Mole concept

1
The Mole Concept
2.1 The Mole 2.2 Molar Volume and Avogadros
Law 2.3 Ideal Gas Equation 2.4 Determination of
Molar Mass 2.5 Daltons Law of Partial Pressures
2
The Mole
3
Mole
P. 3 / 66
4
This number, known as the Avogadros
constant, can be determined by mass spectrometry.
B is the magnetic field strength. V is the
accelerating potential. k is a constant of the
instrument.
5
At fixed e, k, B and V m can be determined.
6
Q.5
Avogadros number ? 1.992648 ? 10-23 g
6.022 ? 1023 mol?1
7
2.1 The mole (SB p.18)
What is mole?
Item Unit used to count No. of items per unit
Shoes pairs 2
Eggs dozens 12
Paper reams 500
Particles in Chemistry moles 6.022 ? 1023
8
602,200,000,000,000,000,000,000 mol?1
???
1 mole
602.2 sextillions
P. 8 / 66
9
The fastest supercomputer can count 1.759?1015
atoms per second. Calculate the time taken for
the superconductor to count 1 mole of
carbon-12 atoms.
? 10.85 years
P. 9 / 66
10
We can count the number of coins by weighing if
the mass of one coin is known.
Similarly, we can count the number of 12C by
weighing if the mass of one 12C is known.
P. 10 / 66
11
Molar mass is the mass, in grams, of 1 mole of a
substance
12
Q.6
Relative isotopic mass Relative intensity
12.000 100.00
13.003 1.12
13
Q.6
Relative isotopic mass Relative intensity
12.000 100.00
13.003 1.12
Molar mass of carbon 12.01 g mol?1
14
Q.6
Relative isotopic mass Relative intensity
12.000 100.00
13.003 1.12
Relative isotopic mass is not exactly equal to
mass number of the isotope
15
Number of moles of a substance
Q.7
Number of moles of oxygen atoms
16
Q.7
Number of moles of oxygen atoms
3.011 ? 1019
17
2.1 The mole (SB p.20)
Molar mass is the same as the relative atomic
mass in grams.
Molar mass is the same as the relative molecular
mass in grams.
Molar mass is the same as the formula mass in
grams.
18
Molar Volume and Avogadros Law
19
What is molar volume of gases?
2.2 Molar volume and Avogadros law (SB p.24)
Volume occupied by one mole of molecules of a gas.
20
What is molar volume of gases?
2.2 Molar volume and Avogadros law (SB p.24)
Depends on T P Two sets of conditions
21
What is molar volume of gases?
2.2 Molar volume and Avogadros law (SB p.24)
at 298 K 1 atm (Room temp pressure / R.T.P.)
22
What is molar volume of gases?
2.2 Molar volume and Avogadros law (SB p.24)
at 273 K 1 atm (Standard temp pressure /
S.T.P.)
23
 
 
NH3
Not constant
24
Avogadros Law
2.2 Molar volume and Avogadros law (SB p.24)
Equal volumes of ALL gases at the same
temperature and pressure contain the same number
of moles of molecules.
At fixed T P, V ? n
If n 1, V molar volume
25
Avogadros Law
2.2 Molar volume and Avogadros law (SB p.24)
V ? n V Vm ? n
26
Interconversions involving number of moles
2.2 Molar volume and Avogadros law (SB p.24)
27
Ideal Gas Equation
28
Boyles law
2.3 Ideal gas equation (SB p.27)
At fixed n and T, PV constant or
n number of moles of gas molecules
29
2.3 Ideal gas equation (SB p.28)
Schematic diagrams explaining Boyles law
30
(No Transcript)
31
2.3 Ideal gas equation (SB p.28)
A graph of volume against the reciprocal of
pressure for a gas at constant temperature
32
Charles law
2.3 Ideal gas equation (SB p.28)
At fixed n and P,
T is the absolute temperature in Kelvin, K
33
2.3 Ideal gas equation (SB p.28)
Schematic diagrams explaining Charles law
34
2.3 Ideal gas equation (SB p.28)
Volume
0oC
-273.15 oC
Temperature / oC
A graph of volume against temperature for a gas
at constant pressure
35
2.3 Ideal gas equation (SB p.28)
/ K
A graph of volume against absolute temperature
for a gas at constant pressure
36
Ideal gas equation
2.3 Ideal gas equation (SB p.27)
R is the same for all gases R is known as the
universal gas constant
PV nRT
Ideal gas equation
37
2.3 Ideal gas equation (SB p.29)
Relationship between the ideal gas equation and
the individual gas laws
38
At fixed n,
Ideal gas behaviour is assumed in all gas laws
39
Gas laws
2.3 Ideal gas equation (SB p.27)
PV nRT
Ideal gas equation
40
2.2 Molar volume and Avogadros law (SB p.24)
Gas laws vs kinetic theory of gases
What is the difference between a theory and a law?
A law describes what happens under a given set of
circumstances.
A theory attempts to explain why that behaviour
occurs.
41
Ideal gas behaviour
Four assumptions as stated in kinetic theory of
gases

1. Gas particles are in a state of constant and
   random motion in all directions, undergoing
   frequent collisions with one another and with
   walls of the container.
2. Gas particles are treated as point masses, i.e.
   they do not occupy volume.

Volume of a gas capacity of the vessel
42
Ideal gas behaviour
3. There is no interaction among gas particles.
4. Collisions between gas particles are perfectly
elastic, i.e. kinetic energy is conserved.
43
(i) low pressure (ii) high temperature (less
deviation from 24 dm3 at R.T.P.)
44

• At low pressure, gas particles are so far apart
  that
• (1) any interaction among them becomes
  negligible (assumption 3)
• (2) the volume occupied by the gaseous
  molecules becomes negligible when compared
  with that of the container (assumption 2)

45
At high temperature, gaseous
molecules possess sufficient energy to overcome
intermolecular interactions readily. (assumption
3)
46
2.3 Ideal gas equation (SB p.31)
Check Point 2-3
(b) A reaction vessel is filled with a gas at 20
oC and 5 atm. If the vessel can withstand a
maximum internal pressure of 10 atm, what is the
highest temperature it can be safely heated to?
T2 586 K
47
2.3 Ideal gas equation (SB p.31)
Check Point 2-3
(c) A balloon is filled with helium at 25 oC.
The pressure exerted and the volume of balloon
are found to be 1.5 atm and 450 cm3 respectively.
How many moles of helium have been introduced
into the balloon?
Or
n 0.0276 mol
n 0.0276 mol
48
2.3 Ideal gas equation (SB p.31)
Check Point 2-3

• 25.8 cm3 sample of a gas has a pressure of 690
  mmHg and a temperature of 17 oC. What is the
  volume of the gas if the pressure is changed to
  1.85 atm and the temperature to 345 K?
• (1 atm 760 mmHg)

V2 15.1 cm3
49
2.3 Ideal gas equation (SB p.29)
Q.8
Calculate the universal gas constant at S.T.P.
For one mole of an ideal gas at S.T.P., P 1 atm
or 101,325 Nm-2 (Pa) V 22.4 dm3 or 0.0224 m3 n
1 mol T 273K
50
2.3 Ideal gas equation (SB p.29)
Or,
8.314 Nm K?1 mol?1 8.314 J K?1 mol?1
51
Q.9 PV nRT
m mass of the gas M molar mass of the gas
52
Determination of Molar Mass
53
Determination of Molar Mass
1. Mass Spectrometry 2. Density Measurement
54
2.4 Determination of molar mass (SB p.32)
(Mass of syringe liquid) before injection (m1)
38.545 g
55
2.4 Determination of molar mass (SB p.32)
(Mass of syringe liquid) after injection (m2)
38.260 g
56
2.4 Determination of molar mass (SB p.32)
Mass of liquid injected (m1 m2) 38.545 g
38.260 g 0.285 g
57
2.4 Determination of molar mass (SB p.32)
Volume of air in syringe before injection (V1)
10.5 cm3
58
2.4 Determination of molar mass (SB p.32)
Volume of air vapour in syringe after injection
(V2) 146.6 cm3
59
2.4 Determination of molar mass (SB p.32)
Volume of vapour in syringe (V2 V1) 146.6
cm3 - 10.5 cm3 136.1 cm3
60
2.4 Determination of molar mass (SB p.32)
Once m and V of the vapour are known,
61
2.4 Determination of molar mass (SB p.32)
Temperature 273 65 338 K
62
2.4 Determination of molar mass (SB p.32)
Pressure 1 atm
63
2.4 Determination of molar mass (SB p.32)
Q.10
Molar mass
58.0 g mol?1
Relative molecular mass 58.0
64
2.5 Daltons law of partial pressures (SB p.35)
Unit conversions -
R 8.314 J K?1mol?1 0.082 atm dm3 K?1mol?1 1m3
103 dm3 106 cm3 1 atm 760 mmHg 101325
Nm?2 101325 Pa
65
2.4 Determination of molar mass (SB p.34)
Check Point 2-4

1. 0.204 g of phosphorus vapour occupies a volume of
   81.0 cm3 at 327 oC and 1.00 atm. Determine the
   molar mass of phosphorus.

124 g mol-1
66
2.4 Determination of molar mass (SB p.34)
Check Point 2-4
(b) A sample of gas has a mass of 12.0 g and
occupies a volume of 4.16 dm3 measured at 97 oC
and 1.62 atm. Calculate the molar mass of the
gas. (1 atm 101325 Nm-2 ideal gas constant
8.314 J K-1 mol-1)
54.1 g mol-1
67
2.4 Determination of molar mass (SB p.34)
Check Point 2-4
(c) A sample of 0.037 g magnesium reacted with
hydrochloric acid to give 38.2 cm3 of hydrogen
gas measured at 25 oC and 740 mmHg. Use this
information to calculate the relative atomic mass
of magnesium.
1.52?10?3 mol
68
2.4 Determination of molar mass (SB p.34)
Check Point 2-4
(c) A sample of 0.037 g magnesium reacted with
hydrochloric acid to give 38.2 cm3 of hydrogen
gas measured at 25 oC and 740 mmHg. Use this
information to calculate the relative atomic mass
of magnesium.
Mg(s) 2HCl(aq) ? MgCl2(aq) H2(g)
1.52?10?3 mol
1.52?10?3 mol
69
Daltons Law of Partial Pressures
70
Experiment 1

• At fixed T n,
• PV constant
• (15 atm)(5 dm3) (PA)(15 dm3)
• PA 5 atm

71
Experiment 2
12 atm
empty
Gas B

• At fixed T n
• PV constant
• (12 atm)(10 dm3) (PB)(15 dm3)
• PB 8 atm

72
Experiment 3
12 atm
The total pressure PT 13 atm 5 atm
8 atm PA PB
73
12 atm
PA 5 atm PB 8 atm
Partial pressure of a constituent gas in a
mixture is the pressure that the gas would exert
if it were present alone under the same conditions
74
2.5 Daltons law of partial pressures (SB p.35)
Daltons Law of Partial Pressures
In a mixture of gases which do not react
chemically, the total pressure of the mixture is
the sum of the partial pressures of the component
gases (the sum of the pressure that each gas
would exert if it were present alone under the
same conditions).
75
2.5 Daltons law of partial pressures (SB p.35)
Derivation from ideal gas equation
Consider a mixture of gases A, B and C at fixed T
V. nA, nB and nC are the numbers of moles of
each gas. The total number of moles of gases in
the mixture nT nA nB nC
Multiply by the constant RT/V
nT(RT/V) nA (RT/V) nB (RT/V) nC (RT/V)
If gases A, B and C obey ideal gas
behaviour Ptotal PA PB PC
76
Partial Pressures and Mole Fractions
Consider a mixture of two gases A and B in a
container of capacity V at temperature T
PA PTXA
PB PTXB
77
Consider a mixture of gases A, B, C, D,
78
Q.11
At fixed T n, PV constant For N2, P1V1
P2V2 (0.20 Pa)(1.0 dm3) P2(4.0 dm3) ? P2 0.05
Pa For O2, P1V1 P2V2 (0.40 Pa)(2.0 dm3)
P2(4.0 dm3) ? P2 0.2 Pa By Daltons law of
partial pressures
79
Q.12
At 40oC, only N2 exists as a gas in the
mixture For a given amount of N2 at fixed V, P ?
T
At 200oC
80
At fixed T V,
81
Q.13(a)
At fixed P T, V ? n
20
1.96 ? 104 Nm?2
82
Q.13(a)
5.39 ? 104 Nm?2
83
Q.13(a)
2.45 ? 104 Nm?2
84
Q.13(b)
5.39 ? 104 Nm?2 2.45 ? 104 Nm?2 7.84 ? 104
Nm?2
but PT changes
85
2.5 Daltons law of partial pressures (SB p.39)
Check Point 2-5

• The valve between a 6 dm3 vessel containing gas A
  at a pressure of 7 atm and an 8 dm3 vessel
  containing gas B at a pressure of 9 atm is
  opened. Assuming that the temperature of the
  system remains constant and there is no reaction
  between the gases, what is the final pressure of
  the system?

86
2.5 Daltons law of partial pressures (SB p.39)
Check Point 2-5
Total pressure PA PB (3 5.1) atm 8.1
atm
87
2.5 Daltons law of partial pressures (SB p.39)

• 2.0 g of helium, 3.0 g of nitrogen and 4.0 g of
  argon are introduced into a 15 dm3 vessel at 100
  oC.
• (i) What are the mole fractions of helium,
  nitrogen and argon in the system?

Total no. of moles (0.50 0.11 0.10) mol
0.71 mol
88
2.5 Daltons law of partial pressures (SB p.39)

• 2.0 g of helium, 3.0 g of nitrogen and 4.0 g of
  argon are introduced into a 15 dm3 vessel at 100
  oC.
• (ii) Calculate the total pressure of the system,
  and hence the partial pressures of helium,
  nitrogen and argon.

89
The END
90
2.1 The mole (SB p.20)
Back
Example 2-1A
What is the mass of 0.2 mol of calcium
carbonate?
Answer
The chemical formula of calcium carbonate is
CaCO3. Molar mass of calcium carbonate (40.1
12.0 16.0 ? 3) g mol-1
100.1 g mol-1 Mass
of calcium carbonate Number of moles ? Molar
mass
0.2 mol ? 100.1 g mol-1
20.02 g
91
2.1 The mole (SB p.21)
Back??? 14
Example 2-1B
Calculate the number of gold atoms in a 20 g gold
pendant.
Answer
92
2.1 The mole (SB p.21)
Example 2-1C

• It is given that the molar mass of water is 18.0
  g mol-1.
• What is the mass of 4 moles of water molecules?
• How many molecules are there?
• How many atoms are there?

Answer
93
2.1 The mole (SB p.21)
Example 2-1C

• Mass of water Number of moles ? Molar mass
• 4 mol ? 18.0 g
  mol-1
• 72.0 g
• There are 4 moles of water molecules.
• Number of water molecules
• Number of moles ? Avogadro constant
• 4 mol ? 6.02 ? 1023 mol-1
• 2.408 ? 1024

94
2.1 The mole (SB p.21)
Back
Example 2-1C

• 1 water molecule has 3 atoms (i.e. 2 hydrogen
  atoms and 1 oxygen atom).
• 1 mole of water molecules has 3 moles of atoms.
• Thus, 4 moles of water molecules have 12 moles
  of atoms.
• Number of atoms 12 mol ? 6.02 ? 1023 mol-1
• 7.224 ? 1024

95
2.1 The mole (SB p.22)
Example 2-1D

• A magnesium chloride solution contains 10 g of
  magnesium chloride solid.
• Calculate the number of moles of magnesium
  chloride in the solution.

Answer
96
2.1 The mole (SB p.22)
Example 2-1D

• (b) Calculate the number of magnesium ions in
  the solution.

Answer

• 1 mole of MgCl2 contains 1 mole of Mg2 ions and
  2 moles of Cl- ions.
• Therefore, 0.105 mol of MgCl2 contains 0.105 mol
  of Mg2 ions.
• Number of Mg2 ions
• Number of moles of Mg2 ions ? Avogadro
  constant
• 0.105 mol ? 6.02 ? 1023 mol-1
• 6.321 ? 1022

97
2.1 The mole (SB p.22)
Example 2-1D

• (c) Calculate the number of chloride ions in
  the solution.

Answer

• 0.105 mol of MgCl2 contains 0.21 mol of Cl- ions.
• Number of Cl- ions
• Number of moles of Cl- ions ? Avogadro
  constant
• 0.21 mol ? 6.02 ? 1023 mol-1
• 1.264 ? 1023

98
2.1 The mole (SB p.22)
Back??? 14
Example 2-1D

• (d) Calculate the total number of ions in the
  solution.

Answer

• Total number of ions
• 6.321 ? 1022 1.264 ? 1023
• 1.896 ? 1023

99
2.1 The mole (SB p.23)
Example 2-1E
Back

• What is the mass of a carbon dioxide molecule?

Answer
100
2.1 The mole (SB p.23)
Check Point 2-1

• Find the mass in grams of 0.01 mol of zinc
  sulphide.

Answer

• Mass No. of moles ? Molar mass
• Mass of ZnS 0.01 mol ? (65.4 32.1) g mol-1
• 0.01 mol ? 97.5 g
  mol-1
• 0.975 g

101
2.1 The mole (SB p.23)
Check Point 2-1

• (b) Find the number of ions in 5.61 g of calcium
  oxide.

Answer
102
2.1 The mole (SB p.23)
Check Point 2-1

• (c) Find the number of atoms in 32.05 g of
  sulphur dioxide.

Answer
103
2.1 The mole (SB p.23)
Check Point 2-1

• There is 4.80 g of ammonium carbonate. Find the
• (i) number of moles of the compound,
• (ii) number of moles of ammonium ions,
• (iii) number of moles of carbonate ions,
• (iv) number of moles of hydrogen atoms, and
• (v) number of hydrogen atoms.

Answer
104
2.1 The mole (SB p.23)
Back
Check Point 2-1
105
2.2 Molar volume and Avogadros law (SB p.24)
Let's Think 1
What is the difference between a theory and a law?
Answer
A law tells what happens under a given set of
circumstances while a theory attempts to explain
why that behaviour occurs.
Back
106
2.2 Molar volume and Avogadros law (SB p.25)
Example 2-2A

• Find the volume occupied by 3.55 g of chlorine
  gas at room temperature and pressure.
• (Molar volume of gas at R.T.P. 24.0 dm3 mol-1)

Answer
Back
107
2.2 Molar volume and Avogadros law (SB p.25)
Example 2-2B

• Find the number of molecules in 4.48 cm3 of
  carbon dioxide gas at standard temperature and
  pressure.
• (Molar volume of gas at S.T.P. 22.4 dm3 mol-1
  Avogadro constant 6.02 ? 1023 mol-1)

Answer
Back
108
2.2 Molar volume and Avogadros law (SB p.26)
Example 2-2C

• The molar volume of nitrogen gas is found to be
  24.0 dm3 mol-1 at room temperature and pressure.
  Find the density of nitrogen gas.

Answer
Back
109
2.2 Molar volume and Avogadros law (SB p.26)
Example 2-2D

• 1.6 g of a gas occupies 1.2 dm3 at room
  temperature and pressure. What is the relative
  molecular mass of the gas?
• (Molar volume of gas at R.T.P. 24.0 dm3 mol-1)

Answer
Back
110
2.2 Molar volume and Avogadros law (SB p.27)
Check Point 2-2

• Find the volume occupied by 0.6 g of hydrogen gas
  at room temperature and pressure.
• (Molar volume of gas at R.T.P. 24.0 dm3 mol-1)

Answer
111
2.2 Molar volume and Avogadros law (SB p.27)
Check Point 2-2
(b) Calculate the number of molecules in 4.48 dm3
of hydrogen gas at standard temperature and
pressure. (Molar volume of gas at S.T.P. 22.4
dm3 mol-1)
Answer
112
2.2 Molar volume and Avogadros law (SB p.27)
Check Point 2-2

• The molar volume of oxygen gas is 22.4 dm3 mol-1
  at standard temperature and pressure. Find the
  density of oxygen gas in g cm-3 at S.T.P.

Answer
113
2.2 Molar volume and Avogadros law (SB p.27)
Check Point 2-2

• (d) What mass of oxygen has the same number of
  moles as that in 3.2 g of sulphur dioxide?

Answer
Back
114
2.3 Ideal gas equation (SB p.30)
Back
Example 2-3A

• A 500 cm3 sample of a gas in a sealed container
  at 700 mmHg and 25 oC is heater to 100 oC. What
  is the final pressure of the gas?

Answer
115
2.3 Ideal gas equation (SB p.30)
Back
Example 2-3B

• A reaction vessel of 500 cm3 is filled with
  oxygen gas at 25 oC and the final pressure
  exerted on it is 101 325 Nm-2. How many moles of
  oxygen gas are there?
• (Ideal gas constant 8.314 J K-1 mol-1)

Answer
PV
nRT 101325 Nm-2 ? 500 ? 10-6 m3 n ? 8.314 J
K-1 mol-1 ? (273 25) K
n 0.02 mol There is 0.02
mol of oxygen gas in the reaction vessel.
116
2.3 Ideal gas equation (SB p.30)
Back
Example 2-3C

• A 5 dm3 vessel can withstand a maximum internal
  pressure of 50 atm. If 2 moles of nitrogen gas
  are pumped into the vessel, what is the highest
  temperature it can be safely heated to?

Answer
117
2.3 Ideal gas equation (SB p.31)
Check Point 2-3
(b) A reaction vessel is filled with a gas at 20
oC and 5 atm. If the vessel can withstand a
maximum internal pressure of 10 atm, what is the
highest temperature it can be safely heated to?
T2 586 K
118
2.3 Ideal gas equation (SB p.31)
Check Point 2-3
(c) A balloon is filled with helium at 25 oC.
The pressure exerted and the volume of balloon
are found to be 1.5 atm and 450 cm3 respectively.
How many moles of helium have been introduced
into the balloon?
Or
n 0.0276 mol
n 0.0276 mol
119
2.3 Ideal gas equation (SB p.31)
Check Point 2-3

• 25.8 cm3 sample of a gas has a pressure of 690
  mmHg and a temperature of 17 oC. What is the
  volume of the gas if the pressure is changed to
  1.85 atm and the temperature to 345 K?
• (1 atm 760 mmHg)

V2 15.1 cm3
120
2.4 Determination of molar mass (SB p.33)
Back
Example 2-4A

• A sample of gas occupying a volume of 50 cm3 at 1
  atm and 25 oC is found to have a mass of 0.0286
  g. Find the molar mass of the gas.
• (Ideal gas constant 8.314 J K-1 mol-1 1 atm
  101325 Nm-2)

Answer
121
2.4 Determination of molar mass (SB p.34)
Back
Example 2-4B

• The density of a gas at 450 oC and 380 mmHg is
  0.0337 g dm-3. What is its molar mass? (1 atm
  760 mmHg 101325 Nm-2 ideal gas constant
  8.314 J K-1 mol-1)

Answer
122
2.4 Determination of molar mass (SB p.34)
Check Point 2-4

• 0.204 g of phosphorus vapour occupies a volume of
  81.0 cm3 at 327 oC and 1 atm. Determine the
  molar mass of phosphorus.
• (1 atm 101325 Nm-2 ideal gas constant 8.314
  J K-1 mol-1)

Answer
123
2.4 Determination of molar mass (SB p.34)
Check Point 2-4

• (b) A sample of gas has a mass of 12.0 g and
  occupies a volume of 4.16 dm3 measured at 97 oC
  and 1.62 atm. Calculate the molar mass of the
  gas.
• (1 atm 101325 Nm-2 ideal gas constant 8.314
  J K-1 mol-1)

Answer
124
2.4 Determination of molar mass (SB p.34)
Check Point 2-4
(c) A sample of 0.037 g magnesium reacted with
hydrochloric acid to give 38.2 cm3 of hydrogen
gas measured at 25 oC and 740 mmHg. Use this
information to calculate the relative atomic mass
of magnesium. (1 atm 760 mmHg 101325 Nm-2
ideal gas constant 8.314 J K-1 mol-1)
Answer
125
2.4 Determination of molar mass (SB p.34)
Check Point 2-4
Back
126
2.5 Daltons law of partial pressures (SB p.36)
Back
Example 2-5A

• Air is composed of 80 nitrogen and 20 oxygen
  by volume. What are the partial pressures of
  nitrogen and oxygen in air at a pressure of 1 atm
  and a temperature of 25 oC?

Answer
127
2.5 Daltons law of partial pressures (SB p.36)
Example 2-5B
The valve between a 5 dm3 vessel containing gas A
at a pressure of 15 atm and a 10 dm3 vessel
containing gas B at a pressure of 12 atm is
opened. (a) Assuming that the temperature
of the system remains constant, what is
the final pressure of the system? (b) What are
the mole fractions of gas A and gas B?
Answer
128
2.5 Daltons law of partial pressures (SB p.36)
Example 2-5B
129
2.5 Daltons law of partial pressures (SB p.37)
Example 2-5B
Back
130
2.5 Daltons law of partial pressures (SB p.37)
Example 2-5C
0.25 mol of nitrogen and 0.30 mol of oxygen are
introduced into a vessel of 12 dm3 at 50 oC.
Calculate the partial pressures of nitrogen and
oxygen and hence the total pressure exerted by
the gases. (1 atm 101325 Nm-2 ideal gas
constant 8.314 J K-1 mol-1)
Answer
131
2.5 Daltons law of partial pressures (SB p.37)
Example 2-5C
Let the partial pressure of nitrogen be PA. Using
the ideal gas equation PV nRT, PA ? 12 ? 10-3
m3 0.25 mol ? 8.314 J K-1 mol-1 ? (273 50)
K PA 55946 Nm-2 (or
0.552 atm) Let the partial pressure of oxygen be
PB. Using the ideal gas equation PV nRT, PB ?
12 ? 10-3 m3 0.30 mol ? 8.314 J K-1 mol-1 ?
(273 50) K PB 67136
Nm-2 (or 0.663 atm)
132
2.5 Daltons law of partial pressures (SB p.37)
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Example 2-5C
Total pressure of gases (55946 67136) Nm-2
123082 Nm-2 Or Total pressure of gases (0.552
0.663) atm 1.215 atm Hence, the partial
pressures of nitrogen and oxygen are 0.552 atm
and 0.663 atm respectively, and the total
pressure exerted by the gases is 1.215 atm.
133
2.5 Daltons law of partial pressures (SB p.38)
Example 2-5D

• 4.0 g of oxygen and 6.0 g of nitrogen are
  introduced into a 5 dm3 vessel at 27 oC.
• What are the mole fraction of oxygen and
  nitrogen in the gas mixture?
• What is the final pressure of the system?
• (1 atm 101325 Nm-2 ideal gas constant
  8.314 J K-1 mol-1)

Answer
134
2.5 Daltons law of partial pressures (SB p.38)
Example 2-5D
135
2.5 Daltons law of partial pressures (SB p.38)
Example 2-5D

• Let P be the final pressure of the system.
• Using the ideal gas equation PV nRT,
• P ? 5 ? 10-3 m3 0.339 mol ? 8.314 J K-1 mol-1
  ? (273 27) K
• P 169 107 Nm-2
  (or 1.67 atm)

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