Transcript Gases - Tripod.com

CHAPTER 12
GASES
2
The Gas Laws
 Describe
HOW gases behave.
 Can be predicted by the theory.
 Amount of change can be calculated
with mathematical equations.
The effect of adding gas.
 When
we blow up a balloon we are
adding gas molecules.
 Doubling the the number of gas
particles doubles the pressure.
(of the same volume at the same
temperature).
Pressure and the number of
molecules are directly related
 More
molecules means more
collisions.
 Fewer molecules means fewer
collisions.
 Gases naturally move from areas of
high pressure to low pressure
because there is empty space to
move in.
 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.
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
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
Kinetic Molecular Theory
 Particles
•
•
•
•
•
in an ideal gas…
have no volume.
have elastic collisions.
are in constant, random, straight-line
motion.
don’t attract or repel each other.
have an avg. KE directly related to
Kelvin temperature.
18
Real Gases
 Particles
in a REAL gas…
• have their own volume
• attract each other
 Gas
behavior is most ideal…
• at low pressures
• at high temperatures
• in nonpolar atoms/molecules
19
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
20
Characteristics of Gases
 Gases
can be compressed.
• no volume = lots of empty space
 Gases
undergo diffusion & effusion.
• random motion
21
Temperature
 Always
use absolute temperature
(Kelvin) when working with gases.
ºF
-459
ºC
-273
K
0
C  59 F  32
32
212
0
100
273
373
K = ºC + 273
22
Pressure
force
pressure 
area
Which shoes create the most pressure?
23
 Barometer
Pressure
• measures atmospheric pressure
Aneroid Barometer
Mercury Barometer
24
Pressure
 Manometer
• measures contained gas pressure
U-tube Manometer
Bourdon-tube gauge 25
Pressure
 KEY
UNITS AT SEA LEVEL
101.325 kPa (kilopascal)
1 atm
760 mm Hg
760 torr
N
kPa  2
m
14.7 psi
26
STP
STP
Standard Temperature & Pressure
0°C
1 atm
273 K
-OR-
101.325 kPa
27
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
28
Boyle’s Law
 The
pressure and volume of
a gas are inversely related
• at constant mass & temp
P
PV = k
V
29
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
30
Charles’ Law
 The
volume and absolute
temperature (K) of a gas are
directly related
• at constant mass & pressure
V
T
V
k
T
31
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
32
Gay-Lussac’s Law
 The
pressure and absolute
temperature (K) of a gas are
directly related
• at constant mass & volume
P
k
T
P
T
33
Combined Gas Law
P
V
PV
PV = k
T
P1V1
P2V2
=
T1
T2
P1V1T2 = P2V2T1
34
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:
P1V1T2 = P2V2T1
(473 cm3)(367 K)=V2(309 K)
V2 = 562 cm3
35
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:
P1V1T2 = P2V2T1
(150.kPa)(100.mL)=(200.kPa)V2
V2 = 75.0 mL
36
Gas
Law
Problems
 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
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
T2 = 273 K
37
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:
P1V1T2 = P2V2T1
(765 torr)T2 = (560. torr)(309K)
T2 = 226 K = -47°C
38
Your Turn
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?
Your Turn
 What
is the temperature 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?
Your Turn
 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?
Your Turn
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?
Avogadro’s Principle
 Equal
volumes of gases
contain equal numbers of
moles
• at constant temp & pressure
• true for any gas
V
k
n
V
n
43
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=8.315 dm3kPa/molK
You don’t need to memorize these values!
44
Ideal Gas Law
PV=nRT
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!
45
The Ideal Gas Law
 Pressure
times Volume equals the
number of moles times the Ideal Gas
Constant (R) times the temperature
in Kelvin.
 R does not depend on anything, it is
really constant
 R = 0.0821 (L atm)/(mol K)
 R = 62.4 (L mm Hg)/(K mol)
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(3.25)=(0.412)(0.0821)(289)
L
mol Latm/molK K
T = 16°C = 289 K
V = 3.25 L
P = 3.01 atm
R = 0.0821Latm/molK
47
Ideal Gas Law Problems
 Find
the volume of 85 g of O2 at 25°C
and 104.5 kPa.
GIVEN:
WORK:
V=?
85 g 1 mol = 2.7 mol
n = 85 g = 2.7 mol
32.00 g
T = 25°C = 298 K PV = nRT
P = 104.5 kPa
(104.5)V=(2.7) (8.315) (298)
kPa
mol
dm3kPa/molK K
R = 8.315 dm3kPa/molK
48
3
V = 64 dm
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?
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.
50
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. 51
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. 52
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 =
m
(PV/RT)
 Molar mass = m RT
V P
 Molar mass = DRT
P
You might need to review
Chapter 8
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
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?
Ideal Gases don’t exist
 Molecules
do take up space
 There are attractive forces
 otherwise there would be no liquids
Real Gases behave like Ideal
Gases
 When
the molecules are
far apart
 The molecules do not
take up as big a
percentage of the space
 We can ignore their
volume.
 This is at low pressure
Real Gases behave like Ideal
gases when
 When
molecules are moving fast.
 Collisions are harder and faster.
 Molecules are not next to each other
very long.
 Attractive forces can’t play a role.
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.
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
67
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
68
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
69
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 Br702.
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
71
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
72
16
2
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.
73
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
= 1.26 mol CO2
Plug this into the Ideal
Gas Law to find liters.
74
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 = 8.315 dm3kPa/molK
PV = nRT
(103 kPa)V
=(1mol)(8.315dm3kPa/molK)(298K)
V = 1.26 dm3 CO2
75
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 = 8.315 dm3kPa/molK
PV = nRT
(97.3 kPa) (15.0 L)
= n (8.315dm3kPa/molK) (294K)
Given liters: Start with
Ideal Gas Law and
calculate moles of O2.
NEXT 
n = 0.597 mol O2
76
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
77