Unit 9:The Gas Laws

Chemistry

~78%

The Atmosphere

an “ocean” of gasesmixed together

Composition

nitrogen (N2)…………..

oxygen (O2)……………

argon (Ar)……………...

carbon dioxide (CO2)…

water vapor (H2O)…….

Trace amounts of:

~21%

~1%

~0.04%

~0.1%He, Ne, Rn, SO2, CH4, NxOx, etc.

Depletion of the Ozone Layer

O3 depletion is caused by chlorofluorocarbons (CFCs).

Uses for CFCs:

refrigerants

Ozone (O3) in upper atmosphereblocks ultraviolet (UV) light from Sun.UV causes skin cancer and cataracts.

O3 is replenished with each strike of lightning.

aerosol propellants -- banned in U.S. in 1996

CFCs

The Greenhouse Effect

CO2 MOLECULES

Upon contact with objects, some of light’s energy

is released as heat.

Lower energy, longer l lightis blocked by CO2 and CH4; energy doesn’t escape into space; atmosphere heats up.

High energy, shortl light passes easilythrough atmosphere.

Energy from Sun has short wavelengths (ls) and high energy. CO2 and methane (CH4) let this light in. Reflected light has longerls and less energy. CO2 and CH4 (“greenhouse gases”) preventreflected light from escaping, thus warming the atmosphere.

Why more CO2 in atmospherenow than 500 years ago?

__________________ _____________

--

--

--

--

--

--

--

burning of fossil fuels deforestation

urban sprawl

wildlifeareas

rain forests

coal

petroleum

natural gas

wood

* The burning of ethanolwon’t slow greenhouse effect.

What can we do?

1. Reduce consumption of fossil fuels.

At home:

On the road:

2. Support environmental organizations.

3. Rely on alternate energy sources.

insulate home; run dishwasher full;avoid temp. extremes (A/C &

furnace);wash clothes on “warm,” not “hot”

bike instead of drive;carpool; energy-efficient vehicles

solar, wind energy,hydroelectric power

The Kinetic Molecular Theory (KMT) --

-- deals w/“ideal” gas particles…1. …are so small that they are

assumed to have zero volume2. …are in constant, straight-line motion

3. …experience elastic collisions in whichno energy is lost

4. …have no attractive or repulsive forces towardeach other

5.…have an average kinetic energy (KE) that isproportional to the absolute temp. of gas(i.e., Kelvin temp.)

explains why gases behave as they do

as Temp. , KE

Theory “works,” except athigh pressures and low temps.

(attractive forcesbecome significant)

N2 can be pumped intotires to increase tire life

KMT “works”

liquid nitrogen (N2);the gas condenses into

a liquid at –196oC

KMT starts to break down

–196oC

H2Ofreezes

37oC 100oC0oC

H2Oboils

bodytemp.

liquidN2

** Two gases w/same # of particles and at sametemp. and pressure have the same kinetic energy.

KE is related to mass and velocity (KE = ½ m v2).

More massive gas particlesare _______ thanless massive gas

particles (on average).

= ½

= ½

m1

m2

v1

v2

KE1

KE2

2

2

sametemp.

To keep same KE, as m , v mustOR as m , v must .

slower

(2 g/mol)

Particle-Velocity Distribution(various gases, same T and P)

# of

parti

cles

Velocity ofparticles (m/s)

(SLOW) (FAST)

CO2

(28 g/mol)

N2

H2

(44 g/mol)CO2

N2

H2

Particle-Velocity Distribution(same gas, same P, various T)

# of

parti

cles

Velocity ofparticles (m/s)

(SLOW) (FAST)

O2 @ 10oC

O2 @ 50oC

O2 @ 100oC

O2 @ 10oC

O2 @ 50oC

O2 @ 100oC

Graham’s Law

Consider two gases at same temp. Gas 1: KE1 = ½ m1 v1

2 Gas 2: KE2 = ½ m2 v22

Since temp. is same, then… KE1 = KE2 ½ m1 v1

2 = ½ m2 v22 m1 v1

2 = m2 v22

Divide both sides by m1 v22…

2

21

222

2112

21 v m1v mv m

v m1

1

22

2

21

mm

vv

Take sq. rt. of both sides to get Graham’s Law:

1

2

2

1

mm

vv

** To use Graham’s Law, both gasesmust be at same temp.

diffusion: effusion: particle

movementfrom high tolow conc.

For gases, rates of diffusion & effusion obey Graham’s law:

diffusion of gas

particles through

an opening

NET MOVEMENT

more massive = slow;less massive = fast

NET MOVEMENT

irrelevant, solong as theyare the same

On avg., carbon dioxide travels at 410 m/s at 25oC.Find avg. speed of chlorine at 25oC.

1

2

2

1

mm

vv

2

2

2

2

Cl

CO

CO

Cl

mm

vv

(CO2)

(Cl2)

7144

410v

2Cl

7144 410 v

2Cl = 320 m/s

**Hint: Put whatever you’re looking for in the numerator.

(the algebrais easier)

At a certain temp., fluorine gas travels at582 m/s and a noble gas travels at394 m/s. What is the noble gas?

He2

4.003

Ne10

20.180

Ar18

39.948

Kr36

83.80

Xe54

131.29

Rn86

(222)

1

2

2

1

mm

vv

(F2)

2

2

F

unk

unk

F

mm

vv

38m

394582 unk

2

mm = 38 g/mol

394582 38 munk = 82.9 g/mol

best guess = Kr

38m

394582 unk

He2

4.003

Ne10

20.180

Ar18

39.948

Kr36

83.80

Xe54

131.29

Rn86

(222)

1

2

2

1

mm

vv

4

4

CH

unk

unk

CH

mm

vv

16

m 1.58 unk2

mm = 16 g/mol

2unk (1.58) 16 m = 39.9 g/mol

Ar

16m

11.58 unk

CH4 moves 1.58 times faster than which noble gas?

“Ar?”

“Aahhrrrr! Buckets o’ blood!Swab de decks, ye scurvy dogs!”

Ne2

or Ar?

mm = 36.5 g/mol

HCl and NH3 are released at same time fromopposite ends of 1.20 m horiz. tube.Where do gases meet?

HCl NH3

more massive

travels slower

mm = 17 g/mol

less massive

travels faster

A B C

A

Gas Pressure

Pressure occurs when a force isdispersed over a given surface area. A

FP

If F acts over a large area…

But if F acts over a small area…

F

A= P

AP=

F

(e.g., your weight)

= 1.44 x 106 in2

At sea level, air pressure is standard pressure:

1 atm = 101.3 kPa = 760 mm Hg = 14.7 lb/in2

= 2 x 107 lb.

lb. 2000ton 1

Find force of air pressureacting on a baseballfield tarp…

100 ft.100 ft.

ft 1in 12

2

F = P A = 14.7 lb/in2 (1.44 x 106 in2)

F = 2 x 107 lb. = 10,000 tons

Key: Gases exert pressurein all directions.

A = 10,000 ft2A = 10,000 ft.ft

Atmospheric pressure changes with altitude:

As altitude , pressure .

barometer: device to measureair pressure

vacuum

airpressure

mercury (Hg)

Bernoulli’s Principle

For a fluid traveling // to a surface:

LIQUIDOR GAS

-- FAST-moving fluidsexert ______ pressure

-- SLOW-moving fluidsexert ______ pressure

FAST

SLOW

LOW

HIGH

HIGH P

LOW P

NETFORCE

roof in hurricane

SLOW

FAST LOW P

HIGH P

airplane wing / helicopter propeller

AIRPARTICLES

FAST

SLOW

Resulting Forces

(BERNOULLI’SPRINCIPLE)

(GRAVITY)

frisbee

HIGH P

LOW P

FAST, LOW P

SLOW, HIGH P

CURTAIN

creeping shower curtain

HIGH PSLOWCOLD

LOW PFAST

WARM

windows andhigh winds

FAST

SLOW windowsburst outwards

TALL BUILDING

LOW P

HIGH P

Pressure and Temperature

STP (Standard Temperature and Pressure)

standard temperature standard pressure 0oC273 K

Equations / Conversion Factors:

K = oC + 273

1 atm = 101.3 kPa = 760 mm Hg

1 atm101.3 kPa760 mm Hg

= 138.8 kPa

Convert 25oC to Kelvin.

How many kPa is 1.37 atm?

K = oC + 273 1 atm = 101.3 kPa = 760 mm Hg

K = oC + 273 = 25 + 273 = 298 K

1.37 atm

atm 1kPa 101.3

How many mm Hg is 231.5 kPa?

= 1737 mm Hg231.5 kPa

kPa 101.3Hg mm 760

measures the pressureof a confined gas

manometer:

CONFINEDGAS

AIRPRESSURE

Hg HEIGHTDIFFERENCE

SMALL + HEIGHT = BIG

differentialmanometer

manometers can be filledwith any of various liquids

Atmospheric pressure is 96.5 kPa;mercury height difference is233 mm. Find confined gaspressure, in atm.

X atm

96.5 kPa

233 mm HgB

96.5 kPa

S

Hg mm 760atm 1

SMALL + HEIGHT = BIG

kPa 101.3atm 1

0.953 atm 0.307 atm+ = 1.26 atm

233 mm Hg+ X atm =

= 22.4 L

The Ideal Gas Law P V = n R T

P = pres. (in kPa) V = vol. (in L or dm3)

T = temp. (in K) n = # of moles of gas (mol)

R = universal gas constant = 8.314 L-kPa/mol-K

32 g oxygen at 0oC is under 101.3 kPa of pressure.Find sample’s volume.

P V = n R TT = 0oC + 273 = 273 K

3.101(273) (8.314) mol 1

2O g 32 n

2

2

O g 32O mol 1 mol 1.0 P P

PT R n V

54.0 kPa

0.25 g carbon dioxide fills a 350 mLcontainer at 127oC. Find pressurein mm Hg.

= 54.0

P V = n R T

T = 127oC + 273 = 400 K

35.0(400) (8.314) 0.00568

2CO g 0.25 n

2

2

CO g 44CO mol 1 mol 0.00568

V V

VT R n P

V = 0.350 L

kPa

kPa 101.3Hg mm 760 = 405 mm Hg

P, V, T Relationships

At constant P, as gas T , its V ___ .

At constant P, as gas T , its V ___ .

balloon placed in liquid nitrogen(T decreases from 20oC to –200oC)

P, V, T Relationships (cont.)

At constant V, as gas T , its P ___ .

At constant V, as gas T , its P ___ .

blown-out truck tire

P, V, T Relationships (cont.)

At constant T, as P on gas , its V ___ .

At constant T, as P on gas , its V ___ .

Gases behave a bit like jacks-in-the-box (or little-brothers-in-the-box)

P1V1 = P2V2

The Combined Gas Law

P = pres. (any unit) 1 = initial conditionsV = vol. (any unit) 2 = final conditionsT = temp. (K)

A gas has vol. 4.2 L at 110 kPa. If temp. is constant,find pres. of gas when vol. changes to 11.3 L.

2

22

1

11

T VP

T VP

2

22

1

11

T VP

T VP

110(4.2) = P2(11.3)

P2 = 40.9 kPa

11.311.3

Original temp. and vol. of gas are 150oCand 300 dm3. Final vol. is 100 dm3.Find final temp. in oC, assumingconstant pressure.

2

22

1

11

T VP

T VP

2

2

1

1

TV

TV

423 K

2T100

423300

300(T2) = 423(100)

T2 = 141 K

300300

T2 = –132oC

A sample of methane occupies 126 cm3 at –75oCand 985 mm Hg. Find its vol. at STP.

2

22

1

11

T VP

T VP

198 K

198

(126) 985273

)(V 760 2

985(126)(273) = 198(760)(V2)198(760) 198(760)

V2 = 225 cm3

Researchers at U of AK, Fairbanks, saymethane has surfaced in the Arctic due toglobal warming. This methane couldaccount for up to 87% of the observedspike in atmospheric methane.

Density of Gases

Density formula for any substance:

For a sample of gas, mass is constant, but pres.and/or temp. changes cause gas’s vol. to change.Thus, its density will change, too.

ORIG. VOL.

VmD

NEW VOL. ORIG. VOL. NEW VOL.

If V (due to P or T ),

then… D

If V (due to P or T ),

then… D

Density of GasesEquation: 22

2

11

1

D TP

D TP ** As always,

T’s must be in K.

A sample of gas has density 0.0021 g/cm3 at –18oCand 812 mm Hg. Find density at 113oCand 548 mm Hg.

812(386)(D2) = 255(0.0021)(548)

22

2

11

1

D TP

D TP

255 K386 K

812(0.0021)

=255

548(D2)386

D2 = 9.4 x 10–4 g/cm3

(386)812 (386)812

A gas has density 0.87 g/L at 30oC and 131.2 kPa.Find density at STP. 303 K

131.2(273)(D2) = 303(0.87)(101.3)

22

2

11

1

D TP

D TP 131.2

(0.87)=

303101.3

(D2)273

D2 = 0.75 g/cm3

(273)131.2 (273)131.2

Find density of argon at STP.

Vm D

L 22.4g 39.9

Lg 1.78

Find density of nitrogen dioxideat 75oC and 0.805 atm.

348 K NO2

D of NO2 @ STP… Vm D

L 22.4g 46

Lg 2.05

1(348)(D2) = 273(2.05)(0.805)

22

2

11

1

D TP

D TP 1

(2.05)=

2730.805

(D2)348

D2 = 1.29 g/L

(348)1 (348)1

NO2 participates in reactionsthat result in smog (mostly O3)

A gas has mass 154 g and density 1.25 g/L at 53oCand 0.85 atm. What vol. doessample occupy at STP?

326 K

Find D @ STP…

0.85(273)(D2) = 326(1.25)(1)

22

2

11

1

D TP

D TP 0.85

(1.25)=

3261

(D2)273

D2 = 1.756 g/L(273)0.85 (273)0.85

Find vol. when gas has that density.

22 V

m D 2

2 Dm V = 87.7 L

g/L 1.756g 154

Dalton’s Law of Partial Pressure

In a gaseous mixture, a gas’spartial pressure is the onethe gas would exert if it wereby itself in the container.

The mole ratio in a mixture ofgases determines each gas’spartial pressure. John Dalton (1766–1844)

Since air is ~80% N2, (i.e., 8 out of every 10 air-gas moles isa mole of N2), then the partial pressure of N2 accounts for ~80%of the total air pressure.

At sea level, where P ~100 kPa, N2 accounts for ~80 kPa.

Total pressure of mixture (3.0 mol He and 4.0 mol Ne)is 97.4 kPa. Find partial pressure of each gas.

kPa 97.4 gas mol 7He mol 3 PHe

= 41.7 kPa

kPa 97.4 gas mol 7Ne mol 4 PNe

= 55.7 kPa

Dalton’s Law: the total pressure exerted by a mixture of gases is the sum of all the partial pressures

PZ = PA,Z + PB,Z + …

80.0 g each of He, Ne, and Ar are in a container.The total pressure is 780 mm Hg. Find each gas’spartial pressure.

g 4mol 180 g He = 20 mol He

g 20mol 180 g Ne = 4 mol Ne

g 40mol 180 g Ar = 2 mol Ar

Total:26 mol

PHe = 20/26

of total

PNe = 4/26

of total

PAr = 2/26

of total

PHe = 600 mm HgPNe = 120 mm HgPAr = 60 mm Hg

Total pressure is 780 mm Hg

Two 1.0 L containers, A and B, contain gases under2.0 and 4.0 atm, respectively. Both gases are forcedinto Container Z (w/vol. 2.0 L). Find total pres. ofmixture in Z.

A B Z

PX VX VZ PX,Z

AB

2.0 atm4.0 atm

1.0 L1.0 L

2.0 L1.0 atm

2.0 atm

=

PRESSURESIN ORIG.

CONTAINERS

VOLUMESOF ORIG.

CONTAINERS

VOLUMEOF FINAL

CONTAINER

PARTIALPRESSURES IN

FINAL CONTAINER

TOTAL PRESSURE IN LAST CONTAINER 3.0 atm

1.0 L2.0 atm

1.0 L4.0 atm

2.0 L

Find total pressure of mixture in Z.

PX VX VZ PX,Z

ABC

A B ZC 1.3 L 2.6 L 3.8 L 2.3 L3.2 atm 1.4 atm 2.7 atm X atm

3.2 atm 1.3 L2.3 L

1.81 atm1.4 atm 2.6 L 1.58 atm2.7 atm 3.8 L 4.46 atm

7.85 atm

=

Gas Stoichiometry Find vol. hydrogen gas made when 38.2 g zinc reactw/excess hydrochloric acid. Pres.=107.3 kPa;temp.= 88oC.

Zn (s) + 2 HCl (aq) ZnCl2 (aq) + H2 (g)38.2 g Zn excess V = X L H2

Zn g 65.4Zn mol 1

Zn mol 1H mol 1 238.2 g Zn = 0.584 mol H2

P = 107.3 kPaT = 88oC 361 K

Not at STP!

P V = n R T = 16.3 L3.107

(361) (8.314) 0.584 P

T R n V

Zn H2

What mass solid magnesium is req’d to react w/250 mLcarbon dioxide at 1.5 atm and 77oC to produce solidmagnesium oxide and solid carbon?

2 Mg (s) + CO2 (g) 2 MgO (s) + C (s) X g Mg V = 250 mL

P = 1.5 atm T = 77oC

T R VP n

0.013 mol CO2

P V = n R T

)350(314.8(0.25) 151.95

2CO mol 1Mg mol 2

= 0.63 g Mg

Mg mol 1Mg g 24.3

= 0.013 mol CO2

0.25 L 151.95 kPa350 K CO2 Mg

Vapor Pressure

-- a measure of the tendency for liquid particles to enter gas phase at a given temp.

-- a measure of “stickiness” of liquid particles to each other

more“sticky”

less likely tovaporize

In general:LOW v.p.

not very“sticky”

more likely tovaporize

In general:HIGH v.p.

0 20 40 60 80 1000

20

40

60

80

100

NOT all liquids have same v.p. at same temp.

TEMPERATURE (oC)

PR

ES

SU

RE

(kP

a)CHLOROFORM

ETHANOL

WATER

____________ substances evaporate easily(have high v.p.’s).

BOILING

Volatile

vapor pressure = confining pressure (usually fromatmosphere)

At sea level and 20oC…

WATER

AIR PRESSURE(~100 kPa)

ETHANOL

VAPORPRES.

(~5 kPa)

V. P.(~10 kPa)

NETPRESSURE

(~95 kPa)

NETPRESSURE

(~90 kPa)

h

h

TP V

h

h