Key points For today’s lecture; Chapter 9
•Ideal solutions, Set up Phase Diagrams
•Raoult’s Law:
Pi = xi Pi *
• Raoult’s Law for a Solvent of a Dilute
Solution: μ (l ) = μ * (l , P* ) + RT ln x
i
i
( i)
•Raoult’s Law for solute.
•Henry’s Law for a solute.
1
Chemical Equilibrium for reactions involving ideal dilute solutions
(For solvents)
Raoult’s Law:
Pi = xi Pi *
Vapor pressure
above solution of
component i
(For solutes)
Henry’s Law:
Vapor pressure of
pure component i
Mole fraction of
component i in
solution
Pi = ki xi
Vapor pressure
above solution of
component i
Mole fraction of
component i in
solution
Henry’s Law
constant
2
μ
*
H 2O
μ EtOH ( g ) PEtOH ( g )
*
*
μethanol
( g ), Pethanol
*
H 2O
(g) P
μ H O ( g ) PH O ( g )
2
+
=
μ H* O
*
μethanol
2
At equilibrium:
μ
*
H 2O
(l ) = μ
*
H 2O
(g)
μ
(l ) = μ
μethanolxethanol
μ H O xH O
2
At equilibrium:
*
ethanol
2
*
ethanol
2
At equilibrium:
(g)
soln
soln
μ EtOH
(l ) = μ EtOH
(g)
μ HsolnO (l ) = μ HsolnO ( g )
2
2
Focus on one species as a solvent. The chemical potentials of the
gas and liquid phases must be equal at equilibrium.
3
Consider a solvent, initially pure, at a given T there will be a vapor
pressure above the liquid, or a partial pressure (if there is air above the
liquid). The partial pressure is determined by the balance (equality) of
the chemical potential of the liquid and the vapor (or gas).
*
⎛
⎞
P
0
*
0
0
solvent
μi
(l , pure) = μi (l ) = μi ( g , P ) = μi ( g , P ) + RT ln ⎜ 0 ⎟
⎝P ⎠
This defines the partial pressure of the gas above the liquid P*.
If we add a solute to the solvent, this lowers the chemical potential
of the liquid. Therefore the gas chemical potential must lower as
well to keep the two chemical potentials balanced.
⎛ Pi ⎞
0 ⎟
⎝P ⎠
μisolution (l , χ i ) = μi0 (l ) + RT ln ( χ i ) = μi ( g , P) = μi0 ( g , P 0 ) + RT ln ⎜
Subtracting the top equation from the bottom one shows the
differences due to the presence of the solute.
⎛ Pi ⎞
⎛ Pi * ⎞
⎛ Pi ⎞
RT ln ( χ i ) = RT ln ⎜ 0 ⎟ − RT ln ⎜ 0 ⎟ = RT ln ⎜ * ⎟
⎝P ⎠
⎝P ⎠
⎝ Pi ⎠
χi =
Pi
P*
or
Pi = χ i Pi *
This is Raoult’s Law
4
Applying the Ideal Solution Model to Binary Solutions
Assume both solvent and solute obey Raoult’s Law
*
*
PTotal = χ benzene Pbenzene
+ (1 − χ benzene ) Ptoluene
*
Pbenzene = χ benzene Pbenzene
*
Ptoluene = (1 − χ benzene ) Ptoluene
This is a coexistence line for vapor and liquid. At fixed T
but X is varying.
5
Dalton’s Laws and the components
N
PTotal = ∑ Pi and Pi = yi PTotal (Dalton's Laws)
i =1
Pi = χ i Pi *
( Rauolt's Law )
Pi is the vapor pressure of i.
N
PTotal = ∑ χ i Pi *
i =1
χ i Pi* = yi PTotal
χi
PTotal
N
yi
1
yi
= * or, after summing
=∑ *
Pi
PTotal i =1 Pi
The partial pressure of species i is the same from both
Dalton’s and Rauolt’s Laws. “y” are the mole fractions in the
vapor phase, and “x” are the mole fractions in the solution.
Using both of these laws simultaneously implies that the mole
fraction of i in the gas phase is in general not the same as the
mole fraction of the same species in the liquid phase.
6
What is the mole fraction of each component in the gas phase?
P1
x1 P1*
y1 =
= *
Ptotal P2 + ( P1* − P2* ) x1
P1 = x1 P1*
P2 = x2 P2* = (1 − x1 ) P2*
y1 P2*
x1 = *
P1 + ( P2* − P1* ) y1
Ptotal
P1* P2*
= P + ( P − P ) x1 = *
P1 + ( P2* − P1* ) y1
*
2
*
1
*
2
P1* Ptotal − P1* P2*
y1 =
Ptotal ( P1* − P2* )
We can find the composition in the gas phase based on the composition in
the liquid phase. If the pressure of the pure substance is very larger
(compared to the other pure substance) the mole fraction of that species will
dominate in the gas phase.
7
PTotal = χ
*
benzene benzene
P
+ (1 − χ benzene ) P
*
toluene
x1 P1*
y1 = *
P2 + ( P1* − P2* ) x1
The mole fraction of benzene in the vapor, and the total
Pressure along the coexistence line are both functions of the
mole fraction of benzene in the solution in this benzenetoluene solution.
8
P – x phase diagram
xbenzene
ybenzene
xbenzene ( or ybenzene )
9
Example Problem
An ideal solution is made from 5.00 moles of benzene and 3.25
moles of toluene. At 300 K, the vapor pressures of the pure
*
*
substances are Pbenzene
= 103Torr and Ptoluene
= 32.1Torr .
a) The pressure above this solution is reduced from 760 Torr. At
what pressure does the vapor phase first appear?
b) What is the composition of the vapor under these conditions?
Solution
a) The mole fractions of the components in the solution are
xbenzene = 0.606 and
xtoluene = 0.394. The vapor pressure above this solution is
*
*
Ptotal = xbenzene Pbenzene
+ xtoluene Ptoluene
= 0.606 ⋅103Torr + 0.394 ⋅ 32.1Torr = 75.1Torr
No vapor will be formed until the pressure has been reduced to
this value.
10
b) The composition of the vapor at a total pressure of 75.1 Torr is
given by
P1
x1 P1* 0.606 ⋅103
y1 =
=
=
= 0.83
Ptotal Ptotal
75.1
ytoluene = 1 − ybenzene = 0.17
Note that the vapor is enriched in the more volatile component.
11
Expressing Raoult’s Law for dilute solutes in terms of concentration
ni
concentration of solute i, ci =
Vtotal
where, Vtotal = Vsolvent + ∑ Vi = nsolvent V solvent + ∑ ni V i
solutes
for a dilute solution:
solutes
∑n
i
<< nsolvent
solute
ni
ni
ciVtotal ci nsolvent V solvent
xi =
≈
=
≈
= ci V solvent
ntotal nsolvent nsolvent
nsolvent
Neglecting these extra terms is not as important as it might seem:
The approximations underestimate both the numerator and denominator,
12
The effects tend to cancel.
Chemical potential for a solute in a dilute solution:
μi (l ) = μi0 (l , P 0 ) + RT ln ( xi )
(
μi (l ) = μi0 (l , P 0 ) + RT ln ci V solvent
)
⎛ V solvent ⎞
ci
⎛
⎞
= μ (l , P ) + RT ln ⎜
⎟ + RT ln ⎜
⎟
1
L/mole
1
mole/L
⎝
⎠
⎝
⎠
0
i
0
ci
⎛
⎞
μi (l ) = μ (l , P ) + RT ln ⎜
⎟
1
mole/L
⎝
⎠
0
i ,c
0
Chemical potential of component i in its
standard state at 1 M and P0
13
Ideal dilute solution: A recap
System
Gas
Chemical Potential
⎛ Pi ⎞
μi ( g , Pi ) = μ ( g , P ) + RT ln ⎜ ⎟
⎝ P0 ⎠
0
i
0
Solvent
μi (l ) = μi0 (l ) + RT ln ( xi )
Solute
μi (l ) = μi0 (l , P 0 ) + RT ln ( xi )
ci
⎛
⎞
⎟
1
mole/L
⎝
⎠
μi (l ) = μi0,c (l , P 0 ) + RT ln ⎜
Raoult’s Law:
Pi = xi Pi *
Henry’s Law:
Pi = ki xi
14
What happens with real solutions?
We define an activity of component I, ai, which acts like an
effective mole fraction. It is related to the real mole fraction
by an activity coefficient, γi:
ai
activity of component i
γi = =
xi mole fraction of component i
For solvents, asolvent =γ solvent xsolvent ≅ xsolvent
ci
ci
ci
≅
=
For solutes, ai = γ i
-1
1 mole L
1 M c0
Pi
Pi
For gases, ai = γ i 0 ≅ 0
P
P
For solids, ai ≅ 1 (and pure liquids )
15
This results in:
System Chemical Potential
Gas
Solvent
μi ( g , Pi ) = μi0 ( g , P 0 ) + RT ln ( ai )
μi (l ) = μi* (l , P* ) + RT ln ( ai )
As the solute becomes
more dilute:
Pi → 0, then
ai → Pi / P 0 → 0
xi → 1, then
ai → xi → 1
Solute
μi (l ) = μ (l , P ) + RT ln ( ai )
0
i
0
xi → 0, then
ai → ci /1 mole L-1 → 0
16
Real Solutions Exhibit Deviations from Raoult’s Law
Why we need Henry’s Law (for “ideal dilute solution”)
ΔGmixing < 0
ΔSmixing > 0
ΔVmixing ≠ 0
ΔH mixing ≠ 0
17
Chemical equilibrium of solution phase chemical reactions
vi
⎛
Π products ( ai )
0
ΔG = ΔG + RT ln ⎜
vi
⎜Π
⎝ reactants ( ai )
At equilibrium,
vi
⎛
Π products ( ai )
0
ΔG = − RT ln ⎜
vi
⎜Π
⎝ reactants ( ai )
Example:
⎞
⎟
⎟
⎠
⎞
⎟ = − RT ln K a
⎟
⎠eq
10 A + 7 B → 8C + 9 D
8
9
⎛
aC ) ( aD ) ⎞
0
(
ΔG = ΔG + RT ln ⎜
⎟
⎜ ( a )10 ( a )7 ⎟
B
⎝ A
⎠
⎛ ( aC )8 ( aD )9 ⎞
Ka = ⎜
⎟
⎜ ( a )10 ( a )7 ⎟
B
⎝ A
⎠eq
18
Example: Consider a reaction among dilute solutes at
constant temperature and pressure
A+ B → C + D
⎛ ( aC )( aD ) ⎞
K a = ⎜⎜
⎟⎟
a
a
⎝ ( A )( B ) ⎠eq
⎛ [C ] / moleL−1[ D] / moleL−1 ⎞
=⎜
−1
−1 ⎟
⎝ [ A] / moleL [ B ] / moleL ⎠eq
19
Example: What is the ΔG per mole for the reaction below at
25 °C?
Zn( solid ) + 2 H + ( aq ) → Zn 2 + ( aq ) + H 2 ( g )
PH 2 = 10−5 atm
⎡⎣ Zn 2+ ⎤⎦ = 10−3 M
⎡⎣ H + ⎤⎦ = 5.0 M
K eq = 5.3 x1025
20
This problem concerns dilute solutions of two volatile liquids A and B, At
50 °C, the vapor pressure of pure liquid A is 0.67 atm and the vapor
pressure of pure liquid B is 1.1 atm. You may assume Raoult’s law is
valid for the solvent species in this solution and Henry’s Law is valid for
the solute species in this solution. The gas phase is ideal.
A) At 50 °C, a solution with mole fraction Xa=0.9 and Xb=0.1 has a total
vapor pressure of 0.75 atm. Calculate the mole fraction Yb of B in the
vapor phase that is in equilibrium with the solution.
B) Calculate the total vapor pressure at 50 °C for a solution with mole
fractions Xa=0.95 and Xb=0.05.
Use Henry’s Law for Oxygen to determine the concentration of
oxygen in water and compare that with the concentration (M) of
oxygen in the Air. k H ( O2 / water ) = 50kBar
21
Example: Using Raoult’s Law to determine Molecular Weight of
large molecules with no vapor phase (proteins)
Adding a small amount of solute (x2) decreases the vapor
pressure of the solvent (x1). This can be used to measure
the molecular weight of the solute.
P1 = x1 P1*
P1 = (1 − x2 ) P1*
P1
x2 = 1 − *
P1
n2
n2
w2
x2 =
≈ =
n1 + n2 n1 M 2 n1
P1
w2
1− * =
P1 M 2 n1
⎛ w2 ⎞
⎜ V⎟
w2
⎝
⎠
=
M2 =
⎛
P1 ⎞ ⎛ n1 ⎞ ⎛
P1 ⎞
n1 ⎜1 − * ⎟ ⎜ ⎟ ⎜1 − * ⎟
⎝ P1 ⎠ ⎝ V ⎠ ⎝ P1 ⎠
22
Applications of Chemical Equilibrium
Colligative Properties: depend only on the solute concentration in dilute
solutions of volatile solvents (no specifics needed about the solute, just
moles); sort of assume they have a very low vapor (or partial) pressure.
a) Freezing point depression
b) Boiling point elevation
c) Osmotic pressure; How Trees Can grow so tall
d) Determine the molecular weight of a solute (applies to large molecule)
e) Gasses in our blood (deep sea diving)
23
The dilemmas of thermodynamics (H. Rocha, who
understands this better than this would imply)
Thermodynamics is hard, I foretold.
No doubt a challenge, but I was bold.
Before me all of Gibbs was exposed.
Amidst laughs and jokes I heard all proposed.
I've learned a lot, but more confusion I now behold.
by Heiddy Rocha
There was an isolated system
whose species sat at equilibrium
Since no barrier did tame
Their potentials were the same
And now we don't have to measure'em
by Bo Zhao
Reference:
24
http://pruffle.mit.edu/3.00/Literature_of_Thermodynamics/
• Ideal Solutions
μ EtOH ( g ) PEtOH ( g )
μ H O ( g ) PH O ( g )
2
2
μethanolxethanol
μ H O xH O
2
2
At equilibrium:
soln
soln
μ EtOH
(l ) = μ EtOH
(g)
μ HsolnO (l ) = μ HsolnO ( g )
Raoult’s law:
Pi = xi Pi
2
2
*
25
Non-ideal solutions:
ai
activity of component i
γi = =
xi mole fraction of component i
For solvents, asolvent ≅ xsolvent
ci
ci
For solutes, ai ≅ =
c0 1 mole L-1
Pi
For gases, ai ≅ 0
P
For solids, ai ≅ 1
26
This results in:
System Chemical Potential
Gas
Solvent
μi ( g , Pi ) = μi0 ( g , P 0 ) + RT ln ( ai )
μi (l ) = μi* (l , P* ) + RT ln ( ai )
As the solute becomes
more dilute:
Pi → 0, then
ai → Pi / P 0 → 0
xi → 1, then
ai → xi → 1
Solute
μi (l ) = μ (l , P ) + RT ln ( ai )
0
i
0
xi → 0, then
ai → ci / c0 → 0
27
Example: Using Raoult’s Law to determine Molecular Weight
Adding a small amount of solute (x2) decreases the vapor pressure of
the solvent (x1). This can be used to measure the molecular weight of
the solute. Particularly useful for measuring the molecular weight of
proteins, which have no vapor pressure on their own but dissolve in
water.
P1 = x1 P1*
P1 = (1 − x2 ) P1*
P1
x2 = 1 − *
P1
n2
n2
w2
x2 =
≈ =
n1 + n2 n1 M 2 n1
P1 −ΔP
w2
1− * = * =
P1
P1
M 2 n1
w2 P1*
M2 =
⋅
n1 −ΔP
28
Henry’s Law and dissolution of gases
Henry’s law and deep-sea diving
Henry’s Law:
Pi = ki xi
As pressure increases, the amount of gas dissolved in the blood
stream increases.
Associated phenomena (problems) that are understood by
Henry’s Law, which can be applied to gasses that have no vapor
or partial pressures, they are just gasses
Divers ascend slowly so as to prevent gas-bubble embolisms (bends)
Divers can be poisoned by oxygen
and have narcotic like responses to all gasses
Divers need to change formula in tanks if going deeper than 250 feet.
29
Example Problem:
An average human weighing 70 kg has a blood volume of 5.00 L. The
Henry’s Law constant for the solubility of N2 in water is 9.04 x104 bar
at 298 K. [This is 105 bars not Pascals.] Assume that this is also the
value of the Henry’s Law constant for blood and that the density of
blood is 1.00 kg L-1 (same as water).
a) Calculate the number of moles of N2 absorbed in the blood stream
assuming that air is 80% N2 at sea level with the pressure at 1 bar.
b) Calculate the number of moles of N2 absorbed in the blood stream
assuming that the inhaled air is 80% N2 and the diver is at a depth of
150 meters. This is 500 feet; 250 feet is about the extent of depth
with air based scuba equipment. Beyond that you nee He/O2 (Helox)
based equipment.
c) Now the diver is brought to sea level. What volume of N2 gas is
released in the diver’s blood stream?
30
a) Partial pressure of N 2 = PN2 = y N 2 Ptotal = 0.8 × (1.00 bar)=0.8 bar
Mole fraction of N 2in blood stream = xN2 =
PN 2
k N2
8.0 × 10 −1bar
−6
=
×
=
8.85
10
9.04 × 10 4bar
Moles of N 2in blood stream = nN 2 = xN 2 (nH 2O + nN 2 ) = xN2 (nH 2O )
(5.0 L)(1.00 kg L−1 )
−3
= 8.85 × 10
=
2.5
10
moles
×
−3
−1
(18 ×10 kg mole )
−6
At 150 meters, one is at 16 Atm of pressure so the amount of nitrogen is:
b) N 2 in blood stream = 40 ⋅ 10 −3moles
If this much N2 came out of the blood as gas at one Atm (if a diver were
to surface too fast), the volume of gas (as bubbles in the blood) would
be about 100 cc of gas. This causes the bends.
31
Other features of deep sea diving
• Oxygen Toxicity occurs at ~1 Atm of pure oxygen. To
avoid this how far can one dive using 20% O2 air in the
tank?
• How much oxygen is in the blood at 200 feet?
• Some technical divers can go to 800 feet. What type of
gas would they breath at that depth?
•
If you hold your breath (apnea) then you don’t need to worry as
much about the bends and your can descend and ascend much
faster. Records are around 200 meters and 10 minutes (Yes, on a
single breath, down and back.) Same for whales and dolphins etc.
32
calf
Bends in marine animals
Osteonecrosis in
sperm whale rib
bones
Mature adult
Science, 306, p 2215
33
O2 in water and air
• “Fish gotta swim; and birds gotta fly” – J. Kearn
• What is the concentration of O2 in air and in water?
• In air [O ] = n = .2 = 0.01M = 10mM
2
V 22.4
• In Water χ O
2
nO2
P
.2
−6
=
=
= 4.25 ⋅10 =
= [O2 ]VH 2O
4
k H 4.95 ⋅10
nH 2O
4.25 ⋅10−6
= 4.25 ⋅10−6 ⋅ 55.5M = 0.25 mM
[O2 ] =
VH 2O
• Fish need 4 ppm; extract with gills using counter flow.
• On a common basis, oxygen in water is about 2% of the
concentration in air.
• Contrast this with a diver using air going to 200 feet. This
is about 7 atm, or 1.4 Atm of O2 (more than pure air in
34
the body). Fish have the equivalent of 6 mAtm.
Rapture of the Deep: Gas Narcosis
• All gasses give divers a “nacotic-like” feeling if exposed
to too much for too long.
• The narcotic potency correlates very well with the
Henry’s law Constant.
Gas
N 2 He O2 Ar CO2
Re lative
1
− 1.7 2.3 20 ⋅
Narcotic
KH ( N2 )
K H (Gas )
1
.06 1.8 2.4
55
• Xe is about 25 times more potent than nitrogen.
• He has no known effect; but its Henry’s law Constant
predicts very little He dissolves in blood.
• Why is CO2 not as well correlated?
35
How is CO2 removed from the body?
• CO2 is about 3 times more soluble in water than oxygen.
• Therefore, there must be an active way to transport CO2
out of the body, just as there is an active way to
transport O2 into the body. (For each O2 used one CO2 is
generated.)
• CO2 is permeable to RBCs, so the shuttling alone would
not change the (steady state) gradient between muscle
cells and environmental levels.
• An “active method” is employed to hasten removal.
+
−
• Hemoglobin has an acidic balance: HHb S H + Hb
• Both forms bind O2, but the protonated from releases the
O2 more easily:
*
+
−
CO
+
H
O
S
H
CO
R
H
+
HCO
• In muscle and lungs:
2
2
2
3
3
• * The enzyme, Carbonic Anydrase, facilitates the
reaction to make CO2 far more soluble in water.
36
Why put salt on roads?
• Consider what happens when a molecule (solvent) is
added to a liquid. It dissolves only in liquid but not in the
solid (or the vapor) phase.
• The presence of such extras lowers the chemical
potential (makes more stable) of the liquid form, relative
to the vapor and solid phases.
• The chemical potentials must still be equal at the
equilibrium point, but that point must change.
• The freezing temperature can be set/adjusted by adding
salt to water. (How to make ice cream.)
37
Example: Adding salt on roads during winter
Make Frozen Ice Cream.
ice
Water, o0 C, 1 atm
μwater
μwater = μice
μ salt water
Salt water, 0 oC, 1 atm
μ salt water < μ water
μ salt water < μice
The salt lowers the freezing point
of the ice, so some ice melts to
cool the entire system, until the
new freezing point is reached.
38
Graphical representation of colligative properties
Pressure
Addition of salt lowers the chemical
potential of the liquid phase.
This makes the liquid more stable
than either the solid or the vapor
(which are not affected by the addition
of salt)
Therefore the coexistence lines must
move down and liquid freezes at
lower temperature and boils at a
higher temperature.
Elevated boiling temperature
Lowered
melting
temperature
Normal boiling temperature
Normal melting temperature
39
Freezing Point Depression
•
The two chemical potentials must be equal. The liquid drops, so the
chem. Pot. difference must compensate.
μ ( A ) = μ ( s ) or d μ ( A ) = d μ ( s )
d μ ( A, T ) = d μ 0 ( A, T ) + RTd ln χ s d μ 0 ( A, T ) = − SA dT d μ ( s ) = − S S dT
• Equate the differences of the solid and liquid Chemical Potentials:
ΔH Melt
− S S dT = − SA dT + RTd ln χ s
RTd ln χ s = SA dT − S S dT =
dT
T
χs
T
ΔH Melt
ΔH Melt
ln
χ
d ln χ s =
dT
d
=
dT
s
2
2
∫
∫
RT
RT
1
T0
Predicts Freezing
ΔH Melt ⎛ T − T 0 ⎞
point depression to
ln χ s = 0 ⎜
ln ( χ s ) ≈ −cVs
⎟
T R ⎝ T ⎠
three sig. figs.
0 2 ⎞
⎛
0
R (T ) water
⎛
ΔT
RT ⎞
⎟ = 1.86 deg
= −c ⎜ Vs ⋅
⎟ ΔT = cK f K f = ⎜⎜ Vs ⋅
M
ΔH Melt ⎠
ΔH Melt ⎟
T
⎝
⎝
⎠
T drop is proportional to the concentration of solute (e.g. salt), the
proportionality constant depends only on the solvent. Why is MgCl2
40
better than NaCl or sugar on a per mole basis to reduce freezing?
And what happens to the boiling point?
•
Liquid becomes more stable than vapor, so the boiling temperature
must go up. (Same principle/derivation as Raoult’s Law )
μisolution (l , χ i ) = μi ( g , P)
•
This is the same conditions as Raoult’s Law and leads to the same
relation:
*
•
The derivation repeated here: The solvent’s chemical potential (in
the liquid) goes down, so the chemical potential of the vapor phase
must follow.
P′ = χ sol P
μ
•
solution
i
(l , χ i ) = μ (l ) + RT ln ( χ i )
0
i
Subtract the pure liquid:
⎛ Pi ⎞
μi ( g , P) = μ ( g , P ) + RT ln ⎜ 0 ⎟
⎝P ⎠
0
i
0
⎛ P* ⎞
μ (l ) = μi ( g , P ) = μ ( g , P ) + RT ln ⎜ 0 ⎟
⎝P ⎠
0
i
*
0
i
⎛ P* ⎞
⎛ Ps ⎞
⎛ Ps ⎞
RT ln ( χ s ) = RT ln ⎜ 0 ⎟ − RT ln ⎜ 0 ⎟ = RT ln ⎜ * ⎟
⎝P ⎠
⎝P ⎠
⎝P ⎠
0
41
Cooking and chemical potentials
Boiling
Not boiling!
P
water vapor
< P*water vapor
P*water vapor
μ
salt
*
water
Water, 100 oC, 1 atm
μ *water = μ *vapor
P*
P′ = χ P*
P’
T0 T
Water, 100 oC, 1 atm
μvapor = μwater < μ *water
Adding salt lowers the vapor pressure
from P*, the boiling point at the boiling
temperaute by Raoult’s law. This is the
new reference point for the new C-C line.
The T is now the new T when the solution
42
boils again.
Combine Raoult’s Law & Clausius-Clapyron (C-C) Eqn
Determine how the temperature changes if the external pressure
changes. The original C-C Eqn
⎛ P ⎞ ΔHVap ΔT
T = T0 when P = P * (boiling ) ln ⎜ * ⎟ =
⋅
RT0 T
⎝P ⎠
The new reference pressure, P’, replaces P* above, and the
temperature is still To. Boiling occurs when P=P*; at this pressure
the temperature is the boiling temperature:
⎞ ΔHVap ΔT
⋅
and ln ( χ s ) = −cVs
⎟=
RT0 T
⎠
⎛
⎛
RT0 ⎞
RT0 2 ⎞
ΔT
= c ⎜ Vs ⋅
⇒ ΔT = cK b and K b = ⎜ Vs ⋅
= 0.52 deg
⎟
⎟
M
⎜
⎟
⎜
⎟
Δ
Δ
T
H
H
Vap
Vap
⎝
⎠
⎝
⎠
⎛ 1
⎛ P* ⎞
ln ⎜ ⎟ = ln ⎜
⎝ P′ ⎠
⎝ χs
The temperature increase is proportional to the number of moles (or
concentration) of whatever is added. The effect is independent of what
the molecules are (as long as they are soluble in the solvent), and the
proportionality constant of the drop depends only on the solvent.
43
Osmotic Pressure
Pure solvent (water) can flow freely
between pure region and solution
(contains e.g. sugar which cannot pass
the membrane). The Gibb’s energy of
mixing drives water into the solution.
This swells the membrane, which must
hold against the increased pressure, (π).
The additional pressure eventually
balances the favorable entropy of mixing.
At equilibrium the water chemical
potential in the pure phase must equal
that in the solution:
*
solution
μ solvent
T
,
P
=
μ
( ) solvent (T , P + π , xsolvent )
solution
*
μ solvent
T
,
P
+
π
,
x
=
μ
(
solvent )
solvent ( T , P + π ) + RT ln xsolvent
44
Osmotic Pressure
*
*
− RT ln xsolvent = μ solvent
T
,
P
+
π
−
μ
(
) solvent (T , P )
μ s* (T , P + π , xsolvent ) − μ s* (T , P ) =
P +π
∫
P
*
∂μ solvent
∂P
P +π
dP =
T
∫
Vm*dP′ Vm*π
P
π Vm* = − RT ln xsolvent
ln xsolvent = ln (1 − xsolute ) ≈ − xsolute
π=
nsolute
nsolute
=−
≈−
nsolute + nsolvent
nsolvent
nsolute RT nsolute RT
=
= csolute RT
*
nsolventVm
V
45
Osmotic Pressure and Trees
• Experiment done by Dr. R. Smith (UW, Botany)
• Measure the pressure of the water across the leaves of
trees as a function of height.
• The experiment requires a pressure chamber and a
shotgun.
• How much pressure does water exert as a function of
height?
F gm h
P= =
⋅ = ρ gh
A
A h
1 Atm=32 feet of water
Measure the height of the leave, remove leave from tree
(this is where the shotgun comes in handy). Measure
pressure needed to push water out of leaf (with pressure
chamber), compare with pressure of water at that height.
46
Result of Experiment
So How do trees work?
Hydrostatic Pressure
of the leaves were
about 10% over what
would be needed to
raise water that far.
How concentrated
does the sugar
solution need to be at
320 ft = 10 Atm?
π = csolute RT
10 = c0.082 ⋅ 298
10
c=
= 0.41M
24.4
A can of soda ~0.25M
47
I leave a beaker filled to the top with concentrated acid solution on
the lab bench on a humid day in Seattle. What will happen?
concentrated
acid solution
Lab bench
Example: Calculate the osmotic pressure of the solution in a 12
oz. coke can at 298 K. You may assume that 8 oz = 0.25 L. A
can of coke says that there is 39 g of sucrose in 12 oz of solution
and the molecular mass of sucrose is 342.3 g /mole. How high
can this much sucrose pull water?
48