Chapter 5Chapter 5

The Gaseous State

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Properties of GasesProperties of Gasescan be compressedexert pressure on whatever surrounds themexpand into whatever volume is availableeasily diffuse into one anothercan be described in terms of their

temperatures and pressure,the volume occupied, and the amount (number of molecules or moles) present

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Mercury BarometerMercury Barometer

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Composition of Air at Sea Composition of Air at Sea LevelLevel

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Important Units of PressureImportant Units of Pressure

Conversion factor to learn

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Boyle’s LawBoyle’s LawAt constant temperature and mass of gas:

V1/P

V = a * 1/P

where a is a proportionality constant

thus

VP = a

V1P1 = a = V2P2

V1P1 = V2P2

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Boyle’s LawBoyle’s Law

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Boyle’s LawBoyle’s Law

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Charles’ LawCharles’ Law

At constant pressure and mass of gas:

VT

V = b * T

where b is a proportionality constant

V/T = b

V1/T1 = b = V2/T2

V1/T1 = V2/T2

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Charles’ Law Charles’ Law

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Combined Gas LawCombined Gas LawAt constant mass of gas

VT/P

V = d * (T/P)

where d is a proportionality constant

(VP)/T = d

V1P1 = d = V2P2

T1 T2

V1P1 = V2P2

T1 T2

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Avogadro’s LawAvogadro’s LawAt constant pressure and temperature

Vn

V = c * n

where c is a proportionality constant

V/n = c

V1/n1 = c = V2 /n2

V1/n1 = V2 /n2

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Ideal Gas LawIdeal Gas LawV(n * T)/P

V = R * (n * T)/P

where R is proportionality constant

P * V = n * R * T

(P*V)/(n*T) =R

Thus,

(P1*V1)/(n1*T1) = (P2*V2)/(n2*T2)

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What will be volume of an ideal gas at absolute zero?

- 10 mL/mole

0 mL/mole

10 mL/mole

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Ideal Gas ConstantIdeal Gas Constant

R = 0.08205 L*atm/mol*K

R has other values for other sets of units.

R = 82.05 mL*atm/mol*K

= 8.314 J/mol*K

= 1.987 cal/mol*K

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Molar MassMolar Massfrom Gas Densityfrom Gas Density

gas density = #g/V = d

PV = nRT

where n = #g/MM

PV = (#g/MM)*RT

MM = (#g*R*T)/(P*V)

MM = (#g/V)*((R*T)/P) = (d*R*T)/P

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Dalton’s LawDalton’s Lawof Partial Pressuresof Partial Pressures

The total pressure of a mixture of gases is equal to the sum of the pressures of the individual gases (partial pressures).

PT = P1 + P2 + P3 + P4 + . . . .

where PT => total pressure

P1 => partial pressure of gas 1

P2 => partial pressure of gas 2

P3 => partial pressure of gas 3

P4 => partial pressure of gas 4

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Dalton’s Law of Partial Dalton’s Law of Partial PressurePressure

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Collecting Gases over WaterCollecting Gases over Water

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Example: Example: What volume will 25.0 What volume will 25.0 g Og O22 occupy at 20 occupy at 20ooC and a C and a

pressure of 0.880 atm?pressure of 0.880 atm? (25.0 g)(1 mol)n = ---------------------- = 0.781 mol

(32.0 g)V =?; P = 0.880 atm; T = (20 + 273)K = 293KR = 0.08205 L*atm/mol*K

V = nRT/P= (0.781 mol)(0.08205L*atm/mol*K)(293K)

0.880atm = 21.3 L

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Example: Example: A student generates A student generates oxygen gas and collects it over oxygen gas and collects it over water. If the volume of the gas water. If the volume of the gas is 245 mL and the barometric is 245 mL and the barometric pressure is 758 torr at 25pressure is 758 torr at 25ooC, C, what is the volume of the “dry” what is the volume of the “dry” oxygen gas at STP? oxygen gas at STP? (P(Pwaterwater = =

23.8 torr at 2523.8 torr at 25ooC)C)PO2 = Pbar - Pwater = (758 - 23.8) torr = 734 torr

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ExampleExample A student generates oxygen gas and collects it A student generates oxygen gas and collects it over water. If the volume of the gas is 245 mL and the over water. If the volume of the gas is 245 mL and the barometric pressure is 758 torr at 25barometric pressure is 758 torr at 25ooC, what is the C, what is the volume of the “dry” oxygen gas at STP?volume of the “dry” oxygen gas at STP?

Pwater = 23.8 torr at 25oC; PO2 = Pbar - Pwater = (758 - 23.8) torr = 734 torr

P1= PO2 = 734 torr; P2= SP = 760. torr

V1= 245mL; T1= 298K; T2= 273K; V2= ?

(V1P1/T1) = (V2P2/T2)

V2= (V1P1T2)/(T1P2)

= (245mL)(734torr)(273K) (298K)(760.torr)

= 217mL

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Kinetic Molecular TheoryKinetic Molecular Theory

Matter consists of particles (atoms or molecules) in continuous, random motion.

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Kinetic Molecular Theory: Kinetic Molecular Theory: GasesGases

particles in continuous, random, rapid motion collisions between particles are elastic volume occupied by the particles has a negligibly

small effect on their behavior attractive forces between particles have a

negligible effect on their behavior gases have no fixed volume or shape, take the

volume and shape of the container

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Maxwell’s Distribution of Maxwell’s Distribution of SpeedsSpeeds

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Real GasesReal Gases

have a finite volume at absolute zerohave attractive forces between gas particles

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Van der Waals EquationVan der Waals Equation

(P + a/V2)(V - b) = nRT

where a => attractive forcesb => residual volume

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Real versus Ideal Gases

0

0.5

1

1.5

2

2.5

0 100 200 300 400 500 600 700 800 900

Pressure, atm

Vob

s/V

idea

l

ideal

H2

O2

N2

CH4

CO2

SO2

Cl2

H2O

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Real versus Ideal Gases

0.9820.9840.9860.9880.99

0.992

0.9940.9960.998

11.0021.004

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8

Pressure, atm

Vob

s/V

idea

l

ideal

H2

O2

N2

CH4

CO2

SO2

Cl2

H2O

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Carbon Dioxide and Greenhouse Carbon Dioxide and Greenhouse EffectEffect

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Some Oxides of NitrogenSome Oxides of NitrogenN2O

NONO2

N2O4

2 NO2 = N2O4

brown colorlessNOx

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Air Pollution in Los AnglesAir Pollution in Los Angles