Pressure of the vapor present when equilibrium is achieved between the rate of vaporization and the rate of condensation At the boiling point the P atm P vapor As the vapor pressure on a pot of water is reduced the energy needed to boil that water is also reduced ID: 757013
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Slide1
Vapor PressureSlide2
Vapor Pressure
Pressure of the vapor present when equilibrium is achieved between the rate of vaporization and the rate of condensation.
At the boiling point, the
P
atm
=
P
vapor
As the vapor pressure on a pot of water is reduced, the energy needed to boil that water is also reduced.
Pressure Cooker: By increasing the vapor pressure, additional energy is needed for the water to boil, therefore the water can boil at temperatures above 100C.Slide3
Effect of Pressure on Boiling Point
Boiling Point of Water at Various Locations
Location
Feet above sea level
P
atm
(
kPa
)
Boiling Point (
C)
Top of Mt. Everest, Tibet
29,028
32
70
Top of Mt. Denali, Alaska
20,320
45.3
79
Top of Mt. Whitney, California
14,494
57.3
85
Leadville, Colorado
10,150
68
89
Top of Mt. Washington, N.H.
6,293
78.6
93
Boulder, Colorado
5,430
81.3
94
Madison, Wisconsin
900
97.3
99
New York City, New York
10
101.3
100
Death Valley, California
-282
102.6
100.3Slide4
Vapor Pressure vs. Temperature
As the temperature increases, a greater number of molecules have sufficient kinetic energy to convert from the liquid to the vapor phase.
There is a
nonlinear
relationship between the vapor pressure of a liquid and temperature.Slide5
Vapor Pressure vs. Temperature
Slide6
The
Clausius
–
Clapeyron
Equation
A mathematical expression which relates the variation of vapor pressure to temperature
ln P = (-DHvap/RT) + C
where C is a constant
IMPORTANCE
:
When the
ln
P is plotted
vs
(1/T) you create a line where the slope is equal to the –
D
H
vap
/R
Which means you can calculate the enthalpy of vaporization from the slope of the line.
R = 8.314 J/
Kmol
Convert all Temps to KelvinSlide7
Vapor Pressure of Solutions
A
nonvolatile
solute lowers the vapor pressure of the solution.
The molecules of the solvent
must overcome the force of both the other solvent molecules and the solute molecules.Slide8
Raoult’s Law:
P
soln
=
c
solvent
x PsolventVapor pressure of the solution = m ole fraction of solvent x vapor pressure of the pure solventApplies only to an ideal solution where the solute doesn’t contribute to the vapor pressure.Slide9
Aqueous Solution
Pure water
Water has a higher vapor pressure than a solutionSlide10
Aqueous Solution
Pure water
Water evaporates faster from for water than solutionSlide11
The water condenses faster in the solution so it should all end up there.
Aqueous Solution
Pure waterSlide12
Practice Problem
A solution of cyclopentane with a nonvolatile compound has vapor pressure of 211 torr. If vapor pressure of the pure liquid is 313 torr, what is the mole fraction of the cyclopentane?
P
soln
= X
cp
Pcp211 torr = Xcp (313 torr).674Slide13
Try one on your own
Determine the vapor pressure of a solution at 25
C that has 45 grams of C
6
H
12O6, glucose, dissolved in 72 grams of H2O. The vapor pressure of pure water at 25 C is 23.8 torr. Psolution= X
solvent
P
solvent
P
solution
= .941(23.8 torr)
P
solution
= 22.4 torrSlide14
Liquid-liquid solutions where both are volatile.
Modify
Raoult’s
Law to
P
total
= PA + PB = cAPA0 +
c
B
P
B
0
P
total
= vapor pressure of mixtureIf this equation works then the solution is ideal.Ideal solutionsSlide15
Vapor Pressure of solutionSlide16
Deviations
If solvent has a strong affinity for solute (H bonding).
Lowers solvent’s ability to escape.
Lower vapor pressure than expected.
Negative
deviation from
Raoult’s law.DHsoln
is large and negative exothermic.
Endothermic
D
H
soln
indicates
positive
deviation.Slide17
χ
b
χ
A
Vapor Pressure
Positive deviations-
Weak attraction between solute and solvent
Positive
Δ
H
solnSlide18
χ
b
χ
A
Vapor Pressure
Negative deviations-
Strong attraction between solute and solvent
Negative
Δ
H
solnSlide19
Problem #1
The vapor pressure of a solution containing 53.6g of glycerin C
3
H
8
O
3 in133.7g ethanol C2H5OH is 113 torr at 40C. Calculate the vapor pressure of pure ethanol at 40C assuming that the glycerin is a non volatile, nonelectrolyte solute in ethanol. Slide20
Answer to #1
P
soln
= X
eth
P
eth113torr = 2.90mol/3.48mol (Peth)135.6 torr = PethSlide21
Problem #2
At a certain temperature, the vapor pressure of pure benzene C
6
H
6
is 0.930atm. A solution was prepared by dissolving 10.0g of a nondissociating, nonvolatile solute in 78.11g of benzene at that temperature. The vapor pressure was found to be 0.900atm. Assuming the solution behave ideally, determine the molar mass of the solute.Slide22
Answer #2
P
soln
= X
benzene
P
benzene.900atm = Xbenzene (.930atm)Xbenzene = .9677 X
solute
= .0323
MM = 10.0g/.0323mol = 310g/molSlide23
Problem #3
A solution of NaCl in water has a vapor pressure of 19.6 torr at 25C. What is the mol fraction of solute particle in this solution if the vapor pressure of water is 23.8 torr at 25C?Slide24
Answer #3
P
soln
= X
water
P
water19.6torr = Xwater(23.8torr) .824 = Xwater therefore Xsolute = .176Slide25
Problem #3
For the same problem as #3:
What is the vapor pressure of the solution at 45C if the vapor pressure of water is 71.9 torr at 45C?Slide26
Answer #3
P
soln
= .824(71.9torr)
P
soln
= 59.2 torrSlide27
Problem #4
A solution is made from 0.0300mol CH
2
Cl
2
and 0.0500mol CH
2Br2 at 25C. Assuming that the solution is ideal, calculate the % composition of the vapor at 25C. PCH2Cl2 = 133 torrPCH2Br2
= 11.4 torrSlide28
Answer #4
P
soln
= X
CH2Cl2
P + X
CH2Br2 PPsoln = (.03/.08)(133torr) + (.05/.08)(11.4 torr)Psoln = 57.0 torr
X
CH2Cl2
= 49.9 / 57 = .875 = 87.5%
X
CH2Br2
= 7.13 / 57 = .125 = 12.5%