Chapter 10 Properties of Gases.ppt

8/19/2019 Chapter 10 Properties of Gases.ppt

1/69

Chapter 10

Properties of Gases

Brady and Senese

5th Edition


2/69

10.1 Familiar properties of gases can be explained at the molecular level 2

ndex

10.1 Familiar properties of gases can be explained at the mo

lecular level10.2 !ressure is a measured property of gases

10." #he gas la$s summari%e experimental observations

10.& 'as volumes are used in solving stoichiometry proble

ms10.5 #he ideal gas la$ relates P, V, T,

and the number of moles of gas, n

10.( n a mixture each gas exerts its o$n partial pressure

10.) Effusion and diffusion in gases leads to 'raham*s la$10.+ #he ,inetic molecular theory explains the gas la$s

10.- eal gases don*t obey the ideal gas la$ perfectly


3/69

10.1 Familiar properties of gases can be explained at the molecular level "

!roperties of 'ases

/ hat is the shape of the air in a balloon 'ases have an indefinite shape

/ hat is the volume of the gas in the balloon #hey have an indefinite volume

/ hy do bubbles rise in liuids 3t room temperature4 air has a density of 1." g6 $hile

$ater has a density of 1.0 gm6

/ hy does a hot air balloon rise 7ot air is less dense than cold air


4/69

10.1 Familiar properties of gases can be explained at the molecular level &

7o$ 8oes a 9olecular 9odel Explain #his

/ 'ases completely fill theircontainers:

'ases are in constant random

motion

/ 'ases have lo$ density and are

easy to compress

'as molecules are very far apart

/ 'ases are easy to expand 'as molecules don;t attract one

another strongly


5/69

10.1 Familiar properties of gases can be explained at the molecular level 5

. ?one of these


6/69

10.2 !ressure is a measured property of gases (

hat s !ressure

/ #he force of the collisions of the gas distributed overthe surface area of the container $alls@ P A forcearea

nits : 1 atmosphere CatmD A )(0 mm 7g A )(0 torr A

1.01"25 105 !ascal C!aD A 1&.) psi A 101" millibar CmbD

/ 9easured $ith a barometer

P A g d h

d A density of the liuid

g A gravitational accelerationh A height of the column supported

/ hy use mercury


7/69

10.2 !ressure is a measured property of gases )

6earning >hec,: !ressure nits

Start: ()5 mm 7g #arget: atm

>onversion factor )(0 mm 7g A 1 atm

>onvert ()5 mm 7g to atm

()5 mm 7g ×1 atm

A 0.+++ atm)(0 mm 7g

÷


8/69

10.2 !ressure is a measured property of gases +


9/69

10.2 !ressure is a measured property of gases -

3tmospheric !ressure

/ s the result of the collisions

of the air in the atmosphere

$ith the obects they contact

/ hy is the pressure less inthe mountains than at sea

level

3ir density is greater at sea

level4 hence there are morecollisions.

p

r

e

s

su

r

e


10/69

10.2 !ressure is a measured property of gases 10

6earning >hec,

nder $ater4 the pressure is increased. hy Because the $eight of the air is added to the

$eight of the $ater4 increasing the force acting onobects

#his is $hy deep sea exploration reuires asubmarine: our bodies cannot handle the extreme

pressures at great depths


11/69


GpenHend 9anometer


12/69


>losedIend 9anometer


13/69

10.2 !ressure is a measured property of gases 1"

. )&" mm 7g

8. 12.2 mm 7g

E. ?one of these +)) mm 7g


14/69

10." #he gas la$s summari%e experimental observations 1&

!roportionality

/ Direct proportionality means that 2 variables are:

Opposite the euality from one another

Gn the same level of the fraction on their respective

sides

/ 8irectly proportional variables follo$ each otherH$hen one increases so does the other

/ i.e. P A F A4 or P 1A F A.

Since P and F are both numerators4 they are directly

proportional.

P and A4 ho$ever4 are inversely proportional


15/69

10." #he gas la$s summari%e experimental observations 15

6earning >hec,

hat happens to gas pressure $hen you raise the

temperature

f the container can expandin response to the force

n a rigid $alled container

?o change in pressure is

observed because the area

increased.

!ressure increases

because the faster moving

molecules hit the $alls ofthe container $ith greater

force

Force of >ollisions

3rea= P


16/69

10." #he gas la$s summari%e experimental observations 1(

6earning >hec,

hat happens to gas pressure $hen you increase the number

of molecules in the container

f a container can expand n a rigid $alled container

?o pressure change

is observed.

!ressure increases

because more molecules

hit the $alls of thecontainer4 thus exert a

greater force on the

container

Force of >ollisions

3rea= P


17/69

10." #he gas la$s summari%e experimental observations 1)

. ?o change is observed.

8. ?ot enough information is ,no$n.


18/69

10." #he gas la$s summari%e experimental observations 1+

Boyle;s 6a$

/ here P is pressure and V isvolume

/ 3ssumes: temperature and the

number of moles of gas are

constant

/ 9ay be used to describe t$o

different conditions

#$o gases in separate containers or 3 sample of gas $hose conditions

change

1 2

1 2

1 orµ =

P P P

V V V


19/69

10." #he gas la$s summari%e experimental observations 1-

>harles; 6a$

/ here V is volume and T is absolute temperature/ 3ssumes: the pressure of gas and the number of

moles of gas are constant

/9ay be used to describe t$o different conditions

#$o gases in separate containers or

3 sample of gas $hose conditions change

1 2

1 2

orµ =V V

V T T T


20/69


3bsolute Jero

#emperature of a gas at

$hich pressure and

volume are %ero

t is not possible to

have a gas $ith a V A 0 9olecular volume

doesn;t change but the

total volume decreases

Extrapolation isnecessary due to

condensation


21/69


deal 'ases

/ #heir behavior is predicted by the gas la$s

/ #here are no ideal gases

/ 7o$ever4 most gases behave ideally under most P and

T conditions/


22/69


'ayH6ussac;s 6a$

/ 'as pressure is directly

proportional to absolute

temperature

/ 3ssumes: the volume and

number of moles are

constant

/ #his is $hy $e don;t heatcanned foods on a campfire

$ithout opening them=

µ P T


23/69

10." #he gas la$s summari%e experimental observations 2"

>ombining #his nformation:

/ Boyle;s 6a$

/ >harles; 6a$

/ 'ayH6ussac;s 6a$

/#hus combining this information

/ 3nd therefore4 for any 2 conditions:

µT

P V

µT P

1 µ P V

µT V

1 1 2 2

1 2

= PV P V

T T


24/69

10." #he gas la$s summari%e experimental observations 2&

onsider the follo$ing: 22.& 6 of 7e at 25 K> areheated to 200. K>. hat is the resulting volume

hich is best suited to solving the problem

3. Boyle;s 6a$

B. >harles; 6a$

>. 'ayH6ussac;s 6a$

8. ?one of these


25/69


. 6

8. ?o specific units as long as they cancel


26/69

sed for calculating the effects of changing conditions or,s if the temperature is in Lelvin4 but P and V can

have any units so long as they cancel

6earning >hec,

f a sample of air occupies 500. m6 at S#!M4 $hat is

the volume at +5 K> and 5(0 torr

10." #he gas la$s summari%e experimental observations 2(

>ombined 'as 6a$

25(0 torr )(0 torr 500. m6

2)".15 L "5+ L

××

=

V

MStandard #emperature C2)".15 LD and !ressure C1 atmD

1 1 2 2

1 2

= PV P V

T T

++- m6


27/69

10." #he gas la$s summari%e experimental observations 2)

6earning >hec,

3 sample of oxygen gas occupies 500.0 m6 at )22torr and I25 K>. >alculate the temperature in K> if

the gas has a volume of 2.5" 6 at &-1 mm 7g.

2

)22 torr 500.0 m6 &-1 torr 25"0 m6

2&+ L

× ×=

T

T 2 A 5+0 K>T 2 A +5" L

1 1 2 2

1 2

= PV P V T T


28/69

10." #he gas la$s summari%e experimental observations 2+

6earning >hec,

3 sample of helium gas occupies 500.0 m6 at 1.21atm. >alculate the volume of the gas if the pressure

is reduced to &-1 torr.

-"( m6

1 1 2 2

1 2

= PV P V T T

1.21 atm 500.0 m6 A 0.(&( atm V 2


29/69

10." #he gas la$s summari%e experimental observations 2-

are heated to 200 K>4$hat is the resulting volume

3. 22.& 6

B. 1)- 6

>. "5.( 6

8. ?ot enough information given


30/69

10.& 'as volumes can be used in solving stoichiometry problems "0

9olar Nolume

/#he volume of one mole of any gas at S#! is 22.& 6 dentity of the gas doesn;t matter

9olar mass of the gas doesn;t matter

/ Corollary: Eual volumes of any gas contain the

same number of particles as long as the # and ! arethe same

/ Gay Lussac’s Law of Combining Volumes: For gas

phase reactions4 $e can use volume ratios in place ofmole ratios in stoichiometry problems


31/69


3vogadro;s !rinciple

/ V is volume and n is moles of gas

/3ssumes: the temperature and pressureremain constant

/ >ontainers of eual volume under the

same conditions contain the same number

of particles

1 2

1 2 orµ =

V V

V n n n


32/69


6earning >hec,

>alculate the volume of ammonia formed by the

reaction of 25 6 of hydrogen $ith excess nitrogen.

?2C g D O "72C g D P 2?7"C g D

"2"

2

2 6 ?725 6 71) 6 ?7

1 " 6 7× =


33/69

10.& 'as volumes can be used in solving stoichiometry problems ""

6earning >hec,

?2C g D O "72C g D P 2?7"C g Df 125 6 72 react $ith 50 ?24 $hat volume of ?7" can

be expected

"2"

2

"2"

2

2 6 ?7125 6 7 +"." 6 ?71 " 6 7

2 6 ?750 6 ?100 6 ?7

1 1 6 ?

× =

× =

72 is limiting reagent +"." 6


34/69

10.& 'as volumes can be used in solving stoichiometry problems "&

6earning >hec,

7o$ many liters of ?2C g D at 1.00 atm and 25.0 K> are

produced by the decomposition of 150. g of ?a?"

2?a?"C sD → 2?aC sD O "?2C g D

" " 2

2"

2

1 2 1 2

21 2 1

2

150. g ?a? 1 mol ?a? " mol ? ".&( mol ?

1 (5.02 g 2 mol ?a?

".&( mol ? 22.& 61 mol at S#!1

@

)).5 6 2-+ L +&.( 6

2)" L

× × =

×

= =×

= =

V V V T

V T T T

V


35/69


$ould be reuired to react 2" 6 of B

3 O 5B O "> P 28

3. "+ 6

B. 1& 6

>. ).2 6

8. ?one of these


36/69

10.5 #he ideal gas la$ relates !4 N4 #4 and the number of moles of gas4 n "(

Bringing t #ogether

3vogadro: n directly proportional to V Boyle: P indirectly proportional to V

>harles: T directly proportional to V

'ayH6ussac: T directly proportional to P

>ombining these variables into one euation results in

the deal 'as 6a$.

R is the constant of proportionality Cthe QidealR or QuniversalR

gas constantD its value is 0.0+205) 6/atmmol/L

PV A nRT


37/69

10.5 #he ideal gas la$ relates !4 N4 #4 and the number of moles of gas4 n ")

deal 'as 6a$

/ sed to describe a sample of gas under one set of

conditions

/ #he units have to be:

P in atm

V in 6 n in mol

T in L

/ R

A 0.0+205) 6/atmmol/L

PV A nRT


38/69

10.5 #he ideal gas la$ relates !4 N4 #4 and the number of moles of gas4 n "+

. hat is the pressure of the gas

3. ".0 atm

B. (0. atm

>. 0.25 atm

8. ?one of these


39/69

10.5 #he ideal gas la$ relates !4 N4 #4 and the number of moles of gas4 n "-

>ase Study

3 hardHboiled egg is placed over the opening of an

Erlenmeyer flas,. hat $ill happen to each gas la$variable $hen the flas, is placed in a tub of liuid

nitrogen


40/69

10.5 #he ideal gas la$ relates !4 N4 #4 and the number of moles of gas4 n &0

'as 8ensity

9olar 9ass× = = g

P RT dRT V

/#he number of moles may be related to both the massC g D of the gas sample and the molar mass of the gas

involved

/ #hus $e may re$rite the deal 'as 6a$ as

/ Further4 since d A mV 4 $e can re$rite the euation in

terms of density

9olar 9ass=

g PV RT


41/69


6earning >hec,

hat is the molar mass of a gas $ith a density of (.) g6

at H)" K> and a pressure of "(.) psi

(.) g 0.0+21 6 atm

2.50 atm 9olar 9ass 200 L 6 mol L

× ×

= × ××

&& gmol A 9olar 9ass

P 9olar mass A d R T


42/69


6earning >hec,

hat is the density of ?G2 at 200 K> and (00. torr

&(.01 g 0.0+21 6 atm0.)+- atm d &)" L

mol mol L

×× = × ×

×

0.-"5 g6

P 9olar mass A d R T


43/69

10.5 #he ideal gas la$ relates !4 N4 #4 and the number of moles of gas4 n &"

6earning >hec,

>alculate the volume of 1.00 mol of gas at S#!

PV A nRT

0.0+21 6 atm1 atm 1.00 mol 2)" L mol L

×× = × ××

V

V A 22.& 6


44/69

10.5 #he ideal gas la$ relates !4 N4 #4 and the number of moles of gas4 n &&


45/69


6earning >hec,

3 sample of fluorine gas occupies 2)5 m6 at -&5torr and )2 K>. hat is the mass of the sample

9olar mass=

g PV RT PV A nRT

mol 0.0+21 6 atm1.2& atm 0.2)5 6 m "&5 L

"+.0 g mol L

×× = × × ×

×

0.&5) g A mass


46/69

10.5 #he ideal gas la$ relates !4 N4 #4 and the number of moles of gas4 n &(

6earning >hec,

8etermine the molecular $eight of a gas if 1.05" g ofthe gas occupies a volume of 1.000 6 at 25 K> and )52

mm 7g C#he 8umas 9ethodD.

9olar mass=

g

PV RT PV A nRT

1.05" g 0.0+21 6 atm0.-+- atm 1.000 6 2-+ L

99 mol L

×× = × ×

×

2(.0 gmol A mass


47/69

10.5 #he ideal gas la$ relates !4 N4 #4 and the number of moles of gas4 n &)

. 0.0-0 gmol

8. ?one of these 11 gmol


48/69

10.( n a mixture each gas exerts its o$n partial pressure &+

8alton;s 6a$

#he partial pressure of a gas is the pressure that thegas $ould exert if it $ere in the container by itself


49/69

10.( n a mixture each gas exerts its o$n partial pressure &-

>ollecting a 'as by ater 8isplacement

>ollected gas pressure must be corrected for water vapor

P total A P gas O P $ater Csee #able 10.2D


50/69

10.( n a mixture each gas exerts its o$n partial pressure 50

P #otal A P 1 O P 2 O P " O .

!ump 520 mm 7g ?2 and 250 mm 7g G2 into anempty gas cylinder. hat is the overall pressure ofthe mixture

6earning >hec,

P t A 520 mm 7g O 250 mm 7g A ))0 mm 7g


51/69


!#otal A !1 O !2 O !" O .

3 "2.5 m6 sample of hydrogen gas is collected over$ater at 25 K> and )55 torr. hat is the pressure of dry

hydrogen gas

6oo, up vapor pressure for $ater: P !ater 25 K> A 2".)(

mm 7g

6earning >hec,

>orrect P t to find the P dry gas:

)55 torr H 2".)( torr A )"1 torr

)"1 torr A P hydrogen


52/69


9ole Fraction4 "

/Each gas molecule contributes afraction of the total pressure

" 3Athe mole fraction of substance Q3R

n3 Athe moles of component Q3R

ntA the total number of moles of gas in

the mixture

/ Application: #he partial pressure

contributed by the component gasQ3R is a fraction of the total pressure

3

3

t

= " n

n

P 3 A " 3 P t


53/69

10.( n a mixture each gas exerts its o$n partial pressure 5"

6earning >hec,

hat is the mole fraction of ?2 in the atmosphere

1.000 atm 3ir A 0.)+0+ atm ?2O 0.20-5 atm G2O

0.00-" atm 3r O 0.000"( atm >G2

= A A

t

" n

n0.)+1 A " nitrogen


54/69

10.( n a mixture each gas exerts its o$n partial pressure 5&

6earning >hec,

n a mixture of gases there are 5.0 g each of ?e4 G2 and 72.

hat is the mole fraction of ?e f the partial pressure of

?e in this mix is 1.0 psi. $hat is the total pressure

?e ?e total= × P " P

nneon A 0.2&+ noxygen A 0.15( nhydrogen A 2.&+0

total

1.0 psi0.0+(1 A

P

?e

0.2&+

2.++= "

t

= A

A

n "

n0.0+(1 A " ?e

P total A 12 psi


55/69


P #V A n

# RT

For a mixture of gases4 the total pressure is the sum

of the contributions of all gases. n a mixturecontaining 5.0 g each of ?e4 G2 and 724 $hat is the

total pressure Cin atmD at 50.0 K> in a 2.5 6 vessel

6earning >hec,

n ?eA 0.2&+ mol noxygenA 0.15( mol nhydrogenA 2.&+ mol

nt A 2.++ P

#V An

# RT

( )

( ) ( )H1 H1

tt

2.++ mol 0.0+21 6 atm mol L "2" L

A2.5 6

× × × × ×=n RT P V

P t A "1 atm


56/69

10.( n a mixture each gas exerts its o$n partial pressure 5(

4 5.0 g each of 7e and of ?e are placed into5.0 6 flas,. hat is the total pressure

3. ).+ atm

B. 1.1 atm

>. 52 atm

8. ?one of these


57/69

10.) Effusion and diffusion in gases leads to 'raham;s 6a$ 5)

8iffusion vs. Effusion

/ hen the partition is

removed4 blue

molecules

diffuse to mix

/ #he molecules

effuse through a

pinhole into avacuum


58/69

10.) Effusion and diffusion in gases leads to 'raham;s 6a$ 5+

'raham;s 6a$ of Effusion

/ elates the velocity Crate at $hich the gas movesthrough a given spaceD to the molecular mass of

the gas.

/ #he greater the molecular mass of the gas4 the

slo$er its velocity.

Effusion rate of B 9olar mass of 3A

Effusion rate of 3 9olar mass of B


59/69

10.) Effusion and diffusion in gases leads to 'raham;s 6a$ 5-

ompared to lighter

atoms at the same temperature4 heavier atoms on

average:

3. 9ove faster

B. 9ove slo$er

>. 9ove at the same average velocity


60/69

10.) Effusion and diffusion in gases leads to 'raham;s 6a$ (0

#hree balloons are filled $ith eual volumes of the

gases: >7&4 724 and 7e. 3fter 5 hours the balloonsloo, li,e the picture.

s this effusion or diffusion

3. 8iffusionB. Effusion

hich is the 7e balloon


61/69


6earning >hec,

f it is observed that Br 2 effuses at a rate of 5 cm sH1

4 if asample of an un,no$n gas travels at half the speed4

$hat is the molecular mass of the un,no$n gas

2

2

2

22

2H1

H1

Effusion rate of Br 9olar mass T

Effusion rate of T 9olar mass Br

Effusion rate of Br 9olar mass Br 9olar mass T

Effusion rate of T

5.0 cm s 15-.+0 g 9olar mass Tmol2.5 cm s

=

× = ÷

× = ÷ ÷

99 A (&0 gmol


62/69


6earning >hec,

3 glass tube is 1.0 m long. 3 sample of ?7" gas is

introduced into one end of the tube at the same time

that 7>l is introduced into the other. here the gases

meet4 they form a ring of crystalline ?7&>l. here

does the ring form inside the tube

?7" 7>l

x A distance traveled by ?7"@

C1 H xD A distance by 7>l

H1

H1"

x time "(.&(1 gmol 7>l

1).0"& gmol ?7C1 xD time

x1.&("

1 x

=−

=−

x A 0.5-& m from the ?7" end


63/69

10.) Effusion and diffusion in gases leads to 'raham;s 6a$ ("

G2

25 K>

1 atm

50 6

7e

25 K>

2 atm

50 6

hich flas, has molecules that

are moving faster

hich flas, has molecules that

have a greater average ,inetic

energy

3. >G2

B. 7e

>. ?either

3. >G2

B. 7e

>. ?either


64/69

10.) Effusion and diffusion in gases leads to 'raham;s 6a$ (&

G2

B. 7e

>. ?either

>G2

25 K>

1 atm

50 6

7e

25 K>

2 atm

50 6


65/69


. 1+.) gmol

8. ?ot enough information given


66/69

10.+ #he ,inetic molecular theory explains the gas la$s ((

Linetic 9olecular #heory Explains 'as

Behaviors

'ases consists of an extremely large number of very

tiny particles that:

3re in constant4 random motion

Gccupy a negligible portion of the total volume of thesampleHtheir individual contribution may be ignored

>ollide elastically $ith themselves and the $alls of the

container

9ove in straight lines bet$een collisions4 neitherattracting nor repelling each other


67/69

10.+ #he ,inetic molecular theory explains the gas la$s ()

Linetic 9olecular #heory H rregularities

/ #he volume of a gas molecule is negligible

?G= nder conditions of extremely high pressure4

gases are closer4 their relative si%e is a factor

/ 'as molecules collide elastically ?G= nder conditions of extremely lo$ temperatures4

gases move more slo$ly and intermolecular attractions

are more significant

2an


68/69

10.- eal gases don;t obey the ideal gas la$ perfectly (+

eal 'ases

/ van der aal;s euationaccounts for deviations from

ideal behavior by removing 2

assumptions:

!article volume is negligible !articles do not interact

/ van der aal;s constants4 a and

b4 are specific to the substance

2C DC D

an P V nb nRT

V + − =


69/69

van der aal;s >onstants

TABLE 10.3

Substance a

C62 atm molU2Db

C6 molU1DSubstance

aC62 atm molU2D

bC6 molU1D

#oble gases $ther Gases

7e 0.0"&21 0.02")0 72 0.02&&& 0.02((1

?e 0.210) 0.01)0- G2 1."(0 0.0"1+"

3r 1."&5 0.0"21- ?2 1."-0 0.0"-1"

Lr 2."1+ 0.0"-)+ >7& 2.25" 0.0&2)+

Te &.1-& 0.05105 >G2 ".5-2 0.0&2()

?7" &.1)0 0.0")0)

72G 5.&(& 0.0"0&-

>27

5G7 12.02 0.0+&0)

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Chapter 10 Properties of Gases.ppt