Gases: Chapter 6
• have large spaces between the gas molecules
→ not tightly packed
• expand to fill the container → take the
shape of the container
• diffuse into one another and mix
• 4 properties that affect its physical
behaviour:
amount of gas (in moles)
volume
temperature
pressure
1
2
Pressure
What is pressure?
• it is the force per unit area – (force divided
by the area over which the force is applied)
• The unit is N.m-2 which is one pascal (Pa)
F (N )
P( Pa) =
2
A(m )
3
Barometric Pressure
• The simplest device
used to measure
pressure is the
Toricelli barometer
• Standard atmosphere
(atm) – is the
pressure exerted by a
mercury column of
760 mm (density of
Hg = 13.5951 g.cm-3 (0
0C and gravity =
9.80665 m.s-2)
4
1 atm = 760 mmHg
1 atm = 760 Torr
Example: What is the height of a column of water
that exerts the same pressure as a column of
mercury 76.0 cm (760 mm) high?
PH O = PHg
2
Using equation (g x h x d)
g x hH
2O
x 1.00 g/cm3 = g x 76.0 cm x 13.6 g/cm3
76.0 cm × 13.6 g / cm 3
3
hH 2O =
=
1
.
03
×
10
cm
3
1.00 g / cm
= 10.3 m
5
Units of Pressure
• Pascals
1 Pa = 1 N/m2
• Bar
1 bar = 105 Pa = 100 kPa
• mm Hg or torr
The difference in the heights in mm (h) of two
connected columns of mercury
• Atmosphere
1.00 atm = 760 torr
6
Common Units of Pressure
Atmosphere
atm
Millimeter of mercury mmHg
Torr
Torr
1 atm = 760 mmHg
= 760 Torr
Pascal
Kilopascal
Pa
kPa
= 101 325 Pa
= 101.325 kPa
Bar
Millibar
bar
mb
= 1.01325 bar
= 1013.25 mb
7
Manometers
8
Example: When a manometer is filled with
liquid mercury (d = 13.6 g/cm3), the
barometric pressure is 748.2 mmHg,
and the difference in mercury levels is
8.6 mmHg. What is the gas pressure
Pgas?
Pgas = Pbar + ∆P
= 748.2 mmHg - 8.6 mmHg
= 739.6 mmHg
9
Gas Laws
Boyle’s Law: - for a
fixed amount of gas
at constant
temperature, the
gas volume is
inversely
proportional to the
gas pressure
1
Pα
V
10
Charles’s Law: - The volume of a fixed amount
of gas at constant pressure is directly
proportional to its temperature.
VαT
A plot of V versus T will be a straight line.
11
Standard Temperature and
Pressure
STP conditions:
• standard temperature for gases = 0 0C =
273.15 K
• standard pressure, 1 atm = 760 mmHg
Avogadro’s Law: - at a fixed temperature
and pressure, the volume of a gas is
directly proportional to the amount of gas
Vαn
1 mol gas = 22.4 L gas (at STP
12 )
Example: What is the mass of 1.00 L of
cyclopropane gas, C3H6, when measured at
STP?
1 mol → 22.4 L gas (at STP )
x → 1.00 L
1.00 L × 1 mol
x=
= 0.0446 mol
22.4 L
mass of C 3 H 6 = n × m.m
= 0.0446 mol × 42.08 g .mol −1
= 1.88 g
13
Ideal-Gas Equation
• So far we’ve seen that:
V ∝ 1/P (Boyle’s law)
V ∝ T (Charles’s law)
V ∝ n (Avogadro’s law)
• Combining these, we get, if we call
the proportionality constant, R:
nT
V=R
P
i.e. PV = nRT
14
The Ideal Gas Equation
PV = nRT
Where R = gas constant
R = 8.3145 m3.Pa.mol-1.K-1 (SI unit-use this)
= 8.3145 J.mol-1.K-1(SI unit)(1m3 Pa = 1J)
= 0.082057 L.atm.mol-1.K-1
= 8.314 kPa. dm3 mol-1.K-1
15
Units (1m3 Pa =1J)
R = 8.3145 m3 Pa mol-1 K-1
F (N )
A (m 2 )
P =
m
s2
kg m
2
s
=
m 2
1 N =
Pa
R= 8.3145 J mol-1 K-1
kg
kgm
s2
Pa
=
Pa
= kg
m 3 Pa
×
1
m 2
m − 1s − 2
= kg
m 2 s− 2 = 1J
16
17
Volume conversion
(from milliliters to litres and then
to cubic meters)
3
1L
1m
V = 305 mL ×
×
1000 mL 1000 L
−4
= 3.05 × 10 m
3
18
Pressure conversion
• 1 atm → 101325 Pa
• 1.37 atm → x Pa
• X = (1.37 atm x 101325 Pa) / 1 atm
= 138815 Pa
= 1.38 x 105 Pa
19
Example: What is the pressure, in kPa, exerted by
1.00 x 1020 molecules of N2 in a 305 mL flask at
175 0C?
1 mol → 6.02 × 10 23 molecules
x → 1.00 × 10 20 molecules
1.00 × 10 20 molecules × 1 mol
x=
6.022 × 10 23 molecules
= 0.000166 mol N 2
3
−1
nRT 0.000166 mol × 8.3145 m .Pa.mol .K
P=
=
V
3.05 × 10 − 4 m 3
= 2.03 × 10 3 Pa
= 2.03 kPa
−1
× 448 K
20
General Gas Equation
PiVi Pf V f
=
ni Ti n f T f
For a fixed mass of gas,
PiVi Pf V f
=
Ti
Tf
21
Example: If a fixed amount of gas held at
constant volume goes from STP conditions
to boiling at 100 0C, what is the new
pressure?
PiVi Pf V f
=
ni Ti n f T f
Pi Pf
=
Ti T f
Pf =
Pi × T f
Ti
1.00 atm × 373 K
=
273 K
= 1.37 atm
22
Applications of the Ideal Gas
Equation – Molar mass
Molar mass determination
PV = nRT
m
n=
M
mRT
PV =
M
23
Example: Propylene is used in the production of
plastics. A glass vessel weighs 40.1305 g when
clean, dry and evacuated; it weighs 138.2410 g when
filled with water at 25 0C(density of water =
0.9970 g/ml) and 40.2959 g when filled with
propylene gas at 740.3 mmHg and 24.0 0C. What is
the molar mass of propylene?
Mass of water to fill vessel = 138.2410 g – 40.1305 g
= 98.1105 g
Density = mass/volume
Volume = mass/density
Volume of water (vol. of vessel) = 98.1105 g/0.9970
g/ml
= 98.41 mL
= 0.09841 L
24
Mass of gas = 40.2959 g – 40.1305 g
= 0.1654 g
Temperature (K) = (24 0C + 273.15 0C) x 1K/10C
= 297.15 K
R = 8.3145 J mol-1 K-1 ( 1J= 1 m3 Pa)
Pressure = 740.3 mmHg/760 mmHg
= 0.9741 atm (convert to Pa)
= 98700 Pa
mRT
M =
PV
0.1654 g × 8.3145 m 3 Pa mol −1 K −1 × 297 .15 K
=
98700 Pa × 9.841 × 10 −5 m 3
= 42 .08 g / mol
25
Gas Densities and Molar Mass
If we divide both sides of the ideal-gas
equation by V and by RT, we get:
PV = nRT
n
P
=
V
RT
26
Applications of the Ideal Gas
Equation - Gas Densities
m n× M n
d= =
= ×M
V
V
V
We can then replace n/V with P/RT
m MP
d= =
V
RT
The density of a gas at STP is calculated by
dividing its molar mass by the molar volume
27
(22.4 L/mol)
Example: What is the density of O2 at 298 K
and 0.987 atm? (Convert 0.987 atm to Pa)
R = 8.3145 J mol-1 K-1 ( 1J= 1 m3 Pa)
m MP
d= =
V RT
32.00 g mol −1 × 95954 Pa
=
3
−1
−1
8.3145 m Pa mol K × 298 K
3
12961. 6 g 1 m
=
×
3
1000 L
m
= 1.29 g / L
28
Gases in Chemical Reactions
Air bags use the decomposition of sodium
azide, NaN3 to produce N2 (g) that inflates
the air bag
Example: From the following equation determine
what volume of N2 is produced when 70.0 g of NaN2
is decomposed at 735 mmHg and 26 0C?
2NaN3 (s) →2Na(l) + 3N2 (g)
70.0 g
n( NaN 3 ) =
= 1.08 mol
65.01 g / mol
1.08 mol × 3
n( N 2 ) =
= 1.62 mol
2
29
P = 735 mmHg/760 mmHg = 0.967 atm (convert to Pa)
n = 1.62 mol
R = 8.3145 J mol-1 K-1 ( 1J= 1 m3 Pa)
T = (26 0C + 273 0C) x 1K/10C = 299 K
nRT
V =
P
1.62 mol × 8.3145 m 3 Pa mol −1 K −1 × 299 K
=
97981 Pa
1000 L
= 0.0411 m ×
1 m3
3
= 41.1 L
30
Law of Combining Volumes
Example: For the following reaction,
determine what volume of SO2 is produced
per liter of O2 consumed? Both gases are
measured at 25 0C and 745 mmHg.
2ZnS ( s ) + 3O2
∆
→
2ZnO ( s ) + 2 SO2 ( g )
1 L → 3 mol O2 ( g )
x → 2 mol SO2 ( g )
2 mol × 1 L
x=
3 mol
= 0.667 L SO2 ( g )
31
Mixtures of Gases
32
Dalton’s law of partial pressures:
• The total pressure of a mixture of gases is
the sum of the partial pressures of the
components of the mixture
Ptot = PA + PB + ……
n A RT
VA =
Ptot
n B RT
VB =
Ptot
Vtot = V A + VB
33
n A ( RT / Vtot ) n A
n A ( RT / Ptot ) n A
PA
VA
=
=
and
=
=
Ptot ntot ( RT / Vtot ) ntot
Vtot ntot ( RT / Ptot ) ntot
nA
PA
VA
=
=
= xA
ntot Ptot Vtot
Where XA is the mole fraction (fraction
of all the molecules in a mixture
contributed by that component)
34
Example: What are the partial pressures of
H2 and He in a gaseous mixture containing
1.0 g of H2 and 5.00 g He that are in a
confined volume of 5.0 L at 20 0C? (convert volume)

 

1.00 g
5.00 g
 + 

ntot = 
 2.02 g / mol H 2   4.003 g / mol He 
= 0.50 mol + 1.25 mol = 1.75 mol
ntot RT
P=
V
1.75 mol × 8.3145 m3 Pa mol −1 K −1 × 293 K
=
0.0050 m3
= 8.5 ×105 Pa
35
Partial pressure of H2 and He
nH 2
ntot
=
PH 2 =
PH 2
Ptot
Ptot × n H 2
ntot
0.50 × 8.4 atm
=
= 2.4 atm
1.75
Ptot × n He 1.25 × 8.4 atm
PHe =
=
= 6.0 atm
ntot
1.75
36
Collecting a gas over water
Gas collected over water is “wet” – made up of
the desired gas and water vapour
Ptot = Patm = Pgas + PH2O
or
Pgas = Patm – PH2O
37
Kinetic Molecular Theory of Gases
• A gas is composed of a
large no. of small particles
in constant, random motion
• Gas molecules have large
spaces between them
• Molecules collide with each
other and with the sides of
the container for short while
• No forces between molecules,
except when they collide
• Molecules may gain or loose
energy when they collide but
the total energy remains constant
38
1.
Gas molecules contain translational kinetic energy
(energy of objects moving through space). The
faster the molecules move, the greater is their
translational energies →greater forces exerted
when they collide.
2. Frequency of molecular collisions – no. of collisions
per second. The higher the frequency, the
greater the total force of the collisions
(increases with no. of molecules per unit volume
and with molecular speed).
3. When a molecule hits the walls of a container,
momentum is transferred as the molecule
reverses direction→impulse. The magnitude of
the impulse is directly proportional to the mass
and velocity of a molecule.
39
The Kelvin temperature (T) of a gas is
directly proportional to the average
translational kinetic energy of its
molecules.
Absolute zero of temperature – the
temperature at which the translational
molecular motion should cease.
40
Nonideal (Real) Gases
Compressibility factor:
PV
nRT
For an ideal gas, PV/nRT = 1
For nonideal gas behaviour, at
high pressures the volume
of a gas becomes very small
→ 0 but this is not so as gas
molecules occupy a volume
themselves.
41
• Since molecules are
attracted to each other, the
force of collisions of gas
molecules with the walls of
the container are less than
expected for an ideal gas

n a
 P + 2  (V − nb ) = nRT
V 

2
• Gases behave ideally at high temperatures and low
pressures.
• Gases tend to behave nonideally at low
42
temperatures and high pressures.
Example: Use the van der Waals equation to
calculate the pressure exerted by 1.00 mol Cl2(g)
confined to a volume of 2.00 L at 273 K. The
volume of a = 0.657 (m3)2Pa .mol-2, and that of b =
0.0000562 m3 mol-1.
R = 8.3145 J mol-1 K-1 ( 1J= 1 m3 Pa)
nRT n2a
P=
− 2
V − nb V
1.00 mol × 8.3145 m3Pa mol−1.K −1 × 273K 1.00 mol2 × 0.657(m3 )2 Pa mol−2
=
−
3
3
−1
(0.002 m −1 mol x 0.0000562 m mol )
(0.002)2 (m3 )2
= 1003343Pa
= 1.00 × 106 Pa
43
To simplify calculation :
• R = 8.3145 J mol-1 K-1 ( 1J= 1 m3 Pa)
3
−1
−1
1
.
00
mol
×
8
.
3145
m
Pa
mol
.
K
× 273 K
• nRT =
• n2a = 1 . 00 mol 2 × 0 . 657 ( m 3 ) 2 Pa mol
−2
(
• V-nb = 0.002 m3 − 1 mol x 0.0000562m3mol−1
44
)
• EX. A 1.65 g sample of an unknown metal M reacts with excess
HCl and the liberated hydrogen is collected over water at 25 oC at a
barometric pressure of 744 mmHg. The gas collected has a volume
of 2367 mL. The equation for the reaction is:
• 2M(s) + 6HCl (aq)→ 2MCl3(aq) + 3H2(g)
•
•
•
•
The gas constant: R = 8.3145 m3 Pa mol-1K-1
Vapour pressure of H2O at 25 oC is 23.76 mmHg
1 atm = 760 mmHg
1 atm = 101325 Pa
•
•
•
•
•
•
•
Calculate the following:
(i) partial pressure of dry hydrogen gas in SI units
(ii) moles of dry hydrogen gas in SI units
(iii) moles of M
(iv) molar mass of M
(v) density of hydrogen gas, in g L-1 at STP.
45
(i) partial pressure of dry hydrogen gas using SI units
PH2 = (744 – 23.76) mmHg
= 720 mmHg/760 mmHg atm-1
= 0.947 atm (convert to Pa)
PH2 = 0.947 atm x 101325 Pa atm-1
= 9.60 x 104 Pa
(ii) moles of dry hydrogen gas using SI units
n = PV/RT
= (9.60 x 104 Pa x 2.367 x 10-3 m3 )/8.3145 m3 Pa mol-1K-1 x 298K )
= 0.0917 mol
46
(iii)
moles of M
From the balanced equation:
moles of M = 2/3 mol of H2
= 2/3 x 0.0917
= 0.0611 mol of M
(iv)
molar mass of M
molar mass of M = mass/mol = 1.65 g/0.0611 mol = 27.0 g mol-1
(M = Al)
(v)
density of hydrogen gas, in g L-1 at STP.
Density = PM/RT
= (101325 Pa x 2.02 g mol-1)/8.3145 m3 Pa mol-1 K-1 x 273 K.
= 90.2 g m-3 ( convert volume from cubic metres to litres )
= 0.0902 g L-1
47