CHAPTER 9
GASES: THEIR PROPERTIES AND
BEHAVIOUR
The Air Around Us
Measuring Gases
Any sample of gas has mass, volume and density. Unlike solids and liquids, gases
expand to fill all of the space available. Gas mixture are always homogeneous at
equilibrium.They are also very much more compressible than solids and liquids.
Under the same conditions carbon dioxide, CO2(g) o, is heavier than air (higher
density) and hydrogen, H2 (g)mis lighter (lower density) than air. So, in pouring carbon
dioxide it must be poured down and hydrogen would have to be poured up! Most of a
sample of a gas at atmospheric pressure is empty space.
9.1
Amount of Gas
There are various properties that we can measure for a sample of gas:
m
V
P
T
Mass
Volume
Pressure
Temperature
To be useful in our calculations we need to be able to relate these measurements to
the number of moles present in the sample.
9.2
Measurement of Pressure
Torricelli (1608 - 1647) - Mercury Barometer
The atmospheric pressure varies from day to day
but can normally support a column of mercury about
760 mmHg high.
Consider a column 760 mm tall x 1 cm2
cross-section. What pressure does it exert in
Pascal the SI pressure unit?
The densityof mercury, Hg = 13.534 g cm-3..
Mass of the column = 76.0 cm x 1 cm2 x 13.534 g
cm-3
= 1028.84 g
So, Force
= 1.028584 kg
= mass x acceleration
= 1.029 kg x 2.81 m s-2 = 10.09 kg m
-2
s
The pressure is force per unit area so now we need to convert to the SI base unit Since
this is the force per cm2 we must convert to the force per m2.
1 cm2 = ? m2
1
= ( 100 m)2 =
Pressure =
1
2
1000 m
10.09 kg m s −2
10 −4 m 2
= 10-4 m2
= 1.009 x 105 kg m-1 s-2 or 1.009 x 105 N m-2
= 101 kPa (since 1 N m-2 = 1 Pa pascal)
The Standard Atmosphere
The standard atmosphere is defined as
= 760 mmHg = 101.3 kPa = 1 atm.
1 standard atm = 101.3 kPa = 760 mmHg.
Since Hg expands with temperature, correction must be applied (density changes!)
Since mercury is recognized as a poisonous substance (remember the Mad Hatter!)
mercury barometer are not often seen today. One might ask why water is not used
instead of mercury. However, because the density of water is only 1.0 g cm-3 a water
barometer would have to be about ~ 33.7 feet or ~10 metres tall not ideal for the dining
room!
9.3
Note: Standard pressure for gas measurements has been redefined to be 1 bar, or
100.000 kPa rater than 101.3 kPa (1 atm) .
Most household barometers are aneroid barometers
like the one in the picture opposite.
Measuring Gas Pressure
The Gas Laws
9.4
Early Study 1660
Robert Boyle - Boyle’s Law
How does the volume of a gas change as the pressure changes (keeping the
temperature and the number of moles constant)?
This can be studied quite easily using something a simple as a bicycle pump and give
results similar to the following graphs:
Volume is inversely proportional to pressure
1
V } P or
1
V = constant x P
so
PV = constant
Amount of substance
at constant n and T
Amount of Substance/Volume
V varies directly with n so V = k1n where
k1 is a constant.
V
P & T constant
Double moles and the volume doubles (if
P and T constant)!
n, number of moles
Charles’ Law
Charles, (1746 - 1823) and (Gay-Lussac
1778 - 1850)
V directly proportional to T absolute
V
If we use the Kelvin Scale - add 273.1 to
the Celsius temp. - get new expression
-273
o
O
V = Vo + k t
V is directly proportional to T
9.5
t C
V = kT
So we have
V = k n and V = k T
The laws can be combined
PV
=
k (amount of substance constant)
PV
=
knT
m
The new k is a constant for any sample of gas and this we call
R the gas const.
Ideal Gas Law
PV = nRT
The law fits behaviour of well-behaved gases quite well but there are significant
deviations for most real gases especially near the boiling point.
Avogadro Law (1776 - 1856)
Equal volumes of gases at the same temperature and pressure
contain equal numbers of mole
Consider the reaction
N2(g) + 3 H2(g) → 2 NH3(g)
Say we have 1 volume of N2(g) and 5 volumes of H2(g) and the reaction goes to
completion. What will the final volume be at the same temperature and pressure?
N2(g)
Initial 1 volume
Final 0 volumes
(limiting)
+
3 H2(g)
5 volumes
2 volumes
(excess)
9.6
→
2 NH3(g)
Total
volume
0 volumes 6 volumes
2 volumes 4 volumes
Standard Temperature & Pressure
STP
Standard Pressure 1 atm 760 mmHg (760 torr)
Standard Temperature
101.3 kPa
273.15 K (0 ºC)
At STP, the molar volume for a well behaved (ideal) gas is about 22.4 L mol–1
Using these values, we get R by rearranging PV = nRT
PV
R = nT
=
760 mm Hg x 22.4 L
1 mol x 273.15 K
R = 62.3 L mmHg K–1 mol–1
or if the pressure is measured in atmospheres
1 atm x 22.4 L
R = 1 mol x 273.15 K
R = 0.0820 L atm K-1 mol-1
or if the pressure is measured in kilopascals, kPa
101.3 kPa x 22.4 L
R = 1 mol x 273.15 K
R = 8.31 L kPa K-1 mol-1
but 1 L kPa = 1 N m = 1 J so
R = 8.314 J K-1 mol-1
9.7
Simple Calculations
Example
If 3.66 L of a gas is collected at 0.886 atm and 100 ºC, what would the volume be at
1.000 atm pressure? Keeping n & T constant.
PV
PV = nRT but at constant n and T we can say T
P1V1
=
constant
V2
=
P1V 1
P2
=
0.886 atm x 3.66 L
1.000 atm
Ans:
=
P2V2
= constant
P1V 1
T1
or
=
P2V 2
T2
will expect the volume to decrease!
3.24 L
Example
A rigid car tyre is pumped to 28.0 lbs in-2 at –10.0 ºC. What is the pressure in the tyre
at 20.0 ºC?
Here we have constant V and n
Again PV = nRT and
P1V 1
T1
=
P2V 2
T2
but V is constant
P1
T1
=
P2
T2
This time the pressure should increase....
We must convert the temperatures to Kelvin
T1 = –10.0 + 273.15 K = 263.15 K and T2 = 20.0 + 273.15 K = 293.15 K
P2 =
P1T 2
293.15 K
-2
-2
T 1 = 28.0 lbs in x 263.15 K = 31.2 lbs in .
This is enough to be a problem, so be careful if you plan to drive fast!
Example:
Making O2 in the Laboratory
What volume of “dry” O2 collected at 25.0ºC and 1.0 atm when 50.0 mL of 5.0% H2O2
is decomposed (density 1.0 g mL-1)
9.8
t
2 H2O2
2 H2O
1
H2O2
available moles of
H2O2
mol O2
nRT
P
=
O2(g)
1
Mass
V=
+
=
50.0 mL x 1.0 g mL
=
2.50 g
2.50 g
34.01 g. mol −1
=
=
1/2
-1
½ x 0.07350 mol
5.0%
100%
x
=
0.07350 mol
=
0.0368 mol
0.0368 mol x 0.0821 L atm K −1 mol−1 x 298.1K
= 0.900 L
1.0 atm
Ans: The volume of dry O2 = 0.90 L
Stoichiometry in Gas Reactions
Once we can calculate moles of gases, we can use gas volumes in stoichiometry
calculations. Consider the reaction:
N2(g) + 3 H2(g) → 2 NH3(g)
What volume of hydrogen, H2, measured at 3.00 atm and 28.0 ºC is required to make
1.00 kg of ammonia, NH3?
1.00 kg x 1000 g kg−1
Moles NH3 required =
= 58.719 mol. Divide by 2 to focus on the
17.03 g mol−1
ammonia:
1
2
N2(g) +
3
2
H2(g) → 1 NH3(g)
3
Moles of H2 required = 2 x 58.719 mol = 88.079 mol. Using the ideal gas law PV = nRT
nRT
and rearranging gives V = P and substituting
V=
88.079 mol x 0.0821 L atm K −1 mol−1 x 301.15 K
= 726 L
3.00 atm
Ans: 726 L
9.9
250 mL of dry HCl(g) at 25.0ºC and 790 mmHg is bubbled into 50.0 mL of ice cold
water so that it all dissolves. The solution is then made up to 100.0 mL. What is the
concentration of HCl (molarity) in the final solution?
CHCl
mol
V b
=
PV
= nRT
790 mmHg x 0.250 L
760 mmHg atm −1
n = 0.0821 L atm −1 K −1 mol −1 x 298.15 K
cHCl =
n
PV
RT
=
=0.0106
145
mol, so
= 0.106 mol L-1
0.0106145 mol
100.0 x 10 −3 L
Ans: 0.106 mol L-1
Notice the concentration of the HCl could also be found by titration
What volume of the CO2(g) + H2O(g) gases produced when 1.00 g of octane, C8H18 ,
burns - measured at, say, 130 ºC at a car exhaust at 100.0 kPa?
So first write the balanced equation (complete combustion of octane gives only carbon
dioxide and water)
C8H18(g) + 12½ O2(g) → 8 CO2(g) + 9 H2O(g)
Moles of octane burned =
1.00 g
114.22 g mol −1=
8.755 x 10-3 mol
so total moles of gas formed = 17 x 8.755 x 10-3 mol = 0.1488 mol
now, V =
nRT
P
=
0.1488 mol x 8.314 L kPa K −1 mol−1 x 403.15 K
100.0 kPa
= 4.99 L
Ans: 4.99 L
Mixtures of Gases - Dalton’s Law of Partial Pressures
Recall Equal volumes of gases at the same T & P contain equal moles. This means
that it is the number of moles that matter - not the mass or what the gas is - as
long as they don’t react. Dalton extended this to deal with mixtures of
non-reacting gases.
The total pressure of a mixture of gases is equal to the sum of the partial
pressures of the individual gas.
P total = p A + p B + p C .............
P total =
ntotal RT
V
=
nART
V
+
nB RT
V
+
nB RT
V
.............
9.10
So, if we have a mixture of gases, we can treat it as if it was one pure gas - as long as
they don’t react!
Example
What is the total pressure inside a 2.00 L container that contains 1.00 g of Ne, 1.00 g
of N2 and 1.00 g of CO2? What is the partial pressures of neon in the mixture. All
pressure measured at 25.0º C.
Calculate the number of moles of each gas:
mol
Ne
1.00 g
20.18 g mol −1
=
0.04955 mol
mol
N2
1.00 g
28.01 g mol −1
=
0.03570 mol
mol
CO2
1.00 g
44.01 g mol −1
=
0.02272 mol
Total moles
PV
=
nRT
P
=
nRT
V
Answer: total pressure
0.1080 mol
=
=
0.1080 mol x 0.0821 L atm K −1 mol −1 x 298.15 K
2.00 L
1.32 atm
The partial pressure is the pressure that the neon gas would exert on
=
0.04955 mol x 0.0821 L atm K −1 mol−1 x 298.15 K
= 0.606 atm.
2.00 L
or in proportion to the FRACTION of the moles that are Ne
0.04955 mol
= 0.1080 mol
xNe = 0.459
pNe = 0.459 x 1.322 atm = 0.606 atm
This number is called the MOLE FRACTION and usually given the symbol
that it has no units.
9.11
x.
Notice
To review:
Mole fraction
We have seen that the mole fraction is simply the fraction of the total moles that is the
mol Ne
substance of interest. Thus xNe = total mol and it is not difficult to see that the partial
pressure, pNe = XNe x Ptotal
Since the above is true
XNe
=
x Ne % P total
P total
p Ne
P total
=
x CO 2
=
P CO 2
= 1.32 atm x 0.2103
In the above question
so
.02272 mol
0.1080 mol =
0.2103
= 0.278 atm.
A common circumstance where a gas mixture is obtained - though it might not be
immediately obvious - is the collection of a gas over water.
Zinc is an active metal that reacts with dilute acid to give hydrogen gas:
Zn(s) + 2 HCl(aq) → Zn2+(aq) + 2 Cl−(aq) + H2(g)
Zn(s) + 2 H+(aq) + Cl−(aq) → Zn2+(aq) + 2 Cl−(aq) + H2(g)
Zn(s) + 2 H+(aq) → Zn2+(aq) + H2(g)
The zinc is oxidized and the hydrogen ions are reduced.
The H2(g) obtained is a mixture of H2O(g) and H2(g) so the total pressure is
Ptotal = p H 2 +
p H 2O
p H 2O , the vapour pressure of water depends on temperature:
9.12
The table shows how the vapour pressure (mmHg) changes with temperature.
Something to think about: Why/when does a liquid boil?
If we collect hydrogen over water at 25.0ºC and 770.0 mmHg, the gas collected is
saturated with water vapour (to the saturated vapour pressure).
Ptotal = 770.0 mmHg
= Ptotal = p H 2 +
p H 2O
= p H 2 + 23.76 mmHg
p H 2 = 770.0 mmHg - 23.76 mmHg
so,
If we were to dry the gas, keeping other things constant, the pressure should drop to
this value.
So, if we collect 600 mL of H2 over water, how many moles of H2 do we really have?
PV
=
nRT
Moles of
hydrogen, n
=
PV
RT
=
n
=
0.0821
770.0 mmHg − 23.76 mmHg
x 0.600 L
760.0 mmHg
L atm −1 K −1 mol −1 x (25.0 + 273.15
K)
0.02407 mol
What volume would the gas occupy if collected under the same conditions DRY?
V=
nRT
P
=
0.02407 mol x 0.0821 L atm K −1 mol −1 x 298.15K
770.0 mmHg
760 mmHg atm −1
=
0.5815 L or 582 mL
9.13
So, as expected, the volume collected is a little greater than it should have been
because the gas was wet.
pH
2
Using the mole fraction of H2 = P total =
746.24 mmHg
770.0 mmHg = 0.9691
so fraction of volume = 600 mL x 0.9691
= 581 mL, just as it should be J (notice the effect of the
borderline significant digit)
A 0.4480 g sample of an alkali metal was allowed to react with water and the H2(g)
produced was collected over water at 1.00 atm at 25.0ºC. the volume collected was
246 mL. What was the molar mass of the alkali metal. The vapour pressure of water,
p H2 at 25.0ºC = 0.0313 atm.
First we must write the equation ....
2 M + 2 H2O → 2 MOH + H2(g)
2 M + 2 H2O
p H2
→ 2 M+(aq) + 2 OH−(aq) + H2(g)
= 1.00 atm – 0.0313 atm = 0.9687 atm
PV=nRT
n
Mole of H2 produced =
=
PV
RT
0.987 atm x 0.246 L
0.0821 L atm K −1 mol −1 x 298.15 K
= 9.735 x 10−3 mol
Moles of the alkali metal, M = 2 x 9.735 x 10-3 mol = 0.01947 mol
‹ molar mass =
0.4480g
0.01947mol =
23.0 g mol−1
So what was the alkali metal?
Most gases don’t behave quite as predicted by the gas laws.
However, we can define exactly what a perfect gas should be by considering what we
call Kinetic Theory
Kinetic Theory
The basic postulates of the Kinetic Theory are
9.14
1. Gases consist of particles in continuous random motion. They collide with one another and
with the walls of the container. Gas Pressure is the result of the collisions. (for those doing
physics: it is the change in momentum of the gas particles per unit surface area per second)
2. Collisions must be perfectly elastic. All Kinetic energy is retained (not converted to heat)
3. The particles must be so small that the volume they occupy can be ignored (To be ideal they
should occupy no volume!). A gas is mostly empty space.
4. The gas particles do not react or interact. (attractive forces)
5. Average translational (motion) kinetic energy, Et, of gas particles is directly proportional to the
absolute temperature (ie. T ∝ K) or the
Et = constant x T where T is the absolute temperature.
6. At a given temperature, all gases have the same average kinetic energy , Et and the constant
is the same (universal) for any gas. Since, the kinetic energy of the particles is given by E = ½
mv2, heavy particles move more slowly on average.
Graham’s Law: Diffusion and Effusion of Gases
Since the Kinetic Energy, K.E., of the gas particles depends only on the absolute
temperature (Kelvin) of the gas we can easily guess that heavy particles will travel
more slowly than light particles. The higher the temperature the higher the average
speeds. Graham observed that the rate of effusion of a gas is inversely proportionally
to the square root of its molar mass, M.
Rateº
1
Molar Mass
So that for two gases 1 and 2
Rate 1
Rate 2
º
M2
M1
We can see that this comes from the the observation that the K.E. depends on
temperature. At a constant temperature
9.15
1
2
2 m 1 u 1 gas 1
u1
u2
=
=
1
2
2 m 2 u 2 gas 2
m2
m1
which is similar to Graham’s Law. The faster the gas particles move (on average) the
faster they would be expected to effuse or diffuse.
This property has been use separate mixtures of gases. It can also be used to
determine the molar mass of a gas.
The Behaviour of Real Gases
Real gases have molecules that size and occupy part of the volume of a the gas. At
low pressure the percentage of the volume used in this way would be very small (at
STP about 0.05%) but a high pressure it can become significant (500 atm about 20%).
This leads to deviations from the gas laws at high pressures. Similarly, if there are
attractive forces between the particle the molecular collisions are not quite perfectly
elastic. To get around these problems attempts have been made to correct for these
deviations.
The van der Waals equation.
P +
an 2
V2
(V − nb ) = nRT
The van der Walls equation attempts to do this with corrections for the pressure and
volume.
Gas Density
Since a gas is mostly empty space, gas densities are much lower than those of solids
and liquids.
The density of water at room temperature is 1.0 g mL-1 or 1000 g L-1 so the volume of a
mole is
18.0 g mol −1
1.0 g mL −1 =18.0
mL mol-1 or 0.0180 L mol-1.
The volume of a mole of carbon dioxide, CO2(g), at STP = 22.4 L mol-1 (about) so the
44.0 g mol −1
mass
density is = volume = 22.4 L mol −1 =1.964 g L-1.
Because gas density is so small we normally quote it in g L-1 rather than g mL-1.
9.16
We normally quote gas density as g L-1 rather than g mL-1
The gas law given by PV = nRT can be expressed in terms of density if we recall that
mass
density = volume and that mass = molar mass (M) x number of moles (n) or mass =
MP
nM
n x M, so density, d = nRT = RT
P
Gas density measurements were once used to determine properties of gases. Eg.
What is the molar mass of the gas particles, are the monatomic or diatomic etc.
dRT
molar mass of gas, M = P
Example
The density of sulfur vapour at 445 ºC and 755 mmHg is found to be 4.33 g L-1. Use
this information to find the molecular formula of the sulfur in the vapour.
This is not difficult if we think the problem through carefully.
First we will assume that the vapour is molecules of sulfur with formula SY where Y is a
whole number.
The density tells us that the mass of 1 litre = 4.33 g
so the number of moles of sulfur atoms, n, in 1 litre of the gas is
=
4.33 g
Y x 32.066 g mol −1
=
0.1350 mol
.
Y
Combining the relationships,
= 0.1350 mol %
So, because PV = nRT, and n =
PV
RT
=
0.1350 mol
so
Y
PV
RT
Y = 0.1350 mol %
RT
PV
0.0821 L atm K −1 mol −1 x 718.15 K
755 mm Hg
760 mm Hg atm −1
%1 L
= 8.006(no units, just a number) So the
formula is S8 and we know from other studies the molecule looks like
S
S
S
S
S
S
S
S
9.17