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1 | 805-808 | This blocks capillaries and creates a medical
condition known as bends, which are painful and dangerous to life If N2 gas is bubbled through water at 293 K, how many millimoles of N2
gas would dissolve in 1 litre of water Assume that N2 exerts a partial
pressure of 0 987 bar |
1 | 806-809 | If N2 gas is bubbled through water at 293 K, how many millimoles of N2
gas would dissolve in 1 litre of water Assume that N2 exerts a partial
pressure of 0 987 bar Given that Henry’s law constant for N2 at 293 K is
76 |
1 | 807-810 | Assume that N2 exerts a partial
pressure of 0 987 bar Given that Henry’s law constant for N2 at 293 K is
76 48 kbar |
1 | 808-811 | 987 bar Given that Henry’s law constant for N2 at 293 K is
76 48 kbar The solubility of gas is related to the mole fraction in aqueous solution |
1 | 809-812 | Given that Henry’s law constant for N2 at 293 K is
76 48 kbar The solubility of gas is related to the mole fraction in aqueous solution The mole fraction of the gas in the solution is calculated by applying
Henry’s law |
1 | 810-813 | 48 kbar The solubility of gas is related to the mole fraction in aqueous solution The mole fraction of the gas in the solution is calculated by applying
Henry’s law Thus:
x (Nitrogen) =
H
p (nitrogen)
K
= 0 |
1 | 811-814 | The solubility of gas is related to the mole fraction in aqueous solution The mole fraction of the gas in the solution is calculated by applying
Henry’s law Thus:
x (Nitrogen) =
H
p (nitrogen)
K
= 0 987bar
76,480 bar = 1 |
1 | 812-815 | The mole fraction of the gas in the solution is calculated by applying
Henry’s law Thus:
x (Nitrogen) =
H
p (nitrogen)
K
= 0 987bar
76,480 bar = 1 29 × 10–5
As 1 litre of water contains 55 |
1 | 813-816 | Thus:
x (Nitrogen) =
H
p (nitrogen)
K
= 0 987bar
76,480 bar = 1 29 × 10–5
As 1 litre of water contains 55 5 mol of it, therefore if n represents
number of moles of N2 in solution,
x (Nitrogen) =
mol
mol
n+55 |
1 | 814-817 | 987bar
76,480 bar = 1 29 × 10–5
As 1 litre of water contains 55 5 mol of it, therefore if n represents
number of moles of N2 in solution,
x (Nitrogen) =
mol
mol
n+55 5 mol
n
= 55 |
1 | 815-818 | 29 × 10–5
As 1 litre of water contains 55 5 mol of it, therefore if n represents
number of moles of N2 in solution,
x (Nitrogen) =
mol
mol
n+55 5 mol
n
= 55 5
n
= 1 |
1 | 816-819 | 5 mol of it, therefore if n represents
number of moles of N2 in solution,
x (Nitrogen) =
mol
mol
n+55 5 mol
n
= 55 5
n
= 1 29 × 10–5
(n in denominator is neglected as it is < < 55 |
1 | 817-820 | 5 mol
n
= 55 5
n
= 1 29 × 10–5
(n in denominator is neglected as it is < < 55 5)
Thus n = 1 |
1 | 818-821 | 5
n
= 1 29 × 10–5
(n in denominator is neglected as it is < < 55 5)
Thus n = 1 29 × 10–5 × 55 |
1 | 819-822 | 29 × 10–5
(n in denominator is neglected as it is < < 55 5)
Thus n = 1 29 × 10–5 × 55 5 mol = 7 |
1 | 820-823 | 5)
Thus n = 1 29 × 10–5 × 55 5 mol = 7 16 × 10–4 mol
=
7 |
1 | 821-824 | 29 × 10–5 × 55 5 mol = 7 16 × 10–4 mol
=
7 16×104
mol × 1000 mmol
1 mol
= 0 |
1 | 822-825 | 5 mol = 7 16 × 10–4 mol
=
7 16×104
mol × 1000 mmol
1 mol
= 0 716 mmol
Example 1 |
1 | 823-826 | 16 × 10–4 mol
=
7 16×104
mol × 1000 mmol
1 mol
= 0 716 mmol
Example 1 4
Example 1 |
1 | 824-827 | 16×104
mol × 1000 mmol
1 mol
= 0 716 mmol
Example 1 4
Example 1 4
Example 1 |
1 | 825-828 | 716 mmol
Example 1 4
Example 1 4
Example 1 4
Example 1 |
1 | 826-829 | 4
Example 1 4
Example 1 4
Example 1 4
Example 1 |
1 | 827-830 | 4
Example 1 4
Example 1 4
Example 1 4
Solution
Solution
Solution
Solution
Solution
Gas
Temperature/K
KH /kbar
Gas
Temperature/K
KH/kbar
He
293
144 |
1 | 828-831 | 4
Example 1 4
Example 1 4
Solution
Solution
Solution
Solution
Solution
Gas
Temperature/K
KH /kbar
Gas
Temperature/K
KH/kbar
He
293
144 97
H2
293
69 |
1 | 829-832 | 4
Example 1 4
Solution
Solution
Solution
Solution
Solution
Gas
Temperature/K
KH /kbar
Gas
Temperature/K
KH/kbar
He
293
144 97
H2
293
69 16
N2
293
76 |
1 | 830-833 | 4
Solution
Solution
Solution
Solution
Solution
Gas
Temperature/K
KH /kbar
Gas
Temperature/K
KH/kbar
He
293
144 97
H2
293
69 16
N2
293
76 48
N2
303
88 |
1 | 831-834 | 97
H2
293
69 16
N2
293
76 48
N2
303
88 84
O2
293
34 |
1 | 832-835 | 16
N2
293
76 48
N2
303
88 84
O2
293
34 86
O2
303
46 |
1 | 833-836 | 48
N2
303
88 84
O2
293
34 86
O2
303
46 82
Table 1 |
1 | 834-837 | 84
O2
293
34 86
O2
303
46 82
Table 1 2: Values of Henry's Law Constant for Some Selected Gases in Water
Argon
298
40 |
1 | 835-838 | 86
O2
303
46 82
Table 1 2: Values of Henry's Law Constant for Some Selected Gases in Water
Argon
298
40 3
CO2
298
1 |
1 | 836-839 | 82
Table 1 2: Values of Henry's Law Constant for Some Selected Gases in Water
Argon
298
40 3
CO2
298
1 67
Formaldehyde
298
1 |
1 | 837-840 | 2: Values of Henry's Law Constant for Some Selected Gases in Water
Argon
298
40 3
CO2
298
1 67
Formaldehyde
298
1 83×10-5
Methane
298
0 |
1 | 838-841 | 3
CO2
298
1 67
Formaldehyde
298
1 83×10-5
Methane
298
0 413
Vinyl chloride
298
0 |
1 | 839-842 | 67
Formaldehyde
298
1 83×10-5
Methane
298
0 413
Vinyl chloride
298
0 611
Rationalised 2023-24
9
Solutions
To avoid bends, as well as, the toxic effects of high concentrations
of nitrogen in the blood, the tanks used by scuba divers are filled
with air diluted with helium (11 |
1 | 840-843 | 83×10-5
Methane
298
0 413
Vinyl chloride
298
0 611
Rationalised 2023-24
9
Solutions
To avoid bends, as well as, the toxic effects of high concentrations
of nitrogen in the blood, the tanks used by scuba divers are filled
with air diluted with helium (11 7% helium, 56 |
1 | 841-844 | 413
Vinyl chloride
298
0 611
Rationalised 2023-24
9
Solutions
To avoid bends, as well as, the toxic effects of high concentrations
of nitrogen in the blood, the tanks used by scuba divers are filled
with air diluted with helium (11 7% helium, 56 2% nitrogen and
32 |
1 | 842-845 | 611
Rationalised 2023-24
9
Solutions
To avoid bends, as well as, the toxic effects of high concentrations
of nitrogen in the blood, the tanks used by scuba divers are filled
with air diluted with helium (11 7% helium, 56 2% nitrogen and
32 1% oxygen) |
1 | 843-846 | 7% helium, 56 2% nitrogen and
32 1% oxygen) · At high altitudes the partial pressure of oxygen is less than that at
the ground level |
1 | 844-847 | 2% nitrogen and
32 1% oxygen) · At high altitudes the partial pressure of oxygen is less than that at
the ground level This leads to low concentrations of oxygen in the
blood and tissues of people living at high altitudes or climbers |
1 | 845-848 | 1% oxygen) · At high altitudes the partial pressure of oxygen is less than that at
the ground level This leads to low concentrations of oxygen in the
blood and tissues of people living at high altitudes or climbers Low
blood oxygen causes climbers to become weak and unable to think
clearly, symptoms of a condition known as anoxia |
1 | 846-849 | · At high altitudes the partial pressure of oxygen is less than that at
the ground level This leads to low concentrations of oxygen in the
blood and tissues of people living at high altitudes or climbers Low
blood oxygen causes climbers to become weak and unable to think
clearly, symptoms of a condition known as anoxia Effect of Temperature
Solubility of gases in liquids decreases with rise in temperature |
1 | 847-850 | This leads to low concentrations of oxygen in the
blood and tissues of people living at high altitudes or climbers Low
blood oxygen causes climbers to become weak and unable to think
clearly, symptoms of a condition known as anoxia Effect of Temperature
Solubility of gases in liquids decreases with rise in temperature When
dissolved, the gas molecules are present in liquid phase and the process
of dissolution can be considered similar to condensation and heat
is evolved in this process |
1 | 848-851 | Low
blood oxygen causes climbers to become weak and unable to think
clearly, symptoms of a condition known as anoxia Effect of Temperature
Solubility of gases in liquids decreases with rise in temperature When
dissolved, the gas molecules are present in liquid phase and the process
of dissolution can be considered similar to condensation and heat
is evolved in this process We have learnt in the last Section that
dissolution process involves dynamic equilibrium and thus must
follow Le Chatelier’s Principle |
1 | 849-852 | Effect of Temperature
Solubility of gases in liquids decreases with rise in temperature When
dissolved, the gas molecules are present in liquid phase and the process
of dissolution can be considered similar to condensation and heat
is evolved in this process We have learnt in the last Section that
dissolution process involves dynamic equilibrium and thus must
follow Le Chatelier’s Principle As dissolution is an exothermic
process, the solubility should decrease with increase of
temperature |
1 | 850-853 | When
dissolved, the gas molecules are present in liquid phase and the process
of dissolution can be considered similar to condensation and heat
is evolved in this process We have learnt in the last Section that
dissolution process involves dynamic equilibrium and thus must
follow Le Chatelier’s Principle As dissolution is an exothermic
process, the solubility should decrease with increase of
temperature Liquid solutions are formed when solvent is a liquid |
1 | 851-854 | We have learnt in the last Section that
dissolution process involves dynamic equilibrium and thus must
follow Le Chatelier’s Principle As dissolution is an exothermic
process, the solubility should decrease with increase of
temperature Liquid solutions are formed when solvent is a liquid The solute can be
a gas, a liquid or a solid |
1 | 852-855 | As dissolution is an exothermic
process, the solubility should decrease with increase of
temperature Liquid solutions are formed when solvent is a liquid The solute can be
a gas, a liquid or a solid Solutions of gases in liquids have already
been discussed in Section 1 |
1 | 853-856 | Liquid solutions are formed when solvent is a liquid The solute can be
a gas, a liquid or a solid Solutions of gases in liquids have already
been discussed in Section 1 3 |
1 | 854-857 | The solute can be
a gas, a liquid or a solid Solutions of gases in liquids have already
been discussed in Section 1 3 2 |
1 | 855-858 | Solutions of gases in liquids have already
been discussed in Section 1 3 2 In this Section, we shall discuss the
solutions of liquids and solids in a liquid |
1 | 856-859 | 3 2 In this Section, we shall discuss the
solutions of liquids and solids in a liquid Such solutions may contain
one or more volatile components |
1 | 857-860 | 2 In this Section, we shall discuss the
solutions of liquids and solids in a liquid Such solutions may contain
one or more volatile components Generally, the liquid solvent is volatile |
1 | 858-861 | In this Section, we shall discuss the
solutions of liquids and solids in a liquid Such solutions may contain
one or more volatile components Generally, the liquid solvent is volatile The solute may or may not be volatile |
1 | 859-862 | Such solutions may contain
one or more volatile components Generally, the liquid solvent is volatile The solute may or may not be volatile We shall discuss the properties
of only binary solutions, that is, the solutions containing two
components, namely, the solutions of (i) liquids in liquids and (ii) solids
in liquids |
1 | 860-863 | Generally, the liquid solvent is volatile The solute may or may not be volatile We shall discuss the properties
of only binary solutions, that is, the solutions containing two
components, namely, the solutions of (i) liquids in liquids and (ii) solids
in liquids Let us consider a binary solution of two volatile liquids and denote the
two components as 1 and 2 |
1 | 861-864 | The solute may or may not be volatile We shall discuss the properties
of only binary solutions, that is, the solutions containing two
components, namely, the solutions of (i) liquids in liquids and (ii) solids
in liquids Let us consider a binary solution of two volatile liquids and denote the
two components as 1 and 2 When taken in a closed vessel, both the
components would evaporate and eventually an equilibrium would be
established between vapour phase and the liquid phase |
1 | 862-865 | We shall discuss the properties
of only binary solutions, that is, the solutions containing two
components, namely, the solutions of (i) liquids in liquids and (ii) solids
in liquids Let us consider a binary solution of two volatile liquids and denote the
two components as 1 and 2 When taken in a closed vessel, both the
components would evaporate and eventually an equilibrium would be
established between vapour phase and the liquid phase Let the total
vapour pressure at this stage be ptotal and p1 and p2 be the partial
vapour pressures of the two components 1 and 2 respectively |
1 | 863-866 | Let us consider a binary solution of two volatile liquids and denote the
two components as 1 and 2 When taken in a closed vessel, both the
components would evaporate and eventually an equilibrium would be
established between vapour phase and the liquid phase Let the total
vapour pressure at this stage be ptotal and p1 and p2 be the partial
vapour pressures of the two components 1 and 2 respectively These
partial pressures are related to the mole fractions x1 and x2 of the two
components 1 and 2 respectively |
1 | 864-867 | When taken in a closed vessel, both the
components would evaporate and eventually an equilibrium would be
established between vapour phase and the liquid phase Let the total
vapour pressure at this stage be ptotal and p1 and p2 be the partial
vapour pressures of the two components 1 and 2 respectively These
partial pressures are related to the mole fractions x1 and x2 of the two
components 1 and 2 respectively The French chemist, Francois Marte Raoult (1886) gave the
quantitative relationship between them |
1 | 865-868 | Let the total
vapour pressure at this stage be ptotal and p1 and p2 be the partial
vapour pressures of the two components 1 and 2 respectively These
partial pressures are related to the mole fractions x1 and x2 of the two
components 1 and 2 respectively The French chemist, Francois Marte Raoult (1886) gave the
quantitative relationship between them The relationship is known as
the Raoult’s law which states that for a solution of volatile liquids,
1 |
1 | 866-869 | These
partial pressures are related to the mole fractions x1 and x2 of the two
components 1 and 2 respectively The French chemist, Francois Marte Raoult (1886) gave the
quantitative relationship between them The relationship is known as
the Raoult’s law which states that for a solution of volatile liquids,
1 4
1 |
1 | 867-870 | The French chemist, Francois Marte Raoult (1886) gave the
quantitative relationship between them The relationship is known as
the Raoult’s law which states that for a solution of volatile liquids,
1 4
1 4
1 |
1 | 868-871 | The relationship is known as
the Raoult’s law which states that for a solution of volatile liquids,
1 4
1 4
1 4
1 |
1 | 869-872 | 4
1 4
1 4
1 4
1 |
1 | 870-873 | 4
1 4
1 4
1 4 Vapour
Vapour
Vapour
Vapour
Vapour
Pressure of
Pressure of
Pressure of
Pressure of
Pressure of
Liquid
Liquid
Liquid
Liquid
Liquid
Solutions
Solutions
Solutions
Solutions
Solutions
Intext Questions
Intext Questions
Intext Questions
Intext Questions
Intext Questions
1 |
1 | 871-874 | 4
1 4
1 4 Vapour
Vapour
Vapour
Vapour
Vapour
Pressure of
Pressure of
Pressure of
Pressure of
Pressure of
Liquid
Liquid
Liquid
Liquid
Liquid
Solutions
Solutions
Solutions
Solutions
Solutions
Intext Questions
Intext Questions
Intext Questions
Intext Questions
Intext Questions
1 6 H2S, a toxic gas with rotten egg like smell, is used for the qualitative analysis |
1 | 872-875 | 4
1 4 Vapour
Vapour
Vapour
Vapour
Vapour
Pressure of
Pressure of
Pressure of
Pressure of
Pressure of
Liquid
Liquid
Liquid
Liquid
Liquid
Solutions
Solutions
Solutions
Solutions
Solutions
Intext Questions
Intext Questions
Intext Questions
Intext Questions
Intext Questions
1 6 H2S, a toxic gas with rotten egg like smell, is used for the qualitative analysis If
the solubility of H2S in water at STP is 0 |
1 | 873-876 | 4 Vapour
Vapour
Vapour
Vapour
Vapour
Pressure of
Pressure of
Pressure of
Pressure of
Pressure of
Liquid
Liquid
Liquid
Liquid
Liquid
Solutions
Solutions
Solutions
Solutions
Solutions
Intext Questions
Intext Questions
Intext Questions
Intext Questions
Intext Questions
1 6 H2S, a toxic gas with rotten egg like smell, is used for the qualitative analysis If
the solubility of H2S in water at STP is 0 195 m, calculate Henry’s law constant |
1 | 874-877 | 6 H2S, a toxic gas with rotten egg like smell, is used for the qualitative analysis If
the solubility of H2S in water at STP is 0 195 m, calculate Henry’s law constant 1 |
1 | 875-878 | If
the solubility of H2S in water at STP is 0 195 m, calculate Henry’s law constant 1 7 Henry’s law constant for CO2 in water is 1 |
1 | 876-879 | 195 m, calculate Henry’s law constant 1 7 Henry’s law constant for CO2 in water is 1 67×108 Pa at 298 K |
1 | 877-880 | 1 7 Henry’s law constant for CO2 in water is 1 67×108 Pa at 298 K Calculate
the quantity of CO2 in 500 mL of soda water when packed under 2 |
1 | 878-881 | 7 Henry’s law constant for CO2 in water is 1 67×108 Pa at 298 K Calculate
the quantity of CO2 in 500 mL of soda water when packed under 2 5 atm
CO2 pressure at 298 K |
1 | 879-882 | 67×108 Pa at 298 K Calculate
the quantity of CO2 in 500 mL of soda water when packed under 2 5 atm
CO2 pressure at 298 K 1 |
1 | 880-883 | Calculate
the quantity of CO2 in 500 mL of soda water when packed under 2 5 atm
CO2 pressure at 298 K 1 4 |
1 | 881-884 | 5 atm
CO2 pressure at 298 K 1 4 1 Vapour
Pressure of
Liquid-
Liquid
Solutions
Rationalised 2023-24
10
Chemistry
the partial vapour pressure of each component of the solution
is directly proportional to its mole fraction present in solution |
1 | 882-885 | 1 4 1 Vapour
Pressure of
Liquid-
Liquid
Solutions
Rationalised 2023-24
10
Chemistry
the partial vapour pressure of each component of the solution
is directly proportional to its mole fraction present in solution Thus, for component 1
p1 µ x1
and p1 =
10
p x1
(1 |
1 | 883-886 | 4 1 Vapour
Pressure of
Liquid-
Liquid
Solutions
Rationalised 2023-24
10
Chemistry
the partial vapour pressure of each component of the solution
is directly proportional to its mole fraction present in solution Thus, for component 1
p1 µ x1
and p1 =
10
p x1
(1 12)
where
10
p is the vapour pressure of pure component 1 at the same
temperature |
1 | 884-887 | 1 Vapour
Pressure of
Liquid-
Liquid
Solutions
Rationalised 2023-24
10
Chemistry
the partial vapour pressure of each component of the solution
is directly proportional to its mole fraction present in solution Thus, for component 1
p1 µ x1
and p1 =
10
p x1
(1 12)
where
10
p is the vapour pressure of pure component 1 at the same
temperature Similarly, for component 2
p2 = p2
0 x2
(1 |
1 | 885-888 | Thus, for component 1
p1 µ x1
and p1 =
10
p x1
(1 12)
where
10
p is the vapour pressure of pure component 1 at the same
temperature Similarly, for component 2
p2 = p2
0 x2
(1 13)
where p2
0 represents the vapour pressure of the pure component 2 |
1 | 886-889 | 12)
where
10
p is the vapour pressure of pure component 1 at the same
temperature Similarly, for component 2
p2 = p2
0 x2
(1 13)
where p2
0 represents the vapour pressure of the pure component 2 According to Dalton’s law of partial pressures, the total pressure
(
ptotal
) over the solution phase in the container will be the sum of the
partial pressures of the components of the solution and is given as:
ptotal = p1 + p2
(1 |
1 | 887-890 | Similarly, for component 2
p2 = p2
0 x2
(1 13)
where p2
0 represents the vapour pressure of the pure component 2 According to Dalton’s law of partial pressures, the total pressure
(
ptotal
) over the solution phase in the container will be the sum of the
partial pressures of the components of the solution and is given as:
ptotal = p1 + p2
(1 14)
Substituting the values of p1 and p2, we get
ptotal = x1 p1
0 + x2 p2
0
= (1 – x2) p1
0 + x2 p2
0
(1 |
1 | 888-891 | 13)
where p2
0 represents the vapour pressure of the pure component 2 According to Dalton’s law of partial pressures, the total pressure
(
ptotal
) over the solution phase in the container will be the sum of the
partial pressures of the components of the solution and is given as:
ptotal = p1 + p2
(1 14)
Substituting the values of p1 and p2, we get
ptotal = x1 p1
0 + x2 p2
0
= (1 – x2) p1
0 + x2 p2
0
(1 15)
= p1
0 + (p2
0 – p1
0) x2
(1 |
1 | 889-892 | According to Dalton’s law of partial pressures, the total pressure
(
ptotal
) over the solution phase in the container will be the sum of the
partial pressures of the components of the solution and is given as:
ptotal = p1 + p2
(1 14)
Substituting the values of p1 and p2, we get
ptotal = x1 p1
0 + x2 p2
0
= (1 – x2) p1
0 + x2 p2
0
(1 15)
= p1
0 + (p2
0 – p1
0) x2
(1 16)
Following conclusions can be drawn from equation (1 |
1 | 890-893 | 14)
Substituting the values of p1 and p2, we get
ptotal = x1 p1
0 + x2 p2
0
= (1 – x2) p1
0 + x2 p2
0
(1 15)
= p1
0 + (p2
0 – p1
0) x2
(1 16)
Following conclusions can be drawn from equation (1 16) |
1 | 891-894 | 15)
= p1
0 + (p2
0 – p1
0) x2
(1 16)
Following conclusions can be drawn from equation (1 16) (i) Total vapour pressure over the solution can be related to the mole
fraction of any one component |
1 | 892-895 | 16)
Following conclusions can be drawn from equation (1 16) (i) Total vapour pressure over the solution can be related to the mole
fraction of any one component (ii) Total vapour pressure over the solution varies linearly with the
mole fraction of component 2 |
1 | 893-896 | 16) (i) Total vapour pressure over the solution can be related to the mole
fraction of any one component (ii) Total vapour pressure over the solution varies linearly with the
mole fraction of component 2 (iii) Depending on the vapour pressures
of the pure components 1 and 2,
total vapour pressure over the
solution decreases or increases with
the increase of the mole fraction of
component 1 |
1 | 894-897 | (i) Total vapour pressure over the solution can be related to the mole
fraction of any one component (ii) Total vapour pressure over the solution varies linearly with the
mole fraction of component 2 (iii) Depending on the vapour pressures
of the pure components 1 and 2,
total vapour pressure over the
solution decreases or increases with
the increase of the mole fraction of
component 1 A plot of p1 or p2 versus the mole
fractions x1 and x2 for a solution gives a
linear plot as shown in Fig |
1 | 895-898 | (ii) Total vapour pressure over the solution varies linearly with the
mole fraction of component 2 (iii) Depending on the vapour pressures
of the pure components 1 and 2,
total vapour pressure over the
solution decreases or increases with
the increase of the mole fraction of
component 1 A plot of p1 or p2 versus the mole
fractions x1 and x2 for a solution gives a
linear plot as shown in Fig 1 |
1 | 896-899 | (iii) Depending on the vapour pressures
of the pure components 1 and 2,
total vapour pressure over the
solution decreases or increases with
the increase of the mole fraction of
component 1 A plot of p1 or p2 versus the mole
fractions x1 and x2 for a solution gives a
linear plot as shown in Fig 1 3 |
1 | 897-900 | A plot of p1 or p2 versus the mole
fractions x1 and x2 for a solution gives a
linear plot as shown in Fig 1 3 These
lines (I and II) pass through the points for
which x1 and x2 are equal to unity |
1 | 898-901 | 1 3 These
lines (I and II) pass through the points for
which x1 and x2 are equal to unity Similarly the plot (line III) of ptotal versus
x2 is also linear (Fig |
1 | 899-902 | 3 These
lines (I and II) pass through the points for
which x1 and x2 are equal to unity Similarly the plot (line III) of ptotal versus
x2 is also linear (Fig 1 |
1 | 900-903 | These
lines (I and II) pass through the points for
which x1 and x2 are equal to unity Similarly the plot (line III) of ptotal versus
x2 is also linear (Fig 1 3) |
1 | 901-904 | Similarly the plot (line III) of ptotal versus
x2 is also linear (Fig 1 3) The minimum
value of ptotal is p1
0
and the maximum value
is p2
0, assuming that component 1 is less
volatile than component 2, i |
1 | 902-905 | 1 3) The minimum
value of ptotal is p1
0
and the maximum value
is p2
0, assuming that component 1 is less
volatile than component 2, i e |
1 | 903-906 | 3) The minimum
value of ptotal is p1
0
and the maximum value
is p2
0, assuming that component 1 is less
volatile than component 2, i e , p1
0 < p2
The composition of vapour phase in0 |
1 | 904-907 | The minimum
value of ptotal is p1
0
and the maximum value
is p2
0, assuming that component 1 is less
volatile than component 2, i e , p1
0 < p2
The composition of vapour phase in0 equilibrium with the solution is determined
by the partial pressures of the components |
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