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PRESENTACION PROPIEDADES COLIGATIVAS.pdf
PRESENTACION PROPIEDADES COLIGATIVAS.pdf
josealbertobarrerava1
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Sep 21, 2025
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About This Presentation
Propiedades coligativas deduccion ingenieril
Size:
492.03 KB
Language:
es
Added:
Sep 21, 2025
Slides:
22 pages
Slide Content
Slide 1
PROPIEDADES
COLIGATIVAS
Slide 2
Estudio termodinámico de
las disoluciones
Slide 3
Potencial químico en una solución ideal
μ
solución =
μ
vap
… ………….1
μ
vap
= μ
gas
= μºgas + RT ln P
Ley de Raoult
P = X
1
Pº
μ
vap
= μ°
vap
+ RT ln X
1
Pº
μ
vap
= μ°
vap
+ RT ln X
1
+ ln
Pº
Slide 4
A se le resta B
μ
solucion
- μ
liq
= μ°
vap
- μ°
vap
+ RT ln X
1
+ RT ln
Pº
+ ln
Pº
μ
solucion
- μ
liq
= RT ln X
1
Slide 5
μ
solucion
= μ°
liq
+ RT ln X
1
Slide 6
PROPIEDADES COLIGATIVAS
1.Disminución de la presión de vapor
2.Disminución de la Temperatura de Fusión
3.Aumento del punto de Ebullición
4.Presión Osmótica
Slide 7
DISMINUCION DE LA PRESION DE VAPOR
P = X
1
P°
-P = -X
1
P°
P°-P = P° - X
1
P°
∆P = (1 - X
1
)P°
X
1
+ X
2
= 1
X
2
= 1 - X
1
∆P = X
2
P°
Slide 8
DISMINUCION DE LA TEMPERATURA DE FUSION
µ
sólu
(T,P,X) = µ°
sól
(T,P)
µ
Solución
= µ°
Liq
+ R T ln X1
µ°
liq
+ R T ln X = µ°
sól
R T ln X = µ°
sól
- µ°
Líq
R T lnX = - (µ°
Líq
- µ°
sól
)
Si: µ= G entonces:
∆G
fusion
=G
Liq
- G
Sólido
∆G
fusion
= ∆µ
Fus
= µ
Líq
- µ
Sólido
Slide 9
lnX = -∆G
fusion
/RT
(δlnX/δX) = -(1/R) (δ(∆G
Fus
/T)/δT)( δT/δX)
(δ∆G
Fus
/T/δT) = (-∆H
fusion
/T
2
)
(X)
-1
= -(1/R) (∆H
fusion
/T
2
)( δT/δX)
(dX/X) = (∆H
fusion
/R)(dT/T
2
)
∫ dX/X = ∆H
fusion
/R ∫ dT/T
2
lnX = -(∆H
fusion
/R)(1/T
2
– 1/T
1
)
Slide 10
m – Molalidad total
m= m
1
¬+m
2
+m
3
+…+m
z
m
i
= n
i
/nM
Donde:
n
i
- Número de moles del soluto
n - Número de moles del solvente
M – Peso molecular del solvente
X = n/(n + n
1
+ n
2
+…+n
z
)
X = n/(m
1
nM + m
2
nM + m
3
nM + …+ m
z
nM) entre n
Slide 11
X = (n/n)/(n/n + n/n m
1
M + n/n m
2
M + n/n m
3
M …+ n/n m
z
M)
X = 1 / (1 + M(m
1
+ m
2
+ m
3
))
X = 1 / (1 + Mm) solución binaria
lnX = ln (1/(1+Mm))
lnX = ln1 – ln(1 + Mm)
lnX =- ln(1+Mm)
lnX = -Mdm/1+Mm
Slide 12
dX/X= (∆H
Fusion
/R)(dT/T
2
)
(-Mdm)/(1+Mm) = ∆H
Fusion
/R)(dT/T
2
)
Solución ideal
La molalidad tiende a 0 porque esta muy diluida
(-Mdm)/(1 + Mm) = (-∆)H
Fus
/R)(dT/T
2
)
-Mdm = (∆H
Fus
/R)(dT/T
2
)
-dm/dT = ∆H
Fus
/RT
2
M
-dT/dm = RT
2
M/∆H
Fus
Slide 13
Donde:
T° - Temperatura de fusión del solvente puro
M – Peso molecular del solvente puro
∆HFusion – Calor de fusión del solvente puro
RT
2
M/∆H
Fusion
= K
f
K
f
- Constante Crioscópica del solvente
-dT/dm = K
f
-dT = k
f
(dm)
∫ dT = K
f
∫dm
- (T – T°) = K
f
(m)
Slide 14
(T° - T) = K
f
(m)
Ѳf - Disminución del punto de fusión
Ѳf = T° - T
Ѳf = K
f
m
m = W
2
/(WM
2
)
Ѳf = K
f
W
2
/WM
m
2
= K
f
W
2
/ Ѳf W
Slide 15
AUMENTO DE LA TEMPERATURA DE EBULLICIÓN
µ
Solución
(T,P,X) = µ
Vapor
(T,P)
µ
Solución
= µ°
Líq
+ R T lnX
µ°
Líq
+ RTlnX = µ°
Vapor
lnX = (µ°
Vap
- µ°
Líq
)/RT
ln = G°
vap
– G°
Líq
/RT
∆G
eb
= G°
vap
– G°
Líq
lnX= ∆G
eb
/RT
lnX = ∆H
eb
/R (1/T – 1/T°)
Slide 16
(δT/δU) = MRT°
2
/∆H
eb
Donde:
M - Peso molecular del solvente
T - Temperatura de ebullición del solvente
∆H
eb
– Calor de ebullicion del solvente
MRT°
2
/∆H
eb
= Kb
K
b
- Constante ebulloscópica
dT/dM = Kb
dT = K
b
(dM)
∫ dT = K
b
∫dm
(T – T°) = K
b
(m)
Slide 17
(T° - T) = K
b
(m)
Ѳ
b
- Aumento en el punto de ebullición
Ѳ
b
= T° - T
Ѳ
b
= K
b
(m)
Slide 18
PRESION OSMOTICA
µ°
Líq
(T,P) = µ
Sol
(T,P,X)
En el équilibrions
µ°
Líq
(T,P) = µ
solución
(T,P,X,π) …….1
Slide 20
µ°(T,P +π) = µ°(T,P ) + V π ……..4
Sustituimos 4 en 3
µ°(T,P )
i
+ V
i
π+ RT ln X = µ°(T,P )
V
i
π = -RT ln X …………..5
X = fracción mol de solvente
Slide 21
- RTlnX =V π
X= 1 – X
2
lnX = ln (1 – X
2
) aprox. – lnX
2
*Serie de Taylor
-lnX
2
= -X
2
Slide 22
X
2
= n
2
/nT X
2
= n
2
/n
Solvente
- n
Soluto
X
2
= n
2
/n + n
2
aprox. n
2
/n
V= V°/n
V° = Vn
-RT(-X
2
) = V/n(π)
-RT(n
2
/n) = V/n(π)
RTn
2
= V° π
RT n
2
/V° = π si C=n
2
/V°
RTC = π
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