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Vapor Pressure Vapor Pressure Vapor Pressure Vapor Pressure

Vapor Pressure Vapor Pressure - PowerPoint Presentation

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Uploaded On 2019-03-16

Vapor Pressure Vapor Pressure - PPT Presentation

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

pressure vapor water solution vapor pressure solution water torr soln solute pure solvent temperature problem answer liquid ideal molecules negative 25c top

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Presentation Transcript

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%