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Fall 1999Chapter 7
Gases and Gas Laws
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Some Gases in Our Lives
Air:
oxygen O2 nitrogen N2 ozone O3
argon Ar 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 VolumeExperiment 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
P1V1= 8.0 atm x 2.0 L = 16 atm L
P2V2= 4.0 atm x 4.0 L = 16 atm L
P1V1 = P2V2 = k
Use this equation to calculate how a volume changes when pressure changes, or how pressure changes when volume changes.
new vol. old vol. x Pfactor new P old P x Vfactor
V2 = V1 x P1 P2 = P1 x V1
P2 V2
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P and V Changes
P1
P2
V1 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
P1V1= 8.0 atm x 2.0 L = 16 atm L
P2V2= 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)
170°C 3) - 82°C
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P and T
P (mm Hg) T (°C)
936 100
761 25
691 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 °C to 45.0°C (constant P and n)
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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
P1V1 P2V2
=
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:
27C +273 = 300 K,127C +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 400 K = 1.2 L
300 K 1.18 atm
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Alternate solution
15 L x 0.15 atm x 400 K = 1.2 L
1.18 atm 300 K
P V 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.
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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 < T2
O O O
O O O
OO O O
OO
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The End