Chapter 12:
Avogardro’s Law
Equal volumes of a gas at the same temperature and pressure contain the same number of
molecules.
n1/ V1 = n2/V2
Example: 1: 5.00 L of a gas is known to contain 0.965 mol. If the amount of gas is increased to
1.80 mol, what new volume will result (constant T and P)?
n1/ V1 = n2/V2
0.965 mol/5.00 L = 1.80/V
V=0.326 L
Molar Volume of Gases
One mole of any ideal gas at STP (OoC;
101.3kPa), has a volume of 22.4L
The volume of a real gas is slightly different
than 22.4L at STP, but so close that 22.4L is
an acceptable approximation
SATP: standard ambient temperature and
pressure (25oC;100 kPa) V = 24.8 L/mol
Real gas deviations occur at high pressures
or low temperatures (Van der Waals equation)
12.1-12.2 The Ideal Gas Law
The idea gas law: The relationship that links moles, pressure, volume, and temperature for a gas.
PV = nRT
P = pressure (Units may vary, kPa, bar, atm)
V = Volume (in litres, L)
n = number of moles (in moles)
R = ideal gas constant (values and units may vary based on units of pressure):
-When kPa is used, R = 8.31 kPaLMol-1K-1
-When atm is used, R = 0.0821 atmLmol-1K-1
T = Temperature in Kelvin
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The ideal gas law states that PV=nRT, where P is the pressure of a gas, V is the volume of the gas, n is the
number of moles of gas present, R is the ideal gas constant, and T is the temperature of the gas in Kelvin.
Or kPa
Or (if kPa is used): 8.31 kPaLmol-1K-1
Common mistakes:
 Students express T in degrees Celsius, rather than Kelvin. This can cause huge problems,
especially when the temperature is below freezing.
 Students use the wrong value of R. You need to make sure that you have the right value of R
for the units you’re using. In this worksheet, R = 0.08206 L.atm/mol.K – some teachers prefer
using units of kPa rather than atmospheres, resulting in huge errors if the wrong R is used.
Example 1: A 2.5 L container is filled with sulphur dioxide gas at a pressure of 120 kPa and a
temperature of 26 C. Calculate the mass of sulphur dioxide in the container. The ideal gas constant
is 8.31 kPaLmol-1K-1.
Plan your strategy
Act on your strategy
-Mass is what needs to be calculated
PV = nRT
-Which law: Ideal gas law
-Isolate the variable you are looking for (in this case N = PV/RT
it is n)
-There is no mass variable, but we do know how
number of moles (n) and mass (g) are related!
-Plug the numbers (and units) of the known variables N = (120 kPa)(2.5 L)/( 8.31 kPaLmol-1K-1)(299
into the equation.
K) = 0.121 moles
-Do not forget to convert Temperature in Kelvin
MSO2 = 64.07 g/mol
-Convert Moles to Mass (g) using the formula m =
m = 64.07 g/mol x 0.121 mol = 7.7 g
nM
Conclusion: The mass of sulphur dioxide gas at a pressure of 120 kPa, a volume of 2.5 L, and a
temperature of 26 C is 7.7 g.
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Example 2:Find the volume of 1 g of water in the gas phase at its boiling point (100 C) at 101 kPa.
The ideal gas constant is 8.31 kPaLmol-1K-1.
Plan your strategy
Act on your strategy
-Volume is what needs to be calculated
PV = nRT
-You are given mass
-Which law: Ideal gas law
-Isolate the variable you are looking for (in this case it is V) V= nRT/P
-There is no mass variable, but we do know how number of
moles (n) and mass (g) are related!
-Plug the numbers (and units) of the known variables into
MH2O = 18.02 g/mol
the equation.
n = 1. 00 g/18.02 g/mol = 0.055 mol
-Do not forget to convert Temperature in Kelvin
V = (0.055 mol)(8.31 kPaLmol-1K1
-Convert Mass (g) to moles using the formula n = m/M
)(373 K)/101 kPa = 1.7 L
Conclusion: The volume of 1.00 g of water in the gas phase at its boiling point (100 C) at 101 kPa is
1.7 L.
Example 3:At 100 C, a glass bulb with a volume of 248 ml contains 1.24 g of a compound, X, in
the gas phase. If the pressure of the gas is 101 kPa, what is the molar mass of compound X?
Plan your strategy
Act on your strategy
-Molar Mass is what needs to be calculated
PV = nRT
-Which law: Ideal gas law
-Isolate the variable you are looking for (in this
n = PV/RT
case it is n)
-There is no molar mass variable, but we do know
how number of moles (n) and mass (g) are related!
-Plug the numbers (and units) of the known
n = (101 kPa)(0.248L)/(8.31 kPaLmol-1K1
variables into the equation.
)(373 K) = 0.00808 mol
-Do not forget to convert Temperature in Kelvin
and ml into L.
MX = 1.24 g/0.000808 mol = 153 g/mol
-Convert moles into molar mass using the formula
M = m/n
Conclusion: The molar mass of compound X is 153 g/mol
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Applications of Ideal Gas Law
Example 4: A liquid can be decomposed by electricity into two gases. In one experiment, one of
the gases was collected. The sample had a mass of 1.090 g, a volume of 850 mL, a pressure of 746
torr, and a temperature of 25C. Calculate its molecular mass and identify the gas.
V = 850 mL = 0.850 L
P = 746 torr/760 torr = 0.982 atm
T = 25.0oC + 273.15 = 298.15 K
R= 0.08206 L atm/mol K
PV = nRT
n= PV/RT
n = (0.982 atm)(0.850 L) /(0.0821 L atm mol-1 K-1)(298.15 K)
n = 0.0341 mol
molecular mass = g/mol = 1.090 g/ 0.0341 M= 31.96 g/mol. The gas is oxygen.
12.3 Gas Stoichiometry
We have looked at stoichiometry: 1) using masses & molar masses, & 2) concentrations.
We can use stoichiometry for gas reactions.
As before, we need to consider mole ratios when examining reactions quantitatively.
molar mass of x
mole ratio from
molar mass of y
balanced equation
grams (x)  moles (x)  moles (y)  grams (y)
 PV = nRT 
P, V, T (x)
P, V, T (y)
At times you will be able to use 22.4 L/mol at STP and 24.8 L/mol at SATP as shortcuts.
Steps to solving gas stoichiometry problems
1. Write a balanced equation for the reaction.
2. Convert all amounts to moles.
3. Compare molar amounts using stoichiometry ratios from the balanced equation. Solve for the
unknown molar amount.
4. Convert the new molar amount into the units required, using a set of conditions with the ideal
gas law, PV = nRT.
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Example 1: CH4 burns in O2, producing CO2 and H2O(g). A 1.22 L CH4 cylinder, at 15°C,
registers a pressure of 328 kPa. What volume of O2 at SATP will be required to react completely
with all of the CH4?
Plan your strategy
1) Write a balanced equation for the chemical
reaction.
i) Calculate the amount of methane present using
n=PV/RT
where P = 328 kPa, V = 1.22 L, T = 288 K
Act on your strategy
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
ii) Using this mole value, determine how many
moles of oxygen gas react with methane.
ii) Hydrogen and iron are in a 1:2 ratio.
-The questions asks you to find the volume of
O2(g) produced. Convert moles of O2(g) into V
using PV = nRT.
-V = nRT/P
Use T=298 K
PV = nRT P= 100 kPa, n= 0.334 mol, T=
298 K
i) n=(328 kPa)(1.22 L)
= 0.167 mol
(8.31 kPa•L/K•mol)(288 K)
# mol O2=0.167 mol CH4
x 2 mol O2 = 0.334 mol
1 mol CH4
V=(0.334 mol)(8.31 kPa•L/K•mol)(298 K) =
8.28 L
(100 kPa)
or # L = 0.334 mol x 24.8 L/mol = 8.28 L
Example 2:What volume of hydrogen gas is produced when excess sulfuric acid reacts with 40 g
of iron at 18 °C and 100.3 kPa?
Plan your strategy
1) Write a balanced equation for the chemical
reaction.
i) Calculate the amount of moles of iron present
using n = m/M.
ii) Using this mole value, determine how many
moles of Hydrogen gas are produced by doing a
molar ratio analysis (as seen in the previous
units).
Act on your strategy
H2SO4(aq) + Fe(s)  H2(g) + FeSO4(aq)
-The questions asks you to find the volume of
H2(g) produced. Convert moles of H2(g) into V
using PV = nRT.
-V = nRT/P
-Do not forget to convert temperatures into
Kelvin (18 + 273.15 K = 291.15 K)
VH2(g) =
(0.716 mol)( 8.31 kPaLmol-1K-1)(291.15 K)
100.3 kPa
= 17.27 L
i) nFe = 40g/55.85g/mol = 0.716 mol
ii) Hydrogen and iron are in a 1:1 ratio.
Therefore 0.716 moles of H2 are produced.
Conclusion: Therefore 17.27 L of H2(g) are produced during this reaction.
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Vapour Pressure
- Air breathed contains H2O (ex. Wet surfaces dry up, glass of H2O evaporates)
Vapour pressure is the pressure exerted by a vapour. E.g. the H2O(g) in a sealed container.
- To measure vapour pressure we can heat a sample of liquid on top of a column of Hg and see the
pressure it exerts at different °C.
- A gas can be collected by downward displacement of H2O but its a mix of that gas and water
vapour (ie. gas is “wet”)
- When we collect a gas using the downward displacement of water method, we must SUBTRACT
the pressure exerted by the water vapour from the atmospheric pressure!
Ptot(look up on table) from the total pressure (set it up so that tot press = atmospheric press)
Pgas = Ptotal - PH2O
- Not all molecules in a substance have the same kinetic energy. This is why we speak of “average
kinetic energy”. Some high-energy molecules have enough energy to escape from the surface, and
enter the gas phase to provide vapour pressure.
- At melting point very few molecules have enough nrg to escape to gas phase,  low vap press
- As temp increases, avg kin nrg increases as will the proportion of molecs with nrg high enough
to escape liquid state
Why do different liquids have different boiling points?
- b/c strength of attraction b/w molecs of a liquid is diff’t for diff’t liqs
- strong forces = lower vap press - weak forces = higher vap press
Vapour pressure
Measuring Vapour Pressure
Temperature
When the vapour pressure is equal to the atmospheric pressure (Patm), the push out is enough to
overcome Patm and boiling occurs.
Thus, water will boil at a temperature below 100 °C if the atmospheric pressure is reduced.
Many times gases are collected over H2O
Often we want to know the volume of dry gas at STP (useful for stoichiometry).
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For this we must make 3 corrections:
1. The level of water inside and outside the tube must be level (so pressure inside is equal to
the pressure outside).
2. The water vapour pressure must be subtracted from the total pressure (to get the pressure of
the dry gas).
3. Finally, values are converted to STP using the combined gas law.
Example: A gas was collected over 21°C H2O. After equalizing water levels, the
volume was 325 mL. Give the volume of dry gas at STP (Patm=102.9 kPa).
Step 1: Determine vapour pressure ( ref text pg. 507)
At 21°C vapour pressure is 2.49 kPa
Step 2: Calculate the pressure of dry gas
Pgas = Patm - PH2O = 102.9 - 2.49 = 100.41 kPa
Step 3: List all of the data
T1 = 294 K, V1 = 325 mL, P1 = 100.41 kPa
Step 4: Convert to STP
V2=
(P1)(V1)(T2)
(100.4 kPa)(325 mL)(273 K)
=
(P2)(T1)
= 299 mL
(101.325 kPa)(294 K)
Sample Problem:
For the decomposition of sodium bicarbonate:
2NaHCO3  Na2CO3 + CO2 + H2O
The carbon dioxide was collected by the downward displacement of water.
It was found that 1.00 g of NaHCO3 produced 145 mL of CO2 (after equalizing water levels)
at 25.0 OC and 98.7 kPa.
i) What is the theoretical (or expected) volume of CO2 under these conditions?
ii) Determine the molar volume of CO2 at STP?
Calculations:
i)
nNaHCO3 = m = 1 g = 0.0119 mol
M
84.01
Since NaHCO3 : CO2 = 2 : 1
CO2 = ½ x .0119 = 0.00595 mole
Finding theoretical volume under given conditions:
PCO2 = PT - PH2O
= 98.7 - 3.17 = 95.53 kPa
V = nRT = 0.00595 x 8.31 x 298
P
95.53
= 0.15424 L = 154 mL
(3 sig. fig)
ii) Finding molar volume of CO2 under given conditions
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“Vm” = V = 0.145 = 24.37 L = 24.4 L (3 sig.fig)
n .00595
or:
Since 0.00595 mol CO2 has a volume of 145 mL ( vol from question)
then
1 mol CO2 has a volume of
145 mL = 24369.74 mL = 24.4 L
0.00595 mol
Finding molar volume of CO2 at STP
P1 = 95.53 kPa
V1 = 24.37 L
T1 = 298 K
P2 = 101.3
V2 = ?
T2 = 273 K
V2 = P1V1T2 = 95.53 x 24.37 x 273 = 21.05 L = 21.1 L
T1P2
298 x 101.3
12.4 Atmospheric Reactions and pollution
Atmosphere: The layer of gases
surrounding planet Earth.
The major components of the Earth
atmosphere are Nitrogen (N2), Oxygen
(O2), Argon (Ar), and Carbon Dioxide
(CO2).
Note: water is also a major component;
however, it is naturally more variable than
the rest).
Percentage Composition of clean dry
air in the homosphere (lower part of the
atmosphere): Major components are
bold italics
Atmosphere Gas
Nitrogen, ______
Oxygen, ______
Argon, ______
Carbon dioxide, ______
Neon, ______
Helium, ______
Methane, ______
Krypton, ______
Hydrogen, ______
Dinitrogen monoxide, ______
Xenon, ______
Ozone, ______
Volume Percent
78.084
20.946
0.0340
0.03845
0.00182
0.00052
0.00015
0.00011
0.00005
0.00005
0.000009
0.000007
Natural Greenhouse Gas Effect: Gases in the Earth’s atmosphere trap heat energy from the sun
on Earth in order to heat up the planet. Without this natural greenhouse effect, the Earth’s
average temperature would be -21 °C.
Gases that contribute to the natural
greenhouse effect are called greenhouse
gases. Water vapour is the main contributor
to the natural greenhouse effect (since it is
most abundant). Other natural greenhouse
gases include: N2O, CO2 and CH4.
Anthropogenic (enhanced) Greenhouse
Effect: Due to human activity, the amount
of greenhouse gases (N2O, CO2, CH4) is
increasing in the atmosphere. These gases
trap more and more heat from the sun on
Earth (less is permitted to escape back into
space). The enhancement of the natural
greenhouse effect due to human activities is called the anthropogenic greenhouse effect.
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Activities that increase Greenhouse gases in the atmosphere:
1) Burning fossil fuels: electricity and transportation are major carbon sources.
2) Raising cattle to make my delicious double cheese burgers is bad!
3) Cutting down trees which are a valuable and necessary carbon sink.
This enhanced greenhouse gas effect is causing a rise in the average temperature of the Earth
(global warming) and because of this, climate change (change in average weather patterns over
long periods of time) is occurring.
According to the graph on the left, CO2 levels are
______________.
According the graph on the right, the average temperature of
Earth is ____________.
http://misterguch.brinkster.net/pra_gaslawsworksheets.html
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