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5.60 Thermodynamics & Kinetics
Spring 2008
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5.60 Spring 2008
Lecture #23
page 1
Colligative Properties
These are properties of solutions in the dilute limit, where there is a solvent “A” and a solute “B” where nA >> nB. pure
These properties are a direct result of µ Amix (l, T, p) < µ A
(l, T, p)
Use two measures of concentration:
a. Mole Fraction: xB = nB/(nA+nB) ~ nB/nA
b. Molalility: mB = (moles solute)/(kg solvent) = nB/(nAMA)
Where MA is the mass in kg of one mole of solvent.
There are FOUR Colligative Properties:
1.
V
apor pressure lowering:
2. Boiling point elevation:
∆pA = pA − pA* = −xB pA*
∆Tb = Tb − Tb* = Kb mB
Kb =
3. Freezing point depression:
2
∆Hvap
∆Tf = Tf − Tf* = −Kf mB
Kf =
4. Osmotic pressure:
( )
MA R Tb*
π = RTc~
( )
MA R Tf*
2
∆Hf
n
where c~ = B is concentration of solute
V
5.60 Spring 2008
Lecture #23
page 2
Let’s go through these one at a time:
1. Vapor pressure lowering: This is just Raoult’s Law.
pA = xA pA* = (1 − xB )pA*
2.
So
∆pA = pA − pA* = −xB pA* (<0)
Boiling point elevation:
Let’s derive this. Start with
µ A (l, T, p) = µ A (g, T, p)
So,
µ A* (l, T, p) + RT ln xA = µ A (g, T, p)
And
ln xA =
∆Gvap
1
[µ A (g, T, p) − µ A* (l, T, p)] =
RT
RT
But lnxA = ln(1-xB) ~ -xB = -nB/(nA+nB) ~ -nB/nA = -(MnB)/(MnA)
Where M is the total mass of A,
So, lnxA ~ mBMA, where MA is the molar mass of A.
mB =
Putting it all together then,
But we need ∆T in there!
…
(
)
− ∆Hvap
− 1 ⎛⎜ ∂ ∆Gvap T ⎞⎟
⎛ ∂mB ⎞
=
⎜
⎟ =
2
⎟
∂T
M
RT
⎝ ∂T ⎠ p MAR ⎜⎝
A
⎠p
which gives us δT =
MA RT 2
∆Hvap
δmB
− ∆Gvap
MA RT
5.60 Spring 2008
Lecture #23
page 3
If δmB = mB – 0 (mixed – pure) and mB is very small
δT = Tb −
Then
3.
Tb*
MA R(Tb* ) 2
=
∆Hvap
Freezing point depression:
Same arguments as above, replacing 4.
mB = Kb mB
∆Gvap with –∆Gf
∆Hvap with –∆Hf
Tb with Tf
Kb with Kf
Osmotic Pressure:
The pressures at points:
pext
h
α•
β•
A+B
ℓ
β:
p = pext + ℓρg
α:
p + π = pext + ℓρg + h ρg
A
At equilibrium:
µ A (l, p + π, T) = µ A* (l, p, T)
Using Raoult’s Law:
RT ln xA + µ A* (l, p + π, T) − µ A* (l, p, T) = 0
At constant T:
dG = Vdp, or dµ A* = VA* dp
5.60 Spring 2008
Integrating…
Lecture #23
µ A* (l, p
+ π, T) −
µ A* (l, p, T)
page 4
p+π
=
*
*
∫ VA dp = VA π
p
(this assumed an incompressible liquid, where volume does not depend
on p)
So then RT ln xA + VA* π = 0
Again using lnxA ~ -nB/nA
Then RT(-nB/nA) + (VA/nA)π = 0
But VA ~ VA + VB = V
So finally,
(VB<<VA)
πV = RTnB
This is the Van’t Hoff Equation. It looks like the ideal gas law! If we replace c~ = nB/V in the Van’t Hoff Eq., then we get the osmotic pressure relation:
π = RTc~