BEHAVIOR OF GASES Chapter 12 1. THREE STATES OF MATTER 2.
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Transcript of BEHAVIOR OF GASES Chapter 12 1. THREE STATES OF MATTER 2.
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BEHAVIOR OF GASESChapter 12
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THREE THREE STATES STATES
OF OF MATTERMATTER
THREE THREE STATES STATES
OF OF MATTERMATTER
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ImportanImportance of ce of GasesGases
• Airbags fill with N2 gas in an accident.
• Gas is generated by the decomposition of sodium azide, NaN3.
• 2 NaN3 ---> 2 Na + 3 N2
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General Properties of Gases
• There is a lot of “free” space in a gas.
• Gases can be expanded infinitely.
• Gases occupy containers uniformly and completely.
• Gases diffuse and mix rapidly.4
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KINETIC MOLECULAR THEORY(KMT) Theory used to explain gas laws
• Gases consist of molecules in constant, random motion.
• P arises from collisions with container walls.
• Collisions elastic. No attractive / repulsive forces between molecules. • Volume of molecules is negligible. 5
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Properties of GasesGas properties can be
modeled using math. Model depends on—• V = volume of the gas (L)• T = temperature (K)• n = amount (moles)• P = pressure (atm)
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Pressure
Air Pressure is measured with a
BAROMETER
(developed by Torricelli in 1643)
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PressureHg rises in tube until
force of Hg (down) balances the force of air (pushing up).
P of Hg pushing down related to • Hg density
• column height 8
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PressureColumn height measures P
1 atm= 760 mm Hgor 29.9 inches Hg
= 34 feet of water
SI unit is PASCAL, Pa,
1 atm = 101.3 kPa9
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Avogadro’s Hypothesis
Equal volumes of gases at the same T and P have the same number of molecules.
V = n (RT/P) = n kV and n are directly related.
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twice as many twice as many moleculesmolecules
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Avogadro’s Hypothesis Avogadro’s Hypothesis and Standard Molar and Standard Molar
VolumeVolume
Avogadro’s Hypothesis Avogadro’s Hypothesis and Standard Molar and Standard Molar
VolumeVolume
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1 mol of 1 mol of ANYANY gas occupies 22.4 L gas occupies 22.4 L at STPat STP
++++ ++++
1mol1mol1mol1mol 1mol1mol1mol1mol 1mol1mol1mol1mol
= ? L= ? L= ? L= ? L
V / n = 22.4 L/mol at STP
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Boyle’s LawIf n and T are constant,
then
PV = (nRT) = kThis means, P and V are inversely
related12
Robert Boyle Robert Boyle (1627-1691). (1627-1691).
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BOYLE’S LAB
P inversely proportional to V
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Practice Problem
• If you had a gas that exerted 202 kPa of pressure and took up a space of 3.00 liters. If you decide to expand the tank to 7.00 liters, what would be the new pressure? (Assume constant T)
• P1V1=P2V2
• 202 kPa x 3.00 liters = P2 x 7.00 liters• 606 = P2 x 7.00 liters• P2 = 86.8 kPa
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torr torr
atm
mmHg = torr
V = ft3
nT = n1 +n2 (PV)T = (pv)1 +(pv)2
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Charles’s Law
If n and P are constant, then
V = (nR/P)T = kT
V and T are directly related.
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Jacques Charles Jacques Charles (1746-1823). (1746-1823).
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Charles’s original balloonCharles’s original balloon
Modern long-distance balloonModern long-distance balloon
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V is directly V is directly proportional to Tproportional to T
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Practice Problem• If you took a balloon outside 5.00C that
was originally inside at 20.00C at 2.0 liters, what volume would the balloon occupy when cold? (constant P)
• T1/V1=T2/V2
• (20+273) = (5+273) 2.00 L X liters• V2 = 278*2/293
• V2 = 1.9L
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NO F
20
K = 273 +C
13 C
27 C
21 C
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Gas Volume, Temperature, and Pressure
Gas Volume, Temperature, and Pressure
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COMBINED GAS LAW: COMBINED GAS LAW: Combines Charles and Boyle’s LawCombines Charles and Boyle’s Law
P1V1 = P2V2 T1 T2
P1V1 = P2V2 T1 T2
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STP
STP = S.C. (STANDARD CONDITIONS)
3 atm
35 C
5 C
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27C
C
1 atm = 760 torr = 101 KPa
27C