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Multimedia Chemistry I & II (1996-9-11) [English].img
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chapter6.4c
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à 6.4cèCombïed Gas Law
äèPlease fïd eiêr ê pressure, volume, or temperature ï ê followïg problems ïvolvïg gases.
âè A cylïder contaïs 49.8 L ç argon at 135 atm at 23°C.èWhat
volume will ê argon occupy at 1.02 atm å 500.°C?èThe combïed gas
law is solved for ê new volume, V╖.
èP╢è T╖èèèèèèèèè 135 atmèè773 Kè ┌────────────┐
V╖ = V╢ x ── x ──.èèV╖ = 49.8 L x ──────── x ───── = │ 1.72x10Å L │
èP╖è T╢èèèèèèèèè 1.02 atmè 296 Kè └────────────┘
éSèNot ëo surprisïgly, ê combïed gas law results from a com-
bïation ç Boyle's, Charles', å Amonëns' gas laws.èThe common facër
ï êse laws is that ê mass ç ê gas remaïs fixed.èThe combïed
gas law is
P╢V╢è P╖V╖
──── = ──── (constant mass)
T╢èè T╖
As with ê oêr gas laws, ê subscript "1" could represent ê ori-
gïal state, å ê subscript "2" would ên represent ê fïal state
ç ê gas.èThere are six variables ï this equation.èIf we know five
ç ê variables, ên we can solve for ê sixth variable.èWe need ë
be sure that ê pressure units å ê volume units are ê same å
that ê temperature is ê absolute temperature.
We can apply ê combïed gas law ë changes ï ê conditions imposed on
ê gas as long as ê mass ç ê gas does not change.èThere are three
possibilities: 1) ê pressure responds ë changes ï volume å/or temp-
erature, 2) ê volume responds ë changes ï pressure å/or temperature,
å 3) ê temperature responds ë changes ï pressure å/or volume.
For ê response ç ê pressure ë volume å temperature changes, we
can rearrange ê combïed gas law ë solve for ê fïal pressure, P╖.
èV╢è T╖
P╖ = P╢ x ── x ──
èV╖è T╢
Viewïg ê equation ï this way shows that ê fïal pressure equals ê
origïal pressure multiplied by volume å temperature correction facërs.
If ê volume ïcreases, V╢/V╖ is less than 1 å ê pressure would drop
ï accordance with Boyle's Law.èIf ê temperature ïcreases, T╖/T╢ is
greater than 1 å ê pressure would ïcrease ï agreement with
Amonëns' Law.
Consider ê followïg change ï conditions ç a gas.èAssume that a
reaction produces 0.800 atm ç CO╖ at 150.°C ï a 200. mL reacër.èWhat
will be ê pressure ç ê CO╖ when it is transferred ë a 0.750 L
sërage bulb at 25°C.èWe use ê combïed gas law ë fïd ê new pres-
sure, because both ê volume å ê temperature are changïg but ê
mass ç ê CO╖ is constant.èWe rearrange ê combïed gas law ë fïd
ê new pressure, which we will label P╖.
èV╢è T╖
P╖ = P╢ x ── x ──
èV╖è T╢
Next, we list ê variables:
P╢ = 0.800 atm P╖ = ? atm
V╢ = 200. mL V╖ = 0.750 L x 1000 mL/1 L = 750. mL
T╢ = 150.+273 = 423 K T╖ = 25+273 = 298 K
We make certaï that ê volume units are ê same å that ê temper-
ature is an absolute temperature, eiêr Kelvï or Rankïe.èSubstitute
ê values for ê variables ï ê equation å calculate ê result.
200. mLè 298 K
P╖ = 0.800 atm x ─────── x ───── = 0.150 atm
750. mLè 423 K
The new pressure ç ê CO╖ will be 0.150 atm.èIn this case, ê volume
ïcreased å ê temperature decreased.èBoth changes cause ê pressure
ë decrease.
If we are ïterested ï fïdïg a new volume as a result ç changes ï
ê pressure å ê temperature, we solve ê combïed gas law for V╖.
èP╢è T╖
V╖ = V╢ x ── x ──
èP╖è T╢
If we want ë fïd a new temperature as a result ç changes ï ê
volume å ê pressure, we solve ê combïed gas law for T╖.
èP╖è V╖
T╖ = T╢ x ── x ──
èP╢è V╢
The general procedure is ë solve ê combïed law for ê desired pres-
sure, volume, or temperature.èInsert ê values ç ê variables ïë
ê equation after makïg certaï that ê units ç ê variables agree
with each oêr.è And one more time, temperatures must be ï absolute
degrees, eiêr Kelvï or Rankïe.
Kelvï: T (K)è= t (°C) + 273.15
Rankïe: T (°R) = t (°F) + 459.67
1èA reaction evolved 228 mL ç H½ at 764.2 ërr å 50.0°C.
Calculate ê pressure ç H╖ when it is sëred ï a 250. mL flask at
20.0°C.
A) 760. ërr B) 632 ërr
C) 279 ërr D) 924 ërr
üèListïg ê variables ï ê combïed gas law, we obtaï
P╢ = 764.2 ërr P╖ = ? ërr
V╢ = 228 mL V╖ = 250. mL
T╢ = 50.0+273.2 = 323.2 K T╖ = 20.0+273.2 = 293.2 K
The volume units agree, å ê temperatures are ï Kelvï.
We want ë fïd ê new pressure, P╖.
èV╢è T╖ èèè228 mLè 293.2 K
P╖ = P╢ x ── x ──.èP╖ = 764.2 ërr x ────── x ─────── = 632 ërrèè
èV╖è T╢ èèè250 mLè 323.2 K
The new pressure is 632 ërr.
Ç B
2èA reaction produced 2.37 atm ç N½ at 220.°C ï a 150. mL
vessel.èWhat will be ê pressure ç nitrogen at 20.°C ï a 1.20 L
flask?
A) 11.3 atm B) 0.176 atm
C) 0.0269 atm D) 1.72 atm
üèListïg ê variables ï ê combïed gas law, we obtaï
P╢ = 2.37 atm P╖ = ? atm
V╢ = 150. mL V╖ = 1.20 L x 1000 mL/1 L = 1200 mL
T╢ = 220.+273 = 493 K T╖ = 20.+273 = 293 K
We converted ê fïal volume ïë mL so that ê volume units would
agree.èThe temperatures were converted ë Kelvï.èWe want ë fïd ê
new pressure, P╖.
èV╢è T╖ èè150. mLè 293 K
P╖ = P╢ x ── x ──.èP╖ = 2.37 atm x ─────── x ───── = 0.176 atm
èV╖è T╢ èè1200 mLè 493 K
The new pressure is 0.176 atm.
Ç B
3èA 50.0 cmÄ sërage bulb contaïs SO╖ at 690. ërr å 25°C.
A valve connects ê sërage bulb ë an empty 20.0 cmÄ flask.èWhat is
ê fïal pressure when ê valve is opened å ê system is heated ë
85°C?
A) 2.07x10Ä ërr B) 1.68x10Ä ërr
C) 332 ërr D) 592 ërr
üèListïg ê variables ï ê combïed gas law, we obtaï
P╢ = 690. ërr P╖ = ? ërr
V╢ = 50.0 cmÄ V╖ = 20.0+50.0 = 70.0 cmÄ
T╢ = 25+273 = 298 K T╖ = 85+273 = 358 K
The fïal volume equals ê sum ç ê two volumes because ê gas will
occupy both contaïers.èThe temperatures must be ï Kelvï.èWe want ë
fïd ê new pressure, P╖.
èV╢è T╖ èè 50.0 cmÄè 358 K
P╖ = P╢ x ── x ──.èP╖ = 690. ërr x ──────── x ───── = 592 ërrèè
èV╖è T╢ èè 70.0 cmÄè 298 K
The new pressure is 592 ërr.
Ç D
4èA balloon contaïs 4.00 L ç He at 1.00 atm å 27°C.èWhat
volume will ê helium occupy at 210. ërr å -40.°C?
A) 11.2 L B) 14.9 L
C) 9.77 L D) 0.858 L
üèListïg ê variables ï ê combïed gas law, we obtaï
P╢ = 1.00 atm x 760 ërr/1 atm = 760. ërr P╖ = 210. ërr
V╢ = 4.00 L V╖ = ? L
T╢ = 27+273 = 300. K T╖ = -40+273 = 233 K
The pressure units differed so we arbitrarily converted ê ïitial pres-
sure ïë ërr ë agree with ê fïal pressure unit.èWe rearrange ê
combïed gas law ë fïd ê new volume, V╖.
èP╢è T╖ è760. ërrè 233 K
V╖ = V╢ x ── x ──.èV╖ = 4.00 L x ───────── x ────── = 11.2 Lè
èP╖è T╢ è210. ërrè 300. K
The new volume is 11.2 L.
Ç A
5èA cylïder with a movable pisën contaïs 8.16 L ç air at
14.7 psi å 21°C.èWhat volume will ê air occupy if ê pressure is
ïcreased ë 50.0 psi å ê temperature is ïcreased ë 60.°C?
A) 2.10 L B) 6.85 L
C) 2.72 L D) 4.06 L
üèListïg ê variables ï ê combïed gas law, we obtaï
P╢ = 14.7 psi P╖ = 50.0 psi
V╢ = 8.16 L V╖ = ? L
T╢ = 21+273 = 294 K T╖ = 60.+273 = 333 K
The pressure units agree, å ê temperatures are ï Kelvï.èSolvïg
ê combïed gas law for ê new volume, we get
èP╢è T╖ è14.7 psiè 333 K
V╖ = V╢ x ── x ──.èV╖ = 8.16 L x ──────── x ───── = 2.72 Lè
èP╖è T╢ è50.0 psiè 294 K
The new volume is 2.72 L.
Ç C
6èA balloon contaïs 250. cu. ft. ç He at 762 ërr å 18°C.
What volume will ê helium occupy at 766 ërr å 32°C?
A) 261 cu. ft. B) 442 cu. ft.
C) 263 cu. ft. D) 691 cu. ft.
üèListïg ê variables ï ê combïed gas law, we obtaï
P╢ = 762 ërr P╖ = 766 ërr
V╢ = 250. cu. ft. V╖ = ? cu. ft.
T╢ = 18+273 = 291 K T╖ = 32+273 = 305 K
The pressure å ê temperature units are ï agreement.èWe rearrange
ê combïed gas law ë fïd ê new volume, V╖.
èP╢è T╖ èè762 ërrè 305 K
V╖ = V╢ x ── x ──.èV╖ = 250. ftÄ x ──────── x ───── = 261 ftÄè
èP╖è T╢ èè766 ërrè 291 K
The new volume is 261 cu. ft.
Ç A
7èA cylïder contaïs 2.00 L ç air at 16.2 atm å 45.0°C.
What volume will ê air occupy at 798 ërr å 35.0°C?
A) 24.0 L B) 0.126 L
C) 0.134 L D) 29.9 L
üèListïg ê variables ï ê combïed gas law, we obtaï
P╢ = 16.2 atm P╖ = 798 ërr x 1 atm/760 ërr = 1.05 atm
V╢ = 2.00 L V╖ = ? L
T╢ = 45.0+273.2 = 318.2 K T╖ = 35.0+273.2 = 308.2 K
The pressure units differed so we arbitrarily converted ê fïal pres-
sure ïë atm ë agree with ê ïitial pressure unit.èRearrangïg ê
combïed gas law ë fïd ê new volume, V╖, we obtaï
èP╢è T╖ è16.2 atmè 308.2 K
V╖ = V╢ x ── x ──.èV╖ = 2.00 L x ──────── x ─────── = 29.9 L
èP╖è T╢ è1.05 atmè 318.2 K
The new volume is 29.9 L.
Ç D
8èA gas has a pressure ç 544 ërr ï a volume ç 2.40 L at
62°C.èAt what temperature ï Celsius will ê pressure be 325 ërr ï a
volume ç 2.05 L?
A) -39°C B) 32°C
C) -102°C D) -57°C
üèListïg ê variables ï ê combïed gas law, we obtaï
P╢ = 544 ërr P╖ = 325 ërr
V╢ = 2.40 L V╖ = 2.05 L
T╢ = 62+273 = 335 K T╖ = ? K
The pressure å volume units agree, so we can rearrange ê gas law ë
fïd T╖.
èP╖è V╖ 325 ërrè 2.05 L
T╖ = T╢ x ── x ──.èT╖ = 335 K x ──────── x ────── = 171 K
èP╢è V╢ 544 ërrè 2.40 L
We calculate ê temperature ï Kelvï, but ê problem asks for ê
temperature ï Celsius.èt(°C) = 171 - 273 = -102°C.èThe new temperature
is -102°C.
Ç C
9èA 2.00 L sample ç SF╗ has a pressure ç 1.13 atm at 10.0°C.
The gas is compressed ë 1.40 L å 3.00 atm.èWhat is ê fïal tempera-
ture ç ê SF╗?
A) 231°C B) 253°C
C) 18.6°C D) 60.9°C
üèListïg ê variables ï ê combïed gas law, we obtaï
P╢ = 1.13 atm P╖ = 3.00 atm
V╢ = 2.00 L V╖ = 1.40 L
T╢ = 10.0+273.2 = 283.2 K T╖ = ? K
The pressure å volume units agree, so we can rearrange ê gas law ë
fïd T╖.
èP╖è V╖ è 3.00 atmè 1.40 L
T╖ = T╢ x ── x ──.èT╖ = 283.2 K x ──────── x ────── = 526 K
èP╢è V╢ è 1.13 atmè 2.00 L
We fïd ê temperature ï Kelvï, but ê problem asks for ê tempera-
ture ï Celsius.èt(°C) = 526 - 273 = 253°C.èThe new temperature
is 253°C.
Ç B
10èA gas has a pressure ç 242 psi ï a volume ç 25.0 ftÄ at
68°F.èIf ê gas expås ë 60.0 ftÄ at a pressure ç 118 psi, what is
ê fïal temperature ç ê gas ï Fahrenheit?
A) -8.8°F B) 80.°F
C) 158°F D) 308°F
üèListïg ê variables ï ê combïed gas law, we obtaï
P╢ = 242 psi P╖ = 118 psi
V╢ = 25.0 ftÄ V╖ = 60.0 ftÄ
T╢ = 68+460 = 528°R T╖ = ? °R
The pressure å volume units agree.èThe temperatures are ï Fahrenheit,
so ê easier absolute temperature scale ë use is ê Rankïe scale.
The Rankïe temperature is obtaïed by addïg 460. ë ê Fahrenheit
temperature.èSolvïg for T╖, we get.
èP╖è V╖ 118 psiè 60.0 ftÄ
T╖ = T╢ x ── x ──.èT╖ = 528°R x ─────── x ──────── = 618°R
èP╢è V╢ 242 psiè 25.0 ftÄ
We calculate ê temperature ï Rankïe, but ê problem asks for ê
temperature ï Fahrenheit.èt(°F) = 618 - 460 = 158°F.èThe new temper-
ature is 158°F.
Ç C