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CENTRO DE CICIAS EXATAS, AMBIENTAIS E DE TECNOLOGIACURSO: ENGENHARIA AMBIENTAL NOTURNO

DISCIPLINA: POLUIO ATMOSFRICADOCENTE: Prof. Dr. Aro!"#$o %. D" S#!"

RESOLUO DA 1. LISTA DE EXERCCIOS

1(a) Converter 32 molculas por 109 molculas de ar em n molculas por 106

molculas dear (32 ppb para n ppm)

ppmouppmardemolec

ardemolecmolecc 2

9

6

12!332!.1

.1.32

=

=

1(b) Converter 32 ppb em n molculas por centmetro cbico (cm 3)

"a##$ 1% D&'&rmnar $ nm&r$ *& m$+ ,$rrp$n*&n'& a 19(1 b+-$) m$+/,0+a# *& ar

molmolmolec

ardemolecmoln 1

123

9

166!1.12!6

.1)(

=

=

"a##$ 2% D&'&rmnar $ $+0m& $,0pa*$ p$r 1166!1 m$+ *& ar (0'+ar a +& 4&ra+ *$#4a#)

nRTPV = ! $0%

Latm

KKmolatmLmol

P

nRTV 15

111

1!51

29....2!166!1

=

==

C$n&r'&n*$ a 0n*a*& *& $+0m& &m L para ,m3 r+'a% 3111!5 cmV ="a##$ 3% D*r $ nm&r$ *& m$+/,0+a# *$ 47# p&+$ $+0m& *& ar &m ,m3

311

311 .1!81!5

32

=

= cmmolculascm

molec

c

1(c) Converter 32 ppb em n mol de gs por litro de ar

"a##$ 1% D&'&rmnar $ nm&r$ *& m$+ ,$rrp$n*&n'& a 32 m$+/,0+a# *& 47#

molmolmolec

molecn 23

123132!

.12!6

.32

=

=

"a##$ 2% D*r a 0an'*a*& *& m$+ *& 47# p&+$ $+0m& $,0pa*$ p$r 19m$+/,0+a# *& ar!&m +'r$ (L)

19

15

23

.131!11!5

132!

=

= Lmol

L

molM

2) Converter 60 ! 101"molculas por centmetro cbico em ppm e mol#$ %1

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a) "ara ppm

"a##$ 1% Ca+,0+ar a 0an'*a*& *& m$+ *& ar 0& $,0pa 1,m3

nRTPV = ! $0!RT

PVn =

molKkmolatmL

cmLcmatmn

11

333

19!529....2!

.1.1.1

==

"a##$ 2% D&'&rmnar $ nm&r$ *& m$+/,0+a# ,$n'*a# &m 5!9 : 1;m$+ *& armolculasmolmolecmolnmolec

19123156!2.12!619!5 ==

"a##$ 3 D&'&rmnar a ,$n,&n'ra

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I'&m a% C$n&r'&r 5 ppb *& O3&m m$+/,0+a#.,m;3

"a##$ 1% D&'&rmnar $ $+0m& $,0pa*$ p$r 1 b+-$ *& m$+/,0+a# *& ar.

P

nRTV =

molmolmolec

molculasn 1

123

9

166!112!6

1

=

=

1113315111

13!5.113!59!

3...2!166!1cmLcmL

atm

KKmolLatmmolV

==

=

"a##$ 2% D*r $ nm&r$ *& m$+/,0+a# *& O3 (5) p&+$ $+0m& *& ar (,m3) 0&$,0pam.

311

311.13!9

13!5

5

=

= cmmoleccm

molecc

I'&m b% C$n&r'&r 5 ppb *& O3&m 4O3.m;3ar"a##$1% D&'&rmnar $ nm&r$ *& m$+ ,$rrp$n*&n'& a 5 m$+/,0+a# *& O3

molmolmolec

molecn 23

123

16!6.12!6

5

=

=

"a##$ 2% Ca+,0+ar a ma##a ,$rrp$n*&n'& a mol2316!6 *& O3

3

1123 118!3.516!6 )molmolm ==

"a##$ 3% D*r a ma##a *& O3p&+$ $+0m& *& ar (m3)

arm)LmL

c 3313315

1

.8!83.113!5

118!3

=

=

5) C$n,&n'ra

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molKKmolLatm

Latm

RT

PVn 8!53

29...2!

15!111

3

=

==

"a##$ 2% D&'&rmnar a 0an'*a*& *& m$+ &0a+&n'& a 1=4 *& CO

molmol

n*)

1

163

18!3.2

.11

=

=

"a##$ 3% D&'&rmnar a 0an'*a*& *& m$+ *& CO &m 16 m$+ *& ar

2!8!53

118!3 6=

=

*)n

E:pr#ar a ,$n,&n'ra

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@4 @ra

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B2S% 333363 .1!.1.12!52!

= mmolmcmcmmol

CO2% 323363 .1!2.1.12!56!

= mmolmcmcmmol

CO% 333363 .12!5.1.12!51! = mmolmcmcmmol2% 33363 .6!16.1.12!5399! = mmolmcmcmmol

"a##$ 2% Ca+,0+ar a ma##a &m m,r$4rama! 0'+an*$ $ p$ m$+&,0+ar *& ,a*a,$mp$n&n'& *$ b$47#.

CB5% 31613 .115!5.1.16.9!2

= mmolmmol

B2S% 316133 .19!2.1.35.1!

= mmolmmol

CO2% 3616132 .112!1.1.55.1!2

= mmolmmol

CO% 316133 .119!1.1.2.12!5 = mmolmmol 2% 31613 .16!5.1.2.6!16

= mmolmmol