calculating the density, the known average molecular weight of the gas mixture plays a major role. The ideal gas equation for a gas mixture is PV = nRT ----- (1)
P RT
=
n V
− − − − − − − − ( 2)
Where n= kmol of gas mixture 3 V = volume of gas mixture, m 3. . R = universal gas constant 8.31451 m kPa/(kmol K), T = temperature K, P = pressure in kPa, moles of gas mixture =
kg of gas mixture M avg
− − − − − − − − (3)
∴ From equation (1) and (2) can be written as
wt of gas mixture M avg . V wt of gas mixture V
=
=
P RT PM avg . RT
ρmix = Density of gas mixture =
ρ mix =
wt of gas mixture Volume
PM avg RT 3.
.
When R is expressed in m kPa/(kmol K), P in kPa and T in K then the density will 3 have the units of kg/m Specific gravity of a gas = density of the gas =molecular weight of the gas density of air at the same T and P molecular weight of air
(45) Calculate the volume occupied by 20 kg of chlorine gas at a pressure of 100 kPa and 0 298 K (25 C). Solution: Basis: 20 kg Cl 2 gas. Moles of Cl2 gas = 20 = 0.2817 kmol 71 XA is the PV = nRT V = nRT P Where, n = 0.2817 kmol, T = 298 K, P = 100 kPa, 3. . R = 8.31451 m kPa/(kmol K) 3 Volume, V = 0.2817 x 8.31451 x 298 = 6.98 m 100 0 (46) 15 kg of carbon dioxide is compressed at a temperature of 303 K (30 C) to a volume 3 of 0.5 m . Calculate the pressure required for given duty. Assume ideal gas law is applicable . Solution: Basis: 15 kg of carbon dioxide gas Molecular weight CO 2 = 44 ∴
∴ Moles
of CO2= 15 = 0.341 kmol 44
PV = nRT ∴ P = nRT V Where, n = 0.341kmol, T = 303 K,V = 0.5m3, 3. . R = 8.31451 m kPa/(kmol K)
Pressure, P = 0.341 x 8.31451 x 303 =1718.16 kPa 0.5 3 (47) 5 kg of oxygen contained in a closed container of volume 1 m is heated without exceeding a pressure of 709.28 kPa. Calculate the maximum temperature of gas attained. Solution: Basis: 5 kg oxygen. Molecular weight of O 2 = 32, ∴ Moles
of O2= 5 = 0.5162 kmol 32 PV = nRT ∴ T
= PV nR 3. . Where, V = 1 m3, P = 709.28 kPa, n=0.5162 kmol, R = 8.31451 m kPa/(kmol K) Maximum temperature of gas, T = 709.28 x 1 = 546.13 K 0.5162 x 8.31451 0
∴ Temperature
of gas attained is 546.13 K (273.13 C) 3 (48) Calculate the weight of sulphur dioxide in a vessel having 2 m volume, the 0 pressure and temperature being 97.33 kPa and 393 K (120 C). Solution: Basis: 2 m3 volume of SO 2 gas PV = nRT n = PV RT Where, P = 97.33 kPa, V = 2 m3, T = 393 K, 3. . R = 8.31451 m kPa/(kmol K) ∴
∴Moles
of SO2 gas =
97.33x 2 = 0.0596 kmol 8.31451 x 393 Weight of SO2gas = 0.0596 x 64 = 3.8144 kg 3 (49) A certain quantity of gas contained in a closed vessel of volume 1 m at a 0 temperature of 298 K (25 C) and pressure of 131.7 kPa is to be heated such that the pressure should exceed 303.98 kPa. Calculate the temperature of gas attained. 3 Solution: Basis: 1 m volume of gas at 298 K P1V1= P2V2 T1 T2 Where, P1 = 131.7 kPa, P2 = 303.98 kPa, T 1 = 298K, 3 V1=V2 = 1 m (as vessel being closed), T 2 = ? ∴
∴
131.7 x 1 = 303.98 x 1 298 T2
T2 = 687.82 K
∴ Temperature
0
of gas attained = 687.82 K (414.82 C) (50) A gas contained in a closed vessel at a pressure of 121.59 kPa g and 299 K 0 0 (26 C) is heated to a temperature of 1273 K (1000 C). Find the pressure to which a closed vessel should be designed. Solution: Basis: A gas at 299 K in closed vessel P1V1= P2V2 T1 T2
As vessel being closed, V 1 = V2 ∴
P1= P2 T1 T2
∴ P2=
P1T2 T1 Where, P1 = 121.59 kPa g, T1 = 299 K, T2 = 1273 K Absolute pressure = Gauge pressure + Atmospheric pressure ∴ P1 =
121.59 + 101.325 = 222.915 kPa P2 = 222.915 x 1273 = 949.07 kPa
Pressure to which vessel should be designed= 949.07 kPa = 0.95 MPa. 3 (51) A sample of gas having volume of 0.5 m is compressed in such a manner so that pressure is increased by 60%. The operation is done for a fixed mass of a gas at constant temperature. Calculate the final volume of gas. 3 Solution: Basis: 0.5 m of gas sample. Let initial pressure be P 1 kPa. It is given that pressure of gas increases by 60%
∴
∴ Final
pressure = P 2 = 1.6 P1, Final volume = V 2= ? Temperature and mass are constant, therefore P1V1 = P2V2 3 V2 = P1V1 =P1 x 0.5 = 0.3125 m P2 1.6 P1 3
∴ Final
volume of gas = 0.3125 m 3 (52) A sample of gas having volume of 1 m is compressed in such a manner so that its pressure is increased by 85%. The operation is done for a fixed mass of gas at constant temperature. Calculate the final volume of gas. 3 Solution: Basis: 1 m of gas 3 Initial volume = 1m , Initial pressure = P 1, kPa. It is given that pressure increases by 85%. Final pressure = P 2 = 1.85 P1, kPa; Final volume of gas = P 1, kPa. For given mass at constant temperature, we have, P 1V1 = P2V2 3 V2 = P1V1 = P1 x 1 = 0.54 m P2 1.85 P1 ∴ Final
3
volume of gas = 0.54 m .
(53) A certain sample of gas at a pressure of 202.65 kPa pressure is expanded so that the volume is increased by 50%. The operation is done for a fixed mass of gas at constant temperature. Calculate the final pressure of gas. Solution: Basis: a gas at initial pressure of 202.65 kPa. Initial volume = V 1 m3=?, Initial pressure = P 1 = 202.65 kPa. It is given that volume increases by 50%. ∴ Final
2
3
volume = V = 1.5 V1 m . For a given mass at constant temperature, we have, P 1V1 = P2V2
P2 = P1V1 =202.65 x V1= 135.1 kPa. V2 1.5 V1 ∴ Final
pressure of gas = 135.1 kPa. 3 (54) A sample of gas having volume of 1 m is compressed to half of its original volume. The operation is carried for a fixed mass of gas at constant temperature. Calculate the percent increase in pressure. 3 Solution: Basis: 1 m of gas. 3 Initial volume = V 1 = 1m , 3 Final volume = V 2 = 0.5 V1 = 0.5 m . Initial pressure = P 1, kPa, Final pressure = P 2, kPa. For given mass of gas at constant temperature, we have,P 1V1 = P2V2 P2= V1= 1.0 P1 V2 1.5 P2 = 2 P1 % increase in pressure = P 2 – P1 x 100 P1 = 2 P1 – P1x 100 P1 ∴
∴ %
increase in pressure = 100 3 (55) A cylinder contains 15 kg of liquid propane. What volume in m will propane occupy if it is released and brought to NTP conditions? Solution:Basis: 15 kg of propane Molecular weight of propane = 44 Moles of propane = 15 = 0.3409 kmol 44 PV = nRT 3. . 3 P= 101.325 kPa, R = 8.31451 m kPa/(kmol K) , T = 273K, V = volume in m at NTP=? ∴
NTP conditions ⇒ P = 101.325 kPa, T = 273 K (0 C) 3 V = 0.3409 x 8.31451 x 273 = 7.637 m 101.325 0
∴Volume
3
of propane gas at NTP = 7.637 m . 3 (56) A certain quantity of a gas contained in a closed vessel of volume 1 m at a 0 temperature of 298 K (25 C) and pressure of 121.59 kPa is to be heated such that pressure should not exceed 405.3 kPa. Calculate temperature of gas attained. Solution:Basis: 1 m3 of gas at temperature of 298 K and pressure of 121.59 kPa. P1V1= P2V2 T1 T2 T2 = P2V2x T1 P1V1 Where, P1 = initial pressure = 121.59 kPa. T1 = initial pressure = 298 K, P 2 = final pressure =405.3 kPa, T 2 = final temperature of gas = ? 3 As vessel is closed, V 1= V2 = 1 m . ∴
T2 =
∴ ∴Temperature
405.3 x 1 x 298 = 993.33 K 121.59 x 1 0
of gas attained = 993.33 K (720.33 C) (57) A gas mixture contains 0.274 kmol of HCl, 0.337 kmol of N 2 and 0.089 kmol of O2. calculate (a) Average molecular weight of gas and (b) Volume occupied by this 0 mixture at 405.3 kPa and 303 K (30 C) Solution:Basis: A gas mixture containing 0.274 kmol HCl, 0.337 kmol N 2 and 0.089 kmol O2. Total moles of gas mixture= 0.274 +0.337+0.089 = 0.7 kmol Mole fraction of HCl (X HCl) = 0.274 = 0.391 0.7 Mole fraction of N 2 (XN2)=0.337 = 0.481 0.7 Mole fraction of O 2 (XO2) = 0.080 = 0.127 0.7 Molecular weight of HCl = M HCl = 36.5, Molecular weight of N 2 = MN2 = 28, Molecular weight of O 2 = MO2 =32 Mavg = ∑Mi Xi = MHCl. XHCl + MN2.xN2 +MO2.xO2 = 36.5 x 0.391+28x0.481+32x0.127 Mavg = 31.80 PV = RT PV = nRT V = nRT P 3 Where, V = volume in m , n= moles of gas mixture = 0.7 kmol, P= 405.3 kPa, T = 363 3. . K, R = 8.31451 m kPa/(kmol K) 3 V = 0.7 x 8.31451 x 303 = 4.35 m 405.3 (58) In case of gas mixture cited in previous example, calculate the partial pressure of each 0 component gas at 405.3 kPa and 303 K (30 C). Solution:Basis: Given gas mixture Total pressure = 405.3 kPa Mole fraction of HCl = xHCl = 0.391 Mole fraction of N 2 = xN2 = 0.481 Mole fraction of O 2 = xO2 = 0.127 Partial pressure of HCl = p HCl = xHCl.P = 0.391 x 405.3 = 158.5 kPa Partial pressure of N 2 = pN2 = xN2.P= 0.481 x 405.3 = 194.95 kPa Partial pressure of HCl = pO 2 = xO2.P= 0.127 x 405.3 = 51.47 kPa (59) A mixture of H2 and O2 contains 11.1% H 2 by weight. Calculate (a) Average molecular 0 weight of gas mixture and (b) Partial pressure of O 2 and H2 at 100 kPa and 303K(30 C). Sol: Basis: 100 kg of gas mixture. It contains 11.1 kg of H 2 and 88.9 kg of O 2. Molecular weight of O 2= 32, Molecular weight of H 2 = 2 ∴
Moles of H2 = 11.1 = 5.55 kmol 2 and Moles of O2 = 88.9 = 2.78 kmol 32
(60)
Amount of gas mixture = 5.55 + 2.78 = 8.33 kmol Mole fraction of H 2 = xH2 =5.55 = 0.67 8.33 Mole fraction of O 2 = xO2 = 2.78 = 0.33 8.33 Mavg = Average molecular weight of gas mixture= M.H 2.xH2+M.O2.xO2 = 2 x 0.67 + 32 x 0.33 = 11.9 P = Total pressure = 100 kPa Partial pressure of H 2 = pH2 = xH2.P= 0.67 x100 = 67 kPa (502.54 torr) Partial pressure of O 2 = pO2 = xO2.P = 0.33 x 100 = 33 kPa (247.52 torr) 0 A mixture of nitrogen and carbon dioxide at 298 K (25 C) and 101.325 kPa ha an average molecular weight of 31. What is the partial pressure of nitrogen? Solution:Basis: Average molecular weight of 31 of gas mixture. Let xN2 and xCO2 be the mole fractions of N 2 and CO2 respectively. Molecular weight of N 2 = 28, Molecular weight of CO 2 = 44. ���� � ∑���� Mavg = MN2.xN2 + MCO2.xCO2 31 = 28 xN 2 + 44xCO2 ------- (1)
∑xi = 1 xN2 + xCO2 = 1
------ (2)
xCO2 = 1-xN2 ------ (3) Put the value of xN 2 from equation (2) in equation (1) and solve for xN 2 31 = 28xN2 + 44 (1-xN2) 16 xN2 = 13 ∴
∴
xN2 = 0.8125
xCO2= 1-0.8125 = 0.1875 Partial pressure of N 2 = xN2.P = 0.8125 x 101.325 = 82.33 kPa(617.52 torr) A mixture of CH4 and C2H6 has the average molecular weight of 22.4. Find mole % CH4 and C2H6 in the mixture. Solution: Basis: Average molecular weight of 22.4 of gas mixture. Let xCH4 and xC2H6 be the mole fraction of CH 4 and C2H6 respectively. ∴
(61)
Mavg = ∑Mixi Mavg= M.CH4.xCH4+M.C2H6.xC2H6 22.4 = 16 xCH4+30xC2H6
∑xi = 1 xCH4+xC2H6 = 1 xC2H6 = 1-xCH4
(62)
Put the value of xC 2H6 from equation (3) into equation (1) and solve for xCH 4. 22.4 = 16 xCH 4 + 30 (1-xCH4) xCH4 = 0.543 xC2H6 = 1- 0.543 = 0.457 Mole % of CH4 = Mole fraction of CH 4 x 100= 0.543 x 100 = 54.30 Mole % of C2H6 = 0.457 x 100 = 45.70 Assuming air to contain 79% N2 and 21 % O2 by volume, calculate the density of air at NTP. Solution: Basis: Air containing 21% O 2 and 79% N2 by volume For ideal gases, mole % = volume % ∴ Mole
% N2 = 79, Mole % of O2 = 21 Mole fraction of N 2 (xN2) = Mole % N2= 79 = 0.79 100 100 Mole fraction of O 2 (xO2)=Mole % O2= 21 = 0.21 100 100 Mavg = MN2.xN2 + MO2.xO2 = 28 x 0.79 + 82 x 0.21 = 28.84 Density of air, ρ = PMavg RT 3. . Where, Mavg = 28.84, T=273 K, P = 101.325 kPa, R = 8.31451 m kPa/(kmol K). 3
= 101.325 x 28.84 = 1.2874 kg/m 8.31451 x 273 3 Density of air = 1.2874 kg/m . 3 0 A mixture of CH 4 and C2H6 has density 1.0 kg/m at 273 K (0 C) and 101.325 kPa pressure. Calculate the mole % and weight % of CH 4 and C2H6 in the mixture. 3 Solution: Basis: 1 kg/m density of gas mixture at 273 K and 101.325 kPa. ρ
(63)
Density of gas mixture, ρ = PMavg RT Mavg= ρ. RT P
∴
3.
.
Where, ρ = 1 kg/m3, T = 273 K, P = 101.325 kPa, R = 8.31451 m kPa/(kmol K) . ∴ Mavg= 1 x 8.31451 x 273 = 22.4 101.325 Let xCH4 and xC2H6 be the mole fractions of CH 4 and C2H6 respectively. Mavg = ∑Mixi = MCH4.xCH4 + MC2H6.xC2H6 ∴
22.4 = 16xCH4 + 30 xC2H6
---- (1)
∑xi = 1 xCH4 + xC2H6 = 1
---- (2)
xC2H6 = 1 – xCH4 -------- (3) Put the value of xC 2H6 from equation (3) into equation (1) and solve for xCH 4 22.4 = 16xCH4 + 30 (1-xCH4) ∴
∴
xCH4 = 0.543 xC2H6 = 1-0.543 = 0.457
Mole % of CH4 = xCH4 x 100 = 0.543 x 100 = 54.3 Mole % of C 2H6 = xC2H6 x 100 = 0.457 x 100 = 45.7 Weight of CH4 in 1 kmol mixture = 0.543 x 16 = 8.69 kg Weight of C 2H6 in 1 kmol mixture = 0.457 x 30=13.71 kg Weight of gas mixture = 22.4 kg
(64)
Weight of CH4 in mixture = 8.69 x 100 = 38.8 22.4 Weight of C 2H6 in mixture = 100 – 38.8 = 61.2 A natural gas has the following composition by volume: CH 4 = 82%, C2H6 = 12% and 0 N2 = 6%. Calculate the density of gas at 288 K (15 C) and 101.325 kPa and composition In weight percent. Solution: Basis: 100 kmol of gas It contains 82 kmol of CH 4, 12 kmol of C2H6 and 6 kmol of N 2. Mole fraction of CH 4 = xCH4= 82= 0.82 100 Mole fraction of C 2H6 = xC2H6= 12= 0.12 100 Mole fraction of N 2 = xN2 = 6 = 0.06 100 Molecular weight of CH 4 = 16, Molecular weight of C 2H6 = 30, Molecular weight of N 2 = 28 Mavg = Average molecular weight of gas Mavg = 16 x 0.82 + 30 x 0.12+28 x 0.06 = 18.4 Density of gas = ρ= PMavg RT 3. . Where, P = 101.325 kPa, M avg = 18.4, R = 8.31451 m kPa/(kmol K) T= 288 K Density,
ρ=
101.325 x 18.4 = 0.78 kg/m 8.31451 x 288 CH4 in gas = 82 x 16 = 1312 kg C2H6 in gas = 12x30 = 360 kg N2 in gas = 6x28 = 168 kg Weight % CH= = kg of CH4x 100 kg of gas Composition by Weight:
3
Component
Quantity in kg
Weight %
CH4
1312
71.30
C2H6
360
19.57
(65)
N2
168
9.13
Total
1840
100
0
Calculate the density of air containing 21% O 2, 79% N2 by volume at 503 K (230 C) and 1519.875 kPa. Solution: Basis: Air containing 21% O 2 and 79% N2 by volume. For ideal gas, Mole % = Volume % Mole % O2 = 21, Mole % N2 = 79. Mole fraction of O 2 = xO2 =21 = 0.21 100 Mole fraction of N 2 = xN2= 79= 0.79 100 Molecular weight of O 2 = 32, Molecular weight of N 2 = 28 Mavg = MO2.xO2 + MN2.xN2 = 32 x 0.21 + 28x 0.79 = 28.84 Density of air, ρ = Pmavg RT 3. . Where, P = 1519.875 kPa, M avg = 28.84,R = 8.31451 m kPa/(kmol K), T = 503 K 3
ρ = 1519.875
(66)
x 28.84 = 10.481 kg/m 8.31451 x 503 A gas mixture has the following composition by volume: SO 2 = 8.5%, O2=10% and N2 0 = 81.5% . Find (a) the density of gas mixture at a temperature of 473 K (200 C) and 202.65 kPa g and (b) composition by weight. Solution: Basis: 100 kmol of gas mixture For ideal gas, Mole % = Volume % It contains 8.5 kmol of SO 2, 10 kmol of O2 and 81.5 kmol of N 2. Mole fraction of SO 2 = xSO2= 8.5= 0.085 100 Mole fraction of O 2 = xO2= 10= 0.10 100 Mole fraction of N 2 = xN2 = 81.5 =0 .815 100 Molecular weight of SO 2 = MSO2 = 64, Molecular weight of O 2 = MO2 = 32, Molecular weight of N 2 = MN2 = 28. Mavg = MSO2.xSO2+MO2.xO2+MN2.xN2 = 64 x 0.085 + 32 x 0.1 + 28 x 0.815 = 31.46 Absolute pressure = Gauge pressure + Atmospheric pressure ∴
P = 202.65 + 101.325 = 303.975 kPa
Density of gas mixture , ρ = PMavg RT 3. . Where, P = 303.975, Mavg = 31.46, R = 8.31451 m kPa/(kmol K), T = 473 K ρ = 303.975
3
x 31.46 = 2.43 kg/m 8.31451 x 473
SO2 in gas mixture = 8.5 x 64 = 544 kg O2 in gas mixture = 10 x32 = 320 kg N2 in gas mixture = 81.5 x 28 = 2282 kg Amount of gas mixture = 3146 kg Weight % SO2 in gas mixture =kg of SO 2x 100 kg of gas mixture Composition by Weight: Quantity in kg
Weight %
SO2
544
17.30
O2
320
10.17
N2
2282
72.53
Total
3146
100
Component
(67)
0
In one case 28.6 litres of NO 2 at 80 kPa and 298 K (25 C) is allowed to stand until the equilibrium is reached. At equilibrium, the pressure is found to be 66.662 kPa. Calculate the partial pressure of N 2O4 in the final mixture. Solution: Basis: 26.6 l of NO2 at 80 kPa and 298 K. 3 Volume of NO2 = 26.6 lit = 0.0266 m P1V1 = n1RT Initial moles, n1 = P1V1 RT1 3. . 3 Where, P1 = 80kPa, T1 = 298 K, R= 8.31451 m kPa/(kmol K), V1 = 0.0266 m n1 = 80 x 0.0266 = 8.6 x 10-4 kmol = 0.86 mol 8.31451 x 298 2 NO2 = N2O4 Let x be the mol of N 2O4 in final gas mixture NO2 reacted = 2x mol NO2 unreacted = 0.86 – 2x mol Final moles = n 2 = 0.86-2x + x = 0.86 - x mol For initial conditions = P 1V1=n1RT1 For final conditions = P 2V2 = n1RT2
But here, V1 = V2 and T1 = T2 P1= n1 P2 n2 80 = 0.86 = 66.62 0.86 – x Solving we get, Final moles = 0.86 – x = 0.1434 mol Mole fraction of N 2O4 in final gas mixture, xN 2O4 = 0.1434 = 0.20 0.7166 Partial pressure of N 2O4 = xN2O4 .P = 0.20 x 66.662 = 13.33 kPa (99.98 torr) A closed vessel contain a mixture of 40% NO 2 and 60% N 2O4 at a temperature of 311 0 K (38 C) and a pressure of 531.96 kPa. When the temperature is increased to 333 K 0 (60 C), some of N2O4 dissociates to NO 2 and a pressure rises to 679.95 kPa. Calculate 0 the composition of gases at 60 C by weight. Sol: Basis: 100 kg of gas mixture at 311 K. NO2 in gas mixture = 40 kg, N2O4 in gas mixture = 60 kg ∴
(68)
∴ Moles
of NO2= 40 = 0.87 kmol 46 Mole of N2O4 = 60 = 0.652 kmol 92 Initial moles, n1= 0.87 + 0.652 = 1.522 kmol N2O4 = 2 NO2 Let x be the kmol of N 2O4 dissociated at 333 K NO2 formed = 2x kmol N2O= at 333 K = 0.652 – x kmol NO2 at 333 K = 0.87 + 2x kmol ∴ Total
moles at 333 K = (0.652-x) + (0.87+2x) = 1.522 +x kmol P1V1 = n1RT1 P2V2 = n2RT2 But V1 = V2 for it being closed vessel. Taking ratio of equations (1) and (2), we get, P1 = n1.T1 P2 n2 T2 P1 = 531.96 kPa, P 2 = 679.95 kPa, n1= 1.522, n2= 1.522+x, T1= 311 K, T2 = 333 K. ∴531.96
= 1.522 x 311 679.95 1.522 + x 333 Solving we get, x = 0.295 kmol NO2 at 333 K = 0.87 + 2x = 0.87 +2 x 0.295 = 1.46 kmol N2O4 at 333 K = 0.652 – x = 0.652 – 0.295 = 0.357 kmol Amount of NO2 at 333 K = 1.46 x 46 = 67.16 kg
Amount of N2O4 at 333 K = 0.357 x 92 = 32.84 kg Composition by Weight: Component
Quantity in kg
Weight %
NO2
67.16
67.16
N2O4
52.84
32.84
Total
100
100
(69)
3
A volume of moist air 30 m at a total pressure of 101.328 kPa and a temperature of 0 303K (30 C) contain water vapour in such that proportions that its partial pressure is 2.933 kPa. Without total pressure being changed, the temperature is reduced to 288 K 0 (15 C) an some of water vapour is removed by condensation. After cooling, it is found that the partial pressure of water vapour is 1.693 kPa. Calculate (a) volume of air at 288 0 K (15 C) and (b) weight of water condensed. 3 Solution: Basis: 30 m of moist air at 303 K Ideal gas law is : PV = nRT n = PV RT 3. . Where, n= moles of moist air, P = 101.325 kPa, R= 8.31451 m kPa/(kmol K), 3 V = 30 m , T = 303 K Moles of air, n = 101.325 x 30 = 1.2066 kmol 8.31451 x 303 Let n1 be the kmol of air and n 2 be the kmol of moisture/water vapour �� � ������� �������� �� ��� �� ��� �
= 101.325 – 2.933 = 98.392 kPa. P2 = Partial pressure of moisture at 303 K = 2.933 kPa For ideal gas, Pressure % = Mole % Pressure fraction = Mole fraction For air, 98.392 = n2 101.325 n1 ∴n1= 98.392x 1.2066 = 1.172 kmol at 303 K 101.325 For water vapour /moisture, 2.933 = n1 101.325 n n2 = 2.933 x 1.2066 = 0.035 kmol at 303 K 101.325 At 288 K, partial pressure of water vapour = 1.633 kPa Let n3 be the moles of water vapour at 288 K Mole of moist air = 1.172 + n 3
(70)
1.693 = n3 101.325 1.172 + n3 n3 = 0.02 kmol Moles of moist air at 288 K = 1.172 + 0.02 = 1.192 kmol Let n’ = 1.192 kmol PV = n’RT 3. . Where, V = volume of air at 288 K, n’ = 1.192 kmol, R= 8.31451 m kPa/(kmol K), T = 288 K, P = 101.325 kPa 3 V = n’RT= 1.192 x 8.31451 x 288 = 28.27 m P 101.325 Moles of condensed = n 1 – n2 = 0.035 – 0.02 = 0.015 kmol Amount of water condensed = 0.015 x 18 = 0.27 kg In manufacture of hydrochloric acid, gas containing 20% HCl and 80% air to volume 0 enters an absorption tower at a temperature of 323 K (50 C) and pressure of 99.325 kPa. 98 percent of HCl is absorbed in water and remaining gas leaves the tower at a 0 temperature of 293 K (20 C) and a pressure of 97.992 kPa. Calculate (a) the weight of HCl absorbed/ removed per m3 of gas entering the system and (b) the volume of gas leaving per m3 of gas entering the system. Solution: Basis: 100 kmol of gas entering the absorption tower. PV = nRT V = nRT P 3 3. . Where, V = volume of gas entering, m ; n = 100 kmol, R= 8.31451 m kPa/(kmol K), P = 99.325 kPa; T=323 K 3 V = 100 x 8.31451 x 323 = 2704 m . 99.325 Moles of HCl in gas entering = 20 kmol Moles of air in gas entering = 80 kmol Moles of HCl absorbed/removed = 0.98 x 20 = 19.6 kmol Moles of HCl unabsorbed and appearing in gas leaving = 20 – 19.6 = 0.4 kmol HCl absorbed based on 100 kmol gas entering = 19.6 x 36.5 = 715.4 kg 3 Volume of 100 kmol gas entering = 2704 m 3
∴ Amount
of HCl absorbed per 1 m of gas entering = 715.4 x 1 = 0.2646 kg 2704 Moles of gas entering = 80 + 0.4 = 80.4 kmol PV = nRT ∴
V = nRT P
3.
.
Where, V = volume of gas leaving, m3; n = 80.4 kmol, R= 8.31451 m kPa/(kmol K), P = 97.992 kPa, T= 293 K 3 V = 80.4 x 8.31451 x 293 = 1999 m 97.992 3 3 Volume of gas leaving based on 100 kmol or 2704 m of gas entering system = 1999 m 3
Volume of gas leaving per 1 m of gas entering the system 3 = 1999 x 1 = 0.74 m 2704 0 Equal masses of CO and N 2 are mixed together in a container at 300 K (27 C). The total pressure was found to be 405.3 kPa. Find the partial pressure of CO gas. Solution: Basis: Let m 1 be the mass of CO and m 1 be the mass of N 2 (as equal mass of CO and N2). Mole of CO = m1 28 Moles of N2= m1 28 Total moles of gas mixture = 2 m 1 28 Mole fraction of CO (xCO)= m1 /28 2m1 /28 Partial pressure of CO = xCO. P = 0.5x405.3 = 202.65 kPa The analysis of the gas sample is given below (on volume basis): CH 4 = 66%, CO2 = 30%, NH3 = 4%. Find (a) the average molecular weight of the gas and (b) density of the 0 gas at 202.65 kPa g pressure and 303 K (30 C). Solution: Basis: Gas sample containing CH 4, CO2 and NH3 For ideal gases, Volume % = Mole % Mole % of CH4 = 66, Mole % of CO2 = 32, Mole % of NH3 = 4 Mole fraction of CH 4=xCH4= Mole % of CH4=66=0 .66 100 ∴
(71)
(72)
Mole fraction of CO 2 = xCO2 =30 = 0.3 100 Mole fraction of NH 3 = xNH3= 4= 0.04 100 Mavg = Average molecular weight of gas M.CH4 .xCH4 +M.CO2.xCO2 + M.NH3.xNH3= 16 x 0.66 + 44 x 0.30 + 0.04 x 17 = 24.22 ρ =
Density of gas = PM avg RT Absolute pressure = Gauge pressure + Atmospheric pressure P = 202.65 + 101.325 = 303.975 kPa, T = 303 K, M avg = 24.44, 3. . R= 8.31451 m kPa/(kmol K). ρ = 303.975
3
x 24.44 = 2.95 kg/m 8.31451 x 303
(73)
By electrolysing a mixed brine, a gaseous mixture is obtained at the cathode having the following composition by weight: Cl 2 = 67%, Br== 28% and O2 = 5%. Calculate (a) composition of gas by volume, (b) average molecular weight and (c) density of gas 0 mixture at 298 K (25 C) and 101.325 kPa. (Atomic weights : Cl = 35.5, Br = 80, O=16) Solution: Basis: 100 kg gas mixture It contains 67 kg of Cl 2, 28 kg of Br 2, and 5 kg of O 2 Moles of O2= 5= 0.1562 kmol 32 Composition of Gas Mixture by Volume :
Component
Quantity in kmol
Volume % (mole % )
Cl2
0.9437
74.02
Br2
0.175
13.73
O2
0.1562
12.25
Total
1.2749
100
Mole fraction of Cl 2 = xCl2 = 74.02 = 0.7402 100
Mole fraction of Br 2 = xBr2 = 13.73 = 0.1373 100 Mole fraction of O 2 = xO2= 12.25 = 0.1225 100 Mavg = Average molecular weight of gas mixture = M.Cl2 .xCl2 + M.Br2.xBr2 + M.O2.xO2= 71 x 0.7402 + 160 x 0.1373 + 32 x 0.1225 = 78.44 Also, Mavg = kg of gas mixture kmol of gas mixture = 100 = 78.44 1.2749 ρ =
Density of gas mixture = PM avg RT 3. . Where, P = 101.325 kPa, Mavg = 78.44, R= 8.31451 m kPa/(kmol K), T = 298 K ∴
3
ρ=
101.325 x 78.44 = 3.208 kg/m 8.31451 x 298 3 (74) An inert gas (molecular weight 28) is admitted at the rate of 1 m /min at 202.65 kPa and 0 303 K (30 C) to a pipe line in which natural gas is flowing. The analysis of this gas at a very long distance shown 2.9% by volume inert gas. Calculate the flow rate of natural 0 gas through the pipe line per min at 101.325 kPa and 303 K (30 C). 3 Sol: Basis: 1 m / min inert gas Molal flow rate of inert gas is n’ = PV’ RT 3 3. . Where, P = 202.65 kPa, V’ = 1m /min, T = 303 K, R= 8.31451 m kPa/(kmol K) n’ = 202.65 x 1 = 0.0804 kmol/min 8.31451 x 303 Let x be the kmol/min of natural gas flowing through pipe line Mole% inert gas = Volume % inert gas = 2.9 0.0804 . 100 0.0804 + x Solving we get, x = 2.692 kmol/min Volumetric flow rate of natural gas = xRT P 3. . Where, x = 2.692 kmol/min, R= 8.31451 m kPa/(kmol K), T = 303 K, P = 101.325 kPa Volumetric flow rate of natural gas 3 = 2.692 x 8.31451 x 303 = 66.93 m /min 101.325 A producer gas has the following composition by volume . CO= 21%, CO 2 = 5%, O2 = 3 0 3% and balance being N 2. Calculate the volume of gas in m at 298 K (25 C) and 99.325 kPa per kg of carbon present. Solution: Basis: 100 kmol of producer gas It contains 21 kmol of CO, 5 kmol of CO 2, 3 kmol of O 2 and 71 kmol of N 2. ∴
(75)
2.9 =
1 kmol of CO ≡ 1 katom C Carbon from CO = 1 x 21 = 21 katom 1 Carbon in producer gas = 21 + 5 = 26 katom = 26x 12 = 312 kg PV = nRT ∴
V = nRT P 3 Where, n = 100 kmol, V = Volume in m , T = 298 K,P = 99.325 kPa, 3. . R= 8.31451 m kPa/(kmol K). 3 V = 100 x 8.31451 x 298 = 2494.56 m 99.325 3 Volume of producer gas corresponding to 312 kg carbon in it = 2492.56 m . ∴
∴ Volume
of producer gas per kg of carbon present
3
(76)
= 2494.56 x1 = 8.0 m . 312 0 Calculate the number if cubic meters of acetylene gas at temperature of 313 K (40 C) and a pressure of 100 kPa that may be produced from 5 kg of calcium carbide. Solution: Basis: 5 kg of calcium carbide CaC2 + 2H2O C2H2 + Ca(OH)2 1 kmol of CaC 2≡ 1 kmol of C 2H2
(77)
64 kg CaC2≡ 1 kmol of C2H2 5 kg CaC2= ? Moles of C2H2 produced = 1x 5 = 0.0781 kmol 64 PV = nRT V = nRT P 3. . Where, n = 0.0781 kmol, T = 313 K, P = 100 kPa, R= 8.31451 m kPa/(kmol K). Volume of acetylene gas produced is 3 V = 0.0781 x 8.31451 x 313 = 2.03 m . 100 3 Nitrogen is to be marketed in cylinder having volume of 0.08 m each containing 3.5 kg of nitrogen. Calculate the pressure for which cylinders must be designed if they are 0 subjected to a maximum temperature of 323 K (50 C). Solution: Basis: 3.5 kg nitrogen Molecular weight of N 2 = 28 Moles of N2 = 3.5 = 0.125 kmol 28 According to Ideal gas law – PV = nRT 3 Where, P = pressure in kPa, V = 0.08 m , 3. . R= 8.31451 m kPa/(kmol K). ∴
∴
(78)
n = 0.0125 kmol, T = 323 K,
P V = nRT
P =0.125 x 8.31451 x 323 = 4196.23 kPa. 0.08 The pressure to which cylinders must be designed = 4196.23 kPa = 4.2 MPa. 3 It is desired to market oxygen in small cylinders having volumes of 0.015 m and each containing 0.5 kg of oxygen. If the cylinders may be subjected to a maximum 0 temperature of 323 K (50 C), calculate the pressure for which they must be designed assuming applicability of ideal gas law. Solution: Basis: 0.5 kg oxygen Moles of O2 = 0.5 = 0.0156 kmol 32 According to Ideal gas law – ∴
PV = nRT P = nRT V 3. . Where, n= 0.0156 kmol, R= 8.31451 m kPa/(kmol K), V = 0.015 m3, T = 323 K P = 0.0156 x 8.31451 x 323 = 2793 kPa 0.015 ∴ Pressure
for which cylinders must be designed = 2793 kPa = 2.8 MPa. (79) A gaseous mixture has the following composition by volume: CO 2 = 8%, CO = 14%, H2O = 5%, CH4 = 1% and N2 = 66%. Calculate (i) Average molecular weight of gas 0 mixture and (ii) Density of gas mixture at 303 K (30 C) and 101.325 kPa. Solution: Basis: A gas mixture containing CO 2, CO, O2, H2O, CH4 and N2 at 303 K and 101.325 kPa. For ideal gases, Volume % = Mole % Mole fraction of CO 2, xCO2=Mole% of CO2=8= 0.08 100 Mole fraction of CO, xCO =14 =0.14 100 Mole fraction of O 2, xO2= 6=0.06 100 Mole fraction of H 2O, xH2O = 5 = 0.05 100 Mole fraction of CH 4, xcH4 = 1 = 0.01 100 Mole fraction of N 2, xN2 = 66 = 0.66 100 Molecular weight of CO 2 = 44, Molecular weight of CO= 28, Molecular weight of O 2 = 32, Molecular weight of H 2O = 18, Molecular weight of CH 4 = 16, Molecular weight of N2 = 28 Average molecular weight of gas mixture is Mavg = ∑Mixi = 44 x 0.08 + 28 x 0.14 + 32 x 0.06 + 18 x 0.05 + 16 x 0.01 + 28 x 0.66 = 28.90 Density of gas mixture is ρ =
PMavg RT Where, P = 101.325 kPa, Mavg = 28.90, ρ = 101.325
(80)
3.
.
R= 8.31451 m kPa/(kmol K), T = 303 K 3
x 28.90 = 1.162 kg/m . 8.31451 x 303 The gas acetylene is produced according to the following reaction: CaC2 + 2H2O C2H2 + Ca(OH)2. Calculate the number of hours of service that can be 3 derived from 1 kg of calcium carbide in an acetylene lamp burning 0.10 m of gas per 0 hour at temperature of 298 K (25 C) and pressure of 99.325 kPa. Solution: Basis: 1 kg calcium carbide CaC2 + 2H2O C2H2 + Ca(OH)2
Molecular weight of CaC 2 = 64 Moles of CaC 2= 1= 0.01562 kmol 64 From reaction, 1 kmol CaC2 = 1 kmol C2H2 Moles of C2H2 produced =1x 0.01562 = 0.01562 kmol 1 Now, PV = nRT V = nRT P 3. . Where, n = 0.01562 kmol, T = 298 K, P = 99.325 kPa, R= 8.31451 m kPa/(kmol K). Volume of C2H2 gas produced 3 = 0.01562 x 8.31451 x 298 = 0.3896 m 99.325 3 Burning rate of acetylene gas = 0.10 m /h No. of hours of service = Volume of acetylene gas Burning rate of acetylene = 0.3896 = 3.896 h = 3.9 h 0.1 0 (81) When heated to 373 K (100 C) and 95.992 kPa pressure, 17.2 grams of N 2O4 gas occupies a volume of 11.45 litres. Assuming that the ideal gas law applies, calculate the percentage dissociation of N 2O4 to NO2. Solution: Basis: 17.2 g of N 2O4. Molecular weight of N 2O4 = 92 ∴
Initial moles of N2O4 = 17.2 = 0.187 mol 92 N2O4 = 2NO2 Let x be mol of N 2O4 dissociated as per the reaction. ∴
NO2 formed = 2x mol N2O4 undissociated = (0.187 - x) mol Total moles of gas after dissociation = (0.187 – x) + 2x = (0.187 + x) mol Now, PV = nRT -3 -3 3 Where, P = 95.992 kPa, n = (0.187+x) x 10 kmol, V = 11.45 x 10 m , T = 373 K, 3. . R= 8.31451 m kPa/(kmol K) -3 -3 95.992 x 11.45x10 = (0.187+x) x 10 x 8.31451 x 373 Solving we get, x = 0.1674 mol % dissociation of N 2O4 = moles of N2O4 dissociated x 100 initial moles of N 2O4 = 0.1674 = 89.52 0.187 ∴
0
(82) A 40 ml sample of a mixture of H 2 and O2 was placed in a vessel at 291 K (18 C) and 101.325 kPa. A spark was applied so that the formation of water was complete. The remaining pure gas was H 2, What was the initial mole % H 2 in the mixture ? Solution: Basis: 40 ml of sample of mixture of H 2 and O2 H2 + ½ O2 H2O PV = nRT n = PV RT 3. . Where, P = 101.325 kPa, T = 291 K,R= 8.31451 m kPa/(kmol K), -6 3 V = 40 ml = 40x 10 m -6 n = 101.325 x 40x10 8.31451 x 291 -6 -3 = 1.675 x 10 kmol = 1.675 x 10 mol -3 (H2O + O2) gas mixture = 1.675 x 10 kmol H2 gas remained = 10 ml -6 Moles of H2 remained = 10x10 x 101.325 8.31451 x 291 -7 -4 = 4.188 x 10 kmol = 4.188 x 10 mol Let x be the initial moles of H 2 and O2 respectively -3
x + y = 1.675 x 10 O2 reacted = y mol H2 reacted = 1 y = 2y mol 0.5 Material balance of H 2: -4 x moles of H 2 = 2y + 4.188 x 10 -4 x = 2y + 4.188 x 10 From equations (1) and (2), we get, 3y + 4.188 x 10-4 = 1.675 x -3 and x = 1.256x 10 mol -3 Initial mole % of H 2 = 1.256 x 10 x 100 = 75 -3 1.675 x 10 Initial mole % of O 2 = 100 – 75 = 25 (83) The combustion of 4.73 kg of sample of coal yielded 5.30 m3 of carbon dioxide gas measured at NTP. Find the carbon content of the sample. Solution: Basis: 4.73 kg of a coal sample C + O2 = CO2 3 At NTP, the volume of CO 2 gas = 5.30 m 3 At NTP, the volume of 1 kmol of CO 2 gas is 22.4 m . ∴
∴ Amount
of CO2 yielded =1 x 5.3= 0.237 kmol 22.4 = 0.273 x 44 = 10.428 kg
From reaction, 1 katom C ≡ 1 kmol CO2 12 kg C ≡ 44 kg CO2
Carbon in the sample = 12 x 10.248 = 2.844 kg 44 Carbon content of sample in weight % = 2.844 x 100 = 60.13 33.73
0
(84) At a temperature of 299 K (26 C),ethanol exerts a vapour pressure of 8 kPa. Calculate the composition of a saturated mixture of air and ethanol vapour at a temperature of 299 K 0 (26 C) and a pressure of 100 kPa in terms of (i) volume, (ii) weight, (iii) kg of vapour per m3 of mixture and (iv) kg of vapour per kg of vapour free air. 3 Solution: Basis: 1 m of gas mixture 3 Pure component volume of ethanol vapour in 1 m of mixture. 3 = 1 x 8 = 0.08 m 1 Composition by Volume: Ethanol vapour
0.08 m3
8.0%
Air
0.92 m3
92%
Total
1.0 m3
100
Amount of ethanol vapour present in 1 kmol of mixture = 0.08 x 1 = 0.08 kmol = 0.08 x 46 = 3.68 kg
������ �� ��� �� �������
= 0.92 x 1 = 0.92 kmol = 0.92 x 28.84 = 26.53 kg ∴ Amount
of mixture = 3.68 + 26.53 = 30.21 kg Composition by Weight: Ethanol vapour
0.08 m3
8.0%
Air
0.92 m3
92%
Total
1.0 m3
100
Now,
PV = nRT
V = nRT P Volume of 1 kmol of mixture = nRT P 3. . Where, n = 1 kmol, R= 8.31451 m kPa/(kmol K), T = 299 K, P = 100 kPa. 3
V = 1 x 8.31451 x 299 = 24.86 m 100 3 Weight of ethanol vapour present per m of mixture = 3.68= 0.148 kg 24.86 Kg of ethanol vapour per kg of vapour free air 3.68 = 0.1387 26.53 0 (85) The vapour pressure of ether (molecular weight 74) is 58.928 kPa at 293 K (25 C). If 3 g of compound A are introduced and dissolved in 50 g of ether at this temperature, the vapour pressure falls to 56.795 kPa. Calculate the molecular weight of A. Assume that the solution of A in ether is very dilute. Solution: Basis: 3 g of a compound A in 50 g of ether at 293 K V.P. of ether in solution = V.P. of pure ether x Mole fraction of ether in solution . 56.795 = 58.928 xi ∴
x1 = 0.9638 x1 + x2 = 1 x2 = 1-x1 = 1 – 0.9638 = 0.0362 Mole fraction of A in solution = 0.0362 Mole fraction of ether in solution = 0.9638 Mole fraction of ether in solution = 50 = 0.6757 mol 74 Total moles of solution = 0.6757 = 0.7011 mol 0.9638 Mole of solution = 0.0362 x 0.7011 = 0.0254 mol Mol of A = gram of A = Molecular weight of A ∴
∴
Molecular weight of A =
3 = 118 0.254
(86) In the manufacture of formaldehyde; O 2, methanol and steam are mixed in proportion 0 1.5: 2:1.33 (by weight ) at 283 K (10 C). The total pressure is 70.928 kPa g. Calculate the partial pressure of each of the components present in this mixture. Solution: Basis: 100 kg of solution In gas mixture, the proportion of O 2:HCHO: Steam (H2O) is 1.5:2:1.33 by weight. ∴ O2 in the
mixture = 1.5 x 100 = 31.06 kg 4.83 = 31.06 = 0.97 kmol 32 Methanol in mixture = 2 x 100 = 41.4 kg 4.83 = 41.4 = 1.3 kmol 32 Steam in mixture = 1.33 x 100 = 27.54 kg 4.83 = 27.54 = 1.53 kmol 18 Moles of gas mixture = 0.97 + 1.3 + 1.53 = 3.8 kmol Mole fraction of O 2 = xO2 = 0.97 = 0.256 3.8 Mole fraction of methanol = xCH 3OH =1.3 = 0.342 3.8 Mole fraction of Steam = xH 2O = 1.53 = 0.403 3.8 Total pressure = 70.928 kPa g P = 70.928 + 101.325 P = 172.253 kPa . Partial pressure of O 2= xO2 P = 0.256 x 172.253 = 44.10 kPa Partial pressure of methanol = 0.342 x 172.253= 58.91 kPa Partial pressure of steam = 0.403x 172.253= 69.42 kPa. ∴
0
(87) Ammonia under a pressure of 1519.875 kPa and 298 K (25 C) is heated to 620 K 0 (347 C) in a closed vessel in the presence of catalyst. Under these conditions, NH 3 is partially decomposed. The vessel is such that the volume remain effectively constant, whereas the pressure increases to 5066.25 kPa. Calculate the % of NH 3 decomposed. Solution: Basis: 100 kmol NH 3 initially present at 298 K Ideal gas law for initial and final conditions is P1V1= n1RT1 P2V2 = n2RT2 P1V1 =n1. T1 P2V2 n2 T2 But V1 = V2 as vessel being closed ∴
P1 =n1. T1 P2 n2 T2 Where, n1 = initial moles = 100 kmol, P 1 = 1519.875 kPa, T1 = 298 K, n2 = fional moles of gas mixture = ?, P 2 = 5066.250 kPa, T 2 = 620 K ∴
n2 = n1x T1 x P1 T2 P2 = 100 x298 x5066.250= 1690.21 kmol 620 1519.875 Total moles after decomposition = 160.21 kmol Decomposition of NH 3 takes place as : 2 NH3 = N2 + 3 H2 Let x be the kmol NH 3 decomposed NH3undecomposed= (100 - x) kmol N2 produced = 1 .x = 0.5 x kmol 2 H2 produced = 3 . x = 1.5 x kmol 2 Moles of NH3+N2+H2 after decomposition =160.21 100 – x + 0.5 x + 1.5 x = 160.21 x = 60.21 kmol ∴
% decomposition of NH 3 = moles of NH3 decomposed x 100 Initial moles NH3 = 60.21 x 100 = 60.21 100 (88) A gas containing 96% ethylene and 4% butenes by volume is passed through a bed of a activated carbon where 98% of the original butenes are adsorbed and none of the ethylene. In five hours of continuous operation if quantity of butenes removed is 0.5 kmol, find (i) Mole % ethylene in gas leaving carbon bed and (ii) Molar flow rate of the feed gas to the carbon bed. Solution: Basis: Five hours of operation Amount of butenes removed = 0.5 kmol Butenes removed = 0.5 = 0.1 kmol/h 5 Let y be the molar flow rate (kmol/h) of gas to the carbon bed Butenes in gas fed = 0.04 y kmol/h Butenes adsorbed = 0.98 x 0.04 y = 0.0392 y kmol/h = 0.0392 y = 0.1 ∴
y = 2.551 kmol/h Molar flow rate of gas to carbon bed = 2.551 kmol/h Butenes not adsorbed = 0.04 y – 0.1= 0.04 x 2.551 – 0.1 = 0.00204 kmol/h Ethylene in inlet gas leaving carbon bed = 0.96 x 2.551 = 2.449 kmol/h Ethylene in gas leaving carbon bed = Ethylene in inlet gas = 2.449 kmol/h Gas leaving carbon bed = 2.449 + 0.00204 = 2.451 kmol/h ∴
Mole% ethylene in gas leaving carbon bed = 2.449 x 100 = 99.92% 2.451
(89) A sample of gas having volume of 10 l at 101.325 kPa pressure and at temperature of 0 298 K (25 C) is compressed to a high pressure so that its volume reduces by 4.5 l. if the pressure rises by 0.1 MPa, what will be the rise in temperature? Solution: : Basis: 10 l of gas at 298 K PV = nRT -3 3 3. . Where, V = 10 l = 10 x 10 m , R= 8.31451 m kPa/(kmol K), T = 298 K, P = 201.325 kPa. n = PV RT -3 -4 = 101.325 x 10 x 10 = 4.09 x 10 kmol 8.31451 x 298 Final pressure = 101.325 + 100 = 201.325 kPa. (as pressure increases by 0.1 MPa i.e., 100 kPa) Final volume = 10 – 4.5 = 5.5 l = 5.5 x -3 3 10 m PV = nRT -3 3 Where, P=201.325 kPa, V = 5.5 x 10 m -3 -4 201.325 x 5.5 x 10 = 4.09 x 10 x 8.31451 x T ∴ ∴
T = 325.6 K Rise in temperature = 325.6 – 298 = 27.6 K (27.6 deg C )
(90) Exhaust gas having 75% N 2 and 25% CO2 (by volume) is passed through a absorption column. 97% CO2 is absorbed in KOH solution. The gas enters the system at a temperature 0 0 323 K (50 C) and 740 torr and leaves at 303 K (30 C) and 737 torr. Calculate (i) The volume of gases leaving per 100 litres entering. (ii) The weight of CO2 absorbed per 100 litres of gas entering Solution: Basis: 100 mol of gas entering It contains 75 mol N2 and 25 mol of CO2 Amount of CO2 absorbed = 0.97 x 25=24.25 mol Amount of CO2 unabsorbed=25-24.25 =0.75 mol Total gas leaving system = 75 + 0.75 =75.75 mol PV = nRT Where, P = 740 torr = 98.66 kPa ∴
V = 100 x 8.31451 x 323 = 2722 l 98.66
Volume of 100 mol of entering gas = 2722 l Weight of CO2 absorbed per 100 l gas entering = 24.25 x 100 = 0.89 mol 2722 = 0.89 x 44 = 39.16 g ∴
Volume of gas leaving based on 100 mol gas entering is P’ = 75.75 x 8.31451 x 303 = 1942 l 98.26 Volume of gas leaving per 100 l gas entering = 1942x 100 = 71.43 l 2722
V’ = n’ RT’
(91) Cracked gas form a petroleum refinery contains 45% CH 4, 1% C2H6, 25% C2H4, 7% C3H8, 8% C3H6 and 5% C4H10 by volume. Calculate (i) the average molecular weight of gas mixture, (ii) the composition by weight, and (iii) the specific gravity of the mixture taking average molecular weight of air as 28.84 Solution: Basis: 100 kmol of cracked gas. It contains 45 kmol CH 4, 10 kmol C2H6, 25 kmol C 2H4, 7 kmol C3H8, and 5 kmol C4H10 Molecular weight data: CH4= 16, C2H6= 30, C2H4 = 28, C3H6= 42, and C4H10 = 58 Weight of methane = 45 x 16 = 720 kg In the same way, calculate the weight of each components of cracked gas Composition of refinery gas: Average molecular weight of refinery gsa is Mavg = 2854 = 26.54 100 OR Mavg = M.CH4.xCH4 + M.C2H6.xC2H6 + M.C2H4.xC2H4 + M.C3H8.xC3H8 + M.C3H6.xC3H6 + M.C4H10.xC4H10 = 16 x 0.45 + 30 x 0.1 + 28 x 0.25 + 44 x 0.07 + 42 x 0.08 + 58 x 0.05 = 26.54 Specific gravity of gas mixture = 26.54 = 0.92 28.84
Component
Kmol
Mol.Wt
kg
Weight%
CH4
45
16
720
27.13
C2H6
10
30
300
11.30
C2H4
25
28
700
26.37
C3H8
7
44
308
11.61
C3H6
8
42
336
12.66
C4H10
5
58
290
10.93
Total
100
-
2654
100
(92) The composition of gas mixture in manufacture of nitric acid at a pressure of 0.709 0 MPa and 923 K (650 C) is as follows: N2 = 70.5%, O2 = 18.8%, H2O = 1.2% and NH3 = 9.5%. Calculate the density of gas mixture using ideal gas law Sol: Basis: 100 kmol of gas mixture at 923 K It contains 70.5 kmol of N 2, 18.8 kmol O 2, 1.2 kmol H2O and 9.5 kmol NH 3. Mole fraction of N 2 = xN2 = 70.5 = 0.705 100 Mole fraction of O 2 = xO2 = 18.8 = 0.188 100 Mole fraction of H 2O = xH2O = 1.2 = 0.012 100 and Mole fraction of NH 3 = xNH3 = 9.5 = 0.095 100 Molecular weight of N 2 = 28, Molecular weight of O 2 = 32, Molecular weight of H 2O = 18 and Molecular weight of NH 3 = 17 Average molecular weight of gas mixture is Mavg= ∑Mixi = 28 x 0.705 + 32 x0.188 + 18 x 0.012 + 17 x 0.095 = 27.59 3. . P = 0.709 MPa g = 810.325 kPa, T = 923 K, R= 8.31451 m kPa/(kmol K) Density of gas mixture is given by ρ =
PMavg RT 3 = 810.325 x 27.59 = 2.913 kg/m 8.31451 x 923
(93) The Orsat (dry) analysis of a flue gas from a boiler house is as given below: CO 2 = 10%, O2= 7.96%, N2 = 82% and SO2 = 0.04% by volume. The flue gas pressure is 100 kPa 0 (750 torr) and temperature is 463 K (190 C). SO2 is undesirable from the point of view of occupational hazards (environmental pollution). Express the concentration of SO 2 in ppm 3 and mg/m . Sol: Basis: 100 kmol of a flue gas �� �������� �� ���� �� ���� ���� ���� ��� �� ���� ��� ��� ���� ���� ����
Amount of CO2 = 10 x 44 = 440 kg Amount of O2 = 7.96 x 32 = 254.72 kg Amount of N2 = 82 x 28 = 2296 kg Amount of SO2 = 0.04 x 64 = 2.56 kg Amount of flue gas = 440 + 254.72 + 2296 + 2.56 = 2993.28 kg 6 Concentration of SO 2 in ppm= 2.56 x10 2993.28 = 855.25
