Chapter 10 Sept13
Transcript of Chapter 10 Sept13
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CHAPTER 10
GASES
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CONTENT
10.1Characteristics of Gases
10.2Pressure
10.3The Gas Laws
10.4The !ea"#Gas E$uatio%
10.&'urther A(("icatio%s of The !ea" GasE$uatio%
10.)Gas *i+ture a%! Partia" Pressures10.,-i%etic#*o"ecu"ar Theor
10./*o"ecu"ar Eusio% a%! iusio%
10.Rea" Gases eiatio%s fro5 !ea"
6ehaiour
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Learning Outcomes
• Able to apply the Ideal Gas Law incalculations involving gaseous system
•
Able to calculate partial and totalpressure in a mixture of gases (with orwithout reaction)
• Able to dierentiate eusion and
diusion• Able to apply inetic molecular theory
in problem solving
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The properties of a gas depends upon four variables-
• Pressure (P)
– Is equal to force/unit area
– Measured b a baro!eter – "I unit # $e%tons/!eter 2 # 1 Pascal (Pa)
– 1 standard at!osphere # 1&1'32 Pa
– 1 at! # *& !! +g # *& torr
• ,olu!e (,) of the gas # volu!e of the container
• Te!perature (T) !easured in .elvin
• $u!ber of !oles (n)
Properties of gas
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Introduction to !ressure !ressure "nits
12/1/1
1 Pascal (Pa) = 1 kg/m.s2
1 Atmosphere (atm) = 101,325 Pascals = 101.325 kPa
1 Bar = 100 kPa
1 Atmosphere (atm) = 1.013 bar
1 Atmosphere (atm) = 14.7 ps
1 Atmosphere (atm) = 7!0 mm "g = 7!0 torr
1 torr = 1 mm "g = 133.3 Pa
2 2
2 2 2
# mmass(m)$accelerato% kg$
&orce & kgt sP = = = = =Area A area(# ) m m 's
÷
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*
6o%!s
a #G$G%&$
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6o%!s
a #G$G%&$
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0
E+a5("e 1
Convert 0.378 atm to: a) torr; b) pascal
(1atm = 760 torrs = 101325 Pa)
torr atm
torr atm 2)1
1
)*&3()& =×
Paatm
Paatm 3*1)3
1
1&132(3()& =×
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10.2.2 Pressure of E%c"ose!Gases a%! *a%o5eters
'losedtube manometer
a measures pressure below atmospheric
pressurea dierence in height of mercury level e*uals topressure of enclosed gas
Opentube manometer a measures pressure near atmospheric pressure
a dierence in height of mercury level relates topressure of enclosed gas
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1&
6o%!s
a #G$G%&$
>
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E+a5("e 2
A vessel connecte to an open!en merc"r#manometer $s %$lle &$t' as to a press"re o%
0.835 atm. 'e atmosp'er$c press"re $s 755torr.
a) In which arm of the manometer will thelevel of mercury be higher+
b) ,hat is the height dierence between the- arms of the manometer+
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E+a5("e 2 7A%swer8
a) *$nce t'e atmosp'er$c press"re $s reatert'an t'e enclose as+ t'e level attac'e tot'e as &$ll be '$'er.
b) P as = 0.835 atm × 760 torr = 635 torr
1atm
P as , P' = Patm
635 torr , P' = 755
P' = (755 ! 635 ) torrs
= 120 torr = 120 mm -
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#eal ases
Beha*e as #escrbe# b+ the #eal gas e-ato% %o real gas s
act-all+ #eal
th% a e , #eal gas e-ato% #escrbes most real gases at
room temperat-re a%# press-res o 1 atm or less % real gases, partcles attract each other re#-c%g the press-re
eal gases beha*e more lke #eal gases as press-re approaches
ero.
• n equation relating the !acroscopic variables that
describe so!e tpe of !atter
• The ideal gas la% is an equation of state for gases
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6o"e9 Law
A sample o% as 'as a vol"me o% 5 ml at press"re 52
mm-. /'at &$ll t'e vol"me be $% t'e press"re $s
c'ane to 6 mm- &'$le t'e temperat"re $s eptconstant
,ol/ !l Pressure/!!+g
Initial(1) 2 4inal(2) 5 *
P 1V 1 # P 2V 2
(2!!+g)(!l) # (*!!+g)V 2V # 3* !l
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1*
V T 7a:s.8 7n; P
co%sta%t8 V 0T 1 1 constant
V / 0T / 1 V - T -
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E+a5("e 3
A sample of gas has a volume of 456 ml at a
temperature of -7 8' ,hat will the volume be if
the temperature is changed to /999 8' while the
pressure is ept constant+
.ol0ml 3emp08' 3emp0:
Initial(1) 3* 2& 2 6 231 # 201
4inal (2) 5 1&&& 1&& 6 231 #331
ml T T
V
V
T
V
T
V
*2
1
1
2
2
2
1
1
=×=
=
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1
, ∝ n
a Avogadro’s law: For a gas at constant
temperature and pressure, the volume of a
gas is directly proportional to the number
of moles of the gas.
a Note: This relationship isobeyed closely by gases at
low pressure.
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2&
10.3.3 Ao<a!ros9 Law The=ua%tit#>o"u5e
Re"atio%shi(• Avogadro’s hypothesis E$ua" o"u5e of
<ases at the sa5e te5(erature a%! (ressureco%tai% e$ua" %u5:ers of 5o"ecu"es.
Ar ;- #-
.olume --6L --6L --6L
!ressure /atm /atm /atm
3emperature 99' 99' 99' <ass of gas 4==7g ->9/g -9-g
;o of molec 59- x /9-4 59- x /9-4 59- x
/9
-4
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E+a5("e 4
?uppose we have a /-- L sample containing 97 mol oxygen
gas (O-) at a pressure of / atm and a temperature of -78' If
all this O- were converted to o@one (O4) at the same
temperature and pressure2 what would be the volume of
o@one+4 O-(g) - O4
4 mol of O- gives - mol of O4
97 mol of O- 1 97 mol O- × - mol O4 1 944 mol O4
4 mol O-
Lmol
mol
Ln
n
V V
nV
nV
&033&
&
2122
1
12
2
2
1
1
=×=×=
=
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The !ea"#Gas E$uatio%
$rom IdealGas *uation B PV = nRT
/) ( constant n )
-) (constant n C P)
4) (constant n C V )
6) (constant n C T )
2
22
1
11
T
V P
T
V P =
2
2
1
1
T
V
T
V =
2
2
1
1
T P
T P =
2211 V P V P =
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Co%9t The !ea"#GasE$uatio%
• 3he idealgas e*uation does not alwaysaccurately describe real gases3he measuredvolume2V for given P2 n and T might dier from
the volume calculated from PV 1 nRT • 3he standard conditions for gas behaviour
(where R is calculated based on 9 °' and /atm) are not the same as the standardconditions in thermodynamics (-7°' and /atm)
• Note: Very small volume difference is noticed in
calculation involving real and ideal gas.
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2
3he volume occupied by / mol of ideal gas at ?3!2 --6L is
nown as the molar volume of an ideal gas at ?3!
P 1 /atm n 1 / mol
R 1 99>-/ Latm0mol :
3(:) 1 3(°') D -E4/7
1 9°' D -E4/7
1 -E4/7:
LV
atm
K mol K atm Lmol
V
P
nRT V
22
&1
123/&021&&1
=
××=
=
&etermine volume of / mole gas x at ?3!
Standard temperature and pressure
(?3!)B
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E+a5("e &
,hat volume would be occupied by /99g ofoxygen (O-) at /> °' and /97 ;m-+
n 1 /99 g 1 4/-7 mol 3 1 /> D -E41 -=/ :
4- g0mol F 1 >4/
;m0:mol( ) ( ) ( )
( )3
23
&2&&
1&1&
21/310123
mV
Nm
K K mol Nmmol V
P
nRT
V
=
×=
=
−
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2*
E+a5("e )
A metal cylinder holds 799 L of oxygen at/>7 atm and -/°' ,hat volume will thegas occupy if the pressure is reduced to
/99atm and temperature is maintained at-/°'
4ro! PV # nRT
7hen T ' n are constant P 1V 1 # P 2V 2
V 2 # P 1V 1
P 2
V 2 # (10 at!) (&&8) # 2 8
(1&& at!)
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Gas Density• Remember that the density of a gas is
the mass divided by the volume
• Gas density is usually expressed a g/L
d =m
V
= nM
V
and n =PV
RT
d =PV
RT
M
V=
PM
RT
d
=
PM
RT
The higher the !olar !ass' the
higher the densit
• For a given pressure and temperature, themolar concentration should be the same for any
gas
• Two eual volumes of ! different gases at the
same temperature and pressure will contain the
same " of molecules#
• $t doesn%t matter if the gasses are the same or
different
Molarit (M) #
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E+a5("e ,
,hat is the density of ''l6 vapour at E/6
torr and /-7 o'+
<olar mass of ''l62 < 1 /-9 D 6(477) 1/769 g0mol
Pressure9 *& torr # 1 at!
1 torr 9 1 torr × 1 at! # &3 at!
*& torr
( ) ( )
( ) ( )
L g d density
K K mol atm L
mol g atm
RT
PM d density
/3'
30/&021&
/13&'
=
==
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2
E+a5("e /
3he industrial synthesis of nitric acidinvolves the reaction of nitrogen dioxide gaswith water
4;O- (g) D #-O (l) -#;O4 (a*) D ;O(g)
#ow many moles of nitric acid can beprepared using 679 L of ;O- at a pressure of
799 atm and a temperature of -=7 :+
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3&
E+a5("e / 7A%swer8
V ;O- 1 679L2 T 1 -=7:2 P 1 799 atm
3$:2 (g) 6 +2: (l) 2+$:3 (aq) 6 $:(g)
4 mol of ;O- will produce - mol of #;O4
=-= mol ;O- will produce #
( ) ( )( ) ( )
mol n
K K mol atm L
Latm
RT
PV n
2
2/&021&
&&&
=
==
mol
mol mol
mol
*1
23
2
=
×
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E+ercise 10.3
1. A lare %las $s evac"ate an %o"n to &e$' 13.567. t $st'en %$lle to a press"re o% 735 torr at 310C &$t' a as o%"nno&n molar mass an t'en re&e$'e; $ts mass $s 137.56. 'e%las $s t'en %$lle &$t' &ater an aa$n &e$'te: $ts mass no&
1067.. /'at $s t'e molar mass o% t'e "nno&n as 4"se =P+ ans&er 7.7 mol (t'e ens$t# o% &ater at 310C $s 0.7cm3)
- 3he density of a gas measured at /79 atm and -E8' and
found to be /=7 g0L 'alculate the molar mass of this gas 4"se =P+ ans&er 32.0 mol
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;alton<s 8a% of Partial Pressures
• The total pressure of a
!i=ture of gases is the su!
of the partial pressures of
its co!ponents
• In gas !i=tures each gas
acts independentl of the
other gases present
Mathe!aticall
P Total # P 1 6 P 2 6 P 3 6 >
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3
E+a5("e
<ixture of helium and oxygen are used in scuba diving tansto help prevent the bendsH $or particular dive2 65 L #e at -78' and /9 atm and /- L O- at -7 8' and /9 atm were
pumped into a tan with a volume of 79 L 'alculate thepartial pressure of each gas and the total pressure in the tanat -7 8'
P &e ' (#) atm* V &e ' + L * R ' )#)-!) L#atm/.#mol
T &e' ! 0!12 ' !3-.
P 4! ' (#) atm * V 4! ' ! L * T 4! ' ! 0!12 ' !3-.
n = P!"# ( ) ( )( ) ( )
( ) ( )( ) ( )
mol K K mol atm L
Latmn
mol K K mol atm L
Latmn
O
He
&20/&02&*&
12&1
120/&02&*&
*&1
2==
==
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3*
*o"e 'ractio%s a%! Partia"Pressure
a <ole fraction represented in terms ofpressureB ( )
( ) ( ) ( ) 321
11
1
+++==
RT V P RT V P RT V P
RT V P
n
n X
TOTL
( )
( )( )
TOTL P
P
P P P
P
P P P RT V
RT V P
1
321
1
321
1
=+++
=
+++
=
TOTL
i
TOTL
ii
P P
nn X ==
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3
E+a5("e 10
A -9 L tan containing oxygen at a pressure of /99 !a is
connected to a 9/ L tan containing helium at a pressure of
499 <!a and the gases are allowed to mix ,hat is the nal
pressure assuming that the temperature is held constant+
4ro! P $ $ = P % %
+e C V 1 # &1 8 V 2 # 21 8
P 1 # 3&& MPa P 2 # 5
( )( )( )
!Pa L
L!Pa P 1312
1&3&&&2 ==
!Pa P P P HeOt 23013(2
=+=+=
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30
E+a5("e 10 7A%swer8
7olume of &e increases from )#( L to !#( L
7olume of oxygen increases from !#) L to !#( L
Dalton8 The total pressure is the sum of partial pressure#
4! * V ( ' !#) L V ! ' !#( L
P ( ' ()) 56a P ! ' 9
From P 1V
1= P
2 V
2
( ) ( )( )
!Pa L
L!Pa P 12
&21&&2 ==
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3
E+ercise 2
1 gaseous !i=ture !ade fro! *&& g :2 and && g A+ is placed in a 1& 8
vessel at &&A 7hat is the partial pressure of each gas and %hat is the total
pressure in the vessel5
Note: use Pi # ni( RT /V ) Dns%er9 P:2#&201 at!' PA+#&01 at!'Pt#1122 at!E2 snthetic at!osphere co!posed of 1 !ol percent A:2' 10& !ol percent :2
and 0& !ol percent r
a) Aalculate the partial pressure of :2 in the !i=ture if the total pressure of the
at!osphere is to be torr Dns#13 torrE
b) If this at!osphere is to be held in a 12& 8 space at 2 .' ho% !an !oles
of :2 are needed5 Duse ni#Pi(,/FT)' ans#&0 !olE
.inetic Molecular Theor of Gases
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.inetic-Molecular Theor of Gases• si!ple !odel based on the actions of individual ato!s
– Gases consist of particles in constant !otion
– Pressure derived fro! bo!bard!ent %ith container
– .inetic energ described as B? # H !v2
• Postulates of .inetic Theor
,olu!e of particles is negligible
Particles are in constant !otion
$o inherent attractive or repulsive forces
The average ?inetic energ of a collection of particles is
proportional to the te!perature (.)
&
k k 6 or 6 = c 8
(c s a co%sta%t that s the same or a%+ gas)
<olecular ?peedsB
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<olecular ?peedsB&iusion and usion
3he rootmeans*uare (rms) molecular speed2 u; isa type of average molecular speed2 e*ual to thespeed of a molecule having the average molecularinetic energy
3he s*uare root of the previous e*uation givesB
12/1/11
m
37
9 = :
"nit relationships
F 1 >4/6 gm-0s-0mol•:
3 1 3emperature (:)
<m 1 g0mol (<olar <ass)/ %oule 1 gm-0s-
F 1 >4/6 %0mol•:
(rms) 1 m0s
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2
Co%9t 10./ *o"ecu"arEusio% a%! iusio%a 'onse*uences of the dependence of molecular
speeds on massB
/ EusionB the escape of gas molecules through a tiny hole
into an evacuated space
- !iusionB the spread of one substance throughout aspace or throughout a second substance g molecules of aperfume diuse throughout
a room
EusionB
!iusion
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3
10./.2 iusio% a%! *ea%'ree Path
a &iusion (lie eusion) is faster for light gasmolecules
a <olecular collisions mae diusion more
complicated than eusion
a Average distance of a gas molecule betweencollisions is called "ean #ree path
a 3he higher the density of a gas2 the smaller themean free path 3he more molecules are in agiven volume2 the shorter the average distancetraveled between collisions
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raham;s la o e-so%< The rate of effusion of a gas isinversel proportional to the square root of the !ass of its
particle Felative rates of effusion of t%o gases at the sa!e
te!perature and pressure are given b the inverse ratio ofthe square roots of the !asses of the gas particle9
Fate of effusion for gas 1
Fate of effusion for gas 2
M1 and M2 represents the !olar !asses of the gases
-so%<
;istance traveled b gas 1
;istance traveled b gas 2
= M
M
2
1
=
M
M
2
1
he same raham;s la s e-all+ *al# or e-so% process also.
7hat is the ratio of the average speed of A+ (M7 # 1* g/!ol) !olecules to that
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7hat is the ratio of the average speed of A+ (M7 1* g/!ol) !olecules to that
of ":2 (M7 # * g/!ol) !olecules at 20 .5
F # 031 /(!ol•.) or 031 ?g•!2/s2/!ol•.)
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*
10. Rea" Gases eiatio%sfro5 !ea" 6ehaioura Ideal gas the molecules are assumed to occupy no space
and no attractions for one another
a Feal gas <olecules have nite volumes and they attract oneanother
a At lower pressures (usually below /9 atm)2 the deviation from
ideal behaviour is negligible
C 9t 10 R " G
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Co%9t 10. Rea" Gaseseiatio%s fro5 !ea"
6ehaioura 3emperature determines how eectiveattractive forces between gas moleculesare
a At low temperatureB gases deviate fromidealityB the average inetic energydecreases2 intermolecular attractions
remain constant
a At high temperatureB gas molecules are farapart2 thus the nite volumes of the
molecules predominate
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0
d
F l G
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Feal Gases9
;eviations fro! Idealit
• Feal gases behave ideall at ordinar
te!peratures and pressures
• t lo% te!peratures and high pressures
real gases do not behave ideall
• The reasons for the deviations fro!
idealit are9
1 The !olecules are ver close to one
another' thus their volu!e is
i!portant
2 The !olecular interactions also
beco!e i!portant
van der 7aals' 103-123'Professor of Phsics' !sterda!
$obel PriJe 11&
Feal Gases9
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Feal Gases9
;eviations fro! Idealit
• van der 7aals< equation accounts for the behavior ofreal gases at lo% te!peratures and high pressures
( )P 6n a
,, nb nFT
2
2
− =
The van der 7aals constants Ka< and Kb< ta?e into account t%o things9
1 Ka< accounts for inter!olecular attraction
i 4or nonpolar gases the attractive forces are 8ondon 4orces
ii 4or polar gases the attractive forces are dipole-dipole
attractions or hdrogen bonds
2 Kb< accounts for volu!e of gas !olecules
t large volu!es a and b are relativel s!all and van der 7aal<s
equation reduces to ideal gas la% at high te!peratures and lo%
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B=a!ple 11
a Aalculate the pressure e=erted b 0& g of a!!onia' $+3' in a &&
8 container at 2&& oA using the ideal gas la%
P, # nFTP # nFT/, n # 0&g L 1!ol/1 g T # 2&& 6 23
P # (!ol)(&&02&* 8 at! !ol-1 . -1)(3.)
( 8)
P # 303 at!
bAalculate the pressure e=erted b 0& g of a!!onia' $+3' in a &&
8 container at 2&&o
A using the van der 7aal<s equation The van der7aals constants for a!!onia are9 a # 1 at! 82 !ol-2 b #31=1&-2 8
!ol-1 "olution for part a'
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3
d
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E+a5("e 12
A sample of :'lO4 is partially decomposed2 producing O- gas
that is collected over water 3he volume of gas collected is9-79 l at -58' and E57 torr total pressure
!.:l42;s< !.:l ;s< 0 24! ;g<
a) #ow many moles of O- are collected+
b) #ow many grams of :'lO4 were decomposed+
c) ,hen dry2 what volume would the collected O- gas occupy at
the same temperature and pressure+
!.:l42;s< !.:l ;s< 0 24! ;g<
V total inside ' )#!)L
T total inside ' !=: ;!12#( 0 !< ' !33#(.
P total inside ' 1 torr
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E+a5("e 12 7A%swer8
a< P 4! ' ;1 > !< torr# ' 1+) torr
( ) ( ) ( )( ) ( )
mol n
K K mol atm L
Ltorr atmtorr n
RT
V P n
O
O
O
O
31&1
1(2/&021&
2(&&)*&/1)&
2
2
2
2
−×=
=
=
!.:l42;s< !.:l ;s< 0 24! ;g<
b< ! mols .:l42 ≡ 2 mols of 4!
?olar mass of .:l42 ' (!!# g/mol
gram .:l42
( )
g
molK"lO
gK"lO
molO
molK"lOmolO
011&
1
*122
3
21&1
3
3
2
3
2
3
=
×= −
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*
E+a5("e 12 7A%swer8
c< @se Aoyle%s law8
V (' )#!)L *
P ! ' 1 torr#;assumed as water partial pressure replaced
by 4! <
P ( ' 1+) torr ;from 4! > water vapour<
V ! ' ;assumed dry 4! without water vapour<
2
11
2
P
P V V =
( ) ( )( )
LV
torr
torr LV
22&
*
&2&&
2
2
=
=
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EN of CHAPTER 10