Transcript The Nature of Gases

Gases
Properties of Gases
Gas Pressure
1
Gases
What gases are important for each of the
following: O2, CO2 and/or He?
A.
B.
C.
D.
2
Gases
What gases are important for each of the
following: O2, CO2 and/or He?
A. CO2
B. O2/CO2
C. O2
D. He
3
Some Gases in Our Lives
Air:
oxygen O2
argon Ar
helium He
fluorine F2
nitrogen N2
ozone O3
carbon dioxide CO2
water H2O
Noble gases:
neon Ne krypton Kr
xenon Xe
Other gases:
chlorine Cl2
ammonia NH3
methane CH4
carbon monoxide CO
nitrogen dioxide NO2
sulfur dioxide SO2
4
The Nature of Gases
 Gases are compressible
Why can you put more air in a tire, but
can’t add more water to a glass full of
water?
 Gases have low densities
Dsolid or liquid = 2 g/mL
Dgas 2 g/L
5
Nature of Gases
1. Why does a round balloon
become spherical when filled
with air?
2. Suppose we filled this room halfway
with water. Where would pressure be
exerted?
6
Nature of Gases
 Gases fill a container completely and
uniformly
 Gases exert a uniform pressure on all
inner surfaces of their containers
7
Kinetic Theory of Gases
The particles in gases
•
Are very far apart
•
Move very fast in straight lines until they
collide
•
Have no attraction (or repulsion)
•
Move faster at higher temperatures
8
Barometers
760 mmHg
atm
pressure
9
Learning Check G1
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 (DHg = 13.6 g/mL)
because
1) H2O is less dense
2) Hg is heavier
3) air is more dense than H2O
10
Solution G1
A.The downward pressure of the Hg in a
barometer is 3) the same (as) the weight of
the atmosphere.
B.A water barometer has to be 13.6 times
taller than Hg barometer (DHg = 13.6 g/mL)
because
1) H2O is less dense
11
Unit of Pressure
One atmosphere (1 atm)
Is the average pressure of the atmosphere
at sea level
Is the standard of pressure
 P = Force
Area
1.00 atm = 760 mm Hg = 760 torr
12
Learning Check G2
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?
13
Solution G2
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
14
Types of Pressure Units
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
15
Learning Check G3
A. What is 475 mm Hg expressed in atm?
1) 475 atm
2) 0.625 atm 3) 3.61 x 105 atm
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
16
Solution G3
A. What is 475 mm Hg expressed in 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 1.00 atm x 760 mmHg = 1.52 x 103 mmHg
14.7 psi 1.00 atm
(B)
17
Chapter 7
Gases
Pressure and Volume (Boyle’s Law)
Temperature and Volume
(Charles’ Law)
Temperature and Pressure
(Gay-Lussac’s Law)
Pressure and Volume
Experiment
Pressure Volume
(atm)
(L)
PxV
(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
P x V = k (constant) when
T remains constant
P and V Changes
P1
V1
P2
V2
Boyle's Law
The pressure of a gas is inversely related to
the volume when T does not change
Then the PV product remains constant
P1V1
=
P2V2
P 1 V 1=
8.0 atm x 2.0 L
= 16 atm L
P 2 V 2=
4.0 atm x 4.0 L
= 16 atm L
PV Problem
Freon-12, CCl2F2, is used in refrigeration
systems. What is the new volume (L) of a
1.6 L sample of Freon gas initially at 50
mm Hg after its pressure is changed to
200 mm Hg at constant T?
PV Calculation
Prepare a data table
DATA TABLE
Initial conditions
Final conditions
P1
= 50 mm Hg
P2
= 200 mm Hg
V1
= 1.6 L
V2
= ?
?
Find New Volume (V2)
Solve for V2:
V2 =
P1V2 = P2V2
V1 x P1 /P2
V2 = 1.6 L x 50 mm Hg =
200 mm Hg
0.4 L
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? (T constant) Explain.
1) 3.2 L
2) 6.4 L
3) 12.8 L
Solution 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? (T constant)
6.4 L x 0.70 atm
=
3.2 L (1)
1.40 atm
Volume must decrease to cause an
increase in the pressure
Learning Check GL2
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.
1) 200. mmHg
2) 400. mmHg
3) 1200 mmHg
Solution GL2
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.
Charles’ Law
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?
Charles’ Law: V and T
At constant pressure, the volume of a gas is
directly related to its absolute (K) temperature
V1 = V2
T1
T2
Learning Check GL3
Use Charles’ Law to complete the statements
below:
1. If final T is higher than initial T, final V
is (greater, or less) than the initial V.
2. If final V is less than initial V, final T is
(higher, or lower) than the initial T.
Solution GL3
V1
T1
= V2
T2
1. If final T is higher than initial T, final V
is (greater) than the initial V.
2. If final V is less than initial V, final T is
(lower) than the initial T.
V and T Problem
A balloon has a volume of
785 mL on a Fall day when
the temperature is 21°C. In
the winter, the gas cools to
0°C. What is the new volume
of the balloon?
VT Calculation
Complete the following setup:
Initial conditions
Final conditions
V1 = 785 mL
V2 = ?
T1 = 21°C = 294 K
T2 = 0°C = 273 K
V2 = _______ mL x __
V1
K = _______ mL
K
Check your answer: If temperature decreases,
V should decrease.
Learning Check GL4
A sample of oxygen gas has a volume of
420 mL at a temperature of 18°C. What
temperature (in °C) is needed to change
the volume to 640 mL?
1) 443°C
2) 170°C
3) - 82°C
Solution GL4
A sample of oxygen gas has a volume of
420 mL at a temperature of 18°C. What
temperature (in °C) is needed to change
the volume to 640 mL?
2) 170°C
T2 = 291 K x 640 mL = 443 K
420 mL
= 443 K - 273 K
= 170°C
Gay-Lussac’s Law: P and T
The pressure exerted by a confined gas
is directly related to the temperature
(Kelvin) at constant volume.
P (mm Hg)
T (°C)
936
761
691
100
25
0
Learning Check GL5
Use Gay-Lussac’s law to complete the
statements below:
1. When temperature decreases, the
pressure of a gas (decreases or increases).
2. When temperature increases, the pressure
of a gas (decreases or increases).
Solution GL5
1. When temperature decreases, the
pressure of a gas (decreases).
2. When temperature increases, the
pressure of a gas (increases).
PT Problem
A gas has a pressure at 2.0 atm at 18°C.
What will be the new pressure if the
temperature rises to 62°C? (V constant)
T = 18°C
T = 62°C
PT Calculation
P1 = 2.0 atm
P2 = ? ?
T1 = 18°C + 273 = 291 K
T2 = 62°C + 273 = 335 K
What happens to P when T increases?
P increases (directly related to T)
P 2 = P 1 x T2
T1
P2 =
2.0 atm
x
K =
K
atm
Learning Check GL6
Complete with 1) Increases 2) Decreases
3) Does not change
A. Pressure _____, when V decreases
B. When T decreases, V _____.
C. Pressure _____ when V changes from 12.0 L
to 24.0 L (constant n and T)
D. Volume _____when T changes from 15.0 °C to
45.0°C (constant P and n)
Solution GL6
A. Pressure 1) Increases, when V decreases
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°C (constant P and n)
Chapter 7
Gases
The Combined Gas Law
Volume and Moles
(Avogadro’s Law)
Partial Pressures
LecturePLUS Timberlake
44
Combined Gas Law
P 1V 1
T1
=
P 2V 2
T2
Rearrange the combined gas law to solve for
V2
P 1V 1T 2
V2
=
=
P 2V 2T 1
P 1 V 1T 2
P 2T 1
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45
Combined Gas Law
P 1V 1
T1
=
P 2V 2
T2
Isolate V2
P 1V 1T 2
V2
=
=
P 2V 2T 1
P 1 V 1T 2
P 2T 1
LecturePLUS Timberlake
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Learning Check C1
Solve the combined gas laws for T2.
LecturePLUS Timberlake
47
Solution C1
Solve the combined gas law for T2.
(Hint: cross-multiply first.)
P 1V 1
T1
=
P 2V 2
T2
P 1V 1T 2 = P 2V 2T 1
T2
= P 2V 2T 1
P 1V 1
LecturePLUS Timberlake
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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°C. What is the new
temperature(°C) of the gas at a volume of
90.0 mL and a pressure of 3.20 atm?
LecturePLUS Timberlake
49
Data Table
Set up Data Table
P1 = 0.800 atm
V1 = 0.180 L
T1 = 302 K
P2 = 3.20 atm
V2= 90.0 mL
T2 = ??
??
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50
Solution
Solve for T2
Enter data
T2 = 302 K x
T2 =
atm x
atm
K - 273 =
LecturePLUS Timberlake
mL =
mL
K
°C
51
Calculation
Solve for T2
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
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Learning Check C2
A gas has a volume of 675 mL at 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?
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53
Solution G9
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
646 mm Hg
P inc, T inc
= 178 K - 273
= - Timberlake
95°C
LecturePLUS
x
315 mL
675 mL
V dec, T dec
54
Volume and Moles
How does adding more molecules of a gas
change the volume of the air in a tire?
If a tire has a leak, how does the loss of air
(gas) molecules change the volume?
LecturePLUS Timberlake
55
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.
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 one-half its
original V.
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56
Solution C3
True (1) or False(2)
1. (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
3. (1) At constant T, doubling the P will cause the
V of the gas sample to decrease to one-half its
original V.
LecturePLUS Timberlake
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Avogadro’s Law
When a gas is at constant T and P, the V is
directly proportional to the number of
moles (n) of gas
V1
n1
initial
=
V2
n2
final
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58
STP
The volumes of gases can be compared
when they have the same temperature and
pressure (STP).
Standard temperature
0°C or 273 K
Standard pressure
1 atm (760 mm Hg)
LecturePLUS Timberlake
59
Learning Check C4
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°C?
P1 =
P2 =
V2 = 15 L x
V1 =
V2 = ??
atm
atm
T1 =
T2 =
x
LecturePLUS Timberlake
K
K
K
K
= 6.8 L
60
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
T1 = 273 K
T2 = 248 K
248 K = 6.8 L
273 K
LecturePLUS Timberlake
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Molar Volume
At STP
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
LecturePLUS Timberlake
62
Molar Volume Factor
1 mole of a gas at STP = 22.4 L
22.4 L
1 mole
and
LecturePLUS Timberlake
1 mole
22.4 L
63
Learning Check C5
A.What is the volume at STP of 4.00 g of
CH4?
1) 5.60 L
2) 11.2 L
3) 44.8 L
B. How many grams of He are present in
8.0 L of gas at STP?
1) 25.6 g
2) 0.357 g
LecturePLUS Timberlake
3) 1.43 g
64
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
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
LecturePLUS Timberlake
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Daltons’ Law of Partial
Pressures
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.
PT = P1 +
P2 + P3 + .....
LecturePLUS Timberlake
66
Gases in the Air
The % of gases in air
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
LecturePLUS Timberlake
2
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Learning Check C6
A.If the atmospheric pressure today is 745
mm Hg, what is the partial pressure (mm
Hg) of O2 in the air?
1) 35.6
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
LecturePLUS Timberlake
3) 0.109
68
Solution C6
A.If the atmospheric pressure today is 745
mm Hg, what is the partial pressure (mm
Hg) of O2 in the air?
2) 156
B. At an atmospheric pressure of 714, what is
the partial pressure (mm Hg) N2 in the air?
1) 557
LecturePLUS Timberlake
69
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
LecturePLUS Timberlake
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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.
LecturePLUS Timberlake
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Learning Check C7
A 5.00 L scuba tank contains 1.05 mole of
O2 and 0.418 mole He at 25°C. What is the
partial pressure of each gas, and what is
the total pressure in the tank?
LecturePLUS Timberlake
72
Solution C7
P = nRT
V
PT
=
PT = PO + PHe
2
1.47 mol x 0.0821 L-atm x 298 K
=
5.00 L
7.19 atm
LecturePLUS Timberlake
(K mol)
73
Ideal Gas Law
The equality for the four variables involved
in Boyle’s Law, Charles’ Law, Gay-Lussac’s
Law and Avogadro’s law can be written
PV = nRT
R = ideal gas constant
Lecture PLUS Timberlake 2000
74
Ideal Gases
 Behave as described by the ideal gas
equation; no real gas is actually ideal
 Within a few %, ideal gas equation describes
most real gases at room temperature and
pressures of 1 atm or less
 In real gases, particles attract each other
reducing the pressure
 Real gases behave more like ideal gases as
pressure approaches
zero.
Lecture PLUS Timberlake
2000
75
PV = nRT
R is known as the universal gas constant
Using STP conditions
P
V
R
=
PV
nT
= (1.00 atm)(22.4 L)
(1mol) (273K)
n
T
= 0.0821 L-atm
mol-K
Lecture PLUS Timberlake 2000
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Learning Check G15
What is the value of R when the STP value
for P is 760 mmHg?
Lecture PLUS Timberlake 2000
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Solution G15
What is the value of R when the STP value
for P is 760 mmHg?
R
=
PV
nT
= (760 mm Hg) (22.4 L)
(1mol)
(273K)
= 62.4 L-mm Hg
mol-K
Lecture PLUS Timberlake 2000
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Learning Check G16
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°C,
what is the pressure (mmHg) in the tank in
the dentist office?
Lecture PLUS Timberlake 2000
79
Solution G16
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°C + 273
296 K
n = 2.86 mol
2.86 mol
P =
?
?
Lecture PLUS Timberlake 2000
80
Rearrange ideal gas law for unknown P
P = nRT
V
Substitute values of n, R, T and V and
solve for P
P = (2.86 mol)(62.4L-mmHg)(296 K)
(20.0 L)
(K-mol)
3 mm Hg
=
2.64
x
10
Lecture PLUS Timberlake 2000
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Learning Check G17
A 5.0 L cylinder contains oxygen gas
at 20.0°C and 735 mm Hg. How many
grams of oxygen are in the cylinder?
Lecture PLUS Timberlake 2000
82
Solution G17
Solve ideal gas equation for n (moles)
n = PV
RT
= (735 mmHg)(5.0 L)(mol K)
(62.4 mmHg L)(293 K)
= 0. 20 mol O2 x 32.0 g O2 = 6.4 g O2
1 mol O2
Lecture PLUS Timberlake 2000
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Molar Mass of a gas
What is the molar mass of a gas if 0.250 g of
the gas occupy 215 mL at 0.813 atm and
30.0°C?
n = PV = (0.813 atm) (0.215 L) = 0.00703 mol
RT
(0.0821 L-atm/molK) (303K)
Molar mass =
g
mol
=
0.250 g = 35.6 g/mol
0.00703 mol
Lecture PLUS Timberlake 2000
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Density of a Gas
Calculate the density in g/L of O2 gas at STP.
From STP, we know the P and T.
P = 1.00 atm
T = 273 K
Rearrange the ideal gas equation for moles/L
PV = nRT
PV = nRT
P = n
RTV
RTV
RT
V
Lecture PLUS Timberlake 2000
85
Substitute
(1.00 atm ) mol-K
=
(0.0821 L-atm) (273 K)
0.0446 mol O2/L
Change moles/L to g/L
0.0446 mol O2
1L
x
32.0 g O2
1 mol O2
=
1.43 g/L
Therefore the density of O2 gas at STP is
1.43 grams per liter
Lecture PLUS Timberlake 2000
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Formulas of Gases
A gas has a % composition by mass of
85.7% carbon and 14.3% hydrogen. At
STP the density of the gas is 2.50 g/L.
What is the molecular formula of the
gas?
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Formulas of Gases
Calculate Empirical formula
85.7 g C x 1 mol C = 7.14 mol C/7.14 = 1 C
12.0 g C
14.3 g H x 1 mol H = 14.3 mol H/ 7.14 = 2 H
1.0 g H
Empirical formula = CH2
EF mass = 12.0 + 2(1.0) = 14.0 g/EF
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Using STP and density ( 1 L = 2.50 g)
2.50 g
1L
x
22.4 L
1 mol
n = EF/ mol
=
=
56.0 g/mol
56.0 g/mol = 4
14.0 g/EF
molecular formula
CH2 x 4
=
C 4H 8
Lecture PLUS Timberlake 2000
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Gases in Chemical Equations
On December 1, 1783, Charles used 1.00 x 103
lb of iron filings to make the first ascent in a
balloon filled with hydrogen
Fe(s) + H2SO4(aq)  FeSO4(aq) + H2(g)
At STP, how many liters of hydrogen
gas were generated?
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Solution
lb Fe  g Fe  mol Fe  mol H2  L H2
1.00 x 103 lb x 453.6 g x 1 mol Fe x 1 mol H2
1 lb
55.9 g
1 mol Fe
x 22.4 L H2 = 1.82 x 105 L H2
1 mol H2
Charles generated 182,000 L of hydrogen to fill his
air balloon.
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Learning Check G18
How many L of O2 are need to react 28.0 g
NH3 at 24°C and 0.950 atm?
4 NH3(g) + 5 O2(g)
4 NO(g) + 6 H2O(g)
Lecture PLUS Timberlake 2000
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Solution G18
Find mole of O2
28.0 g NH3 x 1 mol NH3 x 5 mol O2
17.0 g NH3
4 mol NH3
= 2.06 mol O2
V = nRT =
P
(2.06 mol)(0.0821)(297K) = 52.9 L
0.950 atm
Lecture PLUS Timberlake 2000
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Summary of Conversions with
Gases
Volume A
Grams A
Volume B
Moles A
Moles B
Atoms or
molecules A
Grams B
Atoms or
molecules B
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Daltons’ Law of Partial Pressures
The % of gases in air
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
Lecture PLUS Timberlake 2000
2
95
Learning Check G19
A.If the atmospheric pressure today is 745
mm Hg, what is the partial pressure (mm
Hg) of O2 in the air?
1) 35.6
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
Lecture PLUS Timberlake 2000
3) 0.109
96
Solution G19
A.If the atmospheric pressure today is 745
mm Hg, what is the partial pressure (mm
Hg) of O2 in the air?
2) 156
B. At an atmospheric pressure of 714, what is
the partial pressure (mm Hg) N2 in the air?
1) 557
Lecture PLUS Timberlake 2000
97
Partial Pressure
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.
PT = P1 +
P2 + P3 + .....
Lecture PLUS Timberlake 2000
98
Partial Pressures
The total pressure of a gas mixture depends
on the total number of gas particles, not on
the types of particles.
STP
P = 1.00 atm
P = 1.00 atm
1.0 mol He
0.50 mol O2
+ 0.20 mol He
+ 0.30 mol N2
Lecture PLUS Timberlake 2000
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. Lecture PLUS Timberlake 2000
100
Learning Check G20
A 5.00 L scuba tank contains 1.05 mole of
O2 and 0.418 mole He at 25°C. What is the
partial pressure of each gas, and what is
the total pressure in the tank?
Lecture PLUS Timberlake 2000
101
Solution G20
P = nRT
V
PT
=
PT = PO + PHe
2
1.47 mol x 0.0821 L-atm x 298 K
=
5.00 L
7.19 atm
Lecture PLUS Timberlake 2000
(K mol)
102