1a. lista resolvida
TRANSCRIPT
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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