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

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

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

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

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

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

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

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).

BOYLE’S LAB

P inversely proportional to V

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

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

V is directly V is directly proportional to Tproportional to T

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

NO F

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K = 273 +C

13 C

27 C

21 C

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