resultados del 6to informe fisica ii

8/16/2019 RESULTADOS DEL 6TO INFORME FISICA II

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UNIVERSIDAD NACIONAL DE INGENIERÍA

FACULTAD DE INGENIERÍA QUÍMICA Y TEXTIL

ÁREA ACADÉMICA DE CIENCIAS BÁSICAS

INFORME N° 06

PRESION DE VAPOR SATURADO

LABORATORIO DE FÍSICA II

FI – 204B

Realizad !"#

• A• B• C

P"$e%" "e%!&%a'le de la !"()*i)a#

• PROFESOR: I.ALTUNA• PROFESOR: R. REYES

Fe)+a de "ealiza)i,& de la !"()*i)a# -./06/0-.

Fe)+a de e&*"e1a de la !"()*i)a# 0-/02/0-.

INDICE

Re%34e& A'%*"a)*


2/8

I&*"d3))i,&-5 F3&da4e&* *e,"i)5 O'e*i7%85 Me*dl19a:5 Re%3l*ad%

.5 Di%)3%i,& de "e%3l*ad%65 C&)l3%i&e% ;i'li1"a$9a

25 A!


3/8

• Cac!a$(# P) e$ (e*e"+'$a(#% '$%*a$*e%.

h Hg(m) 0.06 0.08 0.099 0.2 0.4 0.7 0.9

h H 2O >4? 0.0 0.0 0.0 0.0 0.02 0.02 0.02

P);Pa< 97199.4292529.86

89461.2

92

85191.

98

82425.

56

78423.

08

77088.

92

• De a %'!'e$*e "eac'1$ α = λvaporizacion

R %e cac!# e ca#" a*e$*e (e

)a"'=ac'1$ , R=1.9872cal /molK

vap=¿9.495 Kcal/mol λ¿

• E "ce$*ae (e e""#" #*e$'(# &a"a e λvap, λvapteo=9.702 Kcal /mol

Error=| λvapteo− λvapexp λvapteo |∗100

Error=2.134

.5 Di%)3%i,& de "e%3l*ad%L#% "e%!*a(#% #*e$'(#% e$ e%*a e>&e"'e$c'a ?!e e -!e (e%e5a+#% #*e$e", e%*# %e"e?ea e$ e a# "ce$*ae (e e""#" e$ e c5c!# (e a e$*a&'a (e )a"'=ac'1$ /

" e$(e e$ a?aα

¿


4/8

• Pa"a -!e #% (a*#% (e a "5?'ca c!+&a$ a ec!ac'1$ (e Ca!%'!%Ca&e/"#$, e%

$ece%a"'# -!e (e c1+# "e%!*a(# a "a?'ca" L$P) )% T !$a "ec*a (e &e$('e$*e

$ea*')a.• La &"e%'1$ (e )a" )a"a e$ "eac'1$ ('"ec*a c#$ a *e+&e"a*!"a.

De a "5?'ca %e c#$c!/e -!e a?a ;@< e% '$(e&e$('e$*e (e a *e+&e"a*!"a / a+!*'&'ca"# " a c#$%*a$*e !$')e"%a (e #% a%e% '(eae% ;.972Ca+# < $#%

(a e )a#" (e ca#" )a"'=ac'1$ (e a!a.

III5 ;i'li1"a$9a

25 A!'%*ee$*"e e a(# (e *!# c#$ +e"c!"'# / e a(# (e *!# e$ e +a*"a=.


5/8

Patm= ρ Hg∗g∗h Hg+ ρ H 2O∗g∗h H 2O+ Pvapor

Patm=101300 Pa

ρ Hg=13600 Kg

m3

g=9.8m/ s

h Hg : (e%$')e (e +e"c!"'#

h H 2O :(e%$')e (e a!a

h Hg(m) 0.06 0.08 0.099 0.2 0.4 0.7 0.9

h H 2O >4? 0.0 0.0 0.0 0.0 0.02 0.02 0.02

P);Pa< 97199.42 92529.86

89461.2

92

85191.

98

82425.

56

78423.

08

77088.

92

• C#$ a/!(a (e a ec!ac'1$: P= P0 e

−α

T &a"a e c5c!# (e α *#+a+#%

#a"'*+# $e&e"'a$# a e%*a e>&"e%'1$.

LnP=ln P0−α

T

D#$(e LnP= y , x=1/T / ln P0=b

y=b−αx


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• Se %ae *a+'3$ -!e α = λvaporizacion

R

• Rea'=a+#% e a!%*e (e c!")a% &a"a #*e$e" α

A!%*a$(# a C!")a " E +3*#(# (e M$'+#% C!a("5*'c#%

LnPvi=¿a0n+a

1∑i=1

n

1/Ti

∑i=1

n

¿

0∗¿7+a1∗0.019067

79.516270=a¿

LnPvi∗1/Ti=¿a0∑i=1

n

1/Ti+a1∗∑

i=1

n

(1/Ti )2

∑i=1

n

¿

0.216580=a0∗0.019067+a1∗5.19369∗10−5

Re%#)'e$(# #% %'%*e+a% %e #*'e$e

' 1Ti

LnPvi LnPvi∗1

Ti ( 1

Ti)2

0.0026954 11.4845200 0.0309556 7.26528∗10−6

2 0.0027100 11.4352867 0.0309899 7.34425∗10−6

6 0.0027174 11.4015613 0.0309825 7.38422∗10

−6

4 0.0027248 11.3526626 0.0309337 7.42451∗10

−6

8 0.0027322 11.3196509 0.0309280 7.46514∗10−6

0.0027397 11.2698736 0.0308764 7.5061∗10

−6

7 0.0027473 11.2527148 0.0309141 7.5474∗10−6

∑i=1

n

0.019067 79.516270 0.216580 5.19369∗10−5


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a0=24.374

a1=−4778.1

P#" # *a$*# a ec!ac'1$ %e"'a

LnPv=a1(1

T )+a

0

y=b−αx

LnPvi=−4778.1( 1Ti )+24.374

D#$(e a &e$('e$*e (e a Rec*a e% e )a#" (e −α

α =4778.1 K

25:5 A&(li%i% de E"""• Te1"'ca+e$*e e ca#" a*e$*e (e )a"'=ac'1$ e% :

λvap=9.702 Kcal/mol

• R c#$%*a$*e !$')e"%a:.972ca+#

α = λvaporizacion

R

vap=¿4778.1 K ∗1.9872cal /molK

λ¿

vap=¿9.495 Kcal/mol λ¿

• Cac!a$(# e "ce$*ae (e e""#"


8/8

Error=| λvapteo− λvapexp λvapteo |∗100

Error=|9.702−9.4959.7020 |∗100 Error=2.134

resultados del 6to informe fisica ii

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