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Copyright 1999, PRENTICE HALL Chapter 11 1 Surface Tension Surface Tension
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Copyright 1999, PRENTICE HALLChapter 111 Surface Tension.

Dec 17, 2015

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Page 1: Copyright 1999, PRENTICE HALLChapter 111 Surface Tension.

Copyright 1999, PRENTICE HALL Chapter 11 1

Surface TensionSurface Tension

Page 2: Copyright 1999, PRENTICE HALLChapter 111 Surface Tension.

Copyright 1999, PRENTICE HALL Chapter 11 2

Intermolecular Forces Bulk and SurfaceIntermolecular Forces Bulk and Surface

Page 3: Copyright 1999, PRENTICE HALLChapter 111 Surface Tension.

Copyright 1999, PRENTICE HALL Chapter 11 3

Phase ChangesPhase Changes

Energy Changes Accompanying Phase ChangesEnergy Changes Accompanying Phase Changes• All phase changes are possible under the right

conditions (e.g. water sublimes when snow disappears without forming puddles).

• The sequence

heat solid melt heat liquid boil heat gas

is endothermic.• The sequence

cool gas condense cool liquid freeze cool solid

is exothermic.

Page 4: Copyright 1999, PRENTICE HALLChapter 111 Surface Tension.

Copyright 1999, PRENTICE HALL Chapter 11 4

Phase ChangesPhase Changes

Heating CurvesHeating Curves

Page 5: Copyright 1999, PRENTICE HALLChapter 111 Surface Tension.

Copyright 1999, PRENTICE HALL Chapter 11 5

Heating Curve IllustratedHeating Curve Illustrated

Page 6: Copyright 1999, PRENTICE HALLChapter 111 Surface Tension.

Copyright 1999, PRENTICE HALL Chapter 11 6

Vapor PressureVapor Pressure

Explaining Vapor Pressure Explaining Vapor Pressure on the Molecular Levelon the Molecular Level

• Dynamic Equilibrium: the point when as many molecules escape the surface as strike the surface.

• Vapor pressure is the pressure exerted when the liquid and vapor are in dynamic equilibrium.

Page 7: Copyright 1999, PRENTICE HALLChapter 111 Surface Tension.

Copyright 1999, PRENTICE HALL Chapter 11 7

Vapor PressureVapor Pressure

Volatility, Vapor Pressure, and TemperatureVolatility, Vapor Pressure, and Temperature• If equilibrium is never established then the liquid

evaporates.• Volatile substances evaporate rapidly.• The higher the temperature, the higher the average

kinetic energy, the faster the liquid evaporates.

Page 8: Copyright 1999, PRENTICE HALLChapter 111 Surface Tension.

Copyright 1999, PRENTICE HALL Chapter 11 8

Liquid Evaporates when no Equilibrium is EstablishedLiquid Evaporates when no Equilibrium is Established

Page 9: Copyright 1999, PRENTICE HALLChapter 111 Surface Tension.

Copyright 1999, PRENTICE HALL Chapter 11 9

Vapor PressureVapor PressureVapor Pressure and Boiling PointVapor Pressure and Boiling Point• Liquids boil when the external pressure equals the vapor

pressure.

• Temperature of boiling point increases as pressure increases.

• Two ways to get a liquid to boil:

• 1. 2.

– Pressure cookers operate at high pressure.

– The higher the pressure the higher the b.p.

– reducing the cooking time required.

• Normal boiling point is the boiling point at 760 mmHg (1 atm).

Page 10: Copyright 1999, PRENTICE HALLChapter 111 Surface Tension.

The Ideal Gas LawThe Ideal Gas Law

PV=nRT• P = Pressure• V = Volume• n = Number of Moles• R = ideal gas constant• T = temperature in Kelvin

Page 11: Copyright 1999, PRENTICE HALLChapter 111 Surface Tension.
Page 12: Copyright 1999, PRENTICE HALLChapter 111 Surface Tension.

R -The ideal gas constant R -The ideal gas constant

• Depends on unit of pressure• 0.0821 L . Atm / K . mol• 62.4 L . mmHg / K . mol (torr is mm Hg)• 8.31 L . kPa / K . mol

Page 13: Copyright 1999, PRENTICE HALLChapter 111 Surface Tension.

Ideal Gas Law Ideal Gas Law example problemexample problem

• Calculate the pressure of 1.65 g of helium gas at 16.0oC and occupying a volume of 3.25 L?

• You will need g to moles and Celsius to Kelvin:• 1.65 g He 1 mole He• 4.0 g He = 0.413 mol He• K = oC + 273 ; 16. 0 + 273 = 289 K• For this problem you will need to pick an R value. For

this problem I will choose to use the R value containing kPa. I picked it. You can’t do anything about it. So; just try and stop me.

• Plug and Chug baby, get ‘R done. Do it. Come on I dare ya.

• Get it - ‘R as in ideal gas constant

Page 14: Copyright 1999, PRENTICE HALLChapter 111 Surface Tension.

Ideal Gas Law Ideal Gas Law example problemexample problem

• P x 3.25 L = 0.413 mol x 8.31 kPa . L x 289 K

• mol . K

• Do the algebra and solve; if you do it right, guess what? You get the answer right. Neat concept, huh? Maybe your mommy will give you a cookie.

• = 305 kPa

• Your turn

• How many moles of gas are present in a sample of Argon at 58oC with a volume of 275 mL and a pressure of 0.987 atm.

Page 15: Copyright 1999, PRENTICE HALLChapter 111 Surface Tension.

Ideal Gas Law Ideal Gas Law example problemexample problem

• Answer• 0.987 atm x 0.275 L = n x 0.0821 L . Atm x 331K• mol . K• Do the dew; oops, I mean the algebra and presto; the

answer with the correct number of sig figs is..• Do you know how to keep a so called chem student in

suspense?• Do ya?• Do ya?• = 0.00999 mol Ar• Congrats - you can plug and chug.

Page 16: Copyright 1999, PRENTICE HALLChapter 111 Surface Tension.

Gas Gas

StoichiometrStoichiometr

yy

You can plug and chug.But, can you do….

For these problems you may need to do tracks firstthen use the ideal gas law or use the gas law first then the tracks.

Page 17: Copyright 1999, PRENTICE HALLChapter 111 Surface Tension.

Sample problem Sample problem Ammonia (NH3) gas can be synthesized from nitrogen gas + hydrogen gas. What volume of ammonia at 450 kPa and 80°C can be obtained from the complete reaction of 7.5 kg hydrogen?

7500 g H2 1 mol H2

2.02 g H2 x = 2500 mol NH3

2 mol NH3

3 mol H2 x

PV = nRT P = 450 kPa, n = 2475 mol, T = 353 K

(450 KPa)

(2500 mol)(8.31)(353 K) = V = 16000 L NH3

First we need a balanced equation:N2(g) + 3H2(g) 2NH3(g)

Page 18: Copyright 1999, PRENTICE HALLChapter 111 Surface Tension.

Your Turn:Your Turn:Hydrogen gas (and NaOH) is produced when sodium metal is added to water. What mass of Na is needed to produce 20.0 L of H2 at STP?

First we need a balanced equation:2Na(s) + 2H2O(l) H2(g) + 2NaOH(aq)

0.893 mol H2 = 41.1 g Na2 mol Na

1 mol H2 x

22.99 g Na

1 mol Nax

PV = nRT

(8.31 kPa•L/K•mol)(273 K)

(101.3 kPa)(20.0 L)= n = 0.893 mol H2

P= 101.3 kPa, V= 20.0 L, T= 273 K

or # mol = 20.0 L x 1 mol / 22.4 L = 0.893 mol