Transcript No Slide Title

BEHAVIOR OF GASES
Chapter 14
1
Importance of
Gases
Airbags fill with N2 gas in an accident.
Gas is generated by the decomposition
of sodium azide, NaN3.
2 NaN3(s) ---> 2 Na(s) + 3 N2(g)
2
THREE
STATES OF
MATTER
Click picture above to view movie on states of matter
Click here for water production
3
General Properties of Gases
There is a lot of "free"
space in a gas.
Gases can be expanded
infinitely.
Gases occupy containers
uniformly and completely.
Gases diffuse and mix
rapidly.
4
Kinetic Theory Revisited
1. Gases consist of hard, spherical
particles (usually molecules or atoms)
2. Small- so the individual volume is
considered to be insignificant
3. Large empty space between them
4. Easily compressed and expanded
5. No attractive or repulsive forces
6. Move rapidly in constant motion
5
Kinetic Theory Revisited
Recall:
that the average kinetic
energy of a collection of gas
particles is directly proportional to
the Kelvin temperature of the gas.
Click for movie
No Kinetic energy lost during
collisions.
All particles have same energy at
same temperature.
6
Properties of Gases
Gas properties can be
modeled using math. Model
depends onV = volume of the gas (L)
T = temperature (K)
n = amount (moles)
P = pressure (atm or
kilopascal)
8
Pressure
Pressure of air is
measured with
a BAROMETER
(developed by
Torricelli in
1643)
Barometer
calibrated for
column width
and pool
width/depth
9
Pressure
Hg rises in tube until
gravitational force
of Hg (down)
balances the force
of atmosphere
(pushing up).
P of Hg pushing down
related to
 Hg density
 column height
10
Pressure
Column height measures
Pressure of atmosphere
1 standard atm
= 760 mm Hg
= 76 cm Hg
= 760 torr
= 29.9 inches
= about 33 feet of water
SI unit is PASCAL, Pa,
where 1 atm = 101.325
kPa
11
Gas -Volume, Temp, & Pressure
 Click
to view movie
12
1. Amount of Gas
When
we inflate a ball, we are
adding gas molecules.
Increasing the number of gas
particles increases the number of
collisions
• thus, the pressure increases
 If
temp. is constant- doubling the
number of particles doubles pressure
Pressure and the Number of
Molecules are Directly Related
Fewer
molecules means fewer
collisions.
Gases naturally move from areas
of high pressure to low pressure
because there is empty space to
move in - spray can is example.
Expanding Gas Uses?
The bombardier
beetle uses
decomposition of
hydrogen peroxide
to defend itself. The
gas acts as a
propellant.
16
The Shuttle
Uses a Solid
Booster and
Uncontrolled
Expanding
Gases Can Be
Disastrous
17
 If
you double the number of
molecules
1 atm
 If
you double the number of
molecules
 You double the pressure.
2 atm
4 atm
 As
you remove
molecules from a
container
2 atm
 As
you remove
molecules from a
container the pressure
decreases
1 atm
 As
you remove
molecules from a
container the pressure
decreases
 Until the pressure inside
equals the pressure
outside
 Molecules naturally
move from high to low
pressure
Changing the Size of the
Container
 In
a smaller container
molecules have less room to
move.
 Hit the sides of the container
more often.
 As volume decreases pressure
increases. Think air pump
1 atm
 As
the
pressure on
a gas
increases
4 Liters
 As
2 atm
2 Liters
the
pressure on
a gas
increases
the volume
decreases
 Pressure and
volume are
inversely
related
What happens to
the air in a diver’s
lungs the deeper
they go?
28
Temperature
Raising the temperature
of a gas increases the
pressure if the volume is
held constant.
 The molecules hit the
walls harder.
 The only way to increase
the temperature at
constant pressure is to
increase the volume.

300 K
 If
you start with 1 liter of gas at 1
atm pressure and 300 K
 and heat it to 600 K one of 2 things
happens
600 K
300 K
 Either
the volume will
increase to 2 liters at 1
atm
300 K
•Or the pressure will increase
to 2 atm.
•Or someplace in between
600 K
The Gas Laws
 Describe
HOW gases behave.
 Can be predicted by theory.
 Amount of change can be calculated
with mathematical equations.
A. Boyle’s Law
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
P
PV = k
V
37
A. Boyle’s Law
 The
pressure and volume of
a gas are inversely related
• at constant mass & temp
P
PV = k
V
38
A. Boyle’s Law
Click to view movie
39
Boyle’s Law
 At
a constant temperature pressure
and volume are inversely related.
 As one goes up the other goes down
P x V = K
(K is some constant)
 Easier
Click
to use P1 x V1=P2 x V2
for movie
Boyle’s Gas Law Problems
A
gas occupies 100. mL at 150. kPa.
Find its volume at 200. kPa.
BOYLE’S LAW
GIVEN: P V
V1 = 100. mL
P1 = 150. kPa
V2 = ?
P2 = 200. kPa
WORK:
P1V1 = P2V2
(150.kPa)(100.mL)=(200.kPa)V2
V2 = 75.0 mL
42
Examples
A
balloon is filled with 25 L of air at
1.0 atm pressure. If the pressure is
change to 1.5 atm what is the new
volume?
 A balloon is filled with 73 L of air at
1.3 atm pressure. What pressure is
needed to change to volume to 43
L?
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
44
B. Charles’ Law
 The
volume and absolute
temperature (K) of a gas are
directly related
• at constant mass &
pressure
V
T
V
k
T
45
Charles’ Law
46
B. Charles’ Law
Click for movie
47
B. Charles’ Law
 The
volume of a gas is directly
proportional to the Kelvin
temperature if the pressure is
held constant.
V = K x T
(K is some constant)
 V/T= K
 V1/T1=
V2/T2
Gas Law Problems
gas occupies 473 cm3 at 36°C. Find
its volume at 94°C.
A
CHARLES’ LAW
GIVEN: T V
V1 = 473 cm3
T1 = 36°C = 309K
V2 = ?
T2 = 94°C = 367K
WORK:
V1T2 = V2T1
(473 cm3)(367 K)=V2(309 K)
V2 = 562 cm3
50
Examples
 What
is the temperature (ºC) of a gas
that is expanded from 2.5 L at 25ºC to
4.1L at constant pressure.
 What is the final volume of a gas that
starts at 8.3 L and 17ºC and is heated
to 96ºC?
C. Gay-Lussac’s Law
P
T
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
52
C. Gay-Lussac’s Law
 The
pressure and absolute
temperature (K) of a gas are
directly related
• at constant mass & volume
P
T
P
k
T
53
C. Gay-Lussac’s Law
 The
temperature and the pressure
of a gas are directly related at
constant volume.
P = K x T
(K is some constant)
 P/T= K
 P1/T1=
P2/T2
E. Gas Law Problems
A
gas’ pressure is 765 torr at 23°C.
At what temperature will the pressure
be 560. torr? GAY-LUSSAC’S LAW
GIVEN: P T
P1 = 765 torr
T1 = 23°C = 296K
P2 = 560. torr
T2 = ?
WORK:
P1T2 = P2T1
(765 torr)T2 = (560. torr)(296K)
T2 = 217 K = -56°C
56
Examples
 What
is the pressure inside a 0.250 L
can of deodorant that starts at 25ºC
and 1.2 atm if the temperature is
raised to 100ºC?
 At what temperature will the can
above have a pressure of 2.2 atm?
Too Much
Pressure
and Not
Enough
Volume!!!!
58
Putting the pieces together
 The
Combined Gas Law Deals with
the situation where only the number
of molecules stays constant.
 (P1 x V1)/T1= (P2 x V2)/T2
 Allows us to figure out one thing
when two of the others change.
 The
combined gas law contains
all the other gas laws!
 If the temperature remains
constant.
P 1 x V1
=
T1Boyle’s Law
P2 x V2
T2
 The
combined gas law contains
all the other gas laws!
 If the pressure remains constant.
P 1 x V1
T1
=
P2 x V2
T2
Charles’ Law
 The
combined gas law contains
all the other gas laws!
 If the volume remains constant.
P 1 x V1
T1
=
P2 x V2
T2
Gay-Lussac’s Law
D. Combined Gas Law
P
V
PV
PV = k
T
P1V1
P2V2
=
T1
T2
P1V1T2 = P2V2T1
63
E. Gas Law Problems
gas occupies 7.84 cm3 at 71.8 kPa &
25°C. Find its volume at STP.
A
COMBINED GAS LAW
GIVEN: P T V WORK:
V1 = 7.84 cm3
P1V1T2 = P2V2T1
P1 = 71.8 kPa
(71.8 kPa)(7.84 cm3)(273 K)
T1 = 25°C = 298 K
=(101.325 kPa) V2 (298 K)
V2 = ?
P2 = 101.325 kPa V2 = 5.09 cm3
64
T2 = 273 K
Examples
A
15 L cylinder of gas at 4.8 atm
pressure at 25ºC is heated to 75ºC
and compressed to 17 atm. What is
the new volume?
 If 6.2 L of gas at 723 mm Hg at 21ºC
is compressed to 2.2 L at 4117 mm
Hg, what is the temperature of the
gas?
The Fourth Part
 Avagadro’s
Hypothesis
click for movie
 V is proportional to number of
molecules at constant T and P
 V is proportional to moles
 One mole = 6.02 x 1023 particles
 One mole = 22.4 L at STP movie
 V = K n ( n ) is the number of moles
 Gets put into the Ideal Gas Law
Ideal Gases
 Remember
in this chapter we assume
the gases behave ideally.
 Ideal gases don’t really exist, but
assuming it makes the math easier. So
we get a close approximation.
 Particles have no volume.
 No attractive forces.
 However, real gases do behave like
ideal gases at high temperature and low
pressure.
The Ideal Gas Law
P
xV=nxRxT
 Pressure times Volume equals the
number of moles times the Ideal Gas
Constant (R) times the temperature
in Kelvin.
 This time R does not depend on
anything, it is really constant
The Ideal Gas Law
R
= 0.0821 (L atm)/(mol K) or
 R = 62.4 (L mm Hg)/(K mol)
 We now have a new way to count
moles. By measuring T, P, and V. We
aren’t restricted to STP.
 n = PV/RT
Examples
 How
many moles of air are there in a
2.0 L bottle at 19ºC and 747 mm Hg?
 What is the pressure exerted by 1.8 g
of H2 gas exert in a 4.3 L balloon at
27ºC?
Density
 The
Molar mass of a gas can be
determined by the density of the gas.
 D=
mass = m
Volume
V
 Molar mass = mass =
m
Moles
n
 n = PV
RT
Molar Mass Formulas
 Molar
Mass =
m
(PV/RT)
 Molar mass = m RT
V P
 Molar mass = DRT
P
Dalton’s Law of Partial Pressures
 The
total pressure inside a container
is equal to the partial pressure due to
each gas.
 The partial pressure is the
contribution by that gas.
 PTotal
= P1 + P2 + P3
 For example
 We
can find out the pressure in the
fourth container.
 By adding up the pressure in the first 3.
2 atm
1 atm
3 atm
6 atm
Examples
 What
is the total pressure in a balloon
filled with air if the pressure of the
oxygen is 170 mm Hg and the
pressure of nitrogen is 620 mm Hg?
 In a second balloon the total pressure
is 1.3 atm. What is the pressure of
oxygen if the pressure of nitrogen is
720 mm Hg?
B. 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 + ...
When a H2 gas is
collected by water
displacement, the gas in
the collection bottle is
actually a mixture of H2
and water vapor.
77
B. Dalton’s Law
 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
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
of decimal places. 78
B. Dalton’s Law

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
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
of decimal places. 79
Diffusion
 Molecules
moving from areas of high
concentration to low concentration.
 Perfume molecules spreading across
the room.
 Effusion Gas escaping through a tiny
hole in a container.
 Depends on the speed of the molecule.
Graham’s Law
 The
rate of effusion and diffusion is
inversely proportional to the square
root of the molar mass of the
molecules.
 Kinetic energy = 1/2 mv2
 m is the mass v is the velocity.
Chem Express
Graham’s Law
 Bigger
molecules move slower at the
same temp. (by Square root)
 Bigger molecules effuse and diffuse
slower
 Helium effuses and diffuses faster
than air - escapes from balloon.
C. Graham’s Law
 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.
vA

vB
mB
mA
86
C. Graham’s Law
 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
159.80g/mol

 1.381
83.80g/mol
Kr diffuses 1.381 times faster than Br2.
87

C. Graham’s Law
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
vA

v O2
mB
mA
mO2
mA
Square both
sides to get rid
of the square
root sign.

32.00 g/mol
g/mol 
32.00
 4.0 



m
A
A


32.00 g/mol
16 
mA
32.00 g/mol
mA 
 2.0 g/mol
88
16
2
C. Graham’s Law

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
Put the gas with
the unknown
speed as
“Gas A”.
12.3 m/s
 3.980
vH2  49.0m/s
89
You may need to review Chapter 12
click for movie
At STP
 At
STP determining the amount of
gas required or produced is easy.
 22.4 L = 1 mole
 For example
How many liters of O2 at STP are
required to produce 20.3 g of
H2O?
Not At STP
 Chemical
reactions happen in
MOLES.
 If you know how much gas - change
it to moles
 Use the Ideal Gas Law
n = PV/RT
 If you want to find how much gas use moles to figure out volume
V = nRT/P
A. Gas Stoichiometry
 Liters of a Gas:
• STP - use 22.4 L/mol
• Non-STP - use ideal gas law
 Non-STP
• Given liters of gas?
 Moles
–start with ideal gas law
• Looking for liters of gas?
–start with stoichiometry
conversions.
97
B. 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
CaCO3
1 mol
CaCO3
1 mol
CO2
100.09g
CaCO3
1 mol
CaCO3
CO2
?L
non-STP
= 0.0525 mol CO2
Plug this into the Ideal
Gas Law to find liters.
98
B. Gas Stoichiometry Problem
 What
volume of CO2 forms from 5.25 g
of CaCO3 at 103 kPa & 25ºC?
GIVEN:
WORK:
P = 103 kPa
PV = nRT
V=?
(103 kPa)V
=(.0525mol)(8.314 LkPa/molK)
n = 0.0525 mol
(298K)
T = 25°C = 298 K
R = 8.314 LkPa/molK
V = 1.26 L CO2
99
B. 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
100
B. Gas Stoichiometry Problem
 How
many grams of Al2O3 are formed
from 15.0 L of O2 at 97.3 kPa & 21°C?
4 Al
+
Use stoich to convert moles
of O2 to grams Al2O3.
0.597
mol O2
3 O2 
15.0L
non-STP
2 mol
Al2O3
101.96 g
Al2O3
3 mol O2
1 mol
Al2O3
2 Al2O3
?g
= 40.6 g Al2O3
101
Example #1
 HCl(g)
can be formed by the
following reaction
 2NaCl(aq) + H2SO4 (aq)
2HCl(g) + Na2SO4(aq)
 What mass of NaCl is needed to
produce 340 mL of HCl at 1.51 atm at
20ºC?
Example #2
 2NaCl(aq)
+ H2SO4 (aq)
2HCl(g) + Na2SO4 (aq)
 What volume of HCl gas at 25ºC and
715 mm Hg will be generated if 10.2 g
of NaCl react?
Too Little Pressure and Too Little
Volume!!!!
104
What is a Turbocharger?
105