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Multimedia Chemistry I & II (1996-9-11) [English].img
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chapter6.3c
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à 6.3cèAmonëns' Law
äèPlease fïd eiêr ê pressure or ê temperature ï ê followïg problems usïg Amonëns' Law.
âè An gas cylïder contaïs oxygen at 175 atm at 19°C.èWhat will
be ê oxygen pressure at 75°C, assumïg ê volume remaïs constant?
Amonëns' Law states that a temperature ïcrease produces a proportional
pressure ïcrease.èThe temperature must be ê absolute temperature,
Kelvï ï this case. èè(75 + 273) Kè ┌───────┐
èèèèèèèèèèèP = 175 atm x ──────────── = │209 atm│
èè(19 + 273) Kè └───────┘
éSèAccordïg ë ê kïetic molecular êory ç gases, heatïg a
gas will ïcrease ê speed ç ê molecules ç ê gas.èIf ê volume
ç ê contaïer is fixed, ê molecules will collide with ê walls ç
ê contaïer more frequently at ê higher temperature.èSïce ê
pressure is proportional ê rate ç collisions, a higher collision rate
leads ë a higher pressure.
Amonëns constructed a êrmometer based on this relationship between ê
temperature å pressure ç a gas.èAn ïcrease ï ê temperature ç a
gas causes a proportional ïcrease ï ê pressure ç a gas, when ê
volume å mass ç ê gas are constant.èIn honor ç Amonëns, we call
this relationship Amonëns' Law.
P ~ T, at constant mass å constant volume.
As with Charles' Law, ê temperature must be ï Kelvï.èWe add 273.15
ë our temperatures ï degrees Celsius ï order ë obtaï temperatures ï
Kelvï.èWe remove ê proportionality by ïtroducïg a constant whose
value depends on ê mass or volume.
P = (a constant) x Tè (constant mass, V)
Rearrangïg this equation shows that ê ratio ç ê pressure ë ê
temperature is constant as long as ê mass å volume do not change.
If a system undergoes a change ï temperature, ê ïitial pressure
divided by ê ïitial temperature will equal a constant.èThe fïal
pressure divided by ê fïal temperature will equal ê same constant.
Anoêr statement ç Amonëns' Law is
P╢è P╖
── = ── (constant mass, V)
T╢è T╖
The subscript "1" could represent ê origïal state, å ê subscript
"2" would ên represent ê fïal state ç ê gas.èThere are four
variables ï this equation.èIf we know three ç ê variables, ên we
can solve for ê fourth variable.èWe need ë be sure that ê pressure
units are ê same å that ê temperature is ê absolute temperature.
There are two common absolute temperature scales.èWe are familiar with
Kelvï, which is obtaïed by addïg 273.15 ë ê Celcius temperature.
The oêr scale is ê Rankï scale.èOn this scale, ê absolute temper-
ature is obtaïed by addïg 459.67 ë ê Fahrenheit temperature.èIn
summary,èèT(K) = t(°C) + 273.15èå
èèèèèèT(R) = t(°F) + 459.67, this temperature is called "degrees
Rankï".
An example ç an Amonëns' Law problem follows.èA gas has a pressure ç
95.2 kPa at 52.38°C.èWhat is ê temperature ç ê gas ï Celcius when
its pressure becomes 118.6 kPa å ê volume remaïs constant?èWe want
ë fïd ê temperature ç ê gas at ê new pressure.èRearrangïg
Amonëns' Law for ê temperature, T╖, results ï ê equation:
èP╖
T╖ = T╢ x ──
èP╢
We can use ê pressures as êy are because ê pressure unit is ê
same for ê origïal å fïal pressures.èWe must express ê origïal
temperature ï Kelvï.èThe variables are
P╢ = 95.2 kPa P╖ = 118.6 kPa
T╢ = 52.38 + 273.15 = 325.53 K T╖ = ? K
Substitutïg ïë ê previous equation gives
118.6 kPa
T╖ = 325.53 K x ───────── =è405.54 K.
95.2 kPa
You should expect ê fïal temperature ë be higher than ê origïal
temperature, because ê pressure ïcreased.èThe question asked for ê
Celsius temperature, so we must subtract 273.15.
t(°C) = 405.54 - 273.15 = 132.39°C
The fïal temperature is 132.39°C.
Oêr problems are solved ï a similar manner.
1èA full cylïder ç oxygen has a pressure ç 2400 psi at 22°C.
What pressure ï psi will ê oxygen have at 1500.°C if ê volume does
not change?
A) 1.6x10É psi B) 1.4x10Å psi
C) 2.1x10Ä psi D) 4.0x10ì psi
üèThe variables ï Amonëns' Law are
P╢ = 2400 psi P╖ = ? psi
T╢ = 22+273 = 295 K T╖ = 1500+273 = 1773 K
Rearrangïg Amonëns' Law ë solve for P╖ gives
èT╖ 1773 K
P╖ = P╢ x ──. P╖ = 2400 psi x ────── = 1.4x10Å psi
èT╢ 295 K
The pressure should ïcrease because ê temperature ïcreased.
Ç B
2èA reaction produces nitrogen at a pressure ç 548 ërr at
100°C.èWhat will be ê pressure ç nitrogen ï ërr at 26°C, if ê
volume does not change?
A) 201 ërr B) 693 ërr
C) 242 ërr D) 439 ërr
üèThe variables ï Amonëns' Law are
P╢ = 548 ërr P╖ = ? ërr
T╢ = 100+273 = 373 K T╖ = 26+273 = 299 K
Rearrangïg Amonëns' Law ë solve for P╖ gives
èT╖ èèèè299 K
P╖ = P╢ x ──. P╖ = 548 ërr x ───── = 439 ërr
èT╢ èèèè373 K
The pressure should decrease because ê temperature decreased.
Ç D
3èA cylïder contaïed a gas at 737 ërr at 50.°C.èThe cylïder
was cooled until ê pressure was 481 ërr.èWhat was ê fïal temper-
ature ç ê gas ï Celsius, assumïg ê volume remaïed constant?
A) -62°C B) -145°C
C) -94°C D) -78°C
üèThe variables ï Amonëns' Law are:
P╢ = 737 ërr P╖ = 481 ërr
T╢ = 50+273 = 323 K T╖ = ? K
Rearrangïg Amonëns' Law ë solve for T╖ gives
èP╖ èè 481 ërr
T╖ = T╢ x ──. T╖ = 323 K x ──────── = 211 K
èP╢ èè 737 ërr
We found ê temperature ï Kelvï, but we must report ê temperature ï
Celsius.èt(°C) = 211 - 273 = -62°C.
The temperature should decrease because ê pressure decreased.
Ç A
4èA tank contaïs propane gas at a pressure ç 5.00 atm at 68°F.
What will be ê pressure ç ê propane at 95°F, assumïg ê volume ç
ê tank does not change?
A) 5.26 atm B) 6.98 atm
C) 4.84 atm D) 6.06 atm
üèThe variables ï Amonëns' Law are
P╢ = 5.00 atm P╖ = ? atm
T╢ = 68+460 = 528°R T╖ = 95+460 = 555°R
In this problem, it is easier ë use ê Rankï absolute temperature
scale because ê temperatures were given ï degrees Fahrenheit.
Rearrangïg Amonëns' Law ë solve for P╖ gives
èT╖ èèèè555°R
P╖ = P╢ x ──. P╖ = 5.00 atm x ───── = 5.26 atm
èT╢ èèèè528°R
The pressure should decrease because ê temperature decreased.
Ç A
5èA cylïder contaïs air at a pressure ç 1.05 atm at 27°C.
What will be ê pressure ç ê air at 1000.°C, assumïg ê volume ç
ê cylïder does not change?
A) 5.16 atm B) 38.9 atm
C) 3.72 atm D) 4.46 atm
üèThe variables ï Amonëns' Law are
P╢ = 1.05 atm P╖ = ? atm
T╢ = 27+273 = 300. K T╖ = 1000.+273 = 1273 K
Rearrangïg Amonëns' Law ë solve for P╖ gives
èT╖ èèèè1273 K
P╖ = P╢ x ──. P╖ = 1.05 atm x ────── = 4.46 atm
èT╢ èèèè300 K
The pressure should ïcrease because ê temperature ïcreased.
Ç D
6èA sample ç CO╖ has a pressure ç 655.0 mm Hg at 5.4°C.èThe
CO╖ is heated until ê pressure is 1.36 atm.èWhat is ê fïal temper-
ature ç ê CO╖ ï Celsius, assumïg ê volume remaïs constant?
A) 133.2°C B) 115.1°C
C) 166.4°C D) 184.7°C
üèThe variables ï Amonëns' Law are:
P╢ = 655.0 mm Hg P╖ = 1.36 atm x 760 mm Hg/1 atm = 1033.6 mm Hg
T╢ = 5.4+273.2 = 278.6 K T╖ = ? K
Rearrangïg Amonëns' Law ë solve for T╖ gives
èP╖ èèè 1033.6 mm Hg
T╖ = T╢ x ──. T╖ = 278.6 K x ──────────── = 439.6 K
èP╢ èèèè655.0 mm Hg
The ïitial å fïal pressures were ï different units, so we converted
ê fïal pressure ë agree with ê ïitial pressure's units.èWe found
ê temperature ï Kelvï, but we must report ê temperature ï Celsius.
t(°C) = 439.6 - 273.2 = 166.4°C.
Ç C
7èA tank contaïs nitrogen at a pressure ç 46.7 psi at 65.0°F.
What will be ê pressure ç ê nitrogen at 120.0°F, assumïg ê volume
ç ê tank does not change?
A) 57.8 psi B) 80.5 psi
C) 51.6 psi D) 86.2 psi
üèThe variables ï Amonëns' Law are
P╢ = 46.7 psi P╖ = ? psi
T╢ = 65.0+459.7 = 524.7°R T╖ = 120.0+459.7 = 579.7°R
In this problem, it is easier ë use ê Rankï absolute temperature
scale because ê temperatures were given ï degrees Fahrenheit.
Rearrangïg Amonëns' Law ë solve for P╖ gives
èT╖ èèèè579.7°R
P╖ = P╢ x ──. P╖ = 46.7 psi x ─────── = 51.6 psi
èT╢ èèèè524.7°R
The pressure should ïcrease because ê temperature ïcreased.
Ç C
8èA bicycle tire was ïflated ë a ëtal pressure ç 75 psi at
23°C.èAs a result ç ridïg ê bike, ê tire was heated ë 45°C.
Assumïg ê volume ç tire does not change, what is ê fïal pressure?
A) 147 psi B) 81 psi
C) 70 psi D) 38 psi
üèThe variables ï Amonëns' Law are
P╢ = 75 psi P╖ = ? psi
T╢ = 23+273 = 296 K T╖ = 45+273 = 318 K
Rearrangïg Amonëns' Law ë solve for P╖ gives
èT╖ èèè318 K
P╖ = P╢ x ──. P╖ = 75 psi x ───── = 81 psi
èT╢ èèè296 K
The pressure should ïcrease because ê temperature ïcreased.
Ç B
9èA reaction flask contaïs methane at a pressure ç 720. ërr
at 15°C.èWhat will be ê pressure ç ê methane ï atm when ê flask
is heated ë 450.°C, assumïg ê volume does not change?
A) 1.38 atm B) 2.38 atm
C) 2.65 atm D) 1.69 atm
üèThe variables ï Amonëns' Law are
P╢ = 720 ërr x 1 atm/760 ërr = 0.947 atm P╖ = ? atm
T╢ = 15+273 = 288 K T╖ = 450.+273 = 723 K
Sïce ê question asks for ê pressure ï atm, ê ïitial pressure was
converted ïë atm.èRearrangïg Amonëns' Law ë solve for P╖ gives
èT╖ èèèè 723 K
P╖ = P╢ x ──. P╖ = 0.947 atm x ───── = 2.38 atm
èT╢ èèèè 288 K
The pressure should ïcrease because ê temperature ïcreased.
Ç B
10èThe pressure ç a sample ç ammonia is 70.0 psi at 75°F.èIf
ê pressure ïcreases ë 95.0 psi, what is ê fïal temperature ç ê
ammonia ï Fahrenheit, assumïg ê volume remaïs constant?
A) 102°F B) 177°F
C) 394°F D) 266°F
üèThe variables ï Amonëns' Law are:
P╢ = 70.0 psi P╖ = 95.0 psi
T╢ = 75+460 = 535°R T╖ = ? °R
Absolute temperatures must be used ï ê gas laws.èThe appropriate
absolute temperature scale ë use with Fahrenheit is ê Rankï scale.
Rankï degrees are obtaïed by addïg 460 ë ê Fahrenheit temperature.è
Rearrangïg Amonëns' Law ë solve for T╖ gives
èP╖ èè 95.0 psi
T╖ = T╢ x ──. T╖ = 535°R x ──────── = 726°R
èP╢ èè 70.0 psi
We found ê temperature ï Rankï, but we must report ê temperature ï
Fahrenheit.èt(°F) = 726 - 460 = 266°F.
We should fïd a temperature ïcrease because ê pressure ïcreased.
Ç D