Transcript Document

Effect of Pressure
•William Henry found that the solubility of a gas increases with
increasing pressure.
C = kPgas
k=
Pgas =
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C
Pgas
C
k
23.54 mL
1.00 atm
=
=
= 23.54 ml N2/atm
100 mL
= 4.25 atm
23.54 ml N2/atm
General Chemistry: Chapter 13
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General Chemistry: Chapter 13
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Vapor Pressures of Solutions
• Raoult, 1880s.
– Dissolved solute lowers vapor pressure of solvent.
– The partial pressure exerted by solvent vapor above an
ideal solution is the product of the mole fraction of
solvent in the solution and the vapor pressure of the
pure solvent at a given temperature.
PA = AP°
A
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General Chemistry: Chapter 13
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Raoult’s Law – Esters?
Raoult’s Law – the Kitchen
• In many cases our enjoyment of foods and
beverages results in part from the presence of
volatile components in what we eat/drink.
Sometimes warming of food is employed
which increases the vapour pressure of all
volatile components.
Mixture with Two (or more) Volatile
Components:
• Solutions can have two (or more) volatile
components. The vapor pressure of the mixture
can be written as
• Ptotal = χAPA0 + χBPB0 + χCPC0 +……..
• In the (unlikely!) event that a mixture
contained only two components (A and B)
with PA = PB the vapor pressure of the mixture
would not vary with the composition (χA etc).
Liquid-Vapor Equilibrium: Ideal
Solutions
•Liquid-vapor equilibrium for benzene-toluene mixtures at 25°C
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General Chemistry: Chapter 13
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Raoult’s Law Example:
• 1. (a) A benzene/toluene mixture is prepared
by mixing 500.0 g benzene, C6H6, and 500.0 g
toluene, C7H8, at 25.0 0C. At this temperature
pure benzene and pure toluene have vapor
pressures of 12.68 kPa and 3.79 kPa
respectively. Calculate the total vapor pressure
of the mixture.
• (b) Find the mole fraction of benzene and
toluene in the gas phase.
Raoult’s Law – Vapor Pressure
Lowering:
• For a mixture of two substances Raoult’s Law
tells us that
• Ptotal = χAPA0 + χBPB0 where PA0 and PB0 are
the vapor pressure of the two pure substances
(at a particular T). If A is totally involatile, PA0
= 0, then a mixture of A and B will have a total
vapor pressure less than that of A. We can
roughly estimate molar masses using this
effect.
Raoult’s Law Example:
• 44.2 g of a simple carbohydrate (known to be
either C12H22O11 or C6H12O6) are dissolved in
99.5 g of water at 24.0 oC. The vapor pressures
of pure water and the solution at 24.0 oC are
2.986 kPa and 2.859 kPa respectively. Find the
molar mass of the carbohydrate and hence its
identity.
Raoult’s Law
• As a mnemonic device can you think of a way
of thinking about Raoult’s Law that involves
“probabilities”?
Osmotic Pressure
• Interesting changes occur when two solutions
of different concentrations are placed in close
contact. The next slide shows what happens
when two aqueous solutions of different
concentrations, each containing a nonvolatile
solute are placed under a transparent
enclosure. Water moves over time from the
dilute solution (A) to the concentrated
solution. Why? (Vapor pressure H2O higher
above the left hand container. Why?)
Osmotic Pressure
FIGURE 13-16
•Observing the direction of flow of water vapor
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General Chemistry: Chapter 13
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Osmotic Pressure – cont’d:
• On the previous slide there is a net transfer of
water (solvent) molecules between the two
containers until both containers have the same
concentration or the vapor pressure of H2O is
the same over both containers. Transfer of
solvent need not occur through the gas phase!
When two solutions of different concentration
are separated by a solvent permeable
membrane there is a surprise!
FIGURE 13-17
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Osmosis
Does pure water or sugar water
have the higher vapor pressure.
Mechanical analogy?
General Chemistry: Chapter 13
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Osmotic Pressure – cont’d:
• On the previous slide the movement of solvent
molecules generates enough pressure to lift
solution through the tube. The size of this
osmotic pressure can be calculated with a
“familiar looking” equation. Here n is the
number of moles of solute (initially a
nonelectrolyte!!!!) and V is the solution
volume (in Liters).
Osmotic Pressure
For dilute solutions of nonelectrolytes:
πV = nRT
n
π=
RT = MRT
V
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General Chemistry: Chapter 13
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