Chapter 11 1 Ch 11 Page 467. STATES OF MATTER CH 11.4-7 CH 11.1-3 CH 5The internet? Phase Change- The transformation from one phase to another upon the.

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Presentation transcript:

Chapter 11 1 Ch 11 Page 467

STATES OF MATTER CH CH CH 5The internet? Phase Change- The transformation from one phase to another upon the addition or release of energy. Temperature does not change during a phase change.

3 Phase Changes Most Order Least Order Ch 11.8 Page 495

4 Phase Changes Outline Liquid ↔ Gas (p ) Solid ↔ Liquid (p ) Solid ↔ Gas (p ) Most Order Least Order

Liquid ↔ Gas Kinetic energy of molecules. “slow” molecules “fast” molecules Some molecules are fast enough to escape intermolecular forces.

Liquid ↔ Gas At time = 0At time > 0At time = ∞ Liquid # of molecules in = # of molecules out At equilibrium! rate of evaporation = rate of condensation

Liquid ↔ Gas Molecular motion causes force. Force creates pressure. We can measure the pressure. Vapor pressure (VP)- is the pressure exerted by a vapor in equilibrium with its condensed phases (solid or liquid) at a given temperature in a closed system. # of molecules in = # of molecules out rate of evaporation = rate of condensation At a given temperature: Weaker intermolecular forces = more molecules out = higher vapor pressure Vapor pressure is temperature dependent!

8 Liquid ↔ Gas Higher temperature (T) = higher average kinetic energy = more molecules escaping solution = higher vapor pressure (P) Lower T = Lower P Higher T = Higher P Ch 11.8 Page 496

9 Relationship Between T and P The experiment- Close container with pressure gauge. Filled with liquid. Increase temperature and measure pressure. Graph pressure vs. temperature. Hot Plate Ch 11.8 Page 497 Each point on the line is the boiling point at that pressure.

10 Relationship Between T and P y = m x + b ln P =  H vap T + C Clausius-Clapeyron Equation R - 1 P = vapor pressure (atm) T = temperature (K) R = gas constant (8.314 J/K mol)  H vap = molar heat of vaporization (J/mol) Ch 11.8 Page 498

ln P =  H vap T + C R - 1 Clausius-Clapeyron Equation P = vapor pressure (atm) T = temperature (K) R = gas constant (8.314 J/K mol)  H vap = molar heat of vaporization (J/mol) Relationship Between T and P Molar heat of vaporization (  H vap ) is the energy required to vaporize 1 mole of a liquid at its boiling point. Dependent on the substance and its intermolecular forces. more intermolecular force = larger  H vap Independent of temperature.

12 Alternate Forms of the Clausius-Clapeyron Equation If we know  H vap, P and T at one temperature we can predict P at another T. or Relationship Between T and P Slope = change in y/the change in x Ch 11.8 Page 499 or T (bp) at another P.

11.7 Diethyl ether is a volatile, highly flammable organic liquid that is used mainly as a solvent. The vapor pressure of diethyl ether is 401 mmHg at 18°C and the  H vap = 26.0 kJ/mol. Calculate its vapor pressure at 32°C. P = vapor pressure (atm) T = temperature (K) R = gas constant (8.314 J/K mol)  H vap = molar heat of vaporization (J/mol)

11.7 Solution Table 11.6 tells us that  H vap = 26.0 kJ/mol. The data are From Equation (11.5) we have Taking the antilog of both sides (see Appendix 4), we obtain P 2 = 656 mmHg Check We expect the vapor pressure to be greater at the higher temperature. Therefore, the answer is reasonable.

15 Closed vs. Open Systems Closed System Open System has a Boiling Point at equilibrium not at equilibrium liquid molecules lost

Boiling Point vapor pressure < external pressure Vapor pressure External pressure vapor pressure > external pressure Vapor pressure External pressure Lower Temp Higher Temp

17 Vapor Pressure and Boiling Point The boiling point is the temperature at which the liquid’s vapor pressure is equal to the applied external pressure. The normal boiling point is the boiling point when the pressure is exactly 1 atm. Decreasing the applied external pressure results in a lower boiling point.

Solid ↔ Liquid Molar heat of vaporization (  H vap ) is the energy required to vaporize 1 mole of liquid to gas. Molar heat of fusion (  H fus ) is the energy required to melt 1 mole of solid. Ch 11.8 Page 501

Solid ↔ Gas Sublimation is the direct transition from the solid to the vapor state “bypassing” the liquid phase During sublimation the solid and gas phases are in equilibrium Iodine (I 2 ) Ice Cubes Ch 11.8 Page 504

Solid ↔ Gas Molar heat of vaporization (  H vap ) is the energy required to vaporize 1 mole of liquid to gas. Molar heat of fusion (  H fus ) is the energy required to melt 1 mole of solid. Molar heat of sublimation (  H sub ) is the energy required to sublime 1 mole of a solid.  H sub =  H fus +  H vap ( Hess’s Law)

Temperature vs. Heat added curve. The temperature of the substance does not rise during the phase change. The heat added to the system at the melting and boiling points goes into breaking intermolecular forces. Phase Change and Heating Curve Ch 11.8 Page 503

Chapter Ch 11 Page 467

Phase Diagram A phase diagram summarizes the conditions at which a substance exists as a solid, liquid, or gas and the places where equilibria exist between phases. Ch 11.9 Page 505

Phase Diagram Important Notes: Each line (AC, AB, and AD): two phases are in equilibrium. B) The critical point: above this critical temperature and critical pressure the liquid and vapor are indistinguishable from each other. Below A: the point below which the substance cannot exist as a liquid. A) The triple point: the point at which all three states are in equilibrium.

25 Phase Diagram of Water Normal melting and boiling points. Solid to liquid with applied pressure. Triple Point (0.006 atm and 0.01 °C)

26 Phase Diagram of CO 2

Phase Diagram of Carbon

28 Actual Phase Diagram of Water I h phase I c phase ICE VII phase

Phase Diagram Review

Chapter Ch 11 Page 467