Think About This…
This is a U-Tube
Manometer. The red
stuff is a liquid that
moves based on the
pressures on each end
of the tube.
Based on the position
of the red liquid, which
has a higher pressure,
the Gas or the
Atmosphere?
How do you know?
Gas
Atmosphere
Think About This…
Gas
Which has a higher
pressure in this UTube, the Gas or the
Atmosphere?
How do you know?
Atmosphere
Think About This…
Gas
What about this UTube? Which has the
higher pressure, the
Gas or the
Atmosphere?
How do you know?
Atmosphere
The Physical
Properties of Gases
Follow along in
your text
Chapter 12
Section 1
Pages 416 422
Kinetic Molecular Theory
 Particles
in an ideal gas…
• have no volume.
• have elastic collisions.
• are in constant, random, straightline motion.
• don’t attract or repel each other.
• have an average kinetic energy
directly related to Kelvin
temperature.
Real Gases
 Particles
in a REAL gas…
• have their own volume
• attract each other
 Gas
behavior is most ideal…
• at high temperatures
• at low pressures
• in nonpolar, basically covalent,
atoms/molecules
Characteristics of Gases
 Gases
expand to fill any container.
• random motion, no attraction
 Gases
are fluids (like liquids).
• no attraction
 Gases
have very low densities.
• no volume = lots of empty space
Characteristics of Gases
 Gases
can be compressed.
• no volume = lots of empty space
 Gases
undergo effusion & diffusion.
• random motion
Temperature
 Always
use absolute temperature
(Kelvin) when working with gases.
ºF
-459
ºC
-273
K
0
32
212
0
100
273
373
K = ºC + 273
What is Pressure?
force
pressure 
area
Which shoes create the most pressure?
Pressure Equipment
 Barometer
• measures atmospheric pressure
Aneroid Barometer
Mercury Barometer
Pressure Equipment
 Manometer
• measures contained gas pressure
U-tube Manometer
Bourdon-tube gauge
Pressure
 KEY
UNITS AT SEA LEVEL
101,325 Pa
101.325 kPa
1 atm
760 mm Hg
760 torr
14.7 psi
1.01 bar
*These are all equal to each other!*
STP is the Norm!
STP
Standard Temperature & Pressure
0°C
1 atm
-OR-
273 K
101.325 kPa
Closure
Example: How many torr equals 5.00 psi?
5.00 psi 760 torr
14.7 psi
= 259 torr
1. How many Pascals equals 7.3 atm?
2. How many mmHg equals 19.3 barr?
3. How many atms equals 96.3 kPa?
Warm-Up :
Complete the following
pressure conversions .
1.
2.
3.
4.
15,650 Pa to mm Hg
23 atm to psi
893,000 torr to kPa
6.5 psi to barr
The Gas Laws
Follow along in
your text
Chapter 12
Section 2
Pages 423 - 432
Meet the Variables
P
= pressure exerted by the gas
T
= temperature in kelvins of the gas
V
= total volume occupied by the gas
n
= number of moles of the gas
k
= symbolizes anything constant
Boyle’s Law
V
Pressure
(kPa)
Volume
(L)
PV
(kPa x L)
150
200
250
300
0.334
0.250
0.200
0.167
50.1
50.0
50.0
50.1
PV = k
P
Boyle’s Law
 The
pressure and volume of a
gas are inversely related at
constant mass & temp
 When
the temp
& number of
particles
remains the
same, this is
the equation
P1V1 = P2V2
Boyle’s Law Problem
A
gas occupies 100. mL at 150.
kPa. Find its volume at 200. kPa.
Remember BOYLE’S LAW!
GIVEN: P V
V1 = 100. mL
P1 = 150. kPa
V2 = ?
P2 = 200. kPa
WORK:
P1V1T2 = P2V2T1
(150.kPa)(100.mL)=(200.kPa)V2
V2 = 75.0 mL
Charles’ Law
V
T
Volume
(mL)
Temperature
(K)
V/T
(mL/K)
545
546
402
199
273
272
200
100
2.00
2.01
2.01
1.99
V
k
T
Charles’ Law
 The
volume and absolute
temperature (K) of a gas are
directly related at constant
mass & pressure
 If
all conditions
are kept
constant, then
the equation
looks like this
V1 V2
=
T1 T2
Charles’ Law Problem
A
gas occupies 473 mL at 36°C.
Find its volume at 94°C.
Remember CHARLES’ LAW!
GIVEN: T V
V1 = 473 mL
T1 = 36°C = 309K
V2 = ?
T2 = 94°C = 367K
WORK:
P1V1T2 = P2V2T1
(473 mL)(367 K)=V2(309 K)
V2 = 562 mL
Gay-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
Gay-Lussac’s Law
 The
pressure and absolute
temperature (K) of a gas
are directly related
At
constant
mass &
volume, the
equation looks
like this
P1
T1
=
P2
T2
Gay-Lussac’s Problem
A
gas’ pressure is 765 torr at 23°C.
At what temperature will the
pressure be 560. torr?
Remember GAY-LUSSAC’S LAW!
GIVEN: P T WORK:
P1 = 765 torr
P1V1T2 = P2V2T1
T1 = 23°C = 296K (765 torr)T2 = (560. torr)(296K)
P2 = 560. torr
T2 = 217 K = -38°C
T2 = ?
Warm Up:
1.
2.
3.
4.
5.

DO NOT SOLVE! Just determine
which Gas Law you should use.
What is the new volume if a 5.0L container of
gas at 23 °C is heated to 70 °C?
If a gas at STP is heated to 900 K, what is the
resulting pressure?
15 mL of a gas at 1 atm is compressed to 5
mL. How much pressure was applied?
A 2.0 L container at 4.0 atm and 0.0 °C is
heated to 50. °C at constant volume. What is
the resulting pressure?
A 35 L container at 23 °C is expanded to 50 L
to obtain a pressure of 4.0 atm. What was the
original pressure if temperature was constant?
Combined Gas Law
 This
is a combination of the 3 main
gas laws: Boyle’s Law, Charles’
Law & Gay-Lussac’s Law
 Pressure
is opposite temperature &
volume
P
T
V
Combined Gas Law
PV
=k
T
P 1V 1 P 2V 2
=
T1
T2
P 1 V 1T 2 = P 2V 2 T 1
Combined Gas Law Problem
A
gas occupies 7.84 mL at 71.8 kPa &
25°C. Find its volume at STP.
COMBINED GAS LAW
GIVEN: P T V WORK:
V1 = 7.84 mL
P1V1T2 = P2V2T1
P1 = 71.8 kPa
(71.8 kPa)(7.84 mL)(273 K)
T1 = 25°C = 298 K
=(101.325 kPa) V2 (298 K)
V2 = ?
P2 = 101.325 kPa V2 = 5.09 mL
T2 = 273 K
Warm Up:
Stoichiometry Quiz #5
How many grams of CO2 are
produced from 75 L of CO at
STP?
___CO + ___O2 → ___CO2
The Ideal Gas Law
Follow along in your text
Chapter 12 Section 3
Pages 433 - 442
Ideal Gas Law
Merge the
Combined Gas Law with Avogadro’s Principle:
PV
V
k
=R
nT
T
n
UNIVERSAL GAS
CONSTANT
R=0.0821 Latm/molK
R=8.314 LkPa/molK
R=62.4 LmmHg/molK
Ideal Gas Law
PV=nRT
UNIVERSAL GAS
CONSTANT
R=0.0821 Latm/molK
R=8.314 LkPa/molK
R=62.4 LmmHg/molK
Ideal Gas Law
PVM=mRT
molar mass
mass
If you are given the mass of a gas,
you can use this equation instead
of converting mass to moles first.
Ideal Gas Law
d=PM/RT
density
molar mass
If you are given the molar mass of
a gas, you can use this equation to
find the density
Ideal Gas Law Problems
 Calculate
the pressure in atmospheres
of 0.412 mol of He at 16°C & occupying
3.25 L.
GIVEN:
WORK:
P = ? atm
PV = nRT
n = 0.412 mol
P=(0.412 mol)(0.0821)(289 K)
T = 16°C = 289 K
(3.25 L)
V = 3.25 L
R = 0.0821Latm/molK P = 3.01 atm
Ideal Gas Law Problems
 Find
the volume of 85 g of O2 at 25°C
and 104.5 kPa.
GIVEN:
V=?
m = 85 g
MO = 32 g/mol
2
T = 25°C = 298 K
P = 104.5 kPa
R = 8.314 LkPa/molK
WORK:
PVM = mRT
V=
(85 g)(8.314)(298 K)
(104.5 kPa)(32 g/mol)
V = 64 L
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
conversion
Gas Stoichiometry Problem
 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.
Gas Stoichiometry Problem
 What
volume of CO2 forms from
5.25 g of CaCO3 at 103 kPa & 25ºC?
GIVEN:
WORK:
P = 103 kPa
V=?
n = 1.26 mol
T = 25°C = 298 K
R = 8.314 LkPa/molK
PV = nRT
(103 kPa)V
=(1mol)(8.314 LkPa/molK)(298K)
V = 1.26 dm3 CO2
Gas Stoichiometry Problem
 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

2 Al2O3
?g
GIVEN:
WORK:
P = 97.3 kPa
V = 15.0 L
n=?
T = 21°C = 294 K
R = 8.314 LkPa/molK
PV = nRT
(97.3 kPa) (15.0 L)
= n (8.314 LkPa/molK) (294K)
Given liters: Start with
Ideal Gas Law and
calculate moles of O2.
NEXT 
n = 0.597 mol O2
Gas Stoichiometry Problem
 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
Closure:
How many grams of CO2 are
produced from 75L of CO at
35 ºC and 96.2 kPa?
___CO + ___O2 → ___CO2
Warm Up:
Solve using the Ideal Gas Law &
Stoichometry
What mass of NH3 is created
when 20.0 L of N2 is combined
with excess H2 at 42 °C and
2.50 atm?
More Gas Laws!
Follow along in
your text
Chapter 12
Sections 2 & 3
Pages 432 - 439
Avogadro’s Law
 Equal
volumes of gases
contain equal numbers of
moles at constant temp &
pressure
• 22.4 L/mole
• 6.02 x 1023 particles/mole
V  kn
V
n
Dalton’s Law of Partial Pressure
 The
total pressure of a mixture
of gases equals the sum of the
partial pressures of the
individual gases.
Ptotal = P1 + P2 + ...
P1 = (Ptotal) (% of Gas1)
Dalton’s Law of Partial Pressure
 Hydrogen
gas is collected over water.
The water vapor has a pressure of 2.72
kPa. 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
Sig Figs: Round to least number of
decimal places.
WORK:
Ptotal = PH2 + PH2O
94.4 kPa = PH2 + 2.72 kPa
PH2 = 91.7 kPa
Graham’s Law of Diffusion
 Diffusion
• Spreading of gas molecules from
high density to low density until
even all over.
 Effusion
• Passing of gas molecules under
pressure through a tiny opening
Graham’s Law of Diffusion
 Speed
of diffusion/effusion
• Kinetic energy is determined by
the temperature of the gas.
• At the same temp & KE, heavier
molecules move more slowly.
 “The smaller the mass, the
faster the gas!”
Graham’s Law of Diffusion
 Graham’s
Law
• Rate of diffusion 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.
v
√m
vA

vB
mB
mA
Graham’s Law of Diffusion
 Determine
the relative rate of diffusion
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 diffuses 1.381 times faster than Br2.
Graham’s Law of Diffusion

vA

vB
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?
mB
mA
vH 2
12.3 m/s

32.00 g/mol
2.02 g/mol
vH 2
vH 2
vO2

mO2
mH 2
Put the gas with
the unknown
speed as
“Gas A”.
12.3 m/s
 3.980
vH2  49.0 m/s