Unit 9: The Gas Laws
description
Transcript of Unit 9: The Gas Laws
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Unit 9:The Gas Laws
Chemistry
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~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.
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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
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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.
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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.
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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
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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
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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
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** 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
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(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
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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
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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.
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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
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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)
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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
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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?
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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
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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)
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= 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
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Atmospheric pressure changes with altitude:
As altitude , pressure .
barometer: device to measureair pressure
vacuum
airpressure
mercury (Hg)
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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
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roof in hurricane
SLOW
FAST LOW P
HIGH P
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airplane wing / helicopter propeller
AIRPARTICLES
FAST
SLOW
Resulting Forces
(BERNOULLI’SPRINCIPLE)
(GRAVITY)
frisbee
HIGH P
LOW P
FAST, LOW P
SLOW, HIGH P
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CURTAIN
creeping shower curtain
HIGH PSLOWCOLD
LOW PFAST
WARM
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windows andhigh winds
FAST
SLOW windowsburst outwards
TALL BUILDING
LOW P
HIGH P
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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
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= 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
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measures the pressureof a confined gas
manometer:
CONFINEDGAS
AIRPRESSURE
Hg HEIGHTDIFFERENCE
SMALL + HEIGHT = BIG
differentialmanometer
manometers can be filledwith any of various liquids
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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 =
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= 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
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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
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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)
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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
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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)
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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
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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
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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.
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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
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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
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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
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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)
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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
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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.
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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 + …
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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
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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
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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
=
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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
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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
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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.
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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)
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At sea level and 20oC…
WATER
AIR PRESSURE(~100 kPa)
ETHANOL
VAPORPRES.
(~5 kPa)
V. P.(~10 kPa)
NETPRESSURE
(~95 kPa)
NETPRESSURE
(~90 kPa)
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h
h
TP V
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h
h