I. Physical
Properties
Real vs. Ideal Gases:
 Ideal
gas = an imaginary gas that
conforms perfectly to all the
assumptions of the kinetic-molecular
theory
• We will assume that the gases
used for the gas law problems are
ideal gases.
Real Gases vs. Ideal Gases:
 Real
gas = a gas that does not
behave completely according to the
assumptions of the kineticmolecular theory.
 All real gases deviate to some
degree from ideal gas behavior.
However, most real gases behave
nearly ideally when their particles
are sufficiently far apart and have
sufficiently high kinetic energy.
Causes of non-ideal behavior:
The kinetic-molecular theory is
more likely to hold true for gases
whose particles have NO attraction
for each other.
Causes of non-ideal behavior:
1. High pressure (low volume):
• Space taken up by gas particles
becomes significant
• Intermolecular forces are more
significant between gas particles
that are closer together
Causes of non-ideal behavior:
2. Low temperature:
• Gas particles move slower
so intermolecular forces
become more important
KMT (Kinetic Molecular Theory)
 Kinetic
Molecular Theory (KMT) =
the idea that particles of matter are
always in motion and that this
motion has consequences.
developed in the late 19th
century to account for the behavior
of the atoms and molecules that
make up matter
 theory
KMT (Kinetic Molecular Theory)

based on the idea that particles in
all forms of matter are always in
motion and that this motion has
consequences
SOLID
LIQUID
GAS
KMT (Kinetic Molecular Theory)

can be used to explain the
properties of solids, liquids, and
gases in terms of the energy of
particles and the forces that act
between them
B. Kinetic Molecular Theory – Ideal gas
 KMT
describing particles in an
IDEAL gas:
• have no volume.
• have elastic collisions.
• are in constant, random motion.
• don’t attract or repel each other.
• have an avg. KE directly related
to Kelvin temperature.
C. Kinetic Molecular Theory - Real Gases
 KMT
describing particles in a REAL
gas:
• have their own volume
• attract each other
behavior is most ideal…
• at low pressures
• at high temperatures
• in nonpolar atoms/molecules
 Gas
D. Characteristics of Gases- using KMT
 Gases
expand to fill any container.
 KMT
- gas particles move rapidly
in all directions without
significant attraction or repulsion
between particles
D. Characteristics of Gases- using KMT
 Gases

are fluids (like liquids).
KMT - No significant attraction or
repulsion between gas particles;
glide past each other
D. Characteristics of Gases- using KMT
 Gases
have very low densities.
• KMT - particles are so much
farther apart in the gas state
sodium in the
solid state:
sodium in the
liquid state:
sodium in the gas
state:
D. Characteristics of Gases- using KMT
 Gases
can be compressed.
• KMT - gas particles are far apart
from one anther with room to be
“squished” together
D. Characteristics of Gases- using KMT
 Gases
undergo diffusion & effusion.
• KMT – gas particles move in
continuous, rapid, random motion
Effusion
E. Temperature
 Always
use Kelvin temperature
when working with gases.
ºF
-459
ºC
-273
K
0
C  F  32
5
9
32
212
0
100
273
373
K = ºC + 273
F. Pressure
force
pressure 
area
Which shoes create the most pressure?
F. Pressure
 Why
do gases exert
pressure?
• Gas particles exert a
pressure on any
surface with which they
collide!
 More collisions =
increase in pressure!
F. Pressure
 Barometer
= measures
atmospheric pressure
 The height of the Hg in
the tube depends on the
pressure
• The pressure of the
atmosphere is proportional to
the height of the Hg column,
so the height of the Hg can be
used to measure atmospheric
pressure!
Mercury Barometer
F. Pressure
 Pressure
UNITS
101.3 kPa (kilopascal)
1 atm
760 mm Hg
760 torr
G. STP
STP
Standard Temperature & Pressure
0°C (exact)
1 atm (exact)
273 K
101.3 kPa
760 mm Hg (exact)
760 torr (exact)
Gases
II. The Gas
Laws
K = ºC + 273
A. Boyle’s Law
P
Volume
(mL)
Pressure
(torr)
P·V
(mL·torr)
10.0
20.0
30.0
40.0
760.0
379.6
253.2
191.0
7.60 x 103
7.59 x 103
7.60 x 103
7.64 x 103
PV = k
V
A. Boyle’s Law
 The
pressure and volume
of a gas are INVERSELY
related
• at constant mass & temp
P
V
A. Boyle’s Law
•Real life application: When you
breathe, your diaphragm moves
downward, increasing the volume of
the lungs. This causes the pressure
inside the lungs to be less than the
outside pressure so air rushes in.
A. Boyle’s Law
•Ex: Halving the volume leads to
twice the rate of collisions and a
doubling of the pressure.
A. Boyle’s Law
• As the
volume
increases,
the
pressure
decreases!
B. Charles’ Law
V
T
Volume
(mL)
Temperature
(K)
V/T
(mL/K)
40.0
44.0
47.7
51.3
273.2
298.2
323.2
348.2
0.146
0.148
0.148
0.147
V
k
T
B. Charles’ Law
 The
volume and
temperature (K) of a gas
are DIRECTLY related
• at constant mass &
pressure
V
T
B. Charles’ Law
 Real
life application: Bread dough
rises because yeast produces carbon
dioxide. When placed in the oven,
the heat causes the gas to expand,
and the bread rises even further.
B. Charles’ Law
 As
temperature decreases, the
volume of the gas decreases
Liquid nitrogen’s temp. is
about 63K or -210 ºC
or -346 ºF!
B. Charles’ Law
 As
temperature
decreases,
the volume
of the gas
decreases
B. Charles’ Law
 NUMBER
OF PARTICLES &
PRESSURE ARE CONSTANT!!!
C. Guy-Lussac’s Law
Temperature
(K)
Pressure
(torr)
P/T
(torr/K)
248
273
298
373
691.6
760.0
828.4
1,041.2
2.79
2.78
2.78
2.79
P
k
T
P
T
C. Guy-Lussac’s Law
 The
pressure and
temperature (K) of a gas
are DIRECTLY RELATED
• at constant mass &
volume
P
T
C. Guy-Lussac’s Law
 When
the temp. of a gas increases
(KE increases) and gas particles
move faster and hit container walls
more frequently and collisions are
more forceful
C. Guy-Lussac’s Law
 Real
life application: The air
pressure inside a tire increases on
a hot summer day.
D. Combined Gas Law
VP
PV
Boyles PV = k
Charles’
Guy
Lussac’sT
T
P 1V 1 P 2V 2
=
T1
T2
E. EXAMPLE Problems
You
need your calculators!
Gas Law Problems
1.) A gas occupies 473 cm3 at 36°C. Find its volume at
94°C. CHARLES’ LAW
GIVEN: T V
V1 = 473 cm3
T1 = 36°C = 309K
V2 = ?
T2 = 94°C = 367K
WORK:
V2= (473 cm3)(367 K)
(309 K)
Johannesson
V2 =C. 562
cm3
Gas Law Problems
2.) A gas occupies 100. mL at 150. kPa. Find its
volume at 200. kPa.
BOYLE’S LAW P1V1 = P2V2
GIVEN: P V WORK:
V2 = P1V1
P2
V1 = 100. mL
V2 = (150.kPa)(100.mL)
P1 = 150. kPa
V2 = ?
P2 = 200. kPa
200.kPa
V2 = 75.0
mL
C. Johannesson
Gas Law Problems
3.) A gas occupies 7.84 cm3 at 71.8 kPa & 25°C. Find its
volume at STP.
COMBINED GAS LAW
GIVEN: P T V WORK:
V1 = 7.84 cm3
V2 = P1V1T2
P1 = 71.8 kPa
P2T1
T1 = 25°C = 298 K
V2 = (71.8 kPa)(7.84 cm3)(273 K)
V2 = ?
(101.3 kPa) (298 K)
P2 = 101.3 kPa
3
V
=
5.09
cm
C. Johannesson
2
T2 = 273 K
Gas Law Problems
4.) A gas’ pressure is 765 torr at 23°C. At what
temperature will the pressure be 560. torr?
GUY LUSSAC’S LAW
GIVEN:
P T WORK:
T2 = P2T1
P1 = 765 torr
P1
T1 = 23°C = 296K
T2 = (560. torr)(296K)
P2 = 560. torr
765 torr
T2 = ?
C. Johannesson
T2 = 217 K
Gases
III. Standard Molar
Volume, Gas
Densities & Molar
Mass
A. Standard Molar Volume
 The
volume occupied by one mole
of a gas at STP
• 22.41410 L/mol or about 22.4
L/mol
 In other words, one mole of any
ideal gas at STP will occupy 22.4 L
B. Example Problems-standard molar volume
1.) A chemical reaction is expected to produce 0.0680 mol of
oxygen gas. What volume of gas in L will be occupied by this
gas sample at STP?
0.068 mol O2 22.4 L O2
=
1 mol O2
1.52 L O2
B. Example Problems-standard molar volume
1.) A chemical reaction is expected to produce 0.0680 mol of
oxygen gas. What volume of gas in L will be occupied by this
gas sample at STP?
0.068 mol O2 22.4 L O2
=
1 mol O2
1.52 L O2
B. Example Problems-standard molar volume
2.) A chemical reaction produced 98.0 mL of SO2 at STP. What
mass (in grams) of the gas was produced?
98.0 mL SO2
1 L SO2
1000 mL SO2
1 mol O2
64.064 g SO2
22.4 L SO2 1 mol SO2
0.280 g SO2
=
C. Density
 The
ratio of an object’s mass to its
volume.
D = M
V
 The volume (and density) of a gas
will change when pressure and
temperature change.
• Densities are usually given in g/L at
STP
C. Density
 For
an ideal gas:
Density (at STP) =
molar mass
standard molar volume
D. Example Problems- density
1.) What is the density of CO2 in g/L at
STP?
44.009 g CO2 1 mol CO2
=
1 mol CO2 22.4 L CO2
1.96 g/L CO2
D. Example Problems- density
2.) What is the molar mass of a gas
whose density at STP is 2.08 g/L.
2.08 g
1L
22.4 L
1 mol
=
46.6 g/mol
Gases
IV. More Gas Laws:
Ideal Gas Law &
Avogadro's Law
A. Avogadro’s Law
 Equal
volumes of gases
contain equal numbers of
moles
• at constant temp &
pressure
V
k
n
V
n
B. Ideal Gas Law
• Boyle’s, Charles, and Avogadro’s laws are contained
within the ideal gas law!
• Merge the Combined Gas Law with Avogadro’s Law:
PV
V
=R
k
T
nT
n
UNIVERSAL GAS
CONSTANT
R=0.0821 Latm/molK
R=8.315 dm3kPa/molK
You don’t need to memorize these values!
B. Ideal Gas Law
PV=nRT
P= pressure
V = volume
n = moles
T = temperature (in K)
R = the ideal gas constant
GAS CONSTANT
R=0.0821 Latm/molK
R=8.315 dm3kPa/molK
You don’t need to memorize these values!
C. Ideal Gas Law Problems
1.) Calculate the pressure in atmospheres of
0.412 mol of He at 16°C & occupying 3.25 L.
PV = nRT
GIVEN:
WORK:
P = ? atm
P = nRT
n = 0.412 mol
V
T = 16°C = 289 K
V = 3.25 L
P =(0.412 mol)(0.0821Latm/molK) (289K)
R = 0.0821Latm/molK
3.25L
C. Johannesson
P = 3.01 atm
C. Ideal Gas Law Problems
2.) Find the volume of 85 g of O2 at 25°C
and 104.5 kPa.
PV = nRT
GIVEN:
WORK:
85 g O2 1 mol O2 = 2.7 mol
31.998 g O2
O2
V=?
n = 85 g = 2.7 mol
V
=
nRT
T = 25°C = 298 K
P
P = 104.5 kPa
3kPa/molK) (298K)
V
=(2.7
mol)(8.315
dm
R = 8.315 dm3kPa/molK
104.5 kPa
C. Johannesson
V = 64 dm3
C. Ideal Gas Law Problems
3.) At 28C and 98.7 kPa, 1.00 L of an unidentified gas
has a mass of 5.16 g. Calculate the number of moles of
gas present and the molar mass of the gas. PV = nRT
GIVEN:
V = 1.00 L
n=?
T = 28°C = 301 K
P = 98.7 kPa
R = 8.314 LkPa/molK
WORK:
n = PV
RT =
(98.7 kPa) (1.00L)
(8.314 LkPa/molK) (301 K)
n = 0.0394 mol
m=
5.16g
= 131 g/mol
C. Johannesson
0.0394
mol
Gases
IV. Two More Laws:
Dalton’s Law &
Graham’s Law
A. Dalton’s Law
 The
total pressure of a mixture
of gases equals the sum of the
partial pressures of the
individual gases.
Ptotal = P1 + P2 + P3 ...
A. Dalton’s Law
Ptotal = P1 + P2 + P3 ...
A. Dalton’s Law
When a H2 gas is collected by water
displacement, the gas in the collection
bottle is actually a mixture of H2 AND water
vapor.
A. Dalton’s Law
1.) Hydrogen gas is collected over water at
22.5°C. Find the pressure of the dry gas
if the atmospheric pressure is 94.4 kPa.
The total pressure in the collection bottle is equal to atmospheric
pressure and is a mixture of H2 and water vapor.
GIVEN:
PH2 = ?
Ptotal = 94.4 kPa
PH2O = 2.72 kPa
Look up water-vapor pressure
on gas formula sheet for 22.5°C.
WORK:
Ptotal = PH2 + PH2O
94.4 kPa = PH2 + 2.72 kPa
PH2 = 91.7 kPa
Sig Figs: Round to least number
C. Johannesson of decimal places.
A. Dalton’s Law
2.) A gas is collected over water at a temp of
35.0°C when the barometric pressure is 742.0
torr. What is the partial pressure of the dry gas?
The total pressure in the collection bottle is equal to barometric
pressure and is a mixture of the “gas” and water vapor.
GIVEN:
Pgas = ?
Ptotal = 742.0 torr
PH2O = 42.2 torr
Look up water-vapor pressure
on gas formula sheet for 35.0°C.
WORK:
Ptotal = Pgas + PH2O
742.0 torr = PH2 + 42.2 torr
Pgas = 699.8 torr
Sig Figs: Round to least number
C. Johannesson of decimal places.
B. Graham’s Law
 Diffusion
=
• Spreading of gas molecules
throughout a container until
evenly distributed.
 Effusion
=
• Passing of gas molecules
through a tiny opening in a
container
B. Graham’s Law
 Speed
of diffusion/effusion
• Kinetic energy is determined by
the temperature of the gas.
• At the same temp & KE, heavier
molecules move more slowly.
 Larger m  smaller v
KE =
2
½mv
B. Graham’s Law
KE
= kinetic energy
m= molar mass
v = velocity
KE =
2
½mv
B. Graham’s Law
 Graham’s
Law = Rate of effusion of
a gas is inversely related to the
square root of its molar mass.
• The equation shows the ratio of
Gas A’s speed to Gas B’s speed.
vA

vB
mB
mA
C. Johannesson
C. Graham’s Law Problems
1.) Determine the relative rate of effusion
for krypton and bromine.
The first gas is “Gas A” and the second gas is “Gas B”.
Relative rate mean find the ratio “vA/vB”.
vA

vB
v Kr

v Br2
m Br2
m Kr

mB
mA
159.80 g/mol
 1.381
83.80 g/mol
Kr effuses 1.381 times faster than Br2.
C. Graham’s Law Problems
2.) A molecule of oxygen gas has an average
speed of 12.3 m/s at a given temp and pressure.
What is the average speed of hydrogen
molecules at the same conditions?
vA

vB
mB
mA
vH 2
12.3 m/s

32.00 g/mol
2.02 g/mol
vH 2
vH 2
vO2

mO2
mH 2
12.3 m/s
Put the gas with
the unknown
speed as
“Gas A”.C. Johannesson
 3.980
vH 2  49.0 m/s
C. Graham’s Law Problems
3.) An unknown gas diffuses 4.0 times faster than
O2. Find its molar mass.
The first gas is “Gas A” and the second gas is “Gas B”.
The ratio “vA/vB” is 4.0.
vA

vB
mB
mA
vA

v O2
mO2
mA

32.00
g/mol
 4.0 

m AA

32.00 g/mol
16 
mA
Square both
sides to get rid
of the square
root sign.




32.00 g/mol
 2.0 g/mol
m A  C. Johannesson
16
2
Gases
VI. Gas Stoichiometry at
Non-STP Conditions
A. Gas Stoichiometry
 Liters of a Gas:
• STP - use 22.4 L/mol
• Non-STP - use ideal gas law
 Moles
 Non-STP
• Given liters of gas?
 start with ideal gas law
• Looking for liters of gas?
 start with stoichiometry conv.
B. Volume  Mass Stoich.
 Thinking:
Gas volume A  moles A  moles B  mass B

If gas A is at STP:
? L “A” 1 mol “A” ? mol “B” molar mass(g) “B”
22.4 L “A” ? mol “A”
1 mol “B”
Mole Ratio
B. Volume  Mass Stoich.
If

gas “A” is NOT at STP:
Use PV = nRT to find moles of “A”
? mol “A” ? mol “B”
? mol “A”
molar mass (g) “B”
1 mol “B”
B. Volume  Mass
1.) How many grams of Al2O3 are formed from
15.0 L of O2 at 97.3 kPa & 21°C?
4 Al
+
3 O2
15.0 L
non-STP
GIVEN:
WORK:

2 Al2O3
?g
Given liters: Start with
Ideal Gas Law and
calculate moles of O2.
n = PV
P = 97.3 kPa
RT
V = 15.0 L
n=
(97.3 kPa) (15.0 L)
n=?
(8.315dm3kPa/molK) (294K)
T = 21°C = 294 K
NEXT 
C. Johannesson
n
=
0.597 mol O2
3
R = 8.315 dm kPa/molK
B. Volume  Mass
1.) How many grams of Al2O3 are formed
from 15.0 L of O2 at 97.3 kPa & 21°C?
3 O2 
15.0L
Use stoich to convert moles
of O to grams Al O .
non-STP
0.597 2 mol 101.96 g
mol O2 Al2O3
Al2O3
4 Al
2
2
+
2 Al2O3
?g
3
3 mol O2
1 mol
Al2O3
= 40.6 g Al2O3
B. Volume  Mass
2.) What mass of sulfur is required to
produce 12.6 L of sulfur dioxide at STP
according to the equation:
S8 (s) + 8 O2 (g) 
?g
12.6
1 mol
L SO2 1 mol SO2 S8
22.4L SO2 8 mol
SO2
8 SO2 (g)
12.6L
256.582
g S8 = 18.0 g
1 mol
S8
S8
C. Mass  Volume
 Thinking:
Mass A  moles A  moles B  gas volume B
 If
gas “B” is at STP:
? g “A”
1 mol “A”
? mol “B”
molar mass (g) “A” ? mol “A”
22.4 L “B”
1 mol “B”
C. Mass  Volume
 If
gas “B” is NOT at STP:
? g “A”
1 mol “A”
molar mass (g) “A”
? mol “B”
? mol “A”
• Then use PV = nRT to find
volume of “B”
C. Mass  Volume
1.) What volume of chlorine gas at STP is needed to react
completely with 10.4 g of sodium to form NaCl?
2 Na +
10.4 g
10.4 g Na
Cl2  2 NaCl
?L
1 mol Cl2
1 mol Na
22.990 g Na 2 mol Na
22.4 L Cl2
1 mol Cl2
= 5. 07 L Cl2
C. Mass  Volume
2.) What volume of CO2 forms from 5.25 g
of CaCO3 at 103 kPa & 25ºC?
CaCO3
5.25 g

CaO
+
Looking for liters: Start with stoich
and calculate moles of CO2.
5.25 g 1 mol
CaCO3 CaCO3
1 mol
CO2
100.09g 1 mol
CaCO3 CaCO3
CO2
?L
non-STP
= 1.26 mol CO2
Plug this into the Ideal
Gas Law to find liters.
C. Mass  Volume
2.) What volume of CO2 forms from
5.25 g of CaCO3 at 103 kPa & 25ºC?
GIVEN:
WORK:
V = nRT
P = 103 kPa
P
V=?
V = (1.26mol)(8.315dm3kPa/molK) (298K)
(103 kPa)
n = 1.26 mol
T = 25°C = 298 K
3 CO
V
=
30.3
dm
2
R = 8.315 dm3kPa/molK
C. Johannesson