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Fall 1999
Chapter 7
Gases and
Gas Laws
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Some Gases in Our Lives
Air:
oxygen O2
argon Ar
nitrogen N2
ozone O3
carbon dioxide CO2
water H2O
Noble gases:
helium He
neon Ne
krypton Kr
xenon Xe
Other gases:
fluorine F2
chlorine Cl2
ammonia NH3
methane CH4
carbon monoxide CO
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Properties of a Gas
• Volume
V
L, mL, cc
• Temperature
T
C,K
• Moles
n
g/mole
• Pressure
P
mmHg, atm, torr
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Units of Pressure
One atmosphere (1 atm)
Is the average pressure of the atmosphere
at sea level
Is a standard of pressure
 P = Force
Area
1.00 atm = 760 mm Hg = 760 torr
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Measuring Pressure
Barometers
760 mmHg
atm
pressure
Hg
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Learning Check
A. What is 475 mm Hg expressed in atm?
1) 475 atm
2) 0.625 atm 3) 361000 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
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Solution
A. What is 475 mm Hg expressed in atm?
475 mm Hg
x
1 atm
= 0.625 atm (2)
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 = 1520 mmHg
14.7 psi 1.00 atm
(2)
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Pressure and Altitude
• As altitude increases,
atmospheric pressure
decreases
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Pressure and Boiling Point
• As P atm decreases, water boils at
lower temperatures and foods cook
more slowly
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Boyle’s Law
Pressure and
Volume
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Pressure and Volume
Experiment
Pressure Volume P x V
(atm)
(L)
(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,n remain constant 13
P and V Changes
P1
V1
P2
V2
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Boyle's Law
The pressure of a gas is inversely related to
the volume when T,n does not change
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
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PV Calculation
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?
( T and n are constant)
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HINT
• The pressure goes from 50. mmHg
to 200. mmHg. Is that an increase
or decrease in pressure ?
• What will happen to the volume?
P
V
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Finding the New
Volume
Take the old volume
and multiply by a factor
of pressures to make
the result bigger.
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Solution
1.6 L x 200 mmHg = 6.4 L
50 mmHg
• Factor greater than 1;
answer is larger
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Learning Check
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 and n constant) Explain.
1) 3.2 L
2) 6.4 L
3) 12.8 L
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Solution
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 and n 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
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Learning Check
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 and n constant) Explain.
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Solution
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.
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Worksheet 7-1
• Do the problems from Worksheet 7-1
• You can work these problems alone
or with others around you.
• You may use your notes and
textbook.
• When you have finished, compare
answers with someone else.
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Charles’ Law
T = 273 K
T = 546 K
Observe the V and T of the balloons. How
does volume change with a temperature
increase ?
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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
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.
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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
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.
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V and T Calculation
A balloon has a volume of 785 mL when
the temperature is 21°C. As the balloon
rises, the gas cools to 0°C. What is the new
volume of the balloon?
Think about what
happens to T;always use K !!!
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Solution
785 mL x 273 K =729 mL
294 K
Factor less than 1; answer is
smaller
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Learning Check
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
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Solution
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?
T2 = 291 K x 640 mL = 443 K
420 mL
= 443 K - 273 K
= 170°C (2)
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P and T
P (mm Hg)
936
761
691
T (°C)
100
25
0
When temperature decreases, the pressure
of a gas (decreases or increases).
When temperature increases, the pressure
of a gas (decreases or increases).
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Gay-Lussac’s Law
• Pressure and Absolute
temperature are directly
proportional
T
P
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P and T Calculation
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,n constant)
T = 18°C
T = 62°C
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Solution
2.0 atm x 335 K =2.3 atm
291 K
Factor more than 1; answer is
larger
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Learning Check
Answer 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
36
°C to 45.0°C (constant P and n)
Solution
Answer with
1) Increases 2) Decreases
3) Does not change
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)
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Worksheet 7-2
• Do the problems from Worksheet 7-2.
• You can work these problems alone or
with others around you.
• You may use your notes and textbook.
• When you have finished, compare
answers with someone else.
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Combined Gas Law
• CGL gives the result of
changing 2 properties
P 1 V1
P2 V2
=
T1
T2
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Problem
• Oxygen gas has a pressure of 0.15 atm
when the volume is 15. L and the
temperature is 27º C. What will the new
volume be if T becomes 127 º C and the
pressure becomes 900. mmHg?
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Answer
Change T to Kelvin:
27C +273 = 300 K,127C +273 = 400 K
Change mmHg to atm:
900. mmHg x 1 atm = 1.18 atm
760 mmHg
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Algebraic solution
15. L x 0.15 atm x
300 K
400 K = 1.2 L
1.18 atm
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Alternate solution
15 L x 0.15 atm
1.18 atm
P
V
x
400 K = 1.2 L
300 K
T
V
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Worksheet 7-3
• Do the problems from Worksheet 7-3.
• You can work these problems alone or
with others around you.
• You may use your notes and textbook.
• When you have finished, compare
answers with someone else.
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Avogadro’s Law
• Volume is directly
related to the number
of moles of gas
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Avogadro’s Law
0.60 moles of O2 gas
has a volume of 50.L.
What is the volume
when 1.0 moles of O2 is
added?
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Does a balloon get bigger or
smaller when air is added?
• Add air
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Does a balloon get bigger or
smaller when air is added?
Add air
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Solution
50 L x 1.6 moles = 133 L
0.6 moles
Factor more than 1; answer is
larger
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STP
• Standard temperature
0 C or 273 K
• Standard pressure
760 mmHg or 1 atm
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Molar Volume
• At STP, 1 mole of gas
has a volume of 22.4 L.
1 mole = 22.4 L (at STP)
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Problem
• What is the mass of 50. L of
CO2 gas at STP? Hint: find
moles first
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•
50. L x
1 mole x
22.4 L
44.0 g
= 98. g
1 mole
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• What is the volume of 100. g
of nitrogen gas N2 at STP?
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100.g
x 1 mole x 22.4 L =
28.0 g
1 mole
80.0 L
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Worksheet 7-4
• Do the problems from Worksheet 7-4.
• You can work these problems alone or
with others around you.
• You may use your notes and textbook.
• When you have finished, compare
answers with someone else.
56
Dalton’s Law
• Total pressure is the sum
of all the partial pressures
• Ptotal =P1 + P2 + P3 +
……..
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• What is the total pressure in
a container with 0.112 atm of
oxygen and 450. mmHg of
nitrogen? Give answer in
mmHg.
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ANSWER
• 0.112 atm x 760 mmHg = 85 mmHg
1 atm
• Pt = P1 + P2 = 85 +450. = 535 mmHg
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Nature of Gases
 Gases fill a container completely and
uniformly
 Gases exert a uniform pressure on all
inner surfaces of their containers
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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
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Question
• Use the KMT to explain why
increasing the temperature
of a gas increases the
pressure. (n and V are
constant)
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•
T1
<
O O O
O
O
O
T2
O
O O
O
O
O
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The End
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