Chapter 10 Sept13

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1

CHAPTER 10

GASES



2

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



3

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



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



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

÷



*

6o%!s

a #G$G%&$



6o%!s

a #G$G%&$



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()& =×



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



1&

6o%!s

a #G$G%&$

>



11

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+



12

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 -



13

#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





1

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



1*

V T 7a:s.8 7n; P

co%sta%t8 V 0T 1 1 constant

V / 0T / 1 V - T -





10

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

=×=

=



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.



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



21

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

=×=×=

=



22

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 =



23

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.



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



2

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

=

×=

=

−



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!)



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) #



20

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&'

=

==



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 :+



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

=

×



31

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



;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 >





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==

==





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 ==



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

=+=+=



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 ==



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



.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



<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



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



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



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



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•.)



*

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



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



0

d

F l G



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



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%



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'





3

d



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



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

=

×= −



*

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

=

=



EN of CHAPTER 10

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Chapter 10 Sept13

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