Chapter 11
Gases
11.1 Gases & Pressure
Defining Gas Pressure
•
•
Pressure Increases
•
Number of Particles Increases
•
How are number of particles and
pressure related?
Pressure –force per unit area
that particles exert on walls of
their container
Gas particles collide with walls =
greater pressure
Pressure is directly proportional
to number of particles.
Temperature & Pressure
•
•
Kelvin Temperature Increases
Pressure of Gas Increases
•
Higher temperature results in more
kinetic energy!
IF the volume of container remains
constant and IF the amount of gas
remains constant:
the pressure of a gas increases in
direct proportion to the Kelvin
temperature.
(Kelvin Temp = Celsius Temp + 273)
Volume of a gas at constant pressure is
directly proportional to Kelvin temp.
Temperature Conversions
• Kelvin & Celsius
Tk = (Tc + 273)
Tc = (Tk - 273)
• Fahrenheit & Celsius
Tf = (9/5 Tc) + 32
Tc = (Tf - 32) 5/9
Devices to Measure Pressure
• Barometer: an instrument that
measures pressure exerted by
the atmosphere.
Invented in 1600’s by an Italian
scientist,
Evangelista Torricelli
• Height of column of
mercury shows the
atmospheric pressure.
(atm)
Atmospheric Pressure
• We live at the bottom of an
ocean of air; highest
pressure occurs at the
lowest altitudes!
• Standard Atmosphere is
pressure that supports a
760 mm column of mercury.
• 1.00 atm = 760 mm Hg
Devices to Measure Pressure
•
Pressure Gauge:
instrument used to
measure pressure
inside a tire or oxygen
tank.
Tire
Pressure
Blood
Pressure
Absolute Pressure
• When measuring tire pressure; you measure
pressure ABOVE atmosphere pressure.
Recommended pressures for tires are gauge
pressures.
• Absolute pressure – the TOTAL pressure of
all gases including the atmosphere.
Q: How would you figure it for an inflated tire?
A: Add barometric pressure to the gauge
pressure.
Pressure Units
•
•
SI unit for measuring pressure is the pascal
(pa) after the French physicist Blaise
Pascal (1600s)
A kilopascal (kPa) is 1000 pascals and is
more commonly used.
Equivalent Pressures
1.00 atm
101300 Pa
101.3 kPa
760 mm Hg
760 Torr
14.7 psi
Sample Calculations
1. Express 1.56 atm in kPa.
2. Convert 801 mm Hg to Pa.
3. How many psi are equivalent to 95.6 kPa?
Answers
1) 1.56 atm X 101.3 kPa
= 158 kPa
1.00 atm
2) 801 mm Hg X 101300 Pa
= 107000 Pa
760 mm Hg
3) 95.6 kPa X
14.7 psi
101.3 kPa
= 13.9 psi
Dalton’s Law of Partial Pressures
• Partial Pressure
The pressure exerted by each gas in a mixture
Dalton’s Law:
PT = P1 + P2 + P3 + …
Practice
• Calculate the partial pressure in mm Hg
exerted by the four main gases in air at 760
mm Hg: nitrogen, oxygen, argon and
carbon dioxide. Their abundance by volume
is 78.08%, 20.95%, 0.934% and 0.035%,
respectively.
N2= 593.4 mm Hg
O2= 159.2 mm Hg
Ar = 7.10 mm Hg
CO2= 0.27 mm Hg
Gases Collected by Water
Displacement
• Gases produced in the lab are often collected by
the displacement of water in a collection bottle
• Water vapor will be present in the collected gas,
and it exerts a pressure
• Water vapor pressure = PH20
• Water vapor pressure increases with temperature
(Appendix A, Table-8)
• Pressure of the dry gas
P atm = P gas + P H20
so… P gas = P atm – P H2O
Practice
• A student has stored 100.0 mL of neon gas
over water on a day when the temperature
if 27.0 °C. If the barometer in the room
reads 743.3 mm Hg, what is the pressure of
the neon gas in its container?
P atm = P Ne + P H2O
P Ne = P atm – P H2O
P Ne = 743.3 mm Hg – 26.7 mm Hg
=716.6 mm Hg
Chapter 11
Gases
11.2 The Gas Laws
Pressure & Volume
• In the 1600s, Robert Boyle did many experiments
involving gases.
• He did these experiments at constant temperature.
Pressure & Volume are
Inversely Proportional!
Boyle’s Law Graph
Boyle’s Law
V1P1=V2P2
Where:
V1 = initial volume
P1 = initial pressure
V2 = final volume
P2 = final pressure
Kinetic Explanation of Boyle’s Law
• As volume is reduced,
number of particles
and temperature
remains constant but
number of collisions
with the walls of the
container increases.
• There is a smaller area of space for the same
number of particles to move around, so pressure
increases.
Practice
• If you have 5.5 L of gas at a pressure of
1.6 atm, and the pressure changes to 1.2
atm, what is your new volume?
V1P1 = V2P2
(5.5 L) x (1.6 atm) = (x L) x (1.2 atm)
x = 7.3 L
Temperature & Volume
• Jacques Charles did experiments concerning
gases held at constant pressure, while varying
temperature.
Temperature & Volume are
Directly Proportional!
Charles’s Law Graph
Charles’s Law
V1 V2
=
T1
T2
Where:
V1 = initial volume
T1 = initial temperature
V2 = final volume
T2 = final temperature
Kinetic Explanation of Charles’s Law
• When a gas is
heated, its
temperature
increases, which
means the kinetic
energy of the
particles has
increased.
• Then the particles begin to move faster, which causes
its volume to increase.
• The reverse occurs as the temperature begins to fall.
Practice
• 3.0 L of Helium gas is in a balloon at 22 C and a
pressure of 760 mm Hg. If the temperature rises to
31 C and the pressure remains constant, what will
the new volume be?
(remember to convert any temperatures to KELVIN!!!)
V1
V2
=
T1
T2
3.0 L
V2
=
(273 + 22C) (273 + 31C )
V2 = (3.0 L x 304 K) / 295 K
V2 = 3.1 L
Pressure & Temperature
• From the prior relationships of volume & pressure,
and temperature & volume, it could be concluded
that a relationship exists between pressure &
temperature.
Pressure & Temperature are
Directly Proportional!
Gay-Lussac’s Law Graph
Gay-Lussac’s Law
P1 P2
=
T1
T2
Where:
P1 = initial pressure
T1 = initial temperature
P2 = final pressure
T2 = final temperature
Practice
• At 27 C, Helium gas is in a balloon at
pressure of 760 mm Hg. If the temperature
rises to 31 C, what will the new pressure be?
(remember to convert any temperatures to KELVIN!!!)
P1
P2
T1 = T2
760 mm Hg =
P2
(273 + 27C) (273 + 31C )
P2 = (760 mm Hg x 304 K) / 300 K
P2 = 770 mm Hg
Combined Gas Law
• All 3 Gas Laws require one variable to be held
constant.
• How can we solve a problem when all 3
variables; volume, pressure & temperature
change?
• Since 2 out of the 3 laws always have a
variable in common, there should be a way to
relate these laws into one formula.
• This new formula is called the Combined Gas
Law.
Combined Gas Law
GayLussac’s
P
V
1
1
Law
Charles’s T1
Law
Where:
= P2 V2
T2
Boyle’s
Law
P1, V1 & T1 are initial values
P2, V2 & T2 are final values
*0C & 1 atm = Standard Temperature &
Pressure, or STP
Practice
• 154 mL of Carbon Dioxide gas is at a pressure of 121
kPa and a temperature of 117C. What volume would
this gas occupy at STP? (Remember to convert your
temps to Kelvin!!!)
1 atm = 101.3 kPa
P1V1/ T1 = P2V2/ T2
(154 mL)(121 kPa) = (101.3 kPa)(V2)
(117C + 273)
(0C + 273)
V2 = (154 mL)(121 kPa)(273 K)
(390 K)(101.3 kPa)
V2 = 129 mL
Chapter 11
Gases
11.3 Gas Volumes & the Ideal
Gas Law
The Law of Combining Gas Volumes
• If one volume of water, H2O, is
decomposed, one volume of
oxygen will be formed and 2
volumes of hydrogen will be
formed.
• How can 3 volumes be formed
from only 1 initial volume?
+
1 L H2
1 L H2O
1 L H2
+ 1LO
2
The Law of
Combining
Gas Volumes
The law of combining
volumes states that in
chemical reactions
involving gases, the ratio
of the gas volumes is a
small whole number.
All of the gases are at the same temperature & pressure,
each of the identical flasks contains the same number of
molecules. Notice how the combining ratio:
2 volumes H2 : 1 volume O2 : 2 volumes H2O leads to a
result in which all the atoms present initially are accounted
for in the product.
The Law of Combining Gas Volumes
• Avogadro was the first to study this and concluded a
water molecule is composed of particles.
• We now know that a water
molecule is composed of 2
hydrogen atoms & 1 oxygen atom.
When a molecule of water breaks
down, it breaks down according to
the ratio of particles that compose
it; 2 volumes of H2 & 1 volume of
O2 from 1 volume of H2O.
The Law of Combining Gas Volumes
• He reasoned that the volume of a
gas depends on the number of gas
particles, provided the temperature
& pressure are constant.
The Law of Combining Gas Volumes
• Under the same conditions of temperature
and pressure, the volumes of reacting gases
and their gaseous products are expressed
in ratios of small whole numbers
2 L H2 + 1 L O2 → 2 L H2O (g)
2 volumes H2 + 1volume O2 → 2 volumes H2O (g)
1 volume H2 + 1 volume Cl2 → 2 volumes HCl
1 volume HCl + 1 volume NH3 → NH4Cl (s)
Avogadro’s Law
For a gas at constant temperature and
pressure, the volume is directly
proportional to the number of moles of gas
(at low pressures).
V = an
a = proportionality constant
V = volume of the gas
n = number of moles of gas
Standard Molar
Volume
Equal volumes of all
gases at the same
temperature and
pressure contain the
same number of
molecules.
- Amedeo Avogadro
Standard Molar Volume
Practice
• You are planning an experiment that requires
0.0580 mol of nitrogen monoxide gas at STP.
What volume would you need?
0.0580 mol x 22.4 L = 1.30 L
1 mol
Gas Stoichiometry
Volume-Volume Calculations
• Assume: All products and reactants are
at the same temp and pressure
• Unless otherwise stated, assume STP
• Solve by normal stoichiometric
processes
• Volume ratios are the same as mole
ratios
Volume-Mass and Mass-Volume
Calculations
• Order of Calculations
•
You are given a gas volume and asked to
find a mass:
gas volume A →moles A →moles B → mass B
•
You are given a mass and asked to find a
gas volume:
mass A → moles A →moles B →gas volume B
Ideal Gas Law
PV = nRT
P = pressure in atm
V = volume in liters
n = moles
R = proportionality constant
= 0.08206 L∙ atm/ mol·K
For units of kPa, L & K:
R = 8.31 kPa ∙ L
Mol ∙ K
T = temperature in Kelvin
Calculate the Value of R
• Use all standard values!
P = 1 atm
V = 22.4 L
n = 1 mole
T = 273 K
• Try substituting different standard
pressures to obtain different values of R
Practice
• A 2.07 L cylinder contains 2.88 mol of helium
gas at 22.0 °C. What is the pressure in
atmospheres of the gas in the cylinder?
PV = nRT
P = nRT
V
= 2.88 mol x 0.0821 (atm∙L/mol∙K) x 295 K
2.07 L
= 33.7 atm
Gas Density
mass
molar mass
Density 

volume molar volume
so at STP…
molar mass
Density 
22.4 L
Variations on the Ideal Gas Law
n = mass (m)
molar mass (M)
If PV = nRT then
PV = mRT
M
M = mRT
PV
So replace n
with m/M
So
rearrange
for M
Variations on the Ideal Gas Law
D = mass (m)
volume (V)
If M = mRT then
VP
M = DRT
P
So replace
m / V with D
Density and the Ideal Gas Law
Combining the formula for density with the Ideal
Gas law, substituting and rearranging algebraically:
MP
D
RT
M = Molar Mass
P = Pressure
R = Gas Constant
T = Temperature in Kelvin
Practice
1) At 28°C and 0.974 atm, 1.00 L of a gas has a mass of
131 g/mol
5.16 g. What is the molar mass of this gas?
2) What is the molar mass of a gas if 0.427g of the gas
83.8 g/mol
occupies a volume of 125 mL at 20.0°C and 0.980 atm?
3) What is the density of a sample of ammonia gas if the
0.572 g/L
pressure is 0.928 atm and the temp is 63.0°C?
4) The density of a gas was found to be 2.0 g/L at 1.50 atm
33 g/mol
and 27°C. What is the molar mass of the gas?
5) What is the density of argon gas at a pressure of 551 torr
1.18 g/L
and a temp of 25°C?
Chapter 11
Gases
11.4 Diffusion & Effusion
Effusion
Effusion: describes the passage of gas into
an evacuated chamber.
Graham’s Law
Rates of Effusion and Diffusion
Effusion:
Rate of effusion for gas 1

Rate of effusion for gas 2
M2
M1
Diffusion:
Distance traveled by gas 1

Distance traveled by gas 2
M2
M1
Graham’s Law
• Density can replace molar mass in
Graham’s formula, since density is
directly proportional to molar mass.
• Isotopes of elements can be separated
by vaporizing the element, and allowing
it to effuse.
• The heavier isotope effuses more
slowly than the lighter isotope
Practice
• At 25 °C, the average velocity of oxygen molecules
is 420 m/s. What is the average velocity of helium
atoms at the same temperature?
Rate of O2 is 420 m/s = √MHe
Rate of He
√MO2
420 m/s = √4 g/mol
x
√32 g/mol
420 m/s = 0.3536
x
1
x = 1188 m/s ≈ 1200 m/s