Chapter 11 Gases
11.1 Gases and Pressure
Pressure and Force: We have learned that pressure is related to the number
of particle collisions and temperature. The collisions cause an outward push,
or force, against the walls of its container. The pressure exerted by a gas is
dependent on the volume, temperature, and number of molecules present.
Pressure (P) is defined as the force per unit area on a surface. The SI unit
for force is the Newton, N.
All objects exert a certain amount of force on the ground based on how
much they weigh. Their weight is compared to the gravitational acceleration
9.8 m/s2. This force is then compared to the area of ground it is resting on.
If the pressure of the same object is compared between two areas we would
find that the pressure exerted by that object is different. The same force is
exerted in 2 different areas by the equation: P= F/A
The Earth also has pressure exerted on it, the atmospheric pressure. This is
made from the individual pressures of the various gases in the atmosphere.
Measuring Pressure: A barometer is used to measure atmospheric pressure.
Torricelli constructed the first barometer in the early 1600’s. He was testing
to see what the maximum height that water could raise to by comparing it to
Hg. He noticed that each time he tested his idea with Hg it would only rise
to a height of 760mm. He concluded that the Hg fell to a certain pt based on
the gravitational force but then was stopped from falling by the atmospheric
pressure that was around it. The maximum height of the Hg in the tube is
based on the surrounding atmospheric pressure.
Units of Pressure: The common unit for the atmospheric pressure is mm of
Hg, (mmHg). A pressure of 1 mmHg can also be referred to as 1 torr.
Pressures can also be measured in atmospheres (atm). This is defined as
being exactly 760 mmHg. The SI unit for pressure is the Pascal (Pa). It is
defined as the pressure exerted by a force of 1 N acting on an area of one
square meter. Pressure can also be expressed as kPa.
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Sample Problem A: The average atmospheric pressure in Denver, CO
is 0.830 atm. Express this pressure in mm Hg and kPa.
P= 0.830 atm
kPa=?
mmHg=?
 760 mmHg 
0.830 atm 
  631 mmHg
 1 atm 
 101.325 kPa 
0.830atm 
  84.1 kPa
1
atm


Dalton’s law of Partial Pressures: The pressure exerted by each gas in an
unreacted mixture is independent of that exerted by other gases present. The
pressure of each gas in a mixture is called the partial pressure of that gas.
Dalton’s law of partial pressures states that the total pressure of a gas
mixture is the sum of the partial pressures of the component gases.
Dalton’s Law of Parital Pressures: P T= P1+P2+P3+P4…
If we think about the partial pressure of each gas along with the KMT we
could understand that as the number of particles increase then the number of
collisions will also increase. Each gas will exert a pressure independently on
the container walls. The total pressure is the result of the total number of
collisions per unit of wall area in a given time.
Gases collected by water displacement: Gases in lab are mostly collected
over water. As the gas enters the container filled with water, the water is
replaced by the gas because it is less dense than the water. This does mean
that the gas that you collect will have added water vapor. This calculation
will always need to be factored vapor pressure from the water.
Sample Problem B: Oxygen gas from the decomposition of potassium
chlorate, KClO3, was collected by water displacement. The barometric
pressure and the temperature during the experiment were 731.0 torr
and 20.0°C, respectively. What was the partial pressure of the oxygen
collected?
PT=Patm= 731.0 torr
PH20= 17.5 torr (found in Table A-8)
Patm= PO2 + PH2O
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Patm PO2  PH2O  PO2  Patm  PT  731atm  17.5 torr  713.5torr
11.2 The Gas Laws
In 1662, Robert Boyle discovered that gas pressure and volume are related
mathematically. The gas laws are simple mathematical relationships
between the volume, temperature, pressure, and amt of gas.
Boyle’s law: Pressure-Volume Relationship: Boyle discovered that
doubling the P on a sample of gas at constant T reduces the volume by ½.
Reducing the P of a gas by ½ will double the V. As one variable increases
the other one decreases.
Boyle’s Law states that the volume of a fixed mass of a gas varies inversely
with the pressure at constant temperature.
PV=k; pressure= P, volume=V, constant=k
Because this expression is equal to the constant we can express this law as
P1V1=P2V2. This allows us to look at the change that a certain gas
undergoes.
Sample Problem C: A sample of gas has a volume of 150 mL when the
pressure if 0.947 atm. What will the volume of the gas be at a pressure
of 0.987 atm if the temperature remains constant?
P1=0.947 atm
V1=150 mL
PV
1 1  PV
2 2 
V2 
P2=0.987 atm
V2=?
PV
1 1
 V2
P2
 0.947atm 150mL 
0.987 atm
V2  144mL O2
Charles’s Law: Volume-Temperature Relationship: We know from the
KMT that if the temp of a gas is increased then the volume will also increase
do to the increased KE of the particles. The volume must expand for the
pressure to remain constant. This relationship was discovered by Jacques
Charles in 1787.
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He found that the volume of a gas will increase by a fraction of 1/273 for
every 1°C the temp is raised in order for the pressure to remain constant.
The Celsius scale can then be related to the Kelvin scale to give it a value of
0K or absolute zero is equal to -273.15°C.
Gas volume and K Temps are directly proportional to each other at constant
pressure. Charles’s law states that the volume of a fixed mass of gas at
constant pressure varies directly with the K.
V=kT or V/T=k T= temp in K, k = constant, V=volume. In the same
manner as before if the ratio of these variables are related to a constant, k we
can show them related to the change of itself.
V1 V2

T1 T2
Sample Problem D: A sample of neon gas occupies a volume of 752 mL
at 25°C. What is the volume the gas will occupy at 50°C if the pressure
remains constant?
V1= 752mL
T1= 25°C
V2=?
T2= 50°C
V1 V2
VT
  V2  1 2
T1 T2
T1
V2 
 752mL  323K 
 298K 
V2  815mL Ne
Gay-Lussac’s law: Pressure-Temp Relationship: the energy and frequency
of gas at constant KE of molecules. At a constant volume the pressure
should be directly proportional to the K temp, which is directly related to the
KE. Joseph Gay- Lussac discovered this relationship in 1802.
Gay-Lussac Law: the pressure of a fixed mass of gas at constant volume
varies directly with the K temp.
P=kT or P/T=k P=pressure, T=Temperature, k=constant
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Sample Problem E: The gas in a container is at a pressure of 3.00 atm
at 25°C. Directions on the container warn the user to keep it in a place
where the temp exceeds 52°C. What would the gas pressure in the
container be at 52°C.
P1=3.00 atm
T1= 25°C
P2 = ?
T2= 52°C
P1 P2
PT
  P2  1 2
T1 T2
T1
P2 
 3.00atm  325K 
 298K 
P2  3.27 atm
The Combined Gas Law: gases can undergo a change of temp, pressure, and
volume all at the same time. We would have to combine the 3 laws that we
just learned about.
PV/T=k
PV
PV
1 1
 2 2
T1
T2
Sample Problem F: A helium-filled balloon has a volume of 50.0L at
25°C and 1.08 atm. What volume will it have at 0.855 atm and 10.0°C.
P1=1.08atm
V1=50.0L
T1=25°C=298K
P2=0.855 atm
V2=?
T2=10°C=283K
PV
PV
PV T
1 1
 2 2  V2  1 1 2
T1
T2
T1 P2
V2 
1.08atm  50.0 L  283K 
 298K  0.855atm 
V2  60.0 L
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11.3 Gas Volumes and the Ideal Gas Law
Measuring and Comparing the Volumes of Reacting Gases:
While Guy-Lussac was studying the relationship between pressure and
temperature he also noticed a simple and definite proportion by volume.
Guy-Lussac’s law of combined volumes of gases states that at constant
temp and pressure with volumes of gaseous rcts and pdts can be
expressed as ratios of small whole numbers.
The ratio of the combo of H and O to form H20 is a ratio of 2:1:2. By
the law it would state that 2mL of H combined with 1mL of O will form
2 mL of H20, or 400 L, 800 L, 400 L…
Avogadro’s Law: Some of the volume relationships observed by GayLussac could not be accounted for by Dalton’s theory. In 1811
Avogadro found a way to satisfy both observations. He thought that
molecules that combine could contain more than one atom.
His law states that equal volumes of gases at the same temp and
pressure contain equal number of molecules. If the temp and pressure
remain constant then the volume will vary directly to the number of
molecules.
Avogadro’s Law: V=kn;
V1 V2

n is the amount of moles and V is the
n1 n2
volume.
From this law we will look at the coefficients of chemical reactions
involving gases as the relative number of moles, molecules, and volume.
Molar Volume of a Gas: Remember that one mole of a substance
contains the number of molecules equal to Avogadro’s number.
According to Avogadro’s law states that one mole of any gas will occupy
the same volume as one mole of any other gas at the same temp and
pressure no matter the difference in masses.
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The volume of gas occupied ay one mole of gas at STP is 22.4 L. The
conversion that can be used for calculations is 1 mol/22.4 L. This can be
used to find the mass or moles of any gas at a particular volume.
Sample Problem G: What volume does a 0.685 mol of gas occupy at STP?
What quantity of gas, in moles, is contained in 2.21 l at STP?
 22.4 L 
0.0685 mol 
  1.53L
 1mol 
 1mol 
2.21L 
  0.0987 mol
 22.4 L 
Gas Stoichiometry:
Sample Problem H: Propane, C3H8, is a gas that is sometimes used as a fuel
for cooking and heating. The complete combustion of propane occurs
according to the following balanced equation.
C3H8 + 5O2  3CO2 + 4H2O
What will the volume, in L, of oxygen required for the complete combustion
of 0.350L of propane? What will the volume of carbon dioxide produced in
the reaction? Assume that all volume measurements are made at the same
temp and pressure.
 5O2 
0.350 L C3 H 8 
  1.75L O2
1
L
C
H
3 8 

 3L CO2 
0.350 L C3 H 8 
  1.05L CO2
1
L
C
H
3 8 

Ideal Gas Law: The mathematical relationship among T, P, V, and n.
This is the equation of state for an ideal gas because the state of a gas
can be defined by its P, V, T, and n. This equation also involves the
constant R.
Ideal Gas Law: PV=nRT
This law shows that the increase in volume keeps that collision rate
constant.
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The Ideal Las Constant: R is the ideal gas constant. The units are
derived from the formula. R= 0.0821 (L atm/mol K)
Sample Problem I: What is the pressure in atm exerted by a 0.500 mol
sample of N2 in a 10.0L container at 298K?
PV  nRT  P 
P
nRT
V
 0.500mol   0.0821
Latm 
  298K 
molK 

 1.22atm
10.0 L
11.4 Diffusion and Effusion
The constant motion of particles causes the random mixing of gas particles
also known as diffusion. Effusion is the process where gas molecules are
confined to a container and randomly pass through a tiny opening.
Rates of diffusion and effusion depend on the velocities of particles. Lighter
molecules will move faster than smaller ones. This gives us the relationship
that the velocities of gas particles vary inversely with the square root of their
mass. Remember that the KE = 1/2mv2 which is also dependent on temp.
We can bring these two equations together to give a mathematical
relationship for the rate of effusion.
MB
rate of effusion gas A

rate of effusion gas B
MA
This formula is known as Graham’s Law. It states that the rates of effusion
of gases at the same temp and pressure are inversely proportional to the
square root of the molar masses.
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Sample Problem J: Compare the rates of effusion of H gas and O gas at
the same temp and pressure.
32.00 g / mol
Rate of Effusion of H 2
32.00 g / mol


 3.98
Rate of Effusion of O2
2.016 g / mol
2.016 g / mol
This tells us that H will effuse almost 4 times faster than O.
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