Vapor Pressure

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 Patm = Pvapor

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.

Effect of Pressure on Boiling PointBoiling Point of Water at Various Locations

Location Feet above sea level

Patm (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 -282 102.6 100.3

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.

Vapor Pressure vs. Temperature

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 –DHvap/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 Kelvin

Vapor Pressure of Solutions• A nonvolatile solute lowers the vapor pressure of the solution.

• The molecules of the solventmust overcome the force of both the other solvent molecules and the solute molecules.

Raoult’s Law:

•Psoln = csolvent x Psolvent

• Vapor pressure of the solution = m ole fraction of solvent x vapor pressure of the pure solvent

• Applies only to an ideal solution where the solute doesn’t contribute to the vapor pressure.

Aqueous Solution

Pure water

• Water has a higher vapor pressure than a solution

Aqueous Solution

Pure water

• Water evaporates faster from for water than solution

• The water condenses faster in the solution so it should all end up there.

Aqueous Solution

Pure water

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?

• Psoln = XcpPcp

• 211 torr = Xcp (313 torr)• .674

Try one on your own

• Determine the vapor pressure of a solution at 25 C that has 45 grams of C6H12O6, glucose, dissolved in 72 grams of H2O. The vapor pressure of pure water at 25 C is 23.8 torr.

• Psolution= Xsolvent Psolvent

• Psolution = .941(23.8 torr)

• Psolution = 22.4 torr

• Liquid-liquid solutions where both are volatile.• Modify Raoult’s Law to

• Ptotal = PA + PB = cAPA0 + cBPB

0

• Ptotal = vapor pressure of mixture

• If this equation works then the solution is ideal.

Ideal solutions

Vapor Pressure of solution

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 DHsoln indicates positive deviation.

χb

χA

Vap

or P

ress

ure Positive deviations-

Weak attraction between solute and solvent

Positive ΔHsoln

χb

χA

Vap

or P

ress

ure Negative deviations-

Strong attraction between solute and solvent

Negative ΔHsoln

Problem #1• The vapor pressure of a solution containing 53.6g of glycerin

C3H8O3 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.

Answer to #1

Psoln = Xeth Peth

113torr = 2.90mol/3.48mol (Peth)

135.6 torr = Peth

Problem #2• At a certain temperature, the vapor pressure of pure benzene

C6H6 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.

Answer #2

Psoln = XbenzenePbenzene

.900atm = Xbenzene (.930atm)

Xbenzene = .9677 Xsolute = .0323

MM = 10.0g/.0323mol = 310g/mol

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?

Answer #3

Psoln = XwaterPwater

19.6torr = Xwater(23.8torr)

.824 = Xwater therefore Xsolute = .176

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?

Answer #3

Psoln = .824(71.9torr)

Psoln = 59.2 torr

Problem #4

A solution is made from 0.0300mol CH2Cl2 and 0.0500mol CH2Br2 at 25C. Assuming that the solution is ideal, calculate the % composition of the vapor at 25C. PCH2Cl2 = 133 torr

PCH2Br2 = 11.4 torr

Answer #4

Psoln = XCH2Cl2 P + XCH2Br2 P

Psoln = (.03/.08)(133torr) + (.05/.08)(11.4 torr)

Psoln = 57.0 torr

XCH2Cl2 = 49.9 / 57 = .875 = 87.5%

XCH2Br2 = 7.13 / 57 = .125 = 12.5%