Gases
Gases
What gases are important for each of the
following: O2, CO2 and/or He?
Properties of Gases
Gas Pressure
A.
1
B.
C.
D.
2
KineticKinetic-Molecular Theory of Matter
Gases
Particles of Matter are always in
motion
Ideal GasesGases-an imaginary gas that
fits all the assumptions of the theory
Kinetic Energy (KE) formula
Physical properties of gases
Real gasesgases- gases in our daily lives
What gases are important for each of the
following: O2, CO2 and/or He?
A. CO2
B. O2/CO2
C. O2
D. He
3
4
Ideal GasGas-imaginary gas that fits all
the following assumptions.
Kinetic Energy of Gases
Particles in gases:
Are very far apart
Have collisions that are elastic (no KE loss)
Move rapidly
Have no attraction (or repulsion)
Have energy increases at higher temperatures
5
At the same temperature, all gas
particles have the same amount of
energy
– Lighter particles move faster than
heavier particles
– KE = mv2 KE =kinetic energy
2
v = velocity (speed)
m = mass
6
1
Physical Properties of Gases
Physical Properties of Gases
Gases are compressible
1. Why does a round balloon
become spherical when filled
with air?
Why can you put more air in a tire, but can’
can’t
add more water to a glass full of water?
Gases have low densities
Dsolid or liquid = 2 g/mL
2. Suppose we filled this room halfway with
water. Where would pressure be exerted?
Dgas 2 g/L
7
8
Real Gases
Physical Properties of Gases
Close to ideal gas at standard
conditions
Have volume
Attraction between particles
Deviation from ideal gas is greater
when
– Particles are close together
Gases fill a container completely and
uniformly
Gases exert a uniform pressure on all
inner surfaces of their containers
Fluidity: gas particles can slide past one
another (gases and liquids are “fluids”
fluids”)
Diffusion: gases move from high
concentration to low concentration
9
» Low temperatures
» High pressures
– Gas is a compound rather than an
element
10
Some Gases in Our Lives
Pressure
Force per unit area
P = force
area
Gas pressure is a result of collisions
of gas particles. Depends on:
– Number of gas particles
– Temperature
– Volume
Air:
oxygen O2
argon Ar
nitrogen N2
ozone O3
carbon dioxide CO2
Noble gases:
gases:
water H2O
helium He
neon Ne krypton Kr
xenon Xe
Other gases:
fluorine F2 chlorine Cl2
ammonia NH3
methane CH4
carbon monoxide CO
nitrogen dioxide NO2
11
sulfur dioxide SO2
12
2
Instruments used to measure pressure
Barometers
Barometer – measures atmospheric
pressure
Manometer – measures gas pressure
of a container
760 mmHg
atm
pressure
13
14
Manometer
Unit of Pressure
One atmosphere (1 atm)
atm)
Is the average pressure of the atmosphere at
sea level
Is the standard of pressure
P = Force
Area
1.00atm=760mmHg=760torr=101.3kPa=14.7psi
15
16
Types of Pressure Units
Conversions
Pressure
Used in
760 mm Hg or 760 torr
Chemistry
14.7 lb/in.2
U.S. pressure gauges
29.9 in. Hg
U.S. weather reports
101.3 kPa (kilopascals)
Weather in all
countries except U.S.
1.013 bars
Physics and
astronomy
17
760.mmHg=
760.mmHg=760.torr=
760.torr=1.00atm=
1.00atm=101.3kPa=
101.3kPa=14.7psi
What is 2.00 atm expressed in torr?
torr?
18
3
Learning Check G1
Conversions
760.mmHg=
760.mmHg=760.torr=
760.torr=1.00atm=
1.00atm=101.3kPa=
101.3kPa=14.7psi
The
pressure of a tire is measured
as 32.0 psi.
What is this pressure in kPa?
kPa?
19
A.The downward pressure of the Hg in a
barometer is _____ than (as) the weight of the
atmosphere.
1) greater
2) less
3) the same
B.A water barometer has to be 13.6 times taller
than Hg barometer (D
(DHg = 13.6 g/mL)
because
1) H2O is less dense
2) Hg is heavier
20
3) air is more dense than H2O
Solution G1
Learning Check G2
A.The downward pressure of the Hg in a
barometer is 3) the same (as) the weight of the
atmosphere.
When you drink through a straw you reduce the
pressure in the straw. Why does the liquid go up
the straw?
1) the weight of the atmosphere pushes it
2) the liquid is at a lower level
3) there is empty space in the straw
Could you drink a soda this way on the moon?
1) yes
2) no 3) maybe
Why or why not?
B.A water barometer has to be 13.6 times taller
than Hg barometer (D
(DHg = 13.6 g/mL)
because
1) H2O is less dense
21
22
Solution G2
Learning Check G3
When you drink through a straw you reduce the
pressure in the straw. Why does the liquid go up
the straw?
1) the weight of the atmosphere pushes it
3) there is empty space in the straw
Could you drink a soda this way on the moon?
2) no
Why or why not? Low atmospheric pressure
A. What is 475 mm Hg expressed in atm?
atm?
1) 475 atm
2) 0.625 atm 3) 3.61 x 105 atm
23
B. The pressure of a tire is measured as 29.4 psi.
What is this pressure in mm Hg?
1) 2.00 mm Hg
2) 1520 mm Hg
3) 22,300 mm Hg
24
4
Gas Laws
Solution G3
Boyle’
Boyle’s Law
Charles’
Charles’ Law
GayGay-Lussac’
Lussac’s Law
Combined Gas Law
Ideal Gas Law
Dalton’
Dalton’s Partial Pressure
Graham’
Graham’s Law of Effusion
A. What is 475 mm Hg expressed in atm?
atm?
485 mm Hg
x
1 atm
= 0.625 atm (B)
760. mm Hg
B. The pressure of a tire is measured as 29.4 psi.
What is this pressure in mm Hg?
29.4 psi x 760. mmHg = 1.52 x 103 mmHg
14.7 psi
(B)
25
26
Boyle’
Boyle’s Law
As the pressure
increases
Reducing the volume by oneone-half
doubles the pressure
The volume of a fixed mass of gas
varies inversely with the pressure at
constant temperature
P1V1 – P2V2
27
Volume
decreases
28
How does Pressure and Volume
of gases relate graphically?
Pressure and Volume
(atm
(atm x L)
1
8.0
2.0
16
2
4.0
4.0
_____
3
2.0
8.0
_____
4
1.0
16
_____
Boyle's Law
29
P x V = k (constant) when
T remains constant
PV = k
Volume
Experiment Pressure Volume P x V
(atm)
(L)
atm)
30
Temperature,
Temperature,
#
# of
of particles
particles
remain
remain constant
constant
Pressure
5
Boyle’
Boyle’s Mathematical Law:
What if we had a change in conditions?
since PV = k
P1V1 = P2V2
Eg:
Eg
Eg:: A
A gas
gas has
has aa volume
volume of
of 3.0
3.0 L
L at
at
atm.
What
is
its
volume
at
4
atm?
??
2
atm
2
atm.
What
is
its
volume
at
4
atm
31
1) determine which variables you
have:
P11 = 2 atm
V11 = 3.0 L
P22 = 4 atm
V22 = ?
2) determine which law is being
represented:
32
3) Rearrange the equation for
the variable you don’
don’t know
Learning Check GL1
A sample of nitrogen gas is 6.4 L at a pressure
of 0.70 atm. What will the new volume be if
the pressure is changed to 1.40 atm?
atm? (T
constant) Explain.
P
P11V
V11 == V
V22
P
P22
4) Plug in the variables and
chug it on a calculator:
1) 3.2 L
2) 6.4 L
(2.0
(2.0 atm)(3.0L)
atm)(3.0L) == V
V22
(4atm)
(4atm)
33
V22 = 1.5L
3) 12.8 L
34
Solution GL1
Learning Check GL2
A sample of nitrogen gas is 6.4 L at a pressure
of 0.70 atm. What will the new volume be if
the pressure is changed to 1.40 atm?
atm? (T
constant)
6.4 L x 0.70 atm
=
A sample of helium gas has a volume of
12.0 L at 600. mm Hg. What new pressure
is needed to change the volume to 36.0 L?
(T constant) Explain.
3.2 L (1)
1) 200. mmHg
1.40 atm
2) 400. mmHg
Volume must decrease to cause an increase
in the pressure
35
P
Boyle
Law
P and
and V
V == Boyle’
Boyle’’ss Law
3) 1200 mmHg
36
6
Volume
Volume of
of balloon
balloon
at
at room
room
temperature
temperature
Solution GL2
Volume
Volume of
of balloon
balloon
at
5°°C
C
at 5°
A sample of helium gas has a volume of
12.0 L at 600. mm Hg. What new pressure
is needed to change the volume to 36.0 L?
(T constant) Explain.
600. mm Hg x 12.0 L = 200. mmHg (1)
36.0 L
Pressure decrease when volume increases.
37
38
Charles’
Charles’ Law
Charles’
Charles’ Law: V and T
At constant pressure, the volume of a gas is
directly related to its absolute (K) temperature
V1 = V2
V1T2 = V2T1
or
T1
T2
V = 125 mL
V = 250 mL
T = 273 K
T = 546 K
Observe the V and T of the balloons. How does
volume change with temperature?
39
40
How
How does
does Temperature
Temperature and
and
Volume
Volume of
of gases
gases relate
relate graphically?
graphically?
Volume
V/T = k
41
K = °C + 273
Pressure,
Pressure,
#
# of
of particles
particles
remain
remain constant
constant
Temp
1) determine which variables you
have:
T11 = 127°
127°C + 273 = 400K
V11 = 3.0 L
T22 = 227°
227°C + 273 = 5ooK
V22 = ?
2) determine which law is being
represented:
42
T
Charles
Law
T and
and V
V == Charles’
Charles’’ss Law
7
4) Plug in the variables:
Learning Check GL3
3.0L
V
3.0L
V22
==
400K
500K
400K
500K
Use Charles’
Charles’ Law to complete the statements
below:
1. If final T is higher than initial T, final V
is (greater,
(greater, or less)
less) than the initial V.
5) Cross multiply and chug
(500K)(3.0L) = V22 (400K)
V22 = 3.8L
43
2. If final V is less than initial V, final T is
(higher, or lower)
lower) than the initial T.
44
Solution GL3
V1
T1
= V2
T2
OR V1T2
=
V and T Problem
V2T1
A balloon has a volume of 785
mL on a Fall day when the
temperature is 21°
21°C. In the
winter, the gas cools to 0°
0°C.
What is the new volume of the
balloon?
1. If final T is higher than initial T, final V
is (greater
(greater)) than the initial V.
2. If final V is less than initial V, final T is (lower
(lower))
than the initial T.
45
46
Learning Check GL4
VT Calculation
Complete the following setup:
Initial conditions
Final conditions
V1 = 785 mL
V2 = ?
T1 = 21°
T2 = 0°
21°C = 294 K
0°C = 273 K
V2 = _______ mL x __
V1
A sample of oxygen gas has a volume of
420 mL at a temperature of 18°
18°C. What
temperature (in °C) is needed to change
the volume to 640 mL?
1) 443°
2) 170°
3) - 82°
443°C
170°C
82°C
K = _______ mL
K
Check your answer: If temperature decreases,
47 V should decrease.
48
8
Solution GL4
GayGay-Lussac’
Lussac’s Law: P and T
A sample of oxygen gas has a volume of
420 mL at a temperature of 18°
18°C. What
temperature (in °C) is needed to change
the volume to 640 mL?
2) 170°
170°C
T2 = 291 K x 640 mL = 443 K
420 mL
= 443 K - 273 K
Doubling the Kelvin temperature doubles the
pressure
The pressure of a fixed mass of gas at constant
volume varies directly with the Kelvin temperature
P1 = P2
T 1 T2
or
P1T2 = P2T1
= 170°
170°C
49
50
Think of a tire...
Car
Car before
before aa trip
trip
Think of a tire...
Car after a long trip
Pressure
Pressure
Gauge
Gauge
Let’
Let’s get on
the road
Dude!
Pressure
Pressure
Gauge
Gauge
WHEW!
51
52
How
How does
does Pressure
Pressure and
and
Temperature
Temperature of
of gases
gases relate
relate
graphically?
graphically?
Lussac’
Lussac’s Mathematical Law:
What if we had a change in conditions?
since P/T = k
P1
P
= 2
T1
T2
Pressure
P/T = k
53
Volume,
Volume,
#
# of
of particles
particles
remain
remain constant
constant
Temp
Eg:
Eg: A gas has a pressure of 3.0 atm at
127ºº C. What is its pressure at 227º
127
227º C?
54
9
1) determine which variables you
have:
T11 = 127°
127°C + 273 = 400K
P11 = 3.0 atm
T22 = 227°
227°C + 273 = 500K
P22 = ?
2) determine which law is
being represented:
55
T
Gay
Lussac
Lussac’’ss Law
Law
T and
and P
P == GayGay--Lussac’
4) Plug in the variables:
3.0atm
P
3.0atm
P22
==
400K
500K
400K
500K
5) Cross multiply and chug
(500K)(3.0atm) = P22 (400K)
P22 = 3.8atm
56
Learning Check GL5
LAW
RELATRELAT-IONSHIP
LAW
CONCON-STANT
Boyle’
Boyle’s
P↑ V↓
P1V1 = P2V2
T, n
Charles’
Charles’
V↑ T↑
V1/T1 = V2/T2
P, n
GayGay-Lussac’
Lussac’s
P↑ T↑
P1/T1 = P2/T2
V, n
57
Use GayGay-Lussac’
Lussac’s law to complete the statements
below:
1. When temperature decreases, the
pressure of a gas (decreases
(decreases or increases).
increases).
2. When temperature increases, the pressure
of a gas (decreases or increases).
increases).
58
Solution GL5
PT Problem
1. When temperature decreases, the
pressure of a gas (decreases
).
(decreases).
A gas has a pressure at 2.0 atm at 18°
18°C.
What will be the new pressure if the
temperature rises to 62°
62°C? (V constant)
2. When temperature increases, the
pressure of a gas (increases
).
(increases).
T = 18°
18°C
59
T = 62°
62°C
60
10
PT Calculation
Learning Check GL6
P1 = 2.0 atm
T1 = 18°
18°C + 273 = 291 K
P2 = ? ?
T2 = 62°
62°C + 273 = 335 K
B. When T decreases, V _____.
P1 x T2
C. Pressure _____ when V changes from 12.0 L to
24.0 L (constant n and T)
T1
P2 =
2.0 atm
1) Increases 2) Decreases
3) Does not change
A. Pressure _____, when V decreases
What happens to P when T increases?
P increases (directly related to T)
P2 =
Complete with
x
K =
atm
K
61
D. Volume _____when T changes from 15.0 °C to
45.0°
45.0°C (constant P and n)
62
Solution GL6
A. Pressure 1) Increases,
Increases, when V decreases
The Combined Gas Law
Volume and Moles
(Avogadro’
(Avogadro’s Law)
Partial Pressures
B. When T decreases, V 2) Decreases
C. Pressure 2) Decreases when V changes
from 12.0 L to 24.0 L (constant n and T)
D. Volume 1) Increases when T changes from 15.0
°C to 45.0°
45.0°C (constant P and n)
63
64
Combined Gas Law
P1V1
T1
=
P2V2
T2
or
P1V1T2 = P2V2T1
Combined Gas Law
P1V1
T1
Rearrange the combined gas law to solve for V2
P1V1T2
V2
=
=
P2V2
T2
or
P1V1T2 = P2V2T1
Isolate V2
P2V2T1
P1V1T2
=
P2V2T1
P1V1T2
V2
P2T1
65
=
66
=
P1V1T2
P2T1
11
Learning Check C1
Solution C1
Solve the combined gas laws for T2.
Solve the combined gas law for T2.
(Hint: crosscross-multiply first.)
P1V1
=
P2V2
T1
T2
P1V1T2 = P2V2T1
T2
67
= P2V2T1
68
P1V1
Data Table
Combined Gas Law Problem
A sample of helium gas has a volume of 0.180
L, a pressure of 0.800 atm and a temperature of
29°
29°C. What is the new temperature(°
temperature(°C) of the
gas at a volume of 90.0 mL and a pressure of
3.20 atm?
atm?
69
Set up Data Table
P1 = 0.800 atm
V1 = 0.180 L
T1 = 302 K
P2 = 3.20 atm
V2= 90.0 mL
T2 = ??
??
70
Solution
Calculation
Solve for T2
Enter data
T2 = 302 K x
T2 =
71
Solve for T2
atm x
atm
K - 273 =
mL =
mL
°C
K
T2 = 302 K x 3.20 atm x 90.0 mL = 604 K
0.800 atm 180.0 mL
T2 = 604 K - 273 = 331 °C
72
12
Solution G9
Learning Check C2
A gas has a volume of 675 mL at 35°
35°C and
0.850 atm pressure. What is the temperature
in °C when the gas has a volume of 0.315 L
and a pressure of 802 mm Hg?
T1 = 308 K
T2 = ?
V1 = 675 mL
V2 = 0.315 L = 315 mL
P1 = 0.850 atm
= 646 mm Hg
P2 = 802 mm Hg
T2 = 308 K x 802 mm Hg x 315 mL
646 mm Hg
675 mL
P inc, T inc
V dec,
dec, T dec
73
74
Volume and Moles
How does adding more molecules of a gas
change the volume of the air in a tire?
= 178 K - 273 = - 95°
95°C
Learning Check C3
True (1) or False(2)
1.___The P exerted by a gas at constant V is not
affected by the T of the gas.
If a tire has a leak, how does the loss of air (gas)
molecules change the volume?
75
2.___ At constant P, the V of a gas is directly
proportional to the absolute T
3.___ At constant T, doubling the P will cause the V
of the gas sample to decrease to oneone-half its
original V.
76
Solution C3
Avogadro’
Avogadro’s Law
True (1) or False(2)
When a gas is at constant T and P, the V is
directly proportional to the number of moles (n)
of gas
1. (2)The
(2)The P exerted by a gas at constant V is not
affected by the T of the gas.
2. (1) At constant P, the V of a gas is directly
proportional to the absolute T
V1
n1
initial
3. (1) At constant T, doubling the P will cause the V
of the gas sample to decrease to oneone-half its
original V.
77
=
V2
n2
final
78
13
Learning Check C4
STP
The volumes of gases can be compared when
they have the same temperature and pressure
(STP).
Standard temperature
0°
0°C or 273 K
Standard pressure 1 atm (760 mm Hg)
A sample of neon gas used in a neon sign has a
volume of 15 L at STP. What is the volume (L) of
the neon gas at 2.0 atm and –25°
25°C?
P1 =
P2 =
V1 =
T1 =
V2 = ?? T2 =
K
V2 = 15 L x
79
atm
atm
x
K
K = 6.8 L
K
80
Solution C4
P1 = 1.0 atm
P2 = 2.0 atm
V1 = 15 L
V2 = ??
V2 = 15 L x 1.0 atm
2.0 atm
x
Ideal Gas Law
The equality for the four variables involved in
Boyle’
Boyle’s Law, Charles’
Charles’ Law, GayGay-Lussac’
Lussac’s Law
and Avogadro’
Avogadro’s law can be written
T1 = 273 K
T2 = 248 K
248 K = 6.8 L
273 K
PV = nRT
R = ideal gas constant
81
82
PV = nRT
Learning Check G15
R is known as the universal gas constant
What is the value of R when the STP value for
P is 760 mmHg?
Using STP conditions
P
V
R
=
PV
nT
= (1.00 atm)(22.4 L)
(1mol) (273K)
n
T
= 0.0821 L-atm
83
molmol-K
84
14
Learning Check G16
Solution G15
Dinitrogen monoxide (N2O), laughing gas, is
used by dentists as an anesthetic. If 2.86 mol
of gas occupies a 20.0 L tank at 23°
23°C, what is
the pressure (mmHg) in the tank in the dentist
office?
What is the value of R when the STP value for
P is 760 mmHg?
R
=
= (760 mm Hg) (22.4 L)
PV
nT
(1mol)
(273K)
= 62.4 LL-mm Hg
molmol-K
85
86
Solution G16
Rearrange ideal gas law for unknown P
P = nRT
V
Set up data for 3 of the 4 gas variables
Adjust to match the units of R
V = 20.0 L
20.0 L
T = 23°
23°C + 273
296 K
Substitute values of n, R, T and V and solve
for P
P = (2.86
(2.86 mol)(62.4L
mol)(62.4L-mmHg)(296
mmHg)(296 K)
n = 2.86 mol 2.86 mol
P =
?
(20.0 L)
?
87
88
Learning Check G17
(K(K-mol)
mol)
= 2.64 x 103 mm Hg
Solution G17
Solve ideal gas equation for n (moles)
n = PV
RT
A 5.0 L cylinder contains oxygen gas at
20.0°
20.0°C and 735 mm Hg. How many
grams of oxygen are in the cylinder?
= (735 mmHg)(5.0 L)(mol
L)(mol K)
K)
(62.4 mmHg L)(293
L)(293 K)
= 0. 20 mol O2 x 32.0 g O2 = 6.4 g O2
89
90
1 mol O2
15
Molar Volume
Molar Volume Factor
At STP
1 mole of a gas at STP = 22.4 L
4.0 g He
1 mole
(STP)
V = 22.4 L
16.0 g CH4
1 mole
(STP)
44.0 g CO2
1mole
(STP)
V = 22.4 L
V = 22.4 L
91
22.4 L
1 mole
1 mole
22.4 L
92
Learning Check C5
A.What is the volume at STP of 4.00 g of
CH4?
1) 5.60 L
2) 11.2 L
1) 25.6 g
Solution C5
A.What is the volume at STP of 4.00 g of CH4?
4.00 g CH4 x 1 mole CH4 x 22.4 L (STP) = 5.60 L
16.0 g CH4
3) 44.8 L
B. How many grams of He are present in 8.0
L of gas at STP?
2) 0.357 g3)
g3) 1.43 g
93
1 mole CH4
B. How many grams of He are present in 8.0 L of gas at
STP?
8.00 L x
1 mole He x
22.4 He
4.00 g He = 1.43 g He
1 mole He
94
Gases in the Air
Daltons’
Daltons’ Law of Partial Pressures
The % of gases in air
Partial Pressure
Pressure each gas in a mixture would exert if it
were the only gas in the container
Dalton's Law of Partial Pressures
The total pressure exerted by a gas mixture is
the sum of the partial pressures of the gases in
that mixture.
95
and
PT = P1 +
P2 + P3 + .....
Partial pressure (STP)
78.08% N2
593.4 mmHg
20.95% O2
159.2 mmHg
0.94% Ar
7.1 mmHg
0.03% CO2
0.2 mmHg
PAIR = PN + PO + PAr + PCO = 760 mmHg
2
2
2
96
16
Learning Check C6
A.If the atmospheric pressure today is 745 mm
Solution C6
A.If the atmospheric pressure today is 745 mm
Hg, what is the partial pressure (mm Hg) of O2
in the air?
Hg, what is the partial pressure (mm Hg) of O2
in the air?
1) 35.6
2) 156
2) 156
3) 760
B. At an atmospheric pressure of 714, what is the
partial pressure (mm Hg) N2 in the air?
1) 557
2) 9.14
B. At an atmospheric pressure of 714, what is the
partial pressure (mm Hg) N2 in the air?
3) 0.109
97
1) 557
98
Partial Pressures
The total pressure of a gas mixture depends
on the total number of gas particles, not on
the types of particles.
P = 1.00 atm
P = 1.00 atm
1 mole H2
0.5 mole O2
+ 0.3 mole He
+ 0.2 mole Ar
99
Health Note
When a scuba diver is several hundred feet
under water, the high pressures cause N2 from the
tank air to dissolve in the blood. If the diver rises
too fast, the dissolved N2 will form bubbles in the
blood, a dangerous and painful condition called
"the bends". Helium, which is inert, less dense,
and does not dissolve in the blood, is mixed with
O2 in scuba tanks used for deep descents.
100
Learning Check C7
A 5.00 L scuba tank contains 1.05 mole of O2
and 0.418 mole He at 25°
25°C. What is the
partial pressure of each gas, and what is the
total pressure in the tank?
Solution C7
P = nRT
V
PT
=
2
1.47 mol x 0.0821 LL-atm x 298 K
=
101
PT = PO + PHe
5.00 L
7.19 atm
(K mol)
102
17
Gases Collected by Displacement of
Water
Table of Vapor Pressures for Water
Atmospheric pressure equals the
pressure of the gas plus the pressure
of the water vapor
Patmosphere = Pgas + PH2O
Water vapor pressure table
103
104
Practice
Find the pressure of oxygen gas
collected over water at 25.0°
25.0°C if the
barometer reading is 752mmHg.
105
Gases and the Mole
106
Gases
the chemicals we deal with
are gases.
They are difficult to weigh, so we’
we’ll
measure volume
Need to know how many moles of gas
we have.
Two things affect the volume of a gas
Temperature and pressure
Compare at the same temp. and
pressure.
Standard Temperature and
Pressure
Many of
107
Avogadro's
Hypothesis - at the same
temperature and pressure equal
volumes of gas have the same number
of particles.
0ºC and 1 atmosphere pressure
Abbreviated atm
273 K and 101.3 kPa
kPa is kiloPascal
108
18
At Standard Temperature and
Pressure
abbreviated
STP
At STP 1 mole of gas occupies 22.4 L
Called the molar volume
Used for conversion factors
Moles to Liter and L to mol
109
Conversion factors
Used to change units.
Three questions
– What can you change the given
into?
– What unit do you want to get rid of?
– Where does it go to cancel out?
110
Examples
Density of a gas
D = m /V
a gas the units will be g / L
We can determine the density of any
gas at STP if we know its formula.
To find the density we need the mass
and the volume.
If you assume you have 1 mole than
the mass is the molar mass (PT)
At STP the volume is 22.4 L.
What
is the volume of 4.59
mole of CO2 gas at STP?
111
for
112
Quizdom
Examples
Find
113
the density of CO2 at STP.
Find the density of CH4 at STP.
114
19
The other way
What is the molar mass of a gas
with a density of 1.964 g/L?
Given
the density, we can find the
molar mass of the gas.
Again, pretend you have a mole at
STP, so V = 22.4 L.
m = D x V
m is the mass of 1 mole, since you
have 22.4 L of the stuff.
115
116
Gases and Stoichiometry
2 H2O2 (l) --->
---> 2 H2O (g) + O2 (g)
Decompose 1.1 g of H2O2 in a flask with a
volume of 2.50 L. What is the volume of O2 at
STP?
Gas Stoichiometry: Practice!
A. What is the volume at STP of 4.00 g of
CH4?
Solution
1.1 g H2O2
1 mol H2O2
1 mol O2
22.4 L O2
34 g H2O2 2 mol H2O2 1 mol O2
= 0.36 L O2 at STP
117
What if it’
it’s NOT at STP?
1.
Do the problem like it was
at STP. (V1)
2. Convert from STP (V1, P1,
T1) to the stated conditions
(P2, T2)
119
B. How many grams of He are present in
8.0 L of gas at STP?
118
GAS DIFFUSION & EFFUSION
diffusion is the
gradual mixing
of molecules of
different gases.
effusion is the
movement of
molecules through
a small hole into an
empty container.
120
20
GAS DIFFUSION & EFFUSION
GAS DIFFUSION & EFFUSION
Graham’
Graham’s law
governs effusion
and diffusion of
gas molecules.
Rate for A
Rate for B
M of B
M of A
Rate
Rate of
of effusion
effusion is
is
inversely
inversely proportional
proportional
to
to its
its molar
molar mass.
mass.
Thomas Graham, 18051805-1869.
Professor in Glasgow and London.
121
122
Gas Diffusion
relation of mass to rate of diffusion
Molecules effuse thru holes
in a rubber balloon, for
example, at a rate (=
moles/time) that is
proportional to T
inversely proportional to
M.
Therefore, He effuses more
rapidly than O2 at same T.
He
If neon travels at 400. m/s, estimate
the average speed of butane (C4H10) at
the same temperature. 235 m/s
HCl
HCland
andNH
NH33diffuse
diffuse
from
fromopposite
oppositeends
endsof
of
tube.
tube.
Gases
Gasesmeet
meetto
toform
form
NH
NH44Cl
Cl
HCl
HClheavier
heavierthan
thanNH
NH33
Therefore,
Therefore,NH
NH44Cl
Cl
forms
formscloser
closerto
toHCl
HCl
end
endof
oftube.
tube.
123
124
Chlorine has a velocity of 0.0380 m/s.
What is the average velocity of sulfur
dioxide under the same conditions?
0.0400 m/s
125
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