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Multimedia Chemistry I & II (1996-9-11) [English].img
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chapter6.6c
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à 6.6cèDalën's Law ç Partial Pressures
äèPlease fïd ê pressure or moles ç gas ï ê followïg gaseous mixtures usïg
Dalën's law ç partial pressures.
âèFïd ê ëtal pressure ç a mixture ç 10.0 g HCl å 90.0 g He
ï a 20.0 L contaïer at 27°C?èFrom Dalën's Law, ê ëtal pressure is
ê sum ç ê partial pressures ç HCl å He.èSïce V å T are const-
ant, P(ëtal) = (mol HCl + mol He)RT/V.
èè P(ëtal) = [(10.0/36.5)mol HCl + (90.0/4.00) mol He]RT/V
èè P(ëtal) = (22.77 mol)(0.08206 L·atm/K·mol)(300. K)/20.0 L
èè P(ëtal) =è28.0 atm
éS1 One ç ê features ç an ideal gas is that ê molecules ç ê
gas do not attract each oêr å, consequently, act completely ïdepend-
ently ç each oêr.èIn a gaseous mixture, each ç ê components ç ê
mixture contributes ë ê ëtal pressure ç ê mixture.èThe partial
pressure ç a gas is ê pressure that ê gas would exert if it occupied
ê contaïer by itself.èThe ëtal pressure ç a gaseous mixture equals
ê sum ç ê partial pressures ç ê components ç ê mixture.èThe
previous statement is Dalën's law ç partial pressures.
What is ê ëtal pressure ç a mixture ç 0.030 mol N╖ å 0.045 mol Cl╖,
at 40.0°C ï a 5.00 L vessel?èThe ëtal pressure equals ê sum ç ê
partial pressures ç N╖, å Cl╖.
P(ëtal) = P(N╖) + P(Cl╖),
where P(N╖),å P(Cl╖) are ê partial pressures ç N╖, å Cl╖, respec-
tively.èWe can fïd ê partial pressures from ê ideal gas law.
P(N╖)è= (mol N╖)RT/V
èèèèP(Cl╖) = (mol Cl╖)RT/V
In this example, ê temperature å ê volume are ê same.èWe can add
êse equations å facër out RT/V.
P(ëtal) = (mol N╖ + moL Cl╖)RT/V.
P(ëtal) = (0.030 mol + 0.045 mol)(0.08206 L·atm/K·mol)(313.2 K)/(5.00 L)
P(ëtal) = 0.39 atm
Frequently ï laboraëry experiments, we collect gases over water.èThe
gas displaces water from a flask such that ê gas is trapped withï ê
flask by ê water remaïïg ï ê flask.èThe pressure ïside ê flask
@fig6601.bmp,5,180,124,80
èèequals ê atmospheric pressure, because ê water
èèoutside ê flask is acted on by ê atmosphere.èThe
èègas pressure ïside ê flask also equals ê sum ç
èèê partial pressures ç ê gases ï ê flask.èOne
èèç ê gases ï ê flask is water vapor.èThe pres-
sure ç ê water vapor is determïed by ê temperature ç ê water.è
Let's consider a typical problem.èIf hydrogen displaced 247 mL ç water
at 23°C å ê atmospheric pressure was 758.4 ërr, what is ê pressure
ç hydrogen ï ê flask?èHow many moles ç hydrogen was collected?
Usïg Dalën's law ç partial pressures, we can write
P(ëtal) = P(H╖) + P(H╖O).
We can fïd ê vapor pressure ç water ï a table that gives ê vapor
pressure as a function ç ê temperature.èThe vapor pressure ç water
is 21.1 ërr at 23°C.èInsertïg ê appropriate values ïë ê above
equation produces
758.4 ërr = P(H╖) + 21.1 ërr
Fïally we obtaï ê partial pressure ç hydrogen ï ê flask,
P(H╖) = 758.4 - 21.1 = 737.3 ërr
This partial pressure ç hydrogen is also called ê pressure ç dry
hydrogen, because this is ê pressure that ê hydrogen would exert if
no water was present.èHavïg found ê partial pressure ç hydrogen, we
can use this pressure ï ê ideal gas law ë fïd ê moles ç hydrogen.
èèPVèèè (737.3 ërr)(0.247 L)
n = ──.èn = ────────────────────────────────────────── = 0.00987 mol H╖
èèRTèèè (760 ërr/atm)(0.08206 L·atm/K·mol)(296 K)
1èA cylïder contaïs N╖, Ar, CH╣, å C╖H╗ at a ëtal pressure
ç 825 ërr.èThe partial pressures ç N╖, Ar, å C╖H╗ are 500. ërr,
120 ërr, å 58 ërr, respectively.èWhat is ê partial pressure ç ê
CH╣?
A) 62 ërr B) 325 ërr C) 678 ërr D) 147 ërr
üèThe ëtal pressure is ê sum ç ê partial pressures.
P(ëtal) = P(N╖) + P(Ar) + P(CH╣) + P(C╖H╗)
Substitutïg ê appropriate pressures ïë ê equation, we obtaï
825 ërr = 500 ërr + 120 ërr + P(CH╣) + 58 ërr.
èèèèP(CH╣) = 825 - 500 - 120 - 58 = 147 ërr
The partial pressure ç methane, CH╣, is 147 ërr.
Ç D
2èWhat is ê ëtal pressure ï a 2.00 L contaïer at 35°C that
contaïs 0.200 g C╖H╗, ethane, å 0.800 g C╕H╜, propane?
R = 0.08206 L·atm/K·mol
A) 12.7 atm B) 0.313 atm
C) 0.0840 atm D) 0.0711 atm
üèThe ëtal pressure is ê sum ç ê partial pressure ç ethane
å propane.
P(ëtal) = P(C╖H╗) + P(C╕H╜)
Usïg ê ideal gas law, ê ëtal pressure is
P(ëtal) = (mol C╖H╗ + mol C╕H╜)RT/V.
The number ç moles ç each gas is found from êir masses.
? mol C╖H╗ = 0.200 g/30.07 g/mol = 6.65x10úÄ mol
? mol C╕H╜ = 0.800 g/44.09 g/mol = 0.0181 mol
P(ëtal) = (0.00665 mol + 0.0181 mol)(0.08206 L·atm/K·mol)(308 K)/2.00 L
P(ëtal) = 0.313 atm
Ç B
3èA 2.00 L flask contaïs nitric oxide å ethane at a ëtal
pressure ç 0.688 atm at 25°C.èIf ê flask contaïs 0.0429 mol ethane,
what is ê partial pressure ç nitric oxide ï ê flask?
R = 0.08206 L·atm/K·mol
A) 0.600 atm B) 0.163 atm C) 0.362 atm D) 0.644 atm
üèThe ëtal pressure is ê sum ç ê partial pressures ç nitric
oxide å ethane.èèP(ëtal) = P(NO) +èP(C╖H╗).èThe partial pressure
ç ethane can be calculated from ê given ïformation.
P(C╖H╗) = (0.0429 mol)(0.08206 L·atm/K·mol)(298 K)/2.00L
P(C╖H╗) = 0.525 atm
Now we can fïd ê partial pressure ç nitric oxide.
P(ëtal) = P(NO) +èP(C╖H╗)
0.688 atm = P(NO) + 0.525 atm
Rearrangïg,èè P(NO) = 0.688 atm - 0.525 atm = 0.163 atm
Ç B
4èA mixture is composed ç 2.40 g CO╖ å 5.06 g CH╕Cl.èWhat
would be ê ëtal pressure ç ê mixture ï a 5.00 L contaïer at 65°C?
R = 0.08206 L·atm/K·mol
A) 7.96 atm B) 0.0303 atm
C) 0.859 atm D) 1.94 atm
üèThe ëtal pressure is ê sum ç ê partial pressures ç CO╖
å CH╕Cl.
P(ëtal) = P(CO╖) + P(CH╕Cl)
Usïg ê ideal gas law, ê ëtal pressure is
P(ëtal) = (mol CO╖ + mol CH╕Cl)RT/V.
The number ç moles ç each gas is found from êir masses.
? mol CO╖è = 2.40 g/44.01 g/mol = 0.054533 mol
? mol CH╕Cl = 5.06 g/50.48 g/mol = 0.100238 mol
We will round-çf ï ê fïal calculation.
èèèèèèè(0.054533 mol + 0.100238 mol)(0.08206 L·atm)(338 K)
è P(ëtal) = ───────────────────────────────────────────────────
èèèèèèèè (5.00 L)èèèèèèèèèèèèè(K·mol)
è P(ëtal) = 0.859 atm
Ç C
5èA 2.00 L bulb contaïïg 600. ërr ç He is connected via a
valve ë a 3.00 L bulb contaïïg 440. ërr ç O╖.èWhat will be ê
ëtal pressure after ê valve is opened å ê gases completely mix?
Assume ê temperature remaïs constant.
A) 504 ërr B) 252 ërr C) 777 ërr D) 1040 ërr
üèThe ëtal pressure will be ê sum ç ê partial pressures ç
ê helium å ê oxygen.èSïce ê gases act ïdependently, each gas
will follow Boyle's Law ïdependently ç ê oêr gas.
èV╢ 2.00 L
P╖ = P╢ x ──.èèP╖(He) = 600 ërr x ─────────────── = 240 ërr
èV╖ èè (2.00 + 3.00) L
3.00L
P╖(O╖) = 440 ërr x ─────────────── = 264 ërr
èè (2.00 + 3.00) L
We added ê volumes because ê gases will occupy both bulbs.
From Dalën's law ç partial pressures,
P(ëtal) =èP╖(He) + P╖(O╖)
P(ëtal) = 240 ërr + 264 ërr = 504 ërr
Ç A
6èA 400. mL bulb contaïïg 375 ërr ç NO╖ is connected via a
valve ë a 150. mL bulb contaïïg 750. ërr ç Ar.èWhat will be ê
ëtal pressure after ê valve is opened å ê gases completely mix?
Assume ê temperature remaïs constant.
A) 563 ërr B) 1125 ërr C) 1281 ërr D) 477 ërr
üèThe ëtal pressure will be ê sum ç ê partial pressures ç
ê NO╖ å ê Ar.èSïce ê gases act ïdependently, each gas will
follow Boyle's Law ïdependently ç ê oêr gas.
èV╢ è400. mL
P╖ = P╢ x ──.èèP╖(NO╖) = 375 ërr x ──────────────── = 272.7 ërr
èV╖ èèè(400. + 150.) mL
è150. mL
P╖(Ar)è= 750 ërr x ──────────────── = 204.5 ërr
èèè(400. + 150.) mL
We added ê volumes because ê gases will occupy both bulbs.
From Dalën's law ç partial pressures,
P(ëtal) =èP╖(NO╖) + P╖(Ar)
P(ëtal) = 272.7 ërr + 204.5 ërr = 477.2 ërr
Roundïg-çf ë ê correct number ç significant figures,
P(ëtal) = 477 ërr
Ç D
7èDecomposition ç a compound produced 422 mL ç N╖ which was
collected over water at 22°C.èThe atmospheric pressure was 764.1 ërr.
How many moles ç nitrogen were produced?èThe vapor pressure ç water at
22°C is 19.8 ërr.
A) 0.0175 mol B) 0.235 mol C) 0.0171 mol D) 0.229 mol
ü To fïd ê number ç moles ç N╖, we need ê partial pressure ç
nitrogen.èFrom Dalën's law, P(N╖) + P(H╖O) = 764.1 ërr.
èèèP(N╖) = 764.1 ërr - P(H╖O).
èèèP(N╖) = 764.1 - 19.8 = 744.3 ërr
Usïg ê partial pressure ï ê ideal gas law, we fïd ê number ç
moles.
èèPVèèè (744.3 ërr)(0.422 L)
n = ──.èn = ────────────────────────────────────────── = 0.0171 mol N╖
èèRTèèè (760 ërr/atm)(0.08206 L·atm/K·mol)(295 K)
Ç C
8èDecomposition ç a compound produced 391 mL ç CO╖ which was
collected over water at 26°C.èThe atmospheric pressure was 757.7 ërr.
How many moles ç CO╖ were produced?èThe vapor pressure ç water at 26°C
is 25.2 ërr.
A) 0.116 mol B) 0.0159 mol C) 0.0154 mol D) 0.0388 mol
ü To fïd ê number ç moles ç CO╖, we need ê partial pressure
ç nitrogen.
From Dalën's law, P(CO╖) + P(H╖O) = 757.7 ërr.
èèèèè P(CO╖) = 757.7 ërr - P(H╖O).
è P(CO╖) = 757.7 - 25.2 = 732.5 ërr
Usïg ê partial pressure ï ê ideal gas law, we fïd ê number ç
moles.
èèPVèèè (732.5 ërr)(0.391 L)
n = ──.èn = ────────────────────────────────────────── = 0.0154 mol CO╖
èèRTèèè (760 ërr/atm)(0.08206 L·atm/K·mol)(299 K)
Ç C
9èA mixture ç 0.0214 mol He å 0.0184 mol Cl╖ has a ëtal
pressure ç 538 ërr.èWhat is ê partial pressure ç Cl╖ ï ê
mixture?
A) 249 ërr B) 115 ërr
C) 269 ërr D) 289 ërr
üèAccordïg ë Dalën's law ç partial pressures
P(ëtal)= P(Cl╖) + P(He).
Sïce we know ê numbers ç moles å T å V are fixed, we can write
è P(ëtal) = [n(Cl╖) + n(He)]RT/V, å
èèèèèèèèP(Cl╖)è = n(Cl╖)RT/V.
Dividïg ê second equation by ê first allows us ë elimïate ê
RT/V term.èèè P(Cl╖)èèèè n(Cl╖)
──────── = ──────────────
èèèèèèèèP(ëtal)è n(Cl╖) + n(He)
We can fïd ê partial pressure ç ê chlorïe because we know ê
ëtal pressure å ê number ç moles ç each gas.
P(Cl╖) = P(ëtal) x n(Cl╖)/[n(Cl╖) + n(He)]
P(Cl╖) = 538 ërr x 0.0184/(0.0184 + 0.0214)
P(Cl╖) = 249 ërr
Ç A
10èA mixture ç 0.0153 mol N╖ å 0.0742 mol CH╣ has a ëtal
pressure ç 755 ërr.èWhat is ê partial pressure ç CH╣ ï ê
mixture?
A) 497 ërr B) 156 ërr
C) 366 ërr D) 626 ërr
üèUsïg Dalën's law ç partial pressures, we can write
P(ëtal)= P(N╖) + P(CH╣).
Sïce we know ê numbers ç moles å T å V are fixed,
è P(ëtal) = [n(N╖) + n(CH╣)]RT/V, å
èèèèèèèèP(CH╣)è = n(CH╣)RT/V.
Dividïg ê second equation by ê first allows us ë elimïate ê
RT/V term.èèè P(CH╣)èèèè n(CH╣)
──────── = ──────────────
èèèèèèèèP(ëtal)è n(CH╣) + n(N╖)
We can fïd ê partial pressure ç ê methane, CH╣, because we know ê
ëtal pressure å ê number ç moles ç each gas.
P(CH╣) = P(ëtal) x n(CH╣)/[n(CH╣) + n(N╖)]
P(Cl╖) = 755 ërr x 0.0742/(0.0742 + 0.0153)
P(Cl╖) = 626 ërr
Ç C